Lp Dirichlet problem for second order elliptic operators having a BMO anti-symmetric part by Linhan Li B.Sc., Shandong University; Jinan, Shandong, China, 2015 M.Sc., Brown University; Providence RI, USA, 2018 A dissertation submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in Department of Mathematics at Brown University PROVIDENCE, RHODE ISLAND May 2019 c Copyright 2019 by Linhan Li This dissertation by Linhan Li is accepted in its present form by Department of Mathematics as satisfying the dissertation requirement for the degree of Doctor of Philosophy. Date Jill Pipher, Ph.D., Advisor Recommended to the Graduate Council Date Hongjie Dong, Ph.D., Reader Date Benoit Pausader, Ph.D., Reader Approved by the Graduate Council Date Andrew G. Campbell, Dean of the Graduate School iii Vitae Linhan Li was born on November 5th, 1992 in the city of Hangzhou, China. She earned a B.Sc. degree in mathematics from the School of Mathematical Sciences at Shandong University in June 2015, and a M.Sc. degree in mathematics from Brown University in May 2018. During her years at Brown, she taught multiple undergraduate courses including Introductory and Advanced Placement Calculus as both a primary lecturer and a teaching assistant. She has given research talks at various conferences such as Geometric and Harmonic Analysis 2019 at University of Connecticut, the 3rd Annual Northeastern Analysis Meeting at State University of New York at New Paltz, and the AMS Graduate Student Conference in Analysis, Probability and PDE at Brown University. Starting from August 2019, she will be appointed as a Dunham Jackson Assistant Professor in the School of Mathematics at University of Minnesota. Below is a list of publications and preprints for the author. 1. Boundary behavior of solutions of elliptic operators in divergence form with a BMO anti-symmetric part, with J.Pipher, communications in Partial Differential Equations (2019): 1-49. DOI: 10.1080/03605302.2018.1542437 2. `2 Decoupling with vanishing curvature, with C. Biswas, M. Gilula, J. Schwend, Y. Xi. Submitted (2018) arXiv:1812.04760. 3. The A∞ condition for elliptic operators having a BMO anti-symmetric part, with S. Hofmann, S. Mayboroda, J. Pipher. In preparation (2019) iv Acknowledgements I could not have gotten to this point alone. There are many people who helped me along the way, and I would like to thank them for their kindness. My deepest gratitude goes to my advisor Jill Pipher. I feel very privileged and honored to be one of her students. Throughout my Ph.D years she has been a great source of inspiration and a generous advisor. Completing this thesis has been a long fight, and I will never forget her enlightening guidance, keen insights, unreserved support, and whole- hearted encouragement. Busy as she is, she always manages to find the time to discuss problems with me, or spend hours thinking of questions even on weekend. I am also be- yond grateful for all the opportunities that she created for me, from attending amazing conferences, to introducing me to amazing people. These valuable experiences have opened my eyes widely. My gratitude also goes to my dissertation committee members, Hongjie Dong and Benoit Pausader, for their patience reading through my dissertation and giving me invalu- able suggestions. Hongjie Dong was the instructor of my PDE course in my first year and has always been so generous to me since then. I would not have known many things without his help. Benoit Pausader has also broadened my mind with his deep and wide knowledge, and given me guidance during my years at Brown. I am also indebted to many mathematicians who have helped me during my Ph.D life. I would like to thank Seick Kim, for introducing to me the operators that this thesis is focused on. Steve Hofmann and Svitlana Mayboroda, who are both full of brilliant ideas and have always been so encouraging and generous to me. Jos´e Conde Alonso, who has v not only given me guidance in math, but also has influenced me with his passion for math and life. I also have learned much through conversations with Martin Dindos and Vladim´ır ˇ ak, for whom I have always been very grateful. Sver´ I would also like to thank my friends, old and new, for making my Ph.D life colorful. I am grateful for all the good times we had and good games we played. Liu Yang, thank you for bringing me so much joy and comfort, and for being so patient and understanding all the time. I would have been much lonelier without you. Finally, I feel extremely fortunate to have parents so understanding and supportive. I would like to thank my dad for guiding and listening to me when I felt lost and frustrated. And I would always remember the first days in Providence, when we explored the sur- roundings together. I would like to thank my mom for reminding me that I always have someone waiting for me at home. You give me the courage to embrace the unknown and move forward bravely. vi Abstract of “ Lp Dirichlet problem for second order elliptic operators having a BMO anti- symmetric part” by Linhan Li, Ph.D., Brown University, May 2019 In this thesis we study the Lp Dirichlet problem for second order divergence-form operators having a BMO antisymmetric part. To be precise, for the divergence-form operator L = − div A∇, we assume that the symmetric part of the matrix A is L∞ and elliptic, and that the anti-symmetric part of A belongs to the space BMO. These operators are relevant to the study of incompressible flows. We show that the weak solution to L is H¨older continuous near the boundary. This enables us to prove the existence of the elliptic measures associated to L in non-tangentially accessible (NTA) domains; these general domains were introduced by Jerison and Kenig and include the class of Lipschitz domains. We also derive pointwise estimates for the elliptic measure, as well as its relation with the Green’s function. Moreover, we are able to prove for these operators that the Lp Dirichlet problem is solvable for p sufficiently large in the upper half space, under the additional assumption that the coefficients are independent of the vertical variable. This result is equivalent to saying that the elliptic measure associated to L belongs to the A∞ class with respect to the Lebesgue measure dx. We prove the A∞ condition by showing a Carelson measure estimate for the weak solution. Our method relies on kernel estimates and off-diagonal estimates for the semigourp e−tL , solution to the Kato problem, and various estimates for the Hardy norms of certain commutators. This result extends the work of Hofmann, Kenig, Mayboroda and Pipher in 2015, which holds for elliptic operators in divergence form with non-symmetric, L∞ coefficients. CONTENTS Vitae iv Acknowledgments v 1 Introduction 1 1.1 More history . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2 Preliminary 10 2.1 Some facts about general domains . . . . . . . . . . . . . . . . . . . . . . . 10 2.2 The Hardy norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 2.3 Weak solutions and Drichlet problem . . . . . . . . . . . . . . . . . . . . . . 16 2.4 Some useful results in PDE . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.5 Theory of weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 3 Estimates for the weak solution 24 3.1 Interior estimates for the solution . . . . . . . . . . . . . . . . . . . . . . . . 24 3.2 Boundary estimates for the solution . . . . . . . . . . . . . . . . . . . . . . 37 3.3 Weak solution of parabolic equations . . . . . . . . . . . . . . . . . . . . . . 53 4 Estimates for the Green functions and elliptic measure 64 4.1 Green’s functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 4.2 Regular points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 4.3 Elliptic measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 5 The Lp Dirichlet problem 79 p 5.1 Dirichlet problem with L (dω) data . . . . . . . . . . . . . . . . . . . . . . 79 5.2 Dirichlet problem with Lp (dσ) data . . . . . . . . . . . . . . . . . . . . . . . 85 6 Characterization of the A∞ condition 89 6.1 Necessary condition for the A∞ in terms of square function and non-tangential maximal function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 6.2 More characterizations of the A∞ condition . . . . . . . . . . . . . . . . . . 105 7 The Gaussian property 107 vii 7.1 Sectorial operators and resolvent estimates . . . . . . . . . . . . . . . . . . 107 7.2 Proof of the Gaussian property . . . . . . . . . . . . . . . . . . . . . . . . . 115 7.3 L2 estimates for ∂tl e−tL and ∇e−tL . . . . . . . . . . . . . . . . . . . . . . . 131 8 The Lp theory for the semigroup and square root 135 8.1 L2 off-diagonal estimates for the semigroup . . . . . . . . . . . . . . . . . . 135 8.2 Lp theory for the semigroup . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 8.3 Lp Theory for the square root . . . . . . . . . . . . . . . . . . . . . . . . . . 152 9 Estimates for Square functions and non-tangential maximal functions involving the heat semigroup 166 9.1 Lp estimates for square functions . . . . . . . . . . . . . . . . . . . . . . . . 166 9.2 Lp estimates for non-tangential maximal functions . . . . . . . . . . . . . . 183 10 The A∞ property for operators with coefficients independent of the transverse variable 189 10.1 Outline of the proof of Theorem 10.0.1 . . . . . . . . . . . . . . . . . . . . . 191 10.2 Hodge decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 10.3 Construction of F and sawtooth domains associated with F . . . . . . . . . 202 10.3.1 The set F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 10.3.2 Sawtooth domains and related estimates . . . . . . . . . . . . . . . . 205 10.3.3 The cut-off function . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 10.4 Proof of the Carleson measure estimate . . . . . . . . . . . . . . . . . . . . 211 viii CHAPTER One Introduction In this thesis, we investigate the Dirichlet problem of divergence-form operators with an elliptic symmetric part and a BM O anti-symmetric part. To be precise, let A be a n × n matrix of real coefficients defined in Rn , such that . 1. The symmetric part a = 12 (A + A| ) = (aij (x)) is L∞ (Rn ), and satisfies the ellipticity condition n X 2 λ0 |ξ| ≤ ha(x)ξ, ξi = aij (x)ξi ξj ∀ ξ ∈ Rn , kak∞ ≤ λ−1 0 (1.0.1) i,j=1 for some 0 < λ0 < 1. . 2. The anti-symmetric part b = 12 (A − A| ) = (bij (x)) is in the space BMO(Rn ), with . kbij kBM O = sup |bij − (bij )Q | dx ≤ Λ0 (1.0.2) Q⊂Rn Q 1 ´ for some Λ0 > 0. Here (f )Q denotes the average |Q| Q f (x)dx. 1 2 When working on a domain Ω ⊂ Rn , we consider BMO coefficients defined in the domain. That is, the anti-symmetric part b ∈ BM O(Ω), with . kbij kBM O(Ω) = sup |bij − (bij )Q | dx ≤ Λ0 , (1.0.3) Q⊂Ω Q where the supremum is taken over all cubes Q inside Ω, with sides parallel to the axes. Note that the anti-symmetry of b implies that the ellipticity condition (1.0.1) on a is actually an ellipticity condition on A. That is, hAξ, ξi = haξ, ξi ≥ λ0 |ξ|2 ∀ ξ ∈ Rn . (1.0.4) We define in Rn a second order divergence form operator L = − div A∇, (1.0.5) which is interpreted in the sense of maximal accretive operators via sesquiliniear form. These operators arise in the study of elliptic equations of the form − ∆u + c · ∇u = f (1.0.6) with a divergence-free drift c (see for example [MyV06]), as well as the parabolic case ∂t u − ∆u + c · ∇u = f with div c = 0 ([Osa87],[Zha04],[LZ04]). Since div c = 0, we can write c = div b for an anti-symmetric tensor b = (bij ), and the equation (1.0.6) becomes − div(I − b)∇u = f. If one requires c ∈ BM O−1 (Rn ), then b ∈ BM O(Rn ), and thus the operator − div(I − b)∇ is a divergence form operator with a BMO anti-symmetric part. We refer to [KT01] for more about the space BM O−1 . 3 ˇ These operators have gained much attention in recent years (see [SSSZ12],[QX18],[EH18], [DK18]), but the study of the boundary value problems was not part of the literature. Our goal is to study the boundary behavior of the weak solutions to the operator L, and the Lp Dirichlet problem in a domain Ω:      Lu = 0 in Ω,   (Dp ) u → g ∈ Lp (∂Ω, dσ) non-tangentially a.e. dσ on ∂Ω (1.0.7)     N α u ∈ Lp (∂Ω).   Here, dσ is the surface measure on ∂Ω, and N α u denotes the non-tangential maximal function of u: . N α u(Q) = sup |u(X)| , (1.0.8) Γα (Q) where Γα (Q) is a “cone” of aperture α at Q ∈ ∂Ω: Γα (Q) = {X ∈ Ω : |X − Q| ≤ (1 + α) dist(X, ∂Ω)}. It is a non-tangential approach region for Q. Note that if ∂Ω is a plane, then Γα (Q) is indeed a cone of aperture α. Since it is well known that Lp norms for N α f are equivalent for any choice of α (see [CMS85]), we shall simply write N for N α when the aperture is unimportant. It is well known that for elliptic operators with L∞ coefficients, the relationship between the elliptic measure ωL associated to L and the surface measure on the boundary of the domain characterizes the solvability of the Lp Dirichlet problem. In fact, the Lp Dirichlet problem is solvable if and only if ωL belongs to the reverse H¨older weight class Bp0 (dσ) (see section 2.5 for the definition of Bp0 (dσ)); and that the Lp Dirichlet problem is solvable for some (sufficiently large) 1 < p < ∞, if and only if ωL ∈ A∞ (dσ). The A∞ condition is a quantitative version of mutual absolute continuity. The precise definition is given in section 2.5. 4 Our first contribution is that we establish the existence of elliptic measures associated to elliptic operators having a BMO anti-symmetric part in non-tangentially accessible (NTA) domains; these domains were introduced by Jerison and Kenig and include Lipschitz domains. The definition of these domains are given in section 2.1. The elliptic measure associated to L is well-defined on the boundary of a domain Ω if one can show that the continuous Dirichlet problem is uniquely solvable in Ω, and that the weak solution has the maximum principle property. We prove these results in section 3.2. In section 4.3, we derive various pointwise estimates for the elliptic measure, which relies on estimates for the Green’s function (see section 4.1). We note that these estimates are the same as the case when the coefficients of the elliptic operators are L∞ . We also observe in section 4.2 that a point on the boundary is a regular point for L = − div A∇ with A satisfying (1.0.1) and (1.0.3) if and only if it is a regular point for the Laplacian. Moreover, we showed that given any function f in Lp (∂Ω, dωL ) on the boundary of an NTA domain Ω, there is a unique u that solves Lu = 0 in Ω such that N u ∈ Lp (∂Ω, dωL ), and that u converges to f on the boundary of Ω except for a set with zero ωL measure. This is proved in Chapter 5. Also in that chapter, we observe that the Lp Dirichlet problem (1.0.7) is uniquely solvable, if one has the estimate kN ukLp (∂Ω,dσ) . kf kLp (∂Ω,dσ) for any weak solution u that solves the continuous Dirichlet problem with data f . This enables us to characterize the solvability of the Lp Dirichlet problem for L using the relationship between ωL and the surface measure. As we pointed earlier, ωL ∈ A∞ (dσ) implies the Lp Dirichlet problem is solvable for some p > 1. And chapter 6 is devoted to the characterization of the A∞ condition. The above results are from the paper [LP19] by the author, coauthored with J. Pipher. Now it is natural to ask, for what kind of operators (and domains), the A∞ condition holds. It turns out that some additional assumption on the operator is required: examples in [CFK81] show that the elliptic measure may be singular to the surface measure for certain operators with merely bounded and measurable coefficients. In fact, this may happen even 5 in smooth domains, which implies that the Lp Dirichlet problem is not solvable for any p. Our second main contribution is that we show in the upper half space Rn+1 + = {(x, t) ∈ Rn × (0, ∞)} with n ≥ 3, under the additional assumption that A is indepen- dent of t, that the elliptic measure associated to L belongs to A∞ (dx) for L = − div A(x)∇. The precise statement and proof are in Chapter 10. This result is new and extends, to elliptic operators with a BMO anti-symmetric part, the result for elliptic operators with bounded, t-independent coefficients obtained in [HKMP15]. We remark that in [HKMP15], the result holds in Rn+1 + for n ≥ 2. And the case when n = 1 was proved previously in [KKPT00] for operators with bounded, t-independent coefficients. We also remark that Theorem 10.0.1 is sharp, in the sense that A∞ is the best possible conclusion. In fact, it was already shown in [KKPT00] that in R2+ , one cannot make precise the exponent p for which one has solvability of the Lp Dirichlet problem for elliptic operators with L∞ , t-independent coefficients. The way we prove the A∞ condition is to estabilish the following Carleson measure es- timate. We note that this characterization of the A∞ condition was obtained in [KKPT16] for elliptic operators with bounded coefficients, and is verified in section 6.2 for elliptic operators with BMO anti-symmetric part. Lemma 1.0.1 (Theorem 6.2.2, [KKPT16] Corollary 3.2). Assume A satisfies (10.0.1) and (10.0.2) in Rn+1 + , and define L as (10.0.3). Assume that there is some uniform constant A < ∞ such that for all Borel sets H ⊂ Rn , the weak solution u to the Dirichlet problem  in Rn+1  Lu = 0  + on ∂Rn+1  u = χH  + satisfies the following Carleson bound ˆ l(Q) ˆ 1 sup |∇u(x, t)|2 t dx ≤ A. (1.0.9) Q⊂Rn |Q| 0 Q Here l(Q) denotes the length of the cube Q. Then ω ∈ A∞ (dx). 6 Our proof relies on some Lp estimates for square functions involving the heat semi- group associated to L = − div A∇, where the anti-symmetric part of A is in BMO. These estimates use ingredients from the Kato problem. Chapter 7 - 9 are devoted to addressing these issues. In chapter 7, we derive the Gaussian estimates for the kernel of the semigroup e−tL as well as the gradient of the kernel; in chapter 8, we obtain Lp estimates for the semi- group and the square root L1/2 . The estimates for the square function and non-tangential function involving the semigroup are obtained in chapter 9. This second main result (Theorem 10.0.1) is a joint work of the author with S. Hofmann, S. Mayboroda, and J. Pipher. 1.1 More history For elliptic operators with bounded coefficients, the investigations on the boundary behav- ior of the solution to Lu = 0 in a domain Ω ⊂ Rn were initiated in the papers of De Giorgi, Nash, and Moser, where sharp regularity of solutions to divergence form elliptic equations with bounded measurable coefficients were obtained. In particular, weak solutions were shown be H¨ older continuous, with a parameter depending only upon ellipticity, in domains satisfying certain exterior cone condition. The boundary regularity can be obtained essen- tially using Moser iteration, see [GT01] for example. The study of regular points, as well as the solvability of the continuous Dirichlet problem for such operators was established in [LSW63]. Then the fundamental work of [HW68] and [CFMS81] showed the existence of non-tangential limits of solutions, paving the way for the study of Lp Dirichlet problem and other boundary value problems for divergence form elliptic operators with bounded and measurable coefficients. In [KKPT00], it was observed that the results of [CFMS81] are valid without the symmetry assumption on A. The book [Ken94] by C. Kenig contains some references to the early work in this literature. For a divergence-form elliptic operators with symmetric, bounded, t-independent (inde- 7 pendent of the transverse variable) coefficients, one can obtain solvability of the L2 Dirichlet problem using an identity of Rellich type [JK81]. Motivated by the study of non-symmetric elliptic equations, Kenig, Koch, Pipher and Toro proved for elliptic operators defined in R2 , with t-independent and bounded coefficients, that the elliptic measure belongs to A∞ (dx), and that the result is sharp [KKPT00]. The result in higher dimensions was proved by Hof- mann, Kenig, Mayboroda and Pipher [HKMP15] using results established in the solution to the Kato problem [HLM02],[AHL+ 02], along with an L-adapted Hodge decomposition of the coefficient vector ~c = (An+1,j )1≤j≤n , and an L-adapted variant of the Dahlberg-Kenig- Stein pullback mapping. They proved the A∞ -property by establishing the Lp bounds for the square function in terms of the non-tangential maximal function of the solution and vice versa. By the result of [KKPT16], one can simplify the argument by showing the Carleson measure estimate we stated earlier. Later, a more streamlined argument was given in [AEN16], which proves the analogous result for the parabolic measure on Rn+1 + . Namely, the authors rely on the reduction to the Carleson measure estimate to obtain the A∞ condition for parabolic operators dependent on time but not dependent on the transverse variable (denoted by λ in their paper, and by t in [HKMP15] and this thesis). Moreover, they avoid the Dahlberg-Kenig-Stein pullback mapping utilized in [HKMP15]. We note that we benefit from the proof of [AEN16] in that the change of variables argument introduced in [HKMP15] does not work for us. This is mainly because that it destroys the structure of the anti-symmetric part of the matrix and thus the BMO coefficients can no longer be controlled. In the recent paper [EH18], the Kato problem was solved for elliptic operators having a BMO anti-symmetric part. Previously, the Kato conjecture was proved for operators having the “Gaussian property” ([HLM02]) and for elliptic operators in divergence form with complex, bounded coefficients ([AHL+ 02]). We note that the Gaussian property for elliptic operators having a BMO anti-symmetric part was obtained in [QX18], and thus one has the Kato estimate 1/2 L f . k∇f kL2 (1.1.1) L2 8 by the arguments of [HLM02]. We present a different proof for the Gaussian property in chapter 7 from the one given in [QX18], and the ideas of the proof are mainly from [Aus96]. The advantage of this method is that it also gives the L2 bound for the gradient of the semigroup, which is essential in estimating the square function involving the semigroup. The Lp estimates for square function involving the semigourp e−tL are essentially from [Aus07], for elliptic operators L with complex, L∞ coefficients in Rn . We note that these estimates exploit not only the L2 Kato estimates (1.1.1), but also the Lp Kato estimate, which are also obtained in [Aus07] for elliptic operators with L∞ coefficients. Finally, we give a brief history of the study of operators having a BMO anti-symmetric ˇ ak, and Zlatoˇs [SSSZ12] part. In an important development, Seregin, Silvestre, Sver´ ˇ dis- covered that a large portion of Moser’s arguments works for this class of operators. In particular, they carried out the Moser iteration, and proved the Liouville theorem and Harnack inequality for solutions to divergence form parabolic operators, including the el- liptic case. Thus, the interior regularity theory of De Giorgi, Nash, and Moser carries over to this setting. In [QX18], the authors showed the existence of the fundamental solution of the parabolic operator L − ∂t , and derived Gaussian estimate for the fundamental solution. H. Dong and S. Kim [DK18] generalized the result to second-order parabolic systems, un- der the assumption that weak solutions of the system satisfy a certain local boundedness estimate. In [EH18], Escauriaza and Hofmann showed that for elliptic operator L having √ a BMO anti-symmetric part, the domain of the square root L contains W 1,2 (Rn ), and ˙ 1,2 (Rn ) without using the Gaussian property of the operator. that (1.1.1) holds over W 1.2 Notation We assume that Ω is an NTA domain in Rn , n ≥ 3, throughout chapter 3 to chapter 6 unless otherwise stated. 9 We use As for the symmetric part of the matrix A and Aa for the anti-symmetric part of A in section 3.3, and throughout chapter 7 to chapter 10. In the rest of the paper, we use a for the symmetric part of the matrix A and b for the anti-symmetric part of A. The meaning should be also clear from the context. In most cases, δ(X) denotes dist(X, ∂Ω). In section 10.3, δ(x) is the distance from x to the set F . f −1,2 the space of bounded semilinear functionals on W 1,2 (Rn ). By We denote by W semilinear we mean, hf, αu + βviW ¯ hf, uiW f −1,2 ,W 1,2 = α ¯ f −1,2 ,W 1,2 + βhf, viW f −1,2 ,W 1,2 , f −1,2 , α for all α, β ∈ C, u, v ∈ W 1,2 (Rn ). Here f ∈ W ¯ is the complex conjugate of α. CHAPTER Two Preliminary 2.1 Some facts about general domains Definition 2.1.1. [Corkscrew condition] A domain Ω ⊂ Rn is said to satisfy the interior corkscrew condition(resp. exterior Corkscrew condition) if there exists M > 1 and R > 0 such that for any Q ∈ ∂Ω and any 0 < r < R, there exists a corkscrew point (or non- tangential point) A = Ar (Q) ∈ Ω (resp. A ∈ Ωc ) such that . r |A − Q| < r and δ(A) = dist(A, ∂Ω) > . (2.1.1) M Definition 2.1.2. [Harnack Chain] A domain Ω ⊂ Rn is said to satisfy the Harnack chain condition if there are universal constants m > 1 and m0 > 0 such that for every pair of points A and A0 in Ω satisfying . |A − A0 | l= > 1, (2.1.2) min{δ(A), δ(A0 )} there is a chain of open Harnack balls B1 , B2 ,. . . , BN in Ω that connects A to A0 . Namely, 10 11 A ∈ B1 , A0 ∈ BN , Bi ∩ Bi+1 6= ∅ and m−1 diam(Bi ) ≤ δ(Bi ) ≤ m diam(Bi ) ∀ i, (2.1.3) and the number of balls N ≤ m0 log2 l. Definition 2.1.3 (NTA domain). We say a domain Ω ⊂ Rn is an NTA domain if it satisfies the interior and exterior Corkscrew condition, and the Harnack chain condition. Definition 2.1.4 (1-sided NTA domain). If Ω ⊂ Rn satisfies only the interior Corkscrew condition, and the Harnack chain condition, then we say that it is a 1-sided NTA domain. 1-sided NTA domains are also called uniform domains. We have the following extension results due to P. Jones: Lemma 2.1.1 ([Jon80]). A domain is uniform if and only if every function b ∈ BM O(Ω) admits an extension to some ˜b ∈ BM O(Rn ). That is, there exists ˜b ∈ BM O(Rn ) such that ˜b|Ω = b and ˜b ≤ C kbkBM O(Ω) , (2.1.4) BM O(Rn ) where the constant C depends only on the domain and dimension. Lemma 2.1.2 ([Jon81]). If the domain Ω is locally uniform (or a (, δ) domain, by the terminology of P. Jones), then there exists a bounded linear extension operator E : W 1,p (Ω) → W 1,p (Rn ) with E|Ω g = g for all g ∈ W 1,p (Ω). We shall also need an extension result that is independent of the domain. Lemma 2.1.3 ([Ste70] p.174 Theorem 3). Let F be any closed set, and let Lip(γ, F ) denote the set consisting of those f defined on F such that |f (x)| ≤ M and |f (x) − f (y)| ≤ M |x − y|γ , ∀ x, y ∈ F. 12 Then there is a linear extension operator E that maps Lip(γ, F ) continuously into Lip(γ, Rn ), if 0 < γ ≤ 1. And the norm of this mapping has a bound independent of the set F . 2.2 The Hardy norm Definition 2.2.1. We say f ∈ L1 (Rn ) is in the real Hardy space H1 (Rn ) if . kf kH1 (Rn ) = sup |ht ∗ f | < ∞, t>0 L1 (Rn ) 1 x , and h is any smooth non-negative function on Rn , with supp h ⊂  where ht (x) = tn h t ´ B1 (0) such that Rn h(x)dx = 1. C. Fefferman [Fef71] showed that BMO is the dual of the Hardy space H1 . We have the following div-curl lemma, also known as the compensated compactness, due to [CLMS93]: 0 1 1 Lemma 2.2.1. Let E ∈ Lp (Rn )n , B ∈ Lp (Rn )n with 1 < p < ∞, p + p0 = 1 and divE = 0, curlB = 0 in D 0 (Rn ). Then E · B ∈ H1 (Rn ), and kE · BkH1 (Rn ) ≤ CkEkLp (Rn ) kBkLp0 (Rn ) . ˇ It was observed in [SSSZ12], that one can obtain the following estimate using the div- curl lemma: ˙ 1,p0 (Rn ). Then for any ˙ 1,p (Rn ), v ∈ W Proposition 2.2.1. Let 1 < p < ∞. Let u ∈ W 13 1 ≤ i, j ≤ n, ∂j u∂i v − ∂i u∂j v ∈ H1 (Rn ) with k∂j u∂i v − ∂i u∂j vkH1 (Rn ) . k∇ukLp k∇vkLp0 , (2.2.1) where the implicit constant depends only on dimension. Proof. For i 6= j, set B = ∇v and E = (0, . . . , ∂j u, 0, . . . , −∂i u, 0, . . . ), whose ith com- 0 ponent is ∂j u and jth component is −∂i u. Observe that E ∈ Lp (Rn )n , B ∈ Lp (Rn )n , div E = 0 and curl B = 0 in D 0 (Rn ). Then the div-curl lemma 2.2.1 gives that E · B ∈ H1 (Rn ), and kE · BkH1 (Rn ) ≤ CkEkLp (Rn ) kBkLp0 (Rn ) . That is, for i 6= j, ∂j u∂i v − ∂i u∂j v ∈ H1 (Rn ) and k∂j u∂i v − ∂i u∂j vkH1 (Rn ) ≤ C k∇ukLp (Rn ) k∇vkLp0 (Rn ) . ˙ 1,p0 (Rn ). Then for any ˙ 1,p (Rn ), v ∈ W Proposition 2.2.2. Let 1 < p < ∞. Let u ∈ W 1 ≤ i ≤ n, ∂i (uv) ∈ H1 (Rn ) with k∂i (uv)kH1 (Rn ) . kukLp k∇vkLp0 + k∇ukLp kvkLp0 , (2.2.2) where the implicit constant depends only on p and dimension. ´ Proof. Let h be a smooth nonnegative compactly supported mollifier with Rn h(x)dx = 1, 14 supp h ⊂ B1 (0). And let ht (x) = t−n h( xt ). Then we have ˆ ˆ ht ∗ ∂i (uv)(x) = ht (x − y)∂i (uv)(y)dy = − ∂i ht (x − y)u(y)v(y)dy Rn Bt (x) ˆ   1 x−y  = n+1 ∂ i h u(y) v(y) − (v) Bt (x) dy Bt (x) t t ˆ   1 x−y . + n+1 i ∂h u(y)(v)Bt (x) dy = I1 + I2 . Bt (x) t t For I1 , we have ˆ v(y) − (v)Bt (x) 1 |I1 | . n |u(y)| dy t Bt (x) t !1/α α0 !1/α0 − v(y) (v)Bt (x) |u|α dy . dy Bt (x) Bt (x) t !1/α !1/β 1  1/β . |u|α dy |∇v|β dy . (M |u|α ) α (x) M |∇v|β (x), Bt (x) Bt (x) 1 1 where α ∈ [1, p), α= 1 + n1 , and M (f ) is the Hardy-Littlewood maximal function of + β ´ f . For I2 , note that I2 = Bt (x) ht (x − y)∂i u(y)(v)Bt (x) dy. So ˆ 1 |I2 | . n |∂i u| (v)Bt (x) dy . |∇u| dy |v| dy t Bt (x) Bt (x) Bt (x) . M (|∇u|)(x)M (v)(x). Combining the estimates for I1 and I2 , and using H¨older inequality, we have ˆ  1/β 1 sup |ht ∗ ∂i (uv)(x)| dx . (M |u|α ) α β M |∇v| + kM (|∇u|)kLp kM (v)kLp0 Rn t>0 Lp Lp0 . kukLp k∇vkLp0 + k∇ukLp kvkLp0 , where in the last inequality we have used that 1 ≤ α < p and 1 < β < p0 . We shall also use the following Hardy norm estimate. Proposition 2.2.3. Let u, v ∈ W 1,2 (Rn ), and ϕ be a Lipschitz function in Rn . Then for 15 any 1 ≤ i, j ≤ n, ∂j (uv)∂i ϕ − ∂i (uv)∂j ϕ ∈ H1 (Rn ) with k∂j (uv)∂i ϕ − ∂i (uv)∂j ϕkH1 (Rn ) . ku |∇ϕ|kL2 k∇vkL2 + kvkL2 k|∇u| |∇ϕ|kL2 , or   k∂j (uv)∂i ϕ − ∂i (uv)∂j ϕkH1 (Rn ) . k∇ϕkL∞ (Rn ) kukL2 k∇vkL2 + kvkL2 k∇ukL2 , (2.2.3) where the implicit constant depends only on dimension. ~ = (0, . . . , 0, ∂j ϕ, 0, . . . , 0, −∂i ϕ, 0, . . . , 0). Then Proof. We can assume ϕ ∈ C 2 (Rn ). Set Φ ~ = 0 and div Φ ~ · ∇(uv) = div(Φuv). ∂i (uv)∂j ϕ − ∂j (uv)∂i ϕ = Φ ~ (2.2.4) ´ Let h be a smooth nonnegative compactly supported mollifier with Rn h(x)dx = 1. ht (x) = t−n h( xt ). We compute ˆ ~ ht ∗ div(Φuv)(x) =− ~ ∇y ht (x − y) · Φ(y)u(y)v(y)dy Bt (x) ˆ =− ~ ∇y ht (x − y) · Φ(y)u(y)(v(y) − (v)Bt (x) )dy Bt (x) ˆ − ~ div(ht Φ)u(y)(v)Bt (x) dy Bt (x) ˆ =− ~ ∇y ht (x − y) · Φ(y)u(y)(v(y) − (v)Bt (x) )dy Bt (x) ˆ + ~ · ∇u(y)(v)B (x) dy ht (x − y)Φ t Bt (x) . = I1 + I2 . (2.2.5) 16 ˆ 1 |I1 | . |∇ϕ| |u| v − (v)B (x) tn+1 Bt (x) t  1/α  v − (v)B (x) 0  α10 α . |u∇ϕ| ( t )α Bt (x) Bt (x) t  1/α  1/β . |u∇ϕ|α |∇v|β . M 1/α (|u∇ϕ|α )(x)M 1/β (|∇v|β )(x), Bt (x) Bt (x) 1 1 1 1 where 1 < α, β < 2, α + α0 = 1, α + β = 1 + n1 . And   |I2 | . |∇ϕ| |∇u| (v)Bt (x) . M (|∇ϕ| |∇u|)(x)M (v)(x). Bt (x) So ~ α β ht ∗ div(Φuv)(x) . M 1/α (|u∇ϕ| )(x)M 1/β (|∇v| )(x) + M (|∇ϕ| |∇u|)(x)M (v)(x) ⇒ ˆ ~ sup ht ∗ div(Φuv)(x) dx . ku |∇ϕ|kL2 k∇vkL2 + kvkL2 k|∇u| |∇ϕ|kL2 . Rn t>0 By (2.2.4) and the definition of Hardy norm, we complete the proof. 2.3 Weak solutions and Drichlet problem Let Ω be an NTA domain in Rn . Let L = − div A∇, with A satisfying (1.0.1) and (1.0.3). We consider the following: 1. the notion of weak solution of Lu = 0 in Ω, (2.3.1) 17 2. the solvability of the classical Dirichlet problem   Lu = 0  in Ω, (2.3.2) u − g ∈ W01,2 (Ω) W 1,2 (Ω),   where g ∈ and 3. the solvability of the continuous Dirichlet problem   Lu = 0  in Ω, (2.3.3)  u = g  on ∂Ω where g ∈ C(∂Ω). Let B[ , ] be the bilinear form associated with L, i.e. ˆ B[u, v] = (a∇u · ∇v + b∇u · ∇v)dx. Ω By the antisymmetry of b, we write ˆ ˆ 1 b∇u · ∇ϕ = bij (x)(∂j u∂i ϕ − ∂i u∂j ϕ)dx (2.3.4) Ω 2 Ω Applying Proposition 2.2.1, we have ˆ ˙ 1,2 (Rn ), b∇u · ∇vdx . Λ0 k∇ukL2 (Rn ) k∇vkL2 (Rn ) , ∀ u, v ∈ W R n ˙ 1,2 (Rn ) is the homogeneous Sobolev space. We claim that the bilinear form is also where W bounded in Ω: |B[u, v]| ≤ C k∇ukL2 (Ω) k∇vkL2 (Ω) (2.3.5) for all u ∈ W01,2 (Ω), v ∈ W01,2 (Ω), and |B[g, v]| ≤ C kgkW 1,2 (Ω) k∇vkL2 (Ω) (2.3.6) 18 for all g ∈ W 1,2 (Ω), v ∈ W01,2 (Ω) To see (2.3.5), it suffices to show ˆ b∇u · ∇v ≤ C k∇uk 2 k∇vk 2 . (2.3.7) L (Ω) L (Ω) Ω ˜ and v˜ be the zero extension to Rn of u and v, respectively. For i 6= j, set B = ∇˜ Let u v ˜, 0, . . . , −∂i u and E = (0, . . . , ∂j u ˜, 0, . . . ), whose ith component is ∂j u ˜ and jth component is −∂i u ˜. Then the div-curl lemma of [CLMS93] gives that for i 6= j, ∂j u ˜∂j v˜ ∈ H1 (Rn ) ˜∂i v˜ −∂i u and k∂j u ˜∂i v˜ − ∂i u ˜∂j v˜kH1 (Rn ) ≤ C k∇˜ ukL2 (Rn ) k∇˜ v kL2 (Rn ) . (2.3.8) Letting ˜b ∈ BM O(Rn ) be the extension of b, we have ˆ ˆ ˜bij (x)(∂j u bij (x)(∂j u∂i v − ∂i u∂j v)dx = ˜∂i v˜ − ∂i u ˜∂j v˜)dx Ω Rn ≤ C ˜b k∇˜ ukL2 (Rn ) k∇˜v kL2 (Rn ) n BM O(R ) ≤ C kbkBM O(Ω) k∇ukL2 (Ω) k∇vkL2 (Ω) , where we have used (2.1.4) in the last inequality. This proves (2.3.7). For (2.3.6), we apply Lemma 2.1.2 to g. That is, find the extension g˜ ∈ W 1,2 (Rn ) such that g˜|Ω = g and k∇˜ g kL2 (Rn ) . kgkW 1,2 (Ω) , with the implicit constant only depends on the domain and dimension. Then by the same argument for (2.3.7) we obtain (2.3.6). The upper bound (2.3.5) of the bilinear form enables us to define weak solutions of problem (2.3.1)–(2.3.3) in the following sense. 1,2 Definition 2.3.1. We say that a function u ∈ Wloc (Ω) is a weak solution of (2.3.1) if ˆ (a∇u · ∇ϕ + b∇u · ∇ϕ)dx = 0 (2.3.9) Ω for all ϕ ∈ W 1,2 (Ω) with supp ϕ ⊂⊂ Ω. 19 1,2 We say that u ∈ Wloc (Ω) is a weak subsolution(supersolution) of (2.3.1) if ˆ (a∇u · ∇ϕ + b∇u · ∇ϕ)dx ≤ (≥)0 (2.3.10) Ω for all ϕ ∈ W 1,2 (Ω) with supp ϕ ⊂⊂ Ω and ϕ ≥ 0 a.e. in Ω. Definition 2.3.2. We say a function u ∈ W 1,2 (Ω) is a weak solution of (2.3.2) if ˆ (a∇u · ∇ϕ + b∇u · ∇ϕ)dx = 0 ∀ ϕ ∈ W01,2 (Ω), Ω and u − g ∈ W01,2 (Ω). (2.3.11) 1,2 Definition 2.3.3. We say a function u ∈ Wloc (Ω) ∩ C(Ω) is a weak solution of (2.3.3) if (2.3.9) holds for all ϕ ∈ W 1,2 (Ω) with supp ϕ ⊂⊂ Ω, and u=g on ∂Ω. The ellipticity of A (1.0.4) immediately gives B[u, u] ≥ λ0 k∇uk2L2 (Ω) ∀u ∈ W01,2 (Ω). (2.3.12) Therefore, together with (2.3.5), we can apply the Lax-Milgram theorem to get Theorem 2.3.1. Given h ∈ W01,2 (Ω)∗ = W −1,2 (Ω), there exists a unique u ∈ W01,2 (Ω) such that Lu = h, in the sense that ˆ (a∇u · ∇ϕ + b∇u · ∇ϕ) = hϕ, hi, ∀ ϕ ∈ W01,2 (Ω). Ω Writing w = u − g ∈ W01,2 (Ω) in the classical Dirichlet problem (2.3.2), the estimate (2.3.6) implies that Lg ∈ W −1,2 (Ω). So we can apply the theorem to conclude that the 20 classical Dirichlet problem (2.3.2) is uniquely solvable. 2.4 Some useful results in PDE We shall frequently use three results from [Gia83]. We include them here for reader’s convenience. The first one is useful in proving reverse H¨older type inequalities. Lemma 2.4.1 ([Gia83] Chapter V Proposition 1.1). Let Q be a cube in Rn . Let g ∈ Lq (Q), q > 1, and f ∈ Ls (Q), s > q, be two nonnegative functions. Suppose !q g q dx ≤ b gdx + f q dx + θ g q dx QR (x0 ) Q2R (x0 ) Q2R (x0 ) Q2R (x0 ) 1 for each x0 ∈ Q and each R < min 2 dist(x 0 , ∂Q), R0 , where R0 , b, θ are constants with b > 1, R0 > 0, 0 ≤ θ < 1. Then g ∈ Lploc (Q) for p ∈ [q, q + ) and  1/p n 1/q  1/p o p q p g dx ≤c g dx + f dx QR Q2R Q2R for Q2R ⊂ Q, R < R0 , where c and  are positive constants depending only on b, θ, q, n (and s). In application, if one can show !1/r 2 2r |∇u| dx ≤ b |∇u| dx + |f |2 dx + θ |∇u|2 dx QR (x0 ) Q2R (x0 ) Q2R (x0 ) Q2R (x0 ) 1 for each x0 ∈ Q and each R < min 2 dist(x 0 , ∂Q), R0 , where b > 1, r ∈ (0, 1) and θ ∈ [0, 1) are some constants. Then by letting g = |∇u|2r , q = 1 r and f be |f |2r in Lemma 21 2.4.1, one obtains that |∇u| ∈ Lploc (Q) for p ∈ [2, 2 + ) and  1/p n 1/2  1/p o p 2 p |∇u| dx ≤c |∇u| dx + |f | dx QR Q2R Q2R for Q2R ⊂ Q, R < R0 , where c and  are positive constants depending only on b, θ, r and n. Lemma 2.4.2 ([Gia83] Chapter V Lemma 3.1). Let f (t) be a nonnegative bounded function defined in [r0 , r1 ], r0 ≥ 0. Suppose that for r0 ≤ t < s ≤ r1 we have f (t) ≤ A(s − t)−α + B + θf (s)  where A, B, α, θ are nonnegative constants with 0 ≤ θ < 1. Then for all r0 ≤ ρ < R ≤ r1 we have f (ρ) ≤ c A(R − ρ)−α + B  where c is a constant depending on α and θ. Lemma 2.4.3 ([Gia83] Chapter III Lemma 2.1). Let φ(t) be a nonnegative and nonde- creasing function. Suppose that  ρ α  φ(ρ) ≤ A + ε φ(R) + BRβ R for all ρ ≤ R ≤ R0 , with A, α, β nonnegative constants, β < α. Then there exists a constant ε0 = ε0 (A, α, β) such that if ε < ε0 , for all ρ ≤ R ≤ R0 we have    ρ β β φ(ρ) ≤ c φ(R) + Bρ R where c = c(α, β, A). 22 2.5 Theory of weights Let ν be any doubling measure on, say ∂Ω. And let µ be another non-negative measure on ∂Ω. Definition 2.5.1. For q ∈ (1, ∞) we say that µ belongs to the reverse H¨ older class Bq (dν) dω if µ << ν, and k = dν ∈ Lq (dν), and verifies  1/q q k dν ≤C kdν ∆ ∆ uniformly for every surface balls ∆ ⊂ ∂Ω. Definition 2.5.2. We say µ belongs to the A∞ class with respect to ν if, given ε > 0, there exists δ = δ(ε) > 0 such that if E ⊂ ∆, where ∆ ⊂ ∂Ω is any surface ball, then ν(E) µ(E) <δ ⇒ < ε. ν(∆) µ(∆) The main properties of weights are summarized in the following lemma. They originates in the work of [Muc72] [Muc74]. Most of the following properties were proved in [CF74] for doubling measures in Rn . The weighted maximal theorem in spaces of homogeneous type is due to A. Calder´ on [Cal76]. Lemma 2.5.1. 1. If µ ∈ A∞ (dν), then µ << ν. 2. If µ ∈ A∞ (dν), then ν ∈ A∞ (dµ). 3. µ ∈ A∞ (dν) if and only if there exist ε, δ ∈ (0, 1) such that µ(E) ν(E) <δ ⇒ < ε. µ(∆) ν(∆) 23 4. µ ∈ A∞ (dν) if and only if there exist C, θ > 0 such that ∀ E ⊂ ∆,  θ µ(E) ν(E) ≤C . µ(∆) ν(∆) S 5. A∞ (dν) = q>1 Bq (dν). 6. If µ ∈ Bq (dν), with q > 1, then there exists ε > 0 such that µ ∈ Bq+ε (dν). 7. µ ∈ Bq (dν) if and only if kMµ f kLq0 (dν) ≤ C kf kLq0 (dν) , . ´ q where Mµ (f )(Q) = supQ∈∆ 1 µ(∆) ∆ |f | dµ, q0 = q−1 . CHAPTER Three Estimates for the weak solution 3.1 Interior estimates for the solution ˇ Almost all the lemmas in this section appeared in [SSSZ12] in the context of parabolic equations, while we specialize to the elliptic case. We include the proofs not merely for the sake of completeness, but because some of the arguments are completely different and lend themselves to the the development of the boundary regularity. 1,2 Lemma 3.1.1 (Caccioppoli inequality). Let u ∈ Wloc (Ω) be a weak solution of (2.3.1). Let BR = BR (x) be a ball centered at x with radius R such that BR ⊂ Ω. Then for any 0 < σ < 1 and c ∈ R, ˆ ˆ ˆ 2 2 1 2 1 |∇u| ϕ . |u − c| + |b − (b)R | |∇uϕ| |u − c| , BR (1 − σ)2 R2 supp ∇ϕ (1 − σ)R supp ∇ϕ (3.1.1) where ϕ ∈ C0∞ (BR ) is nonnegative, satisfying ϕ = 1 on BσR , supp ϕ ⊂ BR , and |∇ϕ| ≤ C (1−σ)R . The implicit constant depends on n and λ0 . 24 25 Proof. Take φ = (u − c)ϕ2 as test function in (2.3.9) to get ˆ ˆ 2 (a + b)∇u · ∇uϕ + (a + b)∇u · ∇ϕ2 (u − c) = 0. (3.1.2) BR BR Using ellipticity of a and anti-symmetry of b, we have ˆ ˆ ˆ ˆ 2 2 2 2 λ0 |∇u| ϕ + a∇u·∇ϕ (u−c)+ (b−(b)R )∇u·∇ϕ (u−c)+ (b)R ∇u·∇ϕ2 (u−c) ≤ 0. BR BR BR BR Since (b)R is an anti-symmetric constant matrix and ϕ is smooth, the divergence theorem gives ˆ ˆ 2 1 (b)R ∇u · ∇ϕ (u − c) = (b)R ∇(u − c)2 · ∇ϕ2 BR 2 BR ˆ ˆ 1  2  2 1 ~ (u − c)2 dσ = div (b)R ∇ϕ (u − c) − ∇ϕ2 · N 2 BR 2 ∂BR = 0. (3.1.3) So ˆ λ0 |∇u|2 ϕ2 BR ˆ ˆ ˆ −1 2 1/2 2 2 1/2 ≤ 2λ0 ( |∇uϕ| ) ( |u − c| |∇ϕ| ) +2 |b − (b)R | |∇uϕ| |u − c| |∇ϕ| . BR BR BR Then the desired result follows from Young’s inequality. The following Caccioppoli type inequality will be used frequently. n Corollary 3.1.1. Let u, BR be as in Lemma 3.1.1. Then for any 1 < s ≤ n−1 , ˆ ˆ n 2s 2−s 2 |∇u| ≤ C(s, n, λ0 )(1 + Λ20 )R2( s0 −1) ( |ˆ u| 2−s ) s , BR/2 BR 26 ˆ = u − (u)R and s0 = where u s s−1 . Equivalently, 2s 2−s |∇u|2 ≤ C(s, n, λ0 )(1 + Λ20 )R−2 ( |ˆ u| 2−s ) s BR/2 BR Proof. Taking c = (u)R and σ = 1/2 in Lemma 3.1.1, we get ˆ ˆ ˆ 2 2 −2 2 −1 |∇u| ϕ ≤ CR |ˆ u| + CR |b − (b)R | |∇uϕ| |ˆ u| BR BR BR . = I1 + I2 . By H¨ older’s inequality, ˆ 2n 2s 2−s −2 I1 ≤ CR s0 ( |ˆ u| 2−s ) s . BR For, I2 , we use H¨ older’s inequality twice and Young’s inequality, as well as the self ffl improving property of functions in BM O: BR |b − (b)R |p ≤ Cp kbkpBM O ∀ p ∈ [1, ∞). We have ˆ n −1 s0 1 1 I2 ≤ CR ( s0 |b − (b)R | ) ( |∇uϕ|s |ˆ u|s ) s s0 BR B ˆ ˆR n 2s 2−s ≤ Cs R s0 −1 Λ0 ( |∇uϕ|2 )1/2 ( |ˆ u| 2−s ) 2s BR BR ˆ ˆ 1 2 n 2( s0 −1) 2 2s 2−s ≤ |∇uϕ| + Cs R Λ0 ( |ˆ u| 2−s ) s . 2 BR BR Combining these two estimates we proved the corollary. Note that by Poincar´e-Sobolev n inequality, kˆ uk 2s . kukW 1,2 for 1 < s ≤ n−1 . L 2−s Remark 3.1.1. The main difference from the usual Caccioppoli’s inequality (3.1.15) is that at this point, the power of u ˆ on the right-hand side is greater than 2. The usual Caccioppoli estimate will hold once we know that  1/p  1/2 |u|p . |u|2 ∀p > 2. BR B2R 27 The recent paper [DK18] contains a somewhat simpler approach to (3.1.15), which is proven there for second-order parabolic systems. Lemma 3.1.2 (Reverse H¨ older inequality of the gradiant). Let u, BR be as in Lemma 3.1.1. Then there exist p > 2 and C depending only on n, λ0 and Λ0 such that 1 1 ( |∇u|p ) p ≤ C( |∇u|2 ) 2 . (3.1.4) BR/2 BR 2s n Proof. Observe that the function s 7→ 2−s is strictly increasing on (1, n−1 ) with range 2n rn 2n (2, n−2 ), and that the function r 7→ n−r is strictly increasing on ( n+2 , 2) with range 2n n 2n 2s rn (2, n−2 ). So for fixed s ∈ (1, n−1 ), there exists a unique r ∈ ( n+2 , 2) that solves 2−s = n−r . Then by Corollary 3.1.1 and Poincar´e-Sobolev inequality, we get ˆ 2 |∇u|2 ≤ C(s, n, λ0 )(1 + kbk2BM O )Rn ( |∇u|r ) r , BR/2 BR that is 1 1 ( |∇u|2 ) 2 ≤ C(s, n, λ0 )(1 + Λ20 )1/2 ( |∇u|r ) r . BR/2 BR Then the result follows from Lemma 2.4.1. ˇ In [SSSZ12], interior regularity of the solution is first established for smooth coefficients case and in the general case via approximation arguments. The following approach gives estimates for solutions directly in the non-smooth case, which is advantageous for boundary estimates. Let us first point out a simple fact: 1,2 Lemma 3.1.3. Let u ∈ Wloc (Ω) be a weak solution of (2.3.1). Let u+ = max{u, 0}, u− = max{−u, 0}. Then u+ and u− are subsolutions of (2.3.1), i.e. ˆ (a∇u± · ∇ϕ + b∇u± · ∇ϕ)dx ≤ 0 (3.1.5) Ω for all ϕ ∈ W 1,2 (Ω) with supp ϕ ⊂⊂ Ω and ϕ ≥ 0 a.e. in Ω. 28 Proof. First consider ϕ ∈ C0∞ (Ω). Consider u+ , and for k > 1, define      0, t ≤ 0,  ηk (t) =  kt, 0 < t ≤ k1 ,    1, t > k1 .  Take ηk (u)ϕ as test function in (2.3.9) and then let k → ∞: ˆ   0= a∇u · ∇(ηk (u)ϕ) + b∇u · ∇(ηk (u)ϕ) dx ˆΩ ˆ ˆ   ≥ a∇u · ∇ϕηk (u) + b∇u · ∇ϕηk (u)dx → a∇u+ · ∇ϕ + b∇u+ · ∇ϕ dx. Ω Ω Ω For ϕ ∈ W 1,2 (Ω) with compact support in Ω, the energy estimates (2.3.5) permits approximation by smooth functions. The argument for u− is similar. 2s n 2n In the rest of the section, let q = 2−s , 1 12 , we have  1 sup u± ≤ C(n, λ0 , Λ0 , q, k0 ) (u± )kq kq . (3.1.6) BR B2R Therefore, 1 sup |u| ≤ C(n, λ0 , Λ0 , q, k0 )( |u|kq ) kq . (3.1.7) BR B2R Proof. We only show (3.1.6) for u+ . 29 1 ∀ > 0, N >> 1 and β ≥ k0 , pick 2 < k1 < min{1, k0 }. Define   tβ ,  t ∈ [, N ], H,N (t) =  Nβ + β β−k1 (tk1 k1 N − N k1 ), t > N.   βtβ−1 , t ∈ (, N ),   0 ⇒ H,N (t) =  βN β−k1 tk1 −1 ,  t > N. Set ˆ w 0 G,N (w) = |H,N (t)|2 dt, w ≥ .  For w ≥ , it is easy to check that H(w) ≤ wβ , (3.1.8) wH 0 (w) ≤ βwβ , (3.1.9) and 1 G(w) ≤ wG0 (w). (3.1.10) 2k1 − 1 Here and in the sequel we omit the subscripts in G,N and H,N . Set u (x) = u+ (x) + . Choose ϕ = G(u )η 2 as test function in (3.1.5), where η = 1 on Br0 , 0 ≤ η ≤ 1, supp η ⊂ Br and |∇η| ≤ C r−r0 for R ≤ r0 < r ≤ 2R. Note that ∀β ≥ k0 , ϕ ∈ W01,2 (Ω), which is important for Moser iteration. In fact, for any β ≥ k0 , H (u ) 2 ∈ L∞ (Ω) 0 ⇒ 30 1 H (u ) 2 u ∈ L2 (Ω). 0 ∇G(u ) ∈ L2 (Ω), |G(u )| ≤ 2k1 − 1 We compute (making use of (3.1.9) and (3.1.10)) ˆ ˆ ˆ a∇u · ∇ϕ = a∇H(u ) · ∇H(u )η 2 + 2 a∇u · ∇ηG(u )η Br Br Br ˆ ˆ 2λ−1 ≥ λ0 |∇H(u )|2 η 2 − 0 |∇u | |∇η| G0 (u )u |η| Br 2k1 − 1 Br ˆ ˆ 2λ−1 = λ0 |∇H(u )|2 η 2 − 0 |∇H(u )η| H 0 (u )u |∇η| B 2k1 − 1 Br ˆr ˆ λ0 C ≥ 2 2 |∇H(u )| η − |u H 0 (u )|2 2 Br ((r − r0 )(2k1 − 1))2 BR ˆ λ0 C(λ0 , n, k0 ) 2 rn 2 ≥ |∇H(u )|2 η 2 − 2 β 0 2 ( |u |βq ) q , (3.1.11) 2 Br (2k0 − 1) (r − r ) Br while ˆ ˆ   (b∇u ) · ∇ϕ = 2 b − (b)Br ∇u · ∇ηG(u )η Br Br ˆ C ≤ |b − (b)Br | |∇u | G(u+ )η (r − r0 ) Br ˆ C ≤ |b − (b)Br | |∇H(u )| H 0 (u )u η (r − r0 )(2k1 − 1) Br ˆ ˆ λ0 C(λ0 , n) 2 |∇H(u )|2 η 2 +  |b − (b)Br |2 H 0 (u )u ≤ 2 8 Br (2k1 − 1)(r − r0 ) Br ˆ λ0 2 2 C(n, q, λ0 )Λ20 rn  0 2 q q ≤ |∇H(u )| η + |H (u )u | 8 Br (2k0 − 1)2 (r − r0 )2 Br ˆ λ0 C(n, q, λ0 , k0 )Λ20 β 2 rn  2 βq q ≤ |∇H(u )|2 η 2 + |u  | , 8 Br (2k0 − 1)2 (r − r0 )2 Br (3.1.12) where in the first equality we have used a similar calculation as in (3.1.3). Then combining (3.1.5), (3.1.11), (3.1.12) and r ∼ r0 , C(n, q, λ0 , k0 )(1 + Λ20 )β 2  2 |∇H(u )|2 ≤ 0 −2 |u |βq q (r − r ) . Br0 (2k0 − 1)2 Br 31 By the Sobolev inequality, 2n n−2 n r02  2 q o ( H(u ) n−2 ) n ≤ C C(λ0 , Λ0 , n, q, k0 )β 2 |u |βq + H 2 (u ) Br0 (r − r0 )2 Br Br0 r02 2 ≤ C(λ0 , Λ0 , n, q, k0 )β 2 ( |u |βq ) q . (r − r0 )2 Br Letting N → ∞, 2n n−2 r0 1 ( |u |β n−2 ) 2n ≤ Cβ ( |u |βq ) q , Br0 r − r0 Br 2n that is, by setting l = (n−2)q (> 1), 1 r0 1 ( |u |βlq ) lq ≤ Cβ ( |u |βq ) q . (3.1.13) Br0 r − r0 Br Now let β = βi = kli , r = ri = R + R 2i and r0 = ri+1 for i = 0, 1, 2, . . . in (3.1.13), one finds i+1 q 1 i 1 ( ukl ) ql ≤ C(λ0 , Λ0 , q, n, k0 )kli 2i ( ukl  ) q q Bri+1 Bri ⇒ i+1 q 1 Pi 1 Pi j 1 ( ukl  ) qli+1 ≤ (Ck) j=0 lj (2l) j=0 lj ( ukq  ) . q (3.1.14) Bri+1 B2R Letting i go to infinity, we get 1 sup u ≤ C(n, λ0 , Λ0 , q, k0 )( ukq  ) . kq BR B2R Finally, by letting  go to 0 we get the desired estimate for u+ . 32 Observe that we now have the usual Caccioppli estimate: 1,2 Corollary 3.1.2 (Caccioppoli). Let u ∈ Wloc (Ω) be a non-negative weak subsolution of (2.3.1). Let BR = BR (X) be a ball centered at X with radius R such that B2R ⊂ Ω. Then |∇u|2 ≤ C(n, λ0 , Λ0 )R−2 u2 . (3.1.15) BR B2R Lemma 3.1.5. Let u ≥ 0 be a weak supersolution of (2.3.1) and B2R ⊂ Ω. Then for any 0 < k < 12 , we have 1 sup uk ≤ C(n, λ0 , Λ0 , q)( ukq ) q . (3.1.16) BR B2R Proof. For any  > 0, let u = u + , so that u ≥ . Set v = uk , φ = ku2k−1  η 2 , where η ∈ C0∞ (B2R ), supp η ⊂ Br , η ≡ 1 on Br0 , |∇η| ≤ C r−r0 and R ≤ r0 < r ≤ 2R. For convenience, we will omit the subscript  in the following. ´ Since u is a supersolution, A∇u · ∇φ ≥ 0, which gives ˆ ˆ 2k − 1 2 A∇v · ∇vη + 2 A∇v · ∇ηηv ≥ 0. (3.1.17) k So ˆ ˆ ˆ ˆ 1 − 2k 2 λ0 |∇vη| ≤ λ−1 0 ( 2 1/2 |∇vη| ) ( |∇ηv| ) 2 1/2 + |b − (b)r | |∇vη| |∇ηv| 2k ˆ ˆ n −1 2 1/2 ≤ (λ0 + Cq Λ0 )( |∇vη| ) ( |v|q )1/q r s0 (r − r0 )−1 Br ⇒ |∇v|2 ≤ C(Λ0 , λ0 , q, n)θ2 (k)(r − r0 )−2 ( v q )2/q , (3.1.18) Br0 Br 2k where θ(k) = 2k−1 . 33 By Sobolev embedding, we have 2n n−2 ( v n−2 ) n ≤ C(r02 |∇v|2 + v2) Br0 Br0 Br0  r0 2  2 ≤ C(Λ0 , λ0 , q, n) θ2 (k)( ) + 1 ( vq ) q . (3.1.19) r − r0 Br 1 5 Recall that l = 2n (n−2)q > 1, so there exists a m ∈ Z+ such that 12 l−m+ 2 ≤ k ≤ 12 l−m+ 2 . 1 Denote k 0 = 12 l−m+ 2 . Then l−2 k ≤ k 0 ≤ k and 1 k 0 < k 0 l < k 0 l2 < · · · < k 0 lm−1 < < k 0 lm < . . . 2 It is easy to check that θ2 (k 0 li ) ≤ (l1/2 − 1)−2 for i = 0, 1, . . . , m − 1. 0 i Letting v = uk l in (3.1.19), i = 0, 1, . . . , m − 1, we get 0 i+1 q 1 r0 0 i 1 ( uk l ) ql ≤ C(n, λ0 , Λ0 , s) ( uk l q ) q . Br0 r − r0 Br Let r = ri = 3R 2 + R 2i+1 and r0 = ri+1 . After m iterations, we have 0 mq 1 0 1 ( uk l ) qlm ≤ C(n, λ0 , Λ0 , q)( uk q ) q . Brm B2R Since k 0 lm > 12 , we can apply Lemma 3.1.4 letting there k = k0 = k 0 lm and get 0 m 0 mq 1 sup uk l ≤ C(n, λ0 , Λ0 , q)( uk l )q BR B3R/2 m 0 lm ≤ C(n, λ0 , Λ0 , q)1+l ( uk q ) q . B2R k Raising to k 0 lm power on both sides and using H¨older inequality, we get −m +1)k/k 0 0 k 1 sup uk ≤ C (l ( uk q ) k0 q ≤ C(n, λ0 , Λ0 , q)( ukq ) q . BR B2R B2R 34 Finally, let  → 0 we finish the proof. By Lemma 3.1.4 and Lemma 3.1.5, we easily obtain the following Corollary 3.1.3. Let u ≥ 0 be a weak solution of (2.3.1) and B2R ⊂ Ω. Then for any p > 0, we have 1 sup u ≤ C(n, λ0 , Λ0 , p)( up ) p . (3.1.20) BR B2R Lemma 3.1.6. Let u ≥ 0 be a weak supersolution of (2.3.1), B2R ⊂ Ω. Then for any k > 0, 1 sup u−k ≤ C(n, Λ0 , λ0 , q)( u−kq ) q . BR B2R Proof. For  > 0, define u = u + , v = u−k −2k−1 η 2 . Then φ ∈ W 1,2 (Ω) for  and φ = ku  0 ´ any k > 0, and u is also a supersolution. So A∇u · ∇φ ≥ 0, which implies ˆ ˆ 2k + 1 A∇v η · ∇v η ≤ −2 A∇v ηv ∇η. k Arguing as in the proof of Lemma 3.1.5, we get 2n n−2  2k r0 2  2 ( vn−2 ) n ≤ C(Λ0 , λ0 , q, n) ( )( ) + 1 ( vq ) q Br 0 2k + 1 r − r0 Br r0 2 ≤ C(Λ0 , λ0 , q, n)( )2 ( vq ) q . r− r0 Br Applying iterations we obtain 1 sup u−k  ≤ C(Λ0 , λ0 , q, n)( u−kq  )q . BR B2R Letting  → 0, one finds the desired inequality for nonnegative solutions. Corollary 3.1.4. Let u ≥ 0 be a weak solution in Ω. B2R ⊂ Ω. Then for any p < 0, 1 inf u ≥ C(n, λ0 , Λ0 , p)( up ) p . (3.1.21) BR B2R 35 Proof. Since supBR u−k = (inf BR u)−k , it follows immediately from the previous Lemma. Indeed, 1 1 inf u ≥ C(n, λ0 , Λ0 , s)− k ( u−kq ) −kq BR B2R for any k > 0. Recall that we have the following result due to F.John and L.Nirenberg: Lemma 3.1.7 ([JN61]). Let B1 be the unit cube. Suppose w ∈ BM O(B1 ), or equivalently, there exists some 0 < C1 < ∞ such that (w − (w)B )2 dx ≤ C12 ∀B ⊂ B1 , B then there exist positive constants α, β depending on n only such that ˆ eα|w−(w)B1 | dx ≤ βC1 . B1 This implies that ˆ ˆ e αw dx e−αw dx ≤ β 2 C12 , B1 B1 and that we can assume α < 12 . We are now ready to show the following. Lemma 3.1.8 (Harnack inequality). Let u ≥ 0 be a solution of (2.3.1). B2R ⊂ Ω. Then sup u ≤ C(λ0 , Λ0 , n) inf u. BR BR Proof. As in Lemma 3.1.6, we first consider u = u +  then let  go to 0. We omit these steps for simplicity. By scaling, we can assume R = 1. Let v = log u. We claim that v ∈ BM O(B3/2 ). To see this, fix Br ⊂ B3/2 . Let η ∈ C0∞ (B3/2 ), supp η ⊂ B4r/3 , η ≡ 1 in Br and |∇η| ≤ C r. 36 Define φ = u1 η 2 . Then ˆ ˆ ˆ 2 0= A∇u · ∇φ = − A∇v · ∇vη + 2 A∇η · ∇vη. ⇒ ˆ ˆ ˆ 2 λ0 |∇v| η ≤ 2 2λ−1 |∇η| |∇vη| + 2 b − (b)B4r/3 |∇η| |∇vη| 0 ˆ 2 ˆ −1 n −1 2 1/2 n −1 2 b − (b)B4r/3 ) ( |∇vη| )1/2 , 1/2 ≤ 2λ0 r ( |∇vη| ) + Cr ( 2 2 B4r/3 ⇒ ˆ |∇v|2 ≤ C(λ0 , Λ0 , n)rn−2 . Br By the Poincar´e inequality, |v − (v)Br |2 ≤ C12 (λ0 , Λ0 , n). Br Since Br ⊂ B3/2 is arbitrary, v ∈ BM O(B3/2 ) with kvkBM O ≤ C1 . So we can apply Lemma 3.1.7 to get ˆ ˆ αv e e−αv ≤ (βC1 )2 , (3.1.22) B3/2 B3/2 for some positive α and β which only depend on n. Then from (3.1.22), (3.1.20) and (3.1.21), Harnack’s inequality follows. Lemma 3.1.9 (interior H¨ older continuity). Let u be a weak solution of (2.3.1), and assume that BR (x) ⊂ Ω. Then 1. For 0 < ρ < r < R, there exists 0 < C0 = C0 (λ0 , Λ0 , n) < 1 and α = − log2 C0 such that ρ ω(ρ) ≤ C0−1 ( )α ω(r), (3.1.23) r where ω(r) = supBr (x) u − inf Br (x) u is the oscillation of u in the ball with radius r. 37 2. For 0 < ρ < r < R/2, ρ ω(ρ) ≤ C(λ0 , Λ0 , n)( )α ( u2 )1/2 . (3.1.24) r B2r Proof. By Harnack’s inequality and a standard argument (see eg. [GT01] Theorem 8.22.), we obtain C −1 ω(r/2) ≤ ω(r) C +1 for 0 < r < R, where C = C(λ0 , Λ0 , n) is the same constant as in Lemma 3.1.8. Let C−1 C0 = C+1 < 1. Iteration gives ω(r2−k ) ≤ C0k ω(r) for k = 1, 2, . . . , which leads to (1). For (2), observe that ω(r) = sup(u+ − u− ) − inf (u+ − u− ) ≤ 2 sup |u| ≤ C( |u|2 )1/2 . Br Br Br B2r 3.2 Boundary estimates for the solution Lemma 3.2.1 (boundary Caccioppoli). Let P ∈ ∂Ω, and let Tr (P ) = Br (P )∩Ω, ∆r (P ) = Br (P ) ∩ ∂Ω. Let u ∈ W 1,2 (TR (P )) be a solution of (2.3.1) in TR (P ), u ≡ 0 on ∆R (P ). 38 Then for any 0 < σ < 1, ˆ ˆ 2 1 2 |∇u| ϕ ≤C(n, λ0 ) |u|2 TR (P ) (1 − σ)2 R2 supp ∇ϕ∩TR (P ) ˆ 1 ˜ ˜ + C(n, λ0 ) b − (b)BR |∇˜ uϕ| |˜ u| , (3.2.1) (1 − σ)R supp ∇ϕ where ϕ ∈ C0∞ (BR ) is nonnegative, satisfying ϕ = 1 on BσR , supp ϕ ⊂ BR , and |∇ϕ| ≤ C (1−σ)R , and ˜b is the extension of b as in (2.1.4), u ˜ is the zero extension of u to BR (P ). Proof. Since u = 0 on ∆R (P ), uϕ2 ∈ W01,2 (TR (P )). Then one can proceed as in the proof of Lemma 3.1.1 and get (denote TR (P ) by TR and BR (P ) by BR ) ˆ λ0 |∇u|2 ϕ2 TR ˆ ˆ ˆ −1 2 2 1/2 2 2 1/2 ≤ 2λ0 ( |∇u| ϕ ) ( |u| |∇ϕ| ) − 2 b∇u · ∇ϕϕu TR TR TR ˆ ˆ ˆ = 2λ−1 0 ( |∇u| 2 2 1/2 ϕ ) ( |u|2 |∇ϕ|2 1/2 ) − 2 (˜b − (˜b)BR )∇˜ u · ∇ϕϕ˜u TR TR BR ˆ ˆ ˆ λ0 2 2 C(n, λ0 ) 2 C(n) ˜ ˜b)B |∇˜ ≤ |∇u| ϕ + |u| + b − ( uϕ| |˜ u| , 2 TR (1 − σ)2 R2 supp ∇ϕ∩TR (1 − σ)R supp ∇ϕ R where we used ˆ (˜b)BR ∇˜ u · ∇ϕϕ˜ u=0 BR as in (3.1.3). The boundary counterpart of Corollary 3.1.1 is the following. n Corollary 3.2.1. Let u, BR (P ), TR (P ) be as in Lemma 3.2.1. Then for any 1 < s ≤ n−1 , ˆ ˆ n 2s 2−s |∇u|2 ≤ C(s, n, λ0 )(1 + Λ20 )R2( s0 −1) ( |u| 2−s ) s , TR/2 (P ) TR (P ) 39 where s0 = s s−1 . Equivalently, 2s 2−s |∇u|2 ≤ C(s, n, λ0 )(1 + Λ20 )R−2 ( |u| 2−s ) s TR/2(P ) TR (P ) Proof. Let ˜b, u ˜ be as in Lemma 3.2.1, and let σ = 1/2. Then we have ˆ ˆ ˆ 2 −2 2 −1 ˜ |∇u| ϕ ≤ CR 2 |u| + CR b − (˜b)R |∇˜ uϕ| |˜ u| TR TR BR . = I1 + I2 . ˆ ˆ 2n 2s 2−s −2 2 −2 I1 = CR |˜ u| ≤ CR s0 ( |˜ u| 2−s ) s BR BR ˆ 2n 2s 2−s −2 = CR s0 ( |u| 2−s ) s TR and n s0 1 ˆ 1 I2 ≤ CR s0 −1 ( uϕ|s |˜ u|s ) s ˜ b − (˜b)R ) s0 ( |∇˜ BR B ˆ ˆ R n 2s 2−s ≤ Cs R s0 −1 Λ0 ( |∇˜uϕ|2 )1/2 ( |˜ u| 2−s ) 2s B B ˆ R ˆ R n 2s 2−s = Cs R s0 −1 Λ0 ( |∇uϕ|2 )1/2 ( |u| 2−s ) 2s TR TR ˆ ˆ 1 2 n 2( s0 −1) 2 2s 2−s ≤ |∇uϕ| + Cs R Λ0 ( |u| 2−s ) s . 2 TR TR Combining these two estimates proves the corollary. We will also need the analog of Lemma 3.1.2 near the boundary: 40 Lemma 3.2.2. Let f ∈ Lip(Ω), and u be the weak solution of   Lu = − div A∇f  in Ω,  u = 0  on ∂Ω. Then there exists a p > 2 and a constant C depending on n, λ0 , Λ0 and the domain, such that  1 n 1 o |∇u|p |∇u|2 p 2 ≤C + k∇f kL∞ (TR (P )) , (3.2.2) T R (P ) TR (P ) 2 where TR (P ) is defined as in Lemma 3.2.1. Proof. For fixed P ∈ ∂Ω and 0 < R < diam Ω, let F ∈ Lip(Rn ) be the extension of f such that k∇F kL∞ (Rn ) ≤ Cn k∇f kL∞ (TR (P )) (3.2.3) where the constant Cn is independent of TR (P ). This is possible due to Lemma 2.1.3. Let ϕ ∈ C0∞ (B2R/3 (P )) be nonnegative, with ϕ ≡ 1 on B R (P ) and |∇ϕ| . R−1 . 2 Let η ∈ C0∞ (BR (P )) be nonnegative, with η ≡ 1 on B 2R (P ) and |∇η| . R−1 . 3 Choose uϕ2 ∈ W01,2 (T2R/3 (P )) as test function to get ˆ ˆ 2 A∇u · ∇(uϕ ) = A∇f · ∇(uϕ2 ). TR TR ⇒ ˆ ˆ ˆ ˆ ˆ a∇u · ∇uϕ2 ≤ −2 A∇u · ∇ϕuϕ + a∇f · ∇uϕ2 + 2 a∇f · ∇ϕuϕ + b∇f · ∇(uϕ2 ) TR TR TR TR TR . = I1 + I2 + I3 + I4 . (3.2.4) I1 can be estimated as in the proof of Lemma 3.2.1 and Corollary 3.2.1. For I2 and I3 , we 41 have ˆ ˆ λ0 |I2 | ≤ |∇u|2 ϕ2 + C(n, λ0 ) |∇f |2 , 8 TR TR and ˆ ˆ |I3 | ≤ λ−1 0 2 |∇f | + C(n, λ0 ) u2 |∇ϕ|2 . TR TR For I4 , let F˜ = F − F (P ). Extending uϕ2 to be zero outside of TR , b to ˜b ∈ BM O(Rn ), we have ˆ I4 = ˜b∇(F˜ η) · ∇(uϕ2 ). Rn ⇒ ˆ  ˆ ˜ 2 1/2 1/2 |I4 | ≤ C(n, Ω)Λ0 ∇(F η) ∇(uϕ2 ) 2 Rn n ˆ 2 ˆR  λ ˆ ˆ  ˜ 2 ˜ 2 0 2 2 2 ≤ C(n, λ0 , Ω)Λ20 ∇F η + F 2 |∇η| + |∇u| ϕ + u2 |∇ϕ| . 8 Rn Rn TR TR Note that by (3.2.3), ˆ ˆ ˆ ˜ 2 2 |BR | ∇ F η ≤ Cn k∇F k2L∞ (BR ) ≤ Cn k∇f k2L∞ (TR ) , Rn BR |TR | TR and ˆ ˆ Cn F˜ 2 |∇η|2 ≤ 2 |F − F (P )|2 Rn R BR ˆ ˆ |BR | ≤ Cn k∇F k2L∞ (BR ) ≤ Cn k∇f k2L∞ (TR ) . BR |TR | TR |BR | By the interior Corkscrew condition, |TR | ≤ M n , where M is the constant in Definition 2.1.1. Therefore, ˆ ˆ ˆ λ0   |I4 | ≤ C(n, λ0 , M, Ω)Λ20 k∇f k2L∞ (TR ) + |∇u| ϕ + 2 2 u2 |∇ϕ|2 . TR 8 TR TR 42 Combining estimates for I1 , I2 , I3 ,I4 , and the inequality (3.2.4), we have ˆ ˆ  2−s ˆ λ0 n 2s 2 C(n, λ0 )(1+Λ20 )R2( s0 −1) k∇f k2L∞ (TR ) , 2 s |∇u| ϕ ≤ |u| 2−s +C(n, λ0 , Ω)(1+Λ20 ) 2 TR TR TR ⇒  2s  2−s |∇u|2 ≤ C(1 + Λ20 )R−2 k∇f k2L∞ (TR ) . s |u| 2−s + C(1 + Λ20 ) TR TR TR 2 1 n By the exterior Corkscrew condition, |BR (P ) \ Ω| ≥ ( 2M ) |BR (P )|. So the assumption that u = 0 on ∂Ω implies  2s  2−s  2 R−2 |∇u|r s r |u| 2−s ≤C TR TR n for some r < 2 (note that 1 < s < n−1 ). The constant depends on n, M, s and r (see eg.[Gia83] p.153 footnote). That is,  2 |∇u|2 ≤ C(1 + Λ20 ) |∇u|r k∇f k2L∞ (TR ) . r + C(1 + Λ20 ) TR TR TR 2 Then the lemma follows Lemma 2.4.1. Remark 3.2.1. Extending u to be zero outside Ω, one can see ˆ 1 n ˆ 1 o p p 2 2 |∇u| ≤C |∇u| + k∇f kL∞ (Ω) Ω Ω for some p > 2, with p and C depending on n, λ0 , Λ0 and the domain. On the other hand, we have ˆ 1 |∇u|2 2 ≤ C(n, λ0 , Λ0 ) k∇f kL2 (Ω) ≤ C(n, λ0 , Λ0 , diam Ω) k∇f kL∞ (Ω) . Ω Therefore, ˆ 1 |∇u|p p ≤ C k∇f kL∞ (Ω) (3.2.5) Ω for some p > 2. 43 Lemma 3.2.3. Let u be a W 1,2 (Ω) subsolution in Ω. Then for any P ∈ Rn , R > 0 and p > 1, we have p 1/p   sup u+ M ≤ C(n, λ0 , Λ0 , p) u+ M , (3.2.6) BR (P ) B2R (P ) where M= sup u, ∂Ω∩B2R (P )   max{u(x), M },  x ∈ Ω, u+ M (x) =  M,  x∈ / Ω. Proof. Note that the truncation uM is also a subsolution in Ω. So if BR (P ) is contained in Ω, then (3.2.6) is obtained by Lemma 3.1.4. Let TR (P ) = BR (P ) ∩ Ω. In the sequel we omit the point P when no confusion is caused. Let   +  u(x) − M  x∈Ω UM (x) =  0  x∈ / Ω. Note that UM = 0 on ∆2R and that u+ M = UM + M. (3.2.7) Moreover, for any ψ ∈ W01,2 (Ω) with ψ ≥ 0 a.e., ˆ (a + b)∇UM · ∇ψ ≤ 0, Ω which can be verified as in Lemma 3.1.3. 44 Claim. ∀ k ≥ k0 > 21 ,  1 kq q sup UM ≤ C(n, λ0 , Λ0 , q, k0 ) UM . TR B2R The same argument in the proof of Lemma 3.1.4 gives this claim. In fact, for any  > 0, define U = UM + . Define H,N , G,N and ϕ as in the proof of Lemma 3.1.4. Then one only need to note that ˆ ˆ a∇U · ∇ϕ = ˜∇U · ∇ϕ, a T2R B2R and ˆ ˆ b∇U · ∇ϕ = ˜b∇U · ∇ϕ, T2R B2R where   aij (x) x ∈ Ω  a ˜ij (x) =  λ0 δij  x∈ / Ω, and ˜b ∈ BM O(Rn ) is the extension of b.  ffl 1  ffl 1  ffl 1 kq kq + kq kq + kq kq By (3.2.7), B2R UM ≤ B2R (uM ) . Also, M ≤ B2R (uM ) . Thus  1 kq sup u+ M ≤ sup UM + M ≤ C (u+ M) kq . BR BR B2R Then (3.2.6) follows. Unlike the interior case, we do not have the result for all p > 0. This is mainly because 1 when 0 < k < 2, we have to modify our test function. However, we have the following result, which is the key to prove boundary H¨older continuity of the solution. 2s n 2k Recall that q = 2−s , 1 0, let u = u− k m +  and v = u . As test function we choose ϕ = k(u2k−1  − (m + )2k−1 )η 2 , where η ∈ C0∞ (B2R ), supp η ⊂ Br , η ≡ 1 in Br0 , |∇η| ≤ C r−r0 and R ≤ r0 < r ≤ 2R. Observe that for k < 21 , ϕ ∈ W01,2 (T4R ), and that 0 ≤ u2k−1  − (m + )2k−1 ≤ u2k−1  . (3.2.9) n Claim 1. For any 0 < p < p1 < n−2 , p p ( u− 1 1/p1 m ) ≤ C( u− m ) 1/p , (3.2.10) BR B2R for some C = C(n, λ0 , Λ0 , p, p1 ). This claim can be justified by considering 0 < k < 21 . Observe that when k is in this range, ϕ ≥ 0 in B4R . Since u is a supersolution, we have ˆ k A∇u ∇(η 2 (u2k−1 − (m + )2k−1 )) ≥ 0. T4R 46 ⇒ ˆ ˆ k(2k − 1) a∇u · ∇u u2k−2 η 2 ≤ (a + b)∇u · ∇η(u2k−1  − (m + )2k−1 )η. (3.2.11) −2k T2R T2R Note that supp η ⊂ Br and that ∇u = 0 in Br \ Tr . We estimate ˆ ˆ 2k − 1 2k − 1 λ0 |∇v|2 η 2 ≤ a∇v · ∇vη 2 −2k Br −2k Br ˆ k(1 − 2k) = a∇u · ∇u u2k−2 η 2 2 Br ˆ ˆ ≤ λ−1 2k−1 ˜ ˜ 2k−1 k |∇u | |∇η| |η| u + k b − (b) |∇u | |∇η| u |η| 0   Br   Br Br ˆ ˆ −1 ˜ = λ0 |∇v| |v| |∇η| |η| + b − (˜b)Br |∇v| |v| |∇η| |η| Br Br ˆ ˆ n −1 2 2 1/2 ≤ (λ0 + Cq Λ0 )r (r − r ) ( s0 0 −1 |∇v| |η| ) ( |v|q )1/q . Br Br ⇒ |∇v|2 ≤ C(n, λ0 , Λ0 , q)θ2 (k)(r − r0 )−2 ( |v|q )2/q , Br0 Br 2k where θ(k) = 2k−1 . By Sobolev embedding we have 1 r0 ( ukql  ) ≤ C(n, λ0 , Λ0 , q)(θ(k) ql + 1)( ukq  ) 1/q , Br0 r − r0 Br 2n where l = (n−2)q > 1. Let ki = kli , r = ri = R + 2Ri and r0 = ri+1 . Then we can iterate as long as 0 < ki < 21 . 2n n Recall that 2 < q < n−2 . Then it is easy to see that for any 0 < p < p1 < n−2 , ( (u− p1 1/p1 m) ) ≤ Cp,p1 ( (u− p 1/p m) ) . (3.2.12) BR B2R 47 Claim 2. For any p < 0, inf u− m ≥ C( (u− p 1/p m) ) (3.2.13) BR B2R for some C = C(n, λ0 , Λ0 , p). This time we consider k < 0. We have ˆ k A∇u ∇(η 2 (u2k−1 − (m + )2k−1 )) ≤ 0. T4R Then it is easy to get ˆ ˆ ˆ 2k − 1 2 n λ0 |∇v| η ≤ 2 (λ−1 0 + Cq Λ0 )r (r − r ) s0 0 −1 ( |∇v| η ) 2 2 1/2 ( |v|q )1/q . 2k Br Br Br Using Sobolev inequality and then iterations, and letting  tend to 0, we obtain   kq sup (u− m ) k ≤ C(n, λ0 , Λ0 , q)( u− m ) 1/q . (3.2.14) BR B2R Since k < 0, this implies kq (inf u− k m ) ≤ C(n, λ0 , Λ0 , q)( u− m ) 1/q . (3.2.15) BR B2R 1 Raise both sides to the power of k we prove the claim. We will show 1 1 ( u−α  ) −α ≥ C( uα ) α , (3.2.16) B2R B2R where α depending on n only is as in Lemma 3.1.7. If this is true, then by Claim 1 and Claim 2, 1 1 p 1 inf u− m ≥ C( (u− m) −α −α ) ≥ C( (u− α α m ) ) ≥ Cp ( u− m ) p BR B2R B2R BR 48 n for any 0 < α < p < n−2 . And trivially, for any 0 < p0 ≤ p 1 1 p p ( u− m ) ≥( p u− 0 p m ) 0. BR BR So it suffices to show (3.2.16). As we did in the proof of Lemma 3.1.8, let w = log u . We will see that w ∈ BM O(B2R ). In fact, fix any Br ⊂ B2R , let φ = (u−1 −1 2  − (m + ) )ζ , where ζ ∈ C0∞ (B4R ), supp ζ ⊂ B2r , ζ ≡ 1 in Br and |∇ζ| ≤ C/r. Then φ ≥ 0 in T4R and φ ∈ W01,2 (T4R ). So ˆ A∇u · ∇φ ≥ 0. T4R Note that ∇u ≡ 0 in B2r \ Ω, and that 0 ≤ u−1  − (m + ) −1 ≤ u−1  . We get ˆ ˆ λ0 2 2 n |∇w| ζ ≤ (λ−1 0 + C(n)Λ0 )( |∇w|2 ζ 2 )1/2 r 2 −1 . (3.2.17) 2 B2r B2r By the Poincar´e inequality we get |w − (w)Br |2 ≤ r2 |∇w|2 ≤ C12 (n, λ0 , Λ0 ). Br Br ⇒ w ∈ BM O(B2R ) with kwkBM O ≤ C1 . Then by Lemma 3.1.7 we proved (3.2.16). 49 |BR (P )\Ω| 1 n Since the exterior Corkscrew condition gives lim inf Rn ≥ ( 2M ) , where P ∈ ∂Ω R→0 and M > 1 is the constant in Definition 2.1.1, we can now prove regularity at the boundary using Lemma 3.2.4 and a standard argument. older continuity). Let u ∈ W 1,2 (Ω) be a solution in Ω and P Lemma 3.2.5 (boundary H¨ be a point on the boundary of Ω. Let BR denote BR (P ), TR denote BR (P ) ∩ Ω and ∆R denote BR (P ) ∩ ∂Ω. Then for any 0 < r ≤ R, we have r √ osc u ≤ C( )α sup |u| + Cσ( rR), (3.2.18) Tr R TR where osc u = supTr u−inf Tr u, σ(r) = osc u = sup∆r u−inf ∆r u, and C = C(n, λ0 , Λ0 , M ), Tr ∆r α = α(n, λ0 , Λ0 , M ) are positive constants. R In particular, if u = 0 on ∆R , 0 < r ≤ 2, then we have r osc u ≤ C( )α ( |u|2 )1/2 . (3.2.19) Tr R TR Proof. It suffices to consider the case supTR |u| < ∞ and r ≤ R/4. Write M4 = supT4r u, m4 = inf T4r u, M1 = supTr u, m1 = inf Tr u, M = sup∆4r u, m = inf ∆4r u. Then M4 − u and u − m4 are solutions in Ω and nonnegative in T4r . Note that inf ∆4r M4 − u = M4 − M and inf ∆4r u − m4 = m − m4 . So applying Lemma 3.2.4 and letting p = 1 there, we get |B2r \ Ω| (M4 − M ) ≤ (M4 − u)− − M4 −M ≤ C(n, λ0 , Λ0 ) inf (M4 − u)M4 −M rn B2r Br ≤ C(n, λ0 , Λ0 )(M4 − M1 ), |B2r \ Ω| (m − m4 ) ≤ (u − m4 )− − m−m4 ≤ C(n, λ0 , Λ0 ) inf (u − m4 )m−m4 rn B2r Br ≤ C(n, λ0 , Λ0 )(m1 − m4 ). 50 Using the exterior Corkscrew condition, we get c(M4 − M ) ≤ C(n, λ0 , Λ0 )(M4 − M1 ) (3.2.20) and c(m − m4 ) ≤ C(n, λ0 , Λ0 )(m1 − m4 ). (3.2.21) By addition we get 1 osc u ≤ (1 − )osc u + osc u, (3.2.22) Tr C T4r ∆4r where C = C(n, λ0 , Λ0 , c). Then we easily get (see eg.[GT01] Lemma 8.23) r √ osc u ≤ C( )α osc u + σ( rR) ∀r ≤ R (3.2.23) Tr R TR where C = C(n, λ0 , Λ0 , c) and α = α(n, λ0 , Λ0 , c) are positive constants. Note that osc TR (u) ≤ 2 supTR |u|, thus (3.2.18). √ If u = 0 on ∆R , then σ( rR) = 0. Moreover, we can apply Lemma 3.2.3 to get sup |u| ≤ ( |u|2 )1/2 . TR/2 TR Thus (3.2.19) holds. Remark 3.2.2. If, in addition to the assumptions of Lemma 3.2.5, osc ∆R u → 0 as R → 0, then (3.2.18) implies that u(P0 ) = lim u(x) x→P0 ,x∈Ω 51 is well-defined. 1,2 Lemma 3.2.6 (positivity). Let u ∈ Wloc (Ω) ∩ C(Ω) be a supersolution of (2.3.1). u ≥ 0 on ∂Ω. Then u ≥ 0 in Ω. Proof. For any  > 0, set v = min{u, −} + . Then v ≤ 0 and   ∇u  u < −, ∇v =  0  u ≥ − ⇒ v ∈ W 1,2 (Ω) with supp v ⊂⊂ Ω. Since u is a supersolution, we have ˆ ˆ ˆ ˆ 0≥ A∇u · ∇v = A∇u · ∇u = A∇v · ∇v ≥ λ0 |∇v |2 , Ω {u<−} Ω Ω ⇒ ∇v = 0 a.e. in Ω, thus u ≥ − in Ω. Letting  → 0, we obtain u ≥ 0 in Ω. This lemma immediately yields the maximum principle: 1,2 Lemma 3.2.7 (maximum principle). Let u ∈ Wloc (Ω) ∩ C(Ω) be a subsolution of (2.3.1). Then sup u ≤ sup u. (3.2.24) Ω ∂Ω 1,2 As a consequence, if u ∈ Wloc (Ω) ∩ C(Ω) is a weak solution of (2.3.1), then sup |u| ≤ sup |u| . (3.2.25) Ω ∂Ω Proof. Let M = sup∂Ω u. We can assume M < ∞ since otherwise (3.2.24) is trivial. Then apply Lemma 3.2.6 to M − u to get M − u ≥ 0 in Ω. Lemma 3.2.8. Let Ω be an NTA domain, and suppose g ∈ Lip(∂Ω). Then there exists a unique u ∈ W 1,2 (Ω) that solves the classical Dirichlet problem with data g. Moreover, 52 u ∈ C 0,β (Ω) for some 0 < β < 1. Proof. Let G be a Lipschitz function with compact support such that G|∂Ω = g. Then by Theorem 2.3.1, there exists a unique solution u ∈ W 1,2 (Ω) to the classical Dirichlet problem   Lu = 0  in Ω, u − G ∈ W01,2 (Ω).   By Remark 3.2.2, lim u(X) = g(P ) ∀P ∈ ∂Ω. Ω3X→P . To see that u ∈ C 0,β (Ω), fix P0 ∈ ∂Ω and X ∈ Ω with r = |X − P0 | < 21 . By Lemma 3.2.5, |u(X) − g(P0 )| ≤ osc u ≤ Crα sup |u| + osc g T2r (P0 ) T1 ∆√2r (P0 ) √ ≤ Crα kgkC 0,1 (∂Ω) + 2[g]C 0,1 (∂Ω) r1/2 0 ≤ C kgkC 0,1 (∂Ω) |X − P0 |α for some 0 < α0 < 1. So together with Lemma 3.1.9, we can show that for any X, Y ∈ Ω, |u(X) − u(Y )| ≤C for some β ∈ (0, 1). |X − Y |β Now we can immediately obtain the following Theorem 3.2.9. Let Ω be an NTA domain. Then the continuous Dirichlet problem (2.3.3) 53 is uniquely solvable. Proof. Let f ∈ C(∂Ω). Then there exists a sequence {fn }∞ n=1 ⊂ Lip(∂Ω) such that fn → f uniformly on ∂Ω. By Lemma 3.2.8, the corresponding solutions un ∈ W 1,2 (Ω) ∩ C 0,β (Ω). By the maximum principle, sup |ui − uj | ≤ sup |fi − fj | . Ω ∂Ω So un converges uniformly to u ∈ C(Ω). By Caccioppoli inequality (Corollary 3.1.2), 1,2 1,2 u ∈ Wloc (Ω), and also Lu = 0 in Ω, u = g on ∂Ω. If v ∈ Wloc (Ω) ∩ C(Ω) is another solution, then u − v = 0 on ∂Ω. By the maximum principle, u ≡ v. 3.3 Weak solution of parabolic equations In this section, we derive some estimates for the weak solution to the parabolic equation ∂t u+Lu = 0 in Rn ×(0, ∞). Here, L = − div(A∇) is the n-dimensional divergence form op- erator, with A = As + Aa , the symmetric part As satisfying (1.0.1) and the anti-symmetric part Aa satisfying (1.0.2). Note that we write div = divx , and ∇ = ∇x throughout this section. Definition 3.3.1. We say u(x, t) ∈ L2loc (0, ∞), W 1,2 (Rn ) ∩ C (0, ∞), L2 is a weak   solution to the parabolic equation ∂t u − div(A∇u) = 0 in Rn × (0, ∞)   f −1,2 (Rn ) and if for any T > 0, any ϕ ∈ L2 [0, T ], W 1,2 (Rn ) with ∂t ϕ ∈ L2 [0, T ], W  ϕ = 0 when 0 ≤ t ≤ ε for some 0 < ε < T , ˆ ˆ T ˆ ˆ T u(x, T )ϕ(x, T )dx + A∇u · ∇ϕdxdt = h∂t ϕ, uiW f −1,2 ,W 1,2 dt. Rn 0 Rn 0 54 Definition 3.3.2. We say u(x, t) ∈ L2loc [0, ∞), W 1,2 (Rn ) ∩ C [0, ∞), L2 is a weak   solution to the initial value problem  in Rn × (0, ∞)  ∂t u − div(A∇u) = 0   u(x, 0) = f (x),    f −1,2 (Rn ) , if for any T > 0, and any ϕ ∈ L2 [0, T ], W 1,2 (Rn ) with ∂t ϕ ∈ L2 [0, T ], W  ˆ ˆ T ˆ ˆ ˆ T u(x, T )ϕ(x, T )dx+ A∇u·∇ϕdxdt = f (x)ϕ(x, 0)dx+ h∂t ϕ, uiW f −1,2 ,W 1,2 . Rn 0 Rn Rn 0   f −1,2 (Rn ) . Lemma 3.3.1. Suppose u, v ∈ L2 (0, T ), W 1,2 (Rn ) with ∂t u, ∂t v ∈ L2 (0, T ), W  Then (i) u ∈ C [0, T ], L2 (Rn ) ;  (ii) The mapping t 7→ ku(·, t)kL2 (Rn ) is absolutely continuous, with d ku(·, t)k2L2 (Rn ) = 20 . Moreover, (e−tL )t>0 is actually analytic, since we show in chapter 7 that L is sectorial. 55 Proposition 3.3.1. For any u0 ∈ L2 (Rn ), the initial value problem  in Rn × (0, ∞)  ∂t u − div(A∇u) = 0  (3.3.1)  u(x, 0) = u0 (x)  has a unique weak solution u(x, t) = e−tL (u0 )(x). Here, div = divx and ∇ = ∇x . Proof. Existence. Let u(x, t) = e−tL (u0 )(x). Since L = − div(A∇) is sectorial, e−tL is an analytic semi- group. Therefore, for any u0 ∈ L2 (Rn ), ∂t u + Lu = 0 in L2 (Rn ) ∀ t ≥ 0. (3.3.2)   f −1,2 (Rn ) For any 0 < τ < T , and any ϕ ∈ L2 (0, T ), W 1,2 (Rn ) , with ∂t ϕ ∈ L2 (0, T ), W  (so ϕ ∈ C [0, T ], L2 (Rn ) by Lemma 3.3.1 (i)), (3.3.2) implies  ˆ T ˆ T (∂t u, ϕ)L2 dt + (Lu, ϕ)L2 dt = 0. τ τ Since by Theorem 7.3.1 ∂t u ∈ L2loc (0, ∞), L2 (Rn ) , we have  ˆ T ˆ T h∂t u, ϕiW f −1,2 ,W 1,2 dt + hL u, ϕiW f −1,2 ,W 1,2 dt = 0. τ τ d By Lemma 3.3.1 (ii), h∂t u, ϕiW f −1,2 ,W 1,2 = dt (u, ϕ)L2 (Rn ) − h∂t ϕ, uiW f −1,2 ,W 1,2 . So ˆ ˆ T ˆ ˆ ˆ T u(x, T )ϕ(x, T )dx+ A∇u·∇ϕdxdt = u(x, τ )ϕ(x, τ )dx+ h∂t ϕ, uiW f −1,2 ,W 1,2 . Rn τ Rn Rn τ (3.3.3) Recall that by Theorem 7.3.1, u ∈ L2loc [0, ∞), W 1,2 (Rn ) ∩ C (0, ∞), W 1,2 (Rn ) and   56 ∂t u ∈ L2loc (0, ∞), L2 (Rn ) , we can choose ϕ = u as test function. Therefore,  ˆ T ˆ ˆ ˆ T ˆ 2 2 2λ0 |∇u| dxdt ≤ |u(x, T )| dx + 2< A∇u · ∇udxdt Rn n Rn τ ˆR τ 2 |u(x, τ )|2 dx = e−τ L (u0 ) L2 (Rn ) ≤ ku0 k2L2 . = Rn ´∞´ Letting τ → 0+ , T → ∞, we obtain 0 Rn |∇u|2 dxdt ≤ (2λ0 )−1 ku0 k2L2 < ∞. This enables us to take limit as τ go to 0+ on both sides of (3.3.3) and get ˆ ˆ T ˆ ˆ ˆ T u(x, T )ϕ(x, T )dx+ A∇u·∇ϕdxdt = u(x, 0)ϕ(x, 0)dx+ h∂t ϕ, uiW f −1,2 ,W 1,2 , Rn 0 Rn Rn 0 i.e. u(x, t) is a weak solution of (3.3.1). Uniqueness Let v be a weak solution of (3.3.1). We first show for any 0 < T < ∞, ∂t v ∈   f −1,2 (Rn ) . L2 (0, T ), W Define a semilinear functional F on L2 [0, T ], W 1,2 (Rn ) as follows: for any ϕ ∈  L2 [0, T ], W 1,2 (Rn ) , let  ˆ T ˆ . hF, ϕi = A∇v · ∇ϕdxdt. 0 Rn Obviously, |hF, ϕi| ≤ C k∇vkL2 ([0,T ],L2 (Rn )) k∇ϕkL2 ([0,T ],L2 (Rn )) . Then by Riesz representation theorem, there exists w(x, t) ∈ L2 [0, T ], W 1,2 (Rn ) such  that ˆ T ˆ hF, ϕi = (∇w · ∇ϕ + wϕ)dxdt 0 Rn ˆ T = h−∆w(·, t) + w(·, t), ϕiW f −1,2 ,W 1,2 dt, 0 57 and k−∆w + wkL2 ([0,T ],W f −1,2 (Rn )) ≤ kwkL2 ([0,T ],W 1,2 (Rn )) ≤ k∇vkL2 ([0,T ],L2 (Rn )) . Choose ϕ(x, t) = Ψ(x)η(t) as test function in (3.3.1), where Ψ ∈ W 1,2 (Rn ), η ∈ C01 ((0, T )). Then since v is a weak solution, we have ˆ T ˆ T ˆ 0 (v(·, t), Ψ)L2 η (t)dt = A∇v · ∇Ψη(t)dxdt 0 0 Rn ˆ T = h−∆w(·, t) + w(·, t), ΨiW f −1,2 ,W 1,2 η(t)dt. 0 Since Ψ ∈ W 1,2 (Rn ) is arbitrary, ˆ T ˆ T 0 v(x, t)η (t)dt = (−∆w + w)η(t)dt 0 0   f −1,2 (Rn ) . Therefore, we can take ϕ = v as test function ⇒ ∂t v = ∆w − w ∈ L2 (0, T ), W in (3.3.1) and get ˆ ˆ T ˆ ˆ T ˆ |v(x, T )|2 + A∇v · ∇vdxdt = h∂t v, viW f −1,2 ,W 1,2 + |v(x, 0)|2 dx. Rn 0 Rn 0 Rn Using this and Lemma 3.3.1 (ii), we have ˆ ˆ T ˆ ˆ T ˆ d 2 |v(x, T )|2 + 2< A∇v · ∇vdxdt = kv(·, t)k2L2 dt + 2 |v(x, 0)|2 dx. Rn 0 Rn 0 dt Rn ⇒ ˆ ˆ T ˆ ˆ ˆ T ˆ 2 2 2 |v(x, T )| + 2λ0 |∇v| dxdt ≤ |v(x, T )| + 2< A∇v · ∇vdxdt Rn Rn Rn Rn 0 ˆ 0 = |v(x, 0)|2 dx, Rn which implies that if v(x, 0) = 0 then v ≡ 0. Remark 3.3.1. Let u(x, t) be the weak solution to (3.3.1). Then since the coefficients 58 are independent of t, a standard argument shows that ∂t u is a weak solution to ∂t v − div(A∇v) = 0 in Rn × (0, ∞). Moreover, since ∂tl u ∈ L2loc (0, ∞), L2 (Rn ) and ∂tl ∇u ∈  L2loc (0, ∞), L2 (Rn ) for any l ∈ N (see Theorem 7.3.3 and Remark 7.3.1), one can show  that for any l ∈ N, ∂tl u is a weak solution to ∂t v − div(A∇v) = 0 in Rn × (0, ∞). Let u(x, t) = e−tL (f )(x), for some f ∈ L2 (Rn ). Then by the remark above, ∂t u is a weak solution to ∂t v − div(A∇v) = 0 in Rn × (0, ∞). Note that by (7.2.52), one has ∂t u ∈ L∞ ([δ0 , ∞) × Rn ) ∀ δ0 > 0. (3.3.4) Now we are ready to prove the following estimates for ∂t u using Moser iteration. Proposition 3.3.2. Let Q ⊂ Rn be a cube with l(Q) = R0 . Then ˆ ˆ (2R0 )2 !1/2 − n+2 sup |∂t u(x, t)| ≤ CR0 2 R02 |∂t u(x, t)|2 dtdx , (3.3.5) 3 Q×(R02 ,(2R0 )2 ] 2 Q 2 for some C = C(n, λ0 , Λ0 ). Proof. Let v(x, t) = ∂t u(x, t). Then by the definition of weak solution and Lemma 3.3.1 (ii), we have ˆ T ˆ ˆ T ˆ ∂t v(x, t)ϕ(x, t)dxdt + A∇v · ∇ϕ = 0, 0 Rn 0 Rn for all ϕ ∈ L2 [0, T ], W 1,2 (Rn ) with supp ϕ ⊂ Rn × (0, T ]. By considering v ± we can  assume v ≥ 0, and that ˆ T ˆ ˆ T ˆ ∂t v(x, t)ϕ(x, t)dxdt + A∇v · ∇ϕ ≤ 0, (3.3.6) 0 Rn 0 Rn for all ϕ ∈ L2 [0, T ], W 1,2 (Rn ) with supp ϕ ⊂ Rn × (0, T ] and ϕ ≥ 0 a.e..  Now for any 0 ≤ s ≤ 1, define Qs = (1+s)Q, Is = ((1−s)R02 , (2R0 )2 ], and Cs = Qs ×Is . n+2 Fix l ∈ N, define ql = 2k0l , where k0 = n . Note that q0 = 2. 59 4 1 3 1 For any fixed 3 2l+2 ≤ s0 < s1 ≤ 2 2l+2 , choose Ψs0 ,s1 ∈ C02 (C s0 +s1 ), Ψ e s ,s ∈ C 2 (Cs ), 0 1 0 1 2 e s ,s = 1 in C s0 +s1 , 0 ≤ Ψs ,s , Ψ with Ψs0 ,s1 = 1 in Cs0 , Ψ e s ,s ≤ 1, and 0 1 0 1 0 1 2 2 R0−2 |∇Ψs0 ,s1 |2 + |∂t Ψs0 ,s1 | + ∇Ψ + ∂ Ψ t s0 ,s1 . . e e s0 ,s1 (s1 − s0 )2 We omit the subscript s0 , s1 in Ψs0 ,s1 and Ψ e s ,s from now on. 0 1 Let t ∈ Is0 . Recalling (3.3.4), one can take ϕ = v ql −1 Ψ2 as test function. Then (3.3.6) gives ˆ tˆ ˆ tˆ ql −1 ∂t vv Ψ + 2 A∇v · ∇(v ql −1 Ψ2 ) ≤ 0. (3.3.7) 0 Rn 0 Rn For the first term, integration by parts gives ˆ tˆ ˆ ˆ ˆ ql −1 2 1 ql 2 1 t ∂t vv Ψ = v (x, t)Ψ (x, t)dx − v ql ∂t (Ψ2 ) 0 Rn ql Rn ql 0 Rn ˆ ˆ 1 CR0−2 ≥ v ql (x, t)dx − v ql . q l Q s0 ql (s1 − s0 )2 C 1 2l+1 ´t´ ´t´ The second term in (3.3.7) consists of 0 Rn As ∇v · ∇(v ql −1 Ψ2 ) and 0 Rn Aa ∇v · ∇(v ql −1 Ψ2 ). We estimate ˆ tˆ As ∇v · ∇(v ql −1 Ψ2 ) 0 Rn ˆ ˆ ˆ ˆ 4(ql − 1) t s ql ql 2 4 t ql ql = 2 A ∇(v 2 ) · ∇(v 2 )Ψ + As ∇(v 2 ) · ∇Ψv 2 Ψ ql 0 Rn ql 0 Rn ˆ tˆ ˆ 2λ0 (ql − 1) q2l 2 C(n, λ0 )R0−2 ≥ ∇v dxdt − v ql . ql2 q (s − s ) 2 0 Qs0 l 1 0 C 1 2l+1 Note that As ∇v · ∇vΨ2 = 0 due to anti-symmetry, ˆ tˆ ˆ tˆ ˆ tˆ a ql −1 2 a 2 ql −1 1 A ∇v · ∇(v Ψ )= A ∇v · ∇(Ψ )v = As ∇(v ql ) · ∇(Ψ2 ) 0 Rn 0 Rn ql 0 Rn ˆ tˆ 1 ql ql = As ∇(v 2 Ψv e · ∇(Ψ2 ). e 2 Ψ) ql 0 Rn 60 By Proposition 2.2.3, ˆ t ˆ ˆ Cn Λ 0 t q  q  a ql −1 2 l e l A ∇v · ∇(v Ψ ) ≤ k∇ΨkL∞ (Rn ) v 2 Ψ 2 n ∇ v 2 Ψ 2 n dt e 0 Rn ql 0 L (R ) L (R ) ˆ ˆ Cθ Λ0 R0−2 θ q l 2 ≤ v ql dxdt + ∇v 2 . ql (s1 − s0 )2 C 1 ql Cs1 2l+1 Combining these estimates with (3.3.7), we have ˆ ˆ tˆ ˆ ˆ ql q 2 l CR0−2 ql  q  2 l v (x, t)dx + ∇v 2 ≤ v dxdt + Cθ ∇ v 2 dxdt, Q s0 0 Qs0 (s1 − s0 )2 C 1 Cs1 2l+1 where C = C(n, λ0 , Λ0 , θ). Choosing θ to be sufficiently small, and then taking supremum in t ∈ Is0 , we obtain ˆ ˆ ˆ ˆ ql q 2 l CR0−2 ql 1  q  2 l sup v (x, t)dx+ ∇v 2 dxdt ≤ v dxdt+ ∇ v 2 dxdt, t∈Is0 Q s0 Cs0 (s1 − s0 )2 C 1 2 Cs1 2l+1 which implies ˆ ˆ q 2 ˆ ql l −2 l sup v (x, t)dx + ∇v dxdt ≤ C(n, λ0 , Λ0 )R0 4 v ql dxdt 2 t∈I 4 1 Q4 1 C4 1 C 1 3 2l+2 3 2l+2 3 2l+2 2l+1 (3.3.8) by Lemma 2.4.2. We insert a cut-off function Ψl (x, t) ∈ C02 (C 4 1 ) into (3.3.8) so that we can use 3 2l+2 embedding theorem. Ψl satisfies 0 ≤ Ψl ≤ 1, Ψl = 1 in C 1 , and 2l+2 |∇Ψl |2 + |∂t Ψl | . R0−2 4l . Then we have ˆ ˆ 2 ˆ ql ql −2 l sup v (x, t)Ψl (x, t)dx + ∇(v 2 Ψl ) dxdt ≤ CR0 4 v ql dxdt. t∈I 4 1 Q4 1 C4 1 C 1 3 2l+2 3 2l+2 3 2l+2 2l+1 (3.3.9) 61 The Sobolev’s inequality in parabolic domains (see e.g. [Lie96] Theorem 6.9) gives ˆ ˆ ˆ  2/n ql ql ql 2 2k0 2 (v Ψl ) ≤ sup (v Ψl ) (x, t)dx ∇(v Ψl ) dxdt, 2  2 2 C4 1 t∈I 4 1 Q4 1 C4 1 3 2l+2 3 2l+2 3 2l+2 3 2l+2 and thus by (3.3.9) ˆ ˆ ˆ  k0 ql ql 2 v ql k0 ≤  sup (v Ψl )2 (x, t)dx + ∇(v 2 Ψl ) dxdt 2 C 1 t∈I 4 1 Q4 1 C4 1 2l+2 3 2l+2 3 2l+2 3 2l+2 ˆ  k0  k0 ≤ C R0−2 4l  v ql dxdt . C 1 2l+1 ⇒ 1 1 ˆ ˆ    ql+1 ql 1 1  v ql+1 dxdt ≤C ql+1 (R0−2 4l ) ql  ql v dxdt ∀ l = 0, 1, 2, . . . . C 1 C 1 2l+2 2l+1 Then (3.3.5) follows from iteration and letting l go to infinity. We also have a reverse H¨ older type estimate for the gradient of u. Proposition 3.3.3. Let Q ⊂ Rn be a cube with l(Q) = R0 . Then for any t > 0,  1/p  1/2  1/p p 2 p |∇u(x, t)| dx ≤C |∇u(x, t)| dx + R0 |∂t u(x, t)| dx Q 2Q 2Q (3.3.10) for all p ∈ [2, 2 + ), where C = C(n, λ0 , Λ0 ) and  = (n, λ0 , Λ0 ) are positive constants. 1 dist(x0 , 4Q), 2R0 . Choose Ψ ∈ C01 (Q 3 R (x0 )), Proof. Let x0 ∈ 4Q and 0 < R < min 2 2 with Ψ = 1 on QR (x0 ) and |∇Ψ| . R−1 . And choose Ψ e ∈ C 1 (Q2R (x0 )), with Ψ = 1 on 0 Q 3 R (x0 ) and ∇Ψ . R−1 . e 2 62 ffl Fix t > 0 and define u ¯= Q2R (x0 ) u(x, t)dx. ¯)Ψ2 (x) as a test function. Take (u(x, t) − u Then ∂t u − div(A∇u) = 0 implies ˆ ˆ ¯)Ψ2 (x) dx = − ¯)Ψ2 (x)dx. (3.3.11)  A∇u(x, t) · ∇ (u(x, t) − u ∂t u(x, t)(u(x, t) − u Rn Rn For the integral involving the symmetric part of A, we have ˆ As ∇u(x, t) · ∇ (u(x, t) − u ¯)Ψ2 (x) dx  Rn ˆ ˆ λ0 C(n, λ0 ) ≥ |∇u|2 dx − ¯)2 . (u − u 2 QR (x0 ) R2 Q 3 R(x0 ) 2 For the integral involving the anti-symmetric part of A, we insert Ψ e and apply Propo- sition 2.2.3: ˆ a 2  n A ∇u(x, t) · ∇ (u(x, t) − u ¯)Ψ (x) dx ˆR ˆ   a 2 2 a 2 e2 2 = A ∇(u − u ¯) · ∇(Ψ ) = A ∇ (u − u ¯) Ψ · ∇(Ψ ) Rn Rn Cn Λ 0   ≤ (u − u ¯ )Ψ 2 n ∇ (u − u ¯)Ψ e e R 2 n L (R ) L (R ) ˆ ˆ C(n, Λ0 , θ) ≤ Cn θ |∇u|2 dx + 2 ¯)2 dx. (u − u Q2R (x0 ) R Q2R (x0 ) We estimate the right-hand side of (3.3.11) by Cauchy-Schwartz. ˆ ˆ ˆ Cn ¯)Ψ (x)dx ≤ Cn R2 2 |∂t u|2 dx+ u)2 dx. ∂t u(x, t)(u(x, t) − u (u(x, t)−¯ R n Q 3 R (x0 ) R2 Q 3 R(x0 ) 2 2 To summarize, we have ˆ ˆ ˆ ˆ 2 2 |∇u| dx . R −2 (u − u 2 ¯) dx + R 2 |∂t u| dx + θ |∇u|2 dx. QR (x0 ) Q2R (x0 ) Q 3 R (x0 ) Q2R (x0 ) 2 63 Choosing θ to be sufficiently small and using Sobolev inequality, we obtain ! n+2 n 2n 1 |∇u|2 dx ≤ C |∇u| n+2 dx +CR02 |∂t u|2 dx+ |∇u|2 dx. QR (x0 ) Q2R (x0 ) Q2R (x0 ) 2 Q2R (x0 ) Then (3.3.10) follows from Lemma 2.4.1. CHAPTER Four Estimates for the Green functions and elliptic measure 4.1 Green’s functions In this section, we collect some results about the Green’s function and regular points. It turns out that the arguments in [GW82] carry forward with only a few modifications. Thus, we will mainly focus on these modifications. ´ Recall that B[u, v] = Ω (a∇u · ∇v + b∇u · ∇v). Theorem 4.1.1. There exists a unique function (the Green’s function) G : Ω × Ω → R ∪ {∞}, such that for each Y ∈ Ω and any r > 0 G(·, Y ) ∈ W 1,2 (Ω \ Br (Y )) ∩ W01,1 (Ω) (4.1.1) and for all φ ∈ W01,p (Ω) ∩ C(Ω) where p > n B[G(·, Y ), φ] = φ(Y ). (4.1.2) 64 65 . The Green’s function enjoys the following properties: for each Y ∈ Ω (G(X) = G(X, Y )) n G ∈ L n−2 ,∞ (Ω) with kGk n ,∞ ≤ C(n)λ−1 0 , (4.1.3) L n−2 n ∇G ∈ L n−1 ,∞ (Ω) with k∇Gk n ,∞ ≤ C(n, λ0 , Λ0 ), (4.1.4) L n−1 n G ∈ W01,k (Ω) for each k ∈ [1, ), and kGkW 1,k (Ω) ≤ C(n, λ0 , Λ0 , k). (4.1.5) n−1 0 For all X, Y ∈ Ω we have G(X, Y ) ≤ C(n, λ0 , Λ0 ) |X − Y |2−n ; (4.1.6) 1 and for all X, Y ∈ Ω satisfying |X − Y | ≤ 2 dist(Y, ∂Ω) G(X, Y ) ≥ C(n, λ0 , Λ0 ) |X − Y |2−n . (4.1.7) Proof. Let Y ∈ Ω be fixed. For fixed ρ > 0, (Bρ = Bρ (Y )), W01,2 (Ω) 3 φ 7→ φ Bρ is a bounded linear functional on W01,2 (Ω). By (2.3.5), (2.3.12) and the Lax-Milgram theorem, there exists a unique function Gρ ∈ W01,2 (Ω), such that for all φ ∈ W01,2 (Ω) B[Gρ , φ] = φ. (4.1.8) Bρ 66 Using the same argument as in [GW82], we obtain Gρ ≥ 0, and kGρ k n ,∞ ≤ C(n)λ−1 0 , (4.1.9) L n−2 keeping in mind that ˆ b∇Gρ ∇Gρ (Gρ )−2 = 0, Ωt where Ωt = {X ∈ Ω : Gρ (X) > t}. We also have a pointwise estimate for Gρ : Gρ (X) ≤ C(n, λ0 , Λ0 )λ−1 0 |X − Y | 2−n if |X − Y | ≥ 2ρ. (4.1.10) Here the only difference is that the constant depends also on Λ0 , since the constant in Corollary 3.1.3 has such dependence. Now we show k∇Gρ k n ,∞ ≤ C(n, λ0 , Λ0 ), (4.1.11) L n−1 (Ω) and it suffices to establish that ˆ |∇Gρ |2 ≤ C(n, λ0 , Λ0 )R2−n if R ≥ 4ρ. (4.1.12) Ω\BR Let η ∈ C ∞ satisfying η ≡ 1 outside of BR , η ≡ 0 in BR/2 and |∇η| ≤ C/R, and take φ = Gρ η 2 in (4.1.8). Then, as in the proof of Corollary 3.1.1, and using (4.1.10), we have ˆ ˆ  2−s n 2s ρ 2 Λ20 )R2( s0 −1) 2 ρ s λ0 |∇G | η ≤ C(n, λ0 , s)(1 + |G | 2−s Ω BR \BR/2 ≤ C(n, λ0 , Λ0 )R2−n . n We now consider the convergence of Gρ . By (4.1.11), we have for each k ∈ [1, n−1 ) a uniform bound on kGρ kW 1,k with respect to ρ. Thus there exists a G ∈ W01,k for all 0 67 n k ∈ [1, n−1 ) such that Gρµ * G weakly in W01,k , (4.1.13) and kGkW 1,k ≤ lim inf ρµ →0 kGρµ kW 1,k ≤ C(n, λ0 , Λ0 , k), which is (4.1.5). 0 0 Let φ ∈ W01,p (Ω) with p > n. We verify that B[·, φ] is a continuous linear functional on 0 W01,p where p0 = p p−1 < n n−1 . Since ˆ ˆ ˆ a∇u · ∇φ ≤ λ−1 ( |∇u|p0 )1/p0 ( |∇φ|p )1/p , 0 (4.1.14) Ω Ω Ω we use the BMO-extension ˜bij of bij ((2.1.4)) and the zero extensions u ˜ and φ˜ of u and φ to Rn to see that ˆ b∇u · ∇φ ≤ C kbk BM O(Ω) kukW 1,p0 (Ω) kφkW 1,p (Ω) . (4.1.15) Ω Combining (4.1.14) and (4.1.15) we obtain 0 |B[u, φ]| ≤ C kukW 1,p0 (Ω) kφkW 1,p (Ω) ∀u ∈ W01,p , (4.1.16) 0 which shows that B[·, φ] ∈ (W01,p )∗ . Since Gρ * G in W01,k for all k ∈ [1, n−1 n ), B[Gρ , φ] → B[G, φ]. (4.1.17) On the other hand, B[Gρ , φ] = φ → φ(y), Bρ so B[G, φ] = φ(y), which is (4.1.2). From here it is not hard to obtain (4.1.5). And (4.1.6) follows from the pointwise estimate (4.1.10) and H¨older continuity of G(·, y) in Ω \ {y}. 68 We now give the proof of (4.1.7): Let X, Y ∈ Ω, r = |X − Y | < 1 2 dist(Y, ∂Ω). Let η ∈ C0∞ (Ω) with supp η ⊂ B3r/2 (Y ) \ Br/4 (Y ) and η ≡ 1 in Br (Y ) \ Br/2 (Y ). Then replacing the test function in the proof of Corollary 3.1.1 by η, we get ˆ ˆ  2−s n 2s |∇G|2 ≤ C(n, λ0 , Λ0 , s)r2( s0 −1) s |G| 2−s . (4.1.18) Br \Br/2 B3r/2 \Br/4 Now consider a cut-off function ϕ ∈ C0∞ with supp ϕ ⊂ Br (Y ), ϕ ≡ 1 on Br/2 (Y ) and |∇ϕ| ≤ C/r. Then inserting ϕ in (4.1.2), we have ˆ ˆ 1= a∇G · ∇ϕ + b∇G · ∇ϕ. (4.1.19) Br Br We claim that ˆ b∇G · ∇ϕ ≤ C(n, λ0 , Λ0 , s) |X − Y |n−2 G(X, Y ). (4.1.20) Br Actually, ˆ ˆ b∇G · ∇ϕ = (b − (b)Br )∇G · ∇ϕ Br Br ˆ n ≤ Cr 2 −1 ( (b − (b)Br ) ) 2 1/2 ( |∇G|2 )1/2 Br Br \Br/2 n ˆ 2s  2−s + sn0 −2 2s ≤ C(n, λ0 , Λ0 , s)r 2 |G| 2−s B3r/2 \Br/4 ≤ C(n, λ0 , Λ0 )rn−2 sup G B3r/2 \Br/4 ≤ C(n, λ0 , Λ0 )rn−2 inf G B3r/2 \Br/4 ≤ C(n, λ0 , Λ0 ) |x − y|n−2 G(X, Y ), where we used (4.1.18) to obtain the second inequality and applied Harnack inequality in 69 B3r/2 \ Br/4 in the fourth inequality. Similarly, ˆ ˆ n a∇G · ∇ϕ ≤ λ−1 0 r 2 −1 ( |∇G|2 )1/2 ≤ C(n, λ0 , Λ0 ) |X − Y |n−2 G(X, Y ). Br Br \Br/2 Now by (4.1.19), 1 ≤ C(n, λ0 , Λ0 ) |X − Y |n−2 G(X, Y ), which gives (4.1.7). Proof of uniqueness: We only give the proof of (1.49) in [GW82], which is the only place something different occurs. But it again follows from a variation of Corollary 3.1.1 and Harnack inequality: ˆ ˆ  2−s 2 2( n0 −1) 2s s |∇u| ≤ C(n, λ0 , Λ0 )ρν s |u| 2−s Ω\B2ρν (Y ) B2ρν \Bρν ≤ C(n, λ0 , Λ0 )ρn−2 ν ( sup |u|)2 B2ρν \Bρν ≤ C(n, λ0 , Λ0 )ρνn−2 m∗ (ρν )2 , . where m∗ (ρ) = inf ∂Bρ (Y ) u. The standard relations hold between Green’s functions of the operator and its adjoint: Theorem 4.1.2. Let L∗ = − div A∗ ∇ be the adjoint operator to L and consider the Green function G and G∗ corresponding to L and L∗ . Then for all points X, Y ∈ Ω, we have G(X, Y ) = G∗ (Y, X). (4.1.21) 70 And the following representation formula. Proposition 4.1.1. For any X, Y ∈ Ω, ρ > 0 such that ρ < dist(Y, ∂Ω) Gρ (X, Y ) = G(X, Z)dZ. (4.1.22) Bρ (Y ) 4.2 Regular points In this section, we apply the method of [GW82] to investigate the regular points for operator L = − div A∇ with A satisfying (1.0.1) and (1.0.3). Definition 4.2.1. A point Q ∈ ∂Ω is said to be regular for L, if for any Lipschitz function h on ∂Ω, the W 1,2 solution u to Lu = 0 in Ω with boundary data h satisfies lim u(X) = h(Q). (4.2.1) X→Q,X∈Ω If every point Q ∈ ∂Ω is a regular point, then we say the domain Ω is regular. The main result is the following. Theorem 4.2.1. A point of ∂Ω is a regular point for L = − div A∇ with A satisfying (1.0.1) and (1.0.3) if and only if it is a regular point for the Laplacian. Let us first define capL of a compact set. Let E ⊂ Ω be compact and denote by KE the following closed convex subset of W01,2 (Ω) . KE = {v ∈ W01,2 (Ω) : v ≥ 1 on E in the sense of W 1,2 }. (4.2.2) Then there exists a unique solution u ∈ KE of the variational inequality B[u, v − u] ≥ 0 ∀v ∈ KE . (4.2.3) 71 The existence of u can be justified by Theorem 2.1 in [KS00], which only requires that the bilinear form B[·, ·] be coercive. This function u is called the equilibrium potential of E. Letting v = {u}1 , the trunca- tion of u at height 1 in (4.2.3), we get u ≤ 1 a.e. in Ω. Therefore, u≡1 on E in the sense of W 1,2 . (4.2.4) From (4.2.3) we get for all φ ∈ C0∞ with φ ≥ 0 on E B[u, φ] ≥ 0. (4.2.5) Then we immediately get Proposition 4.2.1. The equilibrium potential u of the compact subset E of Ω is the solu- tion to   Lu = 0  in Ω \ E, in the sense of W01,2 .  u|∂Ω = 0,  u|∂E = 1 By (4.2.5) and the Riesz representation theorem, there exists a positive regular Borel measure µ with supp µ ⊂ E such that ˆ B[u, φ] = φdµ ∀ φ ∈ C0∞ (Ω). (4.2.6) E (4.2.4) implies that supp µ ⊂ ∂E. The measure µ is called the equilibrium measure of E. Definition 4.2.2. The capacity of E with respect to the operator L is defined as . capL (E) = µ(E) = B[u, u]. (4.2.7) As in the classical elliptic setting, the capacity with respect to L can be compared to 72 the capacity with respect to the Laplacian. Precisely, λ0 cap∆ (E) ≤ capL (E) ≤ λ−1 0 C(n, λ0 , Λ0 ) cap∆ (E), (4.2.8) ´ where cap∆ (E) = inf{ Ω |∇u|2 : u ∈ C0∞ (Ω), u ≥ 1 on E}. To see this, simply observe that ˆ ˆ 12 1 cap∆ (E) ≤ |∇uL | ≤ A∇uL · ∇uL = capL (E), Ω λ0 Ω λ0 where we use uL to denote the equilibrium potential with respect to operator L of set E. On the other hand, we have capL (E) ≤ B[uL , u∆ ] ≤ C(n, λ0 , Λ0 ) k∇uL kL2 (Ω) k∇u∆ kL2 (Ω) 1 ˆ 1/2 ≤ C(n, λ0 , Λ0 ) A∇uL · ∇uL cap∆ (E)1/2 λ0 Ω −1 = C(n, λ0 , Λ0 )λ0 2 capL (E)1/2 cap∆ (E)1/2 , where u∆ is the potential for the Laplacian. The equilibrium potential u with respect to L of E ⊂ Ω can be represented as follows Lemma 4.2.2. Let µ be the equilibrium measure of E. Let G be the Green function on Ω. Then ˆ u(X) = G(X, Y )dµ(Y ). (4.2.9) E This can be proved by taking φ = Gρ (X, ·) in (4.2.6), and then using Fatou’s lemma. One direction of the inequality using Fatou’s lemma is immediate. For the other direction, note that we can show for Y ∈ Ωδ , X ∈ Ω, Gρ (X, Y ) ≤ C(n, λ0 , Λ0 )(G(X, Y ) + 1), where Ωδ = {X ∈ Ω : dist(X, ∂Ω) > δ} and δ = dist(E, ∂Ω)/10. After these preparations, the classical arguments go through and we obtain the follow- ing. 73 . Lemma 4.2.3. Fix Q ∈ ∂Ω (we may assume Q is the origin). For r > 0, set Cr = Ωc ∩ Br (O). Consider the weak solution u ∈ W 1,2 (Ω) of Lu = 0 in Ω, 0 ≤ u ≤ 1 on ∂Ω 1 and u = 0 on ∂Ω ∩ Bρ . There exist constants 0 < α0 (n, λ0 , Λ0 ) < 2 and C(n, λ0 , Λ0 ) > 0 such that for all α ≤ α0 and r < αρ we have n C ˆ αρ cap (C ) o L t sup u ≤ exp dt . (4.2.10) Ω∩Br log α r tn−1 Theorem 4.2.4. The point Q ∈ ∂Ω (we may assume Q is the origin) is a regular point for Lu = 0 if and only if ˆ capL (Cr ) dr = ∞, (4.2.11) 0 rn−1 where Cr = Ωc ∩ Br (O). Then Theorem 4.2.1 is a consequence of Theorem 4.2.4 and (4.2.8). 4.3 Elliptic measure 1,2 Let g ∈ C(∂Ω), X ∈ Ω. We showed in Section 3.2 that there exists a unique u ∈ Wloc (Ω) ∩ C(Ω) such that Lu = 0 in Ω and u = g on ∂Ω. Consider the linear functional on C(∂Ω) T : g 7→ u(X). By Lemma 3.2.6 and Lemma 3.2.7, T is a positive and bounded linear functional. Moreover, if g ≡ 1, u ≡ 1. By the Riesz representation theorem, there exists a family of regular Borel probability measures {ωLX }X∈Ω such that ˆ u(X) = g(P )dωLX (P ). (4.3.1) ∂Ω This family of measures is called the elliptic measure associated to L, or L-harmonic measure. We omit the reference to L when no confusion arises. For a fixed X0 ∈ Ω, let 74 ω = ω X0 . We list some properties of the elliptic measure. Since the main tools used here are Harnack principle and H¨ older continuity of the solution, most of the properties can be proved as in [Ken94]. We only provide a proof when needed. In what follows C is a constant depending on n, λ0 , Λ0 and the constants M, m, m0 in Definition 2.1.1 and 2.1.2 unless otherwise stated. Its value may vary from line to line. Proposition 4.3.1. ∀ X1 , X2 ∈ Ω, ω X1 and ω X2 are mutually absolutely continuous. Recall that for Q ∈ ∂Ω, Ar (Q) is the corkscrew point in Ω such that |Ar (Q) − Q| < r r and δ(Ar (Q)) > M. Proposition 4.3.2. Let Q ∈ ∂Ω. Then ω Ar (Q) (∆r (Q)) ≥ C. Proposition 4.3.3. Suppose that u ≥ 0, Lu = 0, u ∈ W 1,2 (T2r (Q)) ∩ C(T2r (Q)), u ≡ 0 on ∆2r (Q). Then u(X) ≤ Cu(Ar (Q)) ∀ X ∈ T3r/2 (Q). For a proof see [JK82] Lemma 4.4. Proposition 4.3.4. rn−2 G(X, Ar (Q)) ≤ Cω X (∆2r (Q)) ∀ X ∈ Ω \ Br/2 (Ar (Q)), where G is the Green’s function defined in Section 4.1. Proposition 4.3.5. ω X (∆r (Q)) ≤ C(n, λ0 , Λ0 )rn−2 G(X, Ar (Q)) ∀ X ∈ Ω \ B2r (Q). Proof. Let ϕ ∈ C0∞ (Rn ), u be the weak solution of   Lu = 0  in Ω,  u = ϕ  on ∂Ω. Then as we showed in Section 3.2, u ∈ W 1,2 (Ω) ∩ C(Ω), u − ϕ ∈ W01,2 (Ω). 75 Claim. For any Y ∈ / supp ϕ, ˆ u(Y ) = − (a + b)∇ϕ(X) · ∇G∗ (X, Y )dX, (4.3.2) Ω where G∗ (X, Y ) is the Green function of the adjoint operator L∗ = div AT ∇. To justify this claim, consider (G∗ )ρ ∈ W01,2 (Ω). We have (see (4.1.8)) B[φ, (G∗ )ρ ] = φ ∀ φ ∈ W01,2 (Ω), (4.3.3) Bρ (Y ) and (see (4.1.17)) B[φ, (G∗ )ρ ] → B[φ, G∗ ] as ρ → 0, ∀ φ ∈ W01,p (Ω) with p > n. (4.3.4) So ˆ (u − ϕ) = A∇(u − ϕ) · ∇(G∗ )ρ . (4.3.5) Bρ (Y ) Ω On the other hand, since u is a solution, ˆ A∇u · ∇(G∗ )ρ = 0. Ω ´ ∗ ρ ´ Then the right-hand side of (4.3.5) equals − Ω A∇ϕ·∇(G ) , which goes to − Ω A∇ϕ· ∇G∗ as ρ → 0, while the left-hand side of (4.3.5) goes to u(Y ). Thus we obtain (4.3.2). C Now pick ϕ ≡ 1 in Br (Q), with supp ϕ ⊂ B3r/2 (Q), 0 ≤ ϕ ≤ 1 and |∇ϕ| ≤ r. Then by the claim, ˆ ˆ Y ω (∆r (Q)) ≤ Y ϕ(P )dω (P ) ≤ u(Y ) = − (a + b)∇ϕ(X) · ∇G∗ (X, Y )dX. (4.3.6) ∂Ω Ω 76 We estimate ˆ  1/2  1/2 a∇ϕ · ∇G∗ ≤ C(n, λ0 )rn−1 |∇G∗ |2 ≤ Crn−2 |G∗ |2 Ω T3r/2 (Q) T7r/4 (Q) ≤ Crn−2 G∗ (Ar (Q), Y ) = Crn−2 G(Y, Ar (Q)). (4.3.7) Choose η ∈ C0∞ (Rn ) with η ≡ 1 on B3r/2 (Q), supp η ⊂ B 7r (Q), and |∇η| ≤ C r. Then 4 using the properties of η and ϕ, we have ˆ ˆ b∇ϕ · ∇G∗ = ˜b∇ϕ · ∇(G∗ η), Ω Rn where ˜b ∈ BM O(Rn ) is the extension of b, and we identify G∗ (·, Y ) with its zero extension outside of Ω. / B2r (Q), G∗ (·, Y ) ∈ W 1,2 (T3r/2 (Q)). So we can apply Proposition Note that since Y ∈ 2.2.1 and get ˆ b∇ϕ · ∇G∗ ≤ CΛ0 k∇ϕk 2 n k∇(G∗ η)k 2 n L (R ) L (R ) Ω n ≤ CΛ0 r 2 −1 (kη∇G∗ kL2 (Rn ) + kG∗ ∇ηkL2 (Rn ) ) n n ˆ 1 C  ˆ 1 o 2 −1 ∗ |G∗ |2 2 2 ≤ CΛ0 r 2 |∇G | + T7r/4 (Q) r T7r/4 (Q)  1/2 ≤ CΛ0 rn−2 |G∗ |2 ≤ CΛ0 rn−2 G(Y, Ar (Q)), (4.3.8) T2r (Q) where C depending on dimension and Ω. Combining (4.3.7) and (4.3.8) we have proved ω Y (∆r (Q)) ≤ Crn−2 G(Y, Ar (Q)), for any Y ∈ Ω \ B2r (Q), where the constant C depending on n, λ0 , Λ0 and the domain. Corollary 4.3.1. For X ∈ Ω \ B2r (Q), ω X (∆r (Q)) ≈ rn−2 G(X, Ar (Q)). Corollary 4.3.2. For X ∈ Ω \ B2r (Q), ω X is a doubling measure, i.e. ω X (∆2r (Q)) ≤ 77 Cω X (∆r (Q)). Proposition 4.3.6 (Comparison principle). Let u, v ≥ 0, Lu = Lv = 0, u, v ∈ W 1,2 (T2r (Q))∩ C(T2r (Q)), u, v ≡ 0 on ∆2r (Q). Then ∀ X ∈ Tr (Q), u(Ar (Q)) u(X) u(Ar (Q)) C −1 ≤ ≤C v(Ar (Q)) v(X) v(Ar (Q)) Corollary 4.3.3. Let u, v be as in Proposition 4.3.6. Then there exists α = α(n, λ0 , Λ0 ) such that ≤ C u(Ar (Q)) |X − Y | u(Y ) u(X)  α − v(Y ) v(X) ∀ X, Y ∈ Tr (Q). v(Ar (Q)) r Corollary 4.3.4. Let ∆ = ∆r (Q), ∆0 = ∆s (Q0 ) ⊂ ∆r/2 (Q), X ∈ Ω \ T2r (Q). Then ω X (∆0 ) ω Ar (Q) (∆0 ) ≈ . ω X (∆) dω X The kernel function K(X, Q) is defined to be K(X, Q) = dω (Q), the Radon-Nikodym derivative of ω X with respect to ω. It satisfies the following estimates: Proposition 4.3.7. 1. If X is a corkscrew point relative to ∆r (Q), namely, X ∈ Γα (Q) 1 with |X − Q| ≈ dist(X, ∂Ω) ≈ r, then K(X, P ) ≈ ω(∆r (Q)) for all P ∈ ∆r (Q). 2. Let ∆j = ∆2j r (Q), Rj = ∆j \ ∆j−1 , j ≥ 0. Then 2−αj sup K(Ar (Q), P ) ≤ C , P ∈Rj ω(∆j ) where α = α(n, λ0 , Λ0 , M ) is the same as in Lemma 3.2.5. 3. Let ∆ = ∆r (Q) be any boundary disk centered at Q ∈ ∂Ω. Then sup K(X, P ) → 0 as X → Q. P ∈∂Ω\∆ 78 4. K(X, ·) is a H¨ older continuous function. Namely, for all X ∈ Ω, |K(X, Q1 ) − K(X, Q2 )| ≤ CX |Q1 − Q2 |α , where α > 0 is a constant depending on n, λ0 , Λ0 , M, m, m0 , and CX > 0 depends additionally on X. CHAPTER Five The Lp Dirichlet problem 5.1 Dirichlet problem with Lp (dω) data Definition 5.1.1. A cone of aperture α is a non-tangential approach region for Q ∈ ∂Ω of the form Γα (Q) = {X ∈ Ω : |X − Q| ≤ (1 + α) dist(X, ∂Ω)}, and a truncated cone is defined by . Γhα (Q) = Γα (Q) ∩ Bh (Q). Recall that the non-tangential maximal function is defined as N α u(Q) = sup |u(X)| , Γα (Q) 79 80 and we shall imply write N u for N α u when the aperture is unimportant. We define the truncated non-tangential maximal function as . Nh u(Q) = sup |u(X)| . Γh α (Q) We also write . N ∗h u(Q) = sup |u(X)| . Γα (Q)\Bh (Q) By Proposition 4.3.7 (1)–(3), we obtain (see [CFMS81] or [Ken94]) ´ Lemma 5.1.1. Let ν be a finite Borel measure on ∂Ω, and u(X) = ∂Ω K(X, Q)dν(Q). Then for each P ∈ ∂Ω, N u(P ) ≤ Cα Mω (ν)(P ), 1 where Mω (ν)(P ) = sup∆3P ω(∆) |ν| (∆), and Cα = C(n, λ0 , Λ0 , α). Moreover, if ν ≥ 0, then Mω (ν)(P ) ≤ CN u(P ). We consider the Lp (dω) Dirichlet problem in Ω, where ω = ωL is the elliptic measure associated to L = − div A∇ with A satisfying (1.0.1) and (1.0.3).      Lu = 0 in Ω,   u → g ∈ Lp (∂Ω, dω) non-tangentially a.e. dω on ∂Ω    N α u ∈ Lp (∂Ω).   It turns out that the Lp (dω) Dirichlet problem is uniquely solvable. Theorem 5.1.2. Given f ∈ Lp (∂Ω, dω), 1 < p ≤ ∞, there exists a unique u with Lu = 0, N u ∈ Lp (∂Ω, dω), which converges non-tangentially a.e. (dω) to f . Proof. Existence. The existence of the solution can be proved as in the case where A is elliptic, bounded and measurable. Namely, choose {fn }∞ n=1 ⊂ C(∂Ω) such that fn → f in Lp (∂Ω, dω). Let un be the corresponding solution to the Dirichlet problem with boundary 81 data fn . Then ˆ un (X) = fn (Q)K(X, Q)dω, ∂Ω and by Lemma 5.1.1, N (un )(P ) . Mω (fn )(P ) ∀ P ∈ ∂Ω. So we have kN (un − um )kLp (dω) . kMω (fn − fm )kLp (dω) . kfn − fm kLp (dω) → 0, (5.1.1) as n, m → ∞ (we have used the Lp boundedness of the maximal function). Given any compact subset K in Ω, un is uniformly Cauchy in L∞ (K). In fact, by the maximal principle, we can assume that for fixed n and m, the supremum supX∈K |un (X) − um (X)| is attained at some point Y ∈ ∂K. Then there exists a set ∆Y ⊂ ∂Ω such that ∀ P ∈ ∆Y , Y ∈ Γα (P ), and σ(∆Y ) ≈ δ(Y )n−1 . By (5.1.1) and the doubling property of ω, we have ˆ 1 sup |un (X) − um (X)| ≤ N (un − um )dω . kN (un − um )kLp (dω) → 0 X∈K ω(∆Y ) ∆Y as n, m → ∞. Hence u(X) = limn→∞ un (X), for X ∈ Ω is pointwise well-defined. From ´ here it is not hard to see that u is the weak solution to Lu = 0, u(X) = ∂Ω K(X, Q)f (Q)dω(Q), as well as that u converges non-tangentially a.e. (dω) to f . Uniqueness. Assume N u ∈ Lp (∂Ω, dω) with p > 1, and u converges to 0 non- ´ tangentially a.e. (dω). Then ∂Ω Nh udω → 0 as h → 0. Since K(X, ·) is H¨older continuous, ˆ Nh udω X → 0 as h → 0 for any fixed X ∈ Ω. ∂Ω It suffices to show that for any fixed X0 ∈ Ω, u(X0 ) = 0. We first need some geometric observations. 82 Denote Ωδ = {Y ∈ Ω : dist(Y, ∂Ω) > δ}. For δ > 0 small, consider the annulus ΩC1 δ \ Ω(C1 + √1 )δ , where C1 = C1 (n, ∂Ω, α) > 30 is some constant. Cover the annulus by n balls Bδ (Y ) with Y ∈ ∂ΩC1 δ . Since the annulus is compact, there exists N0 = N0 (δ, ∂Ω, n) such that N0 [ ΩC1 δ \ Ω(C1 + √1 )δ ⊂ Bδ (Yk ), n k=1 with Yk ∈ ∂ΩC1 δ , k = 1, 2, . . . , N0 . We can find a C2 = C2 (n, ∂Ω) > 1, such that for any . 1 ≤ k ≤ N0 , there exists a surface ball ∆k = ∆sk (Qk ) with Cδ1 ≤ sk ≤ δ and |Qk − Yk | = dist(Yk , ∂Ω)(= C1 δ), such that B20δ (Yk ) ⊂ Γα (P ), ∀ P ∈ ∆k . It is easy to see that for any X ∈ B20δ (Yk ) and any P ∈ ∆k , |X − P | ≤ 2C1 δ. So B20δ (Yk ) ⊂ Γ2C α 1δ (P ), ∀ P ∈ ∆k . (5.1.2) Choose a subcollection of disjoint balls {Bδ (Ykj )}N 1 j=1 such that N0 [ N1 [ Bδ (Yk ) ⊂ B5δ (Ykj ), (5.1.3) k=1 j=1 for some N1 = N1 (δ, n, ∂Ω). We claim that there exists a kˆ < ∞ depending only on the dimension and the bound- ary of Ω, such that there are at most kˆ overlaps of {∆kj }N1 j=1 . Actually, it is clear that Yk − Yk ≥ 90δ implies Qk − Qk ≥ C1 δ. So by the disjointness of {Bδ (Yk )}N1 , the j l j l j j=1 claim is established. Now we choose δ sufficiently small so that X0 ∈ Ω2C1 δ . Let ϕ ∈ C0∞ (ΩC1 δ ) with ϕ ≡ 1 83 C C in Ω(C1 + √1 )δ , |∇ϕ| ≤ δ and D2 ϕ ≤ δ2 . We establish the following n ˆ   ˆ   u(X0 ) = − A∇u(Y ) · ∇ϕ(Y ) G(X0 , Y )dY + A∇ϕ · ∇G(X0 , Y ) u(Y )dY. (5.1.4) Ω Ω To see this, let Gρ be as in section 5, (4.1.8), and define G∗ρ similarly, corresponding to A∗ . By definition, we have ˆ ˆ u(Y )ϕ(Y )dY = ρ A∇G (Y, X0 ) · ∇(uϕ)dY = A∇(uϕ) · ∇G∗ρ (Y, X0 )dY Bρ (X0 ) Ω Ω ˆ   ˆ   =− A∇u(Y ) · ∇ϕ(Y ) G∗ρ (Y, X0 )dY + A∇ϕ(Y ) · ∇G∗ρ (Y, X0 ) u(Y )dY, Ω Ω ´ where in the last inequality we have used Ω A∇u · ∇(ϕG∗ρ ) = 0. Note that for any fixed 0 < r < dist(X0 , ∂Ω), ∇G∗ρ * ∇G∗ weakly in L2 (Ω \ Br (X0 )), G∗ρ * G∗ weakly in Lp (Ω \ Br (X0 )) for some p > 2, as ρ → 0. Now by the support of ∇ϕ, b ∈ Lp (Ω) for any 1 ≤ p < ∞, and that u ∈ C(Ω), letting ρ → 0 gives (5.1.4). Using (5.1.4), we calculate ˆ ˆ u(X0 ) = − A∇u · ∇ϕG + A∇ϕ · ∇Gu ˆ ˆ ˆ ˆ  =− a∇u · ∇ϕG + a∇ϕ · ∇Gu + b∇ϕ · ∇uG + b∇ϕ · ∇Gu ˆ ˆ ˆ =− a∇u · ∇ϕG + a∇ϕ · ∇Gu + b∇ϕ · ∇(uG) . = I1 + I2 + I3 . (5.1.5) ´ X0 . We show that the above expression can be bounded by ∂Ω N2C1 δ (u)dω Let {ηj }N N1 j=1 be a partition of unity subordinate to {B6δ (Ykj )}j=1 , with |∇ηj | ≤ 1 C δ, N1 X ηj (X) = 1 ∀ X ∈ ΩC1 δ \ Ω(C1 + √1 )δ (5.1.6) n j=1 84 Using Caccioppli’s inequality, Harnack principle, (5.1.2), Corollary 4.3.1, and the bounded overlaps of {∆kj }N1 j=1 , we estimate N1 ˆ X |I1 | = ηj (X)a(X)∇u(X) · ∇ϕ(X)G(X0 , X)dX j=1 Ω N1 ˆ X . δ −1 |∇u(X)| G(X0 , X)dX j=1 B6δ (Ykj ) N1  ˆ 1/2 n X . δ 2 −1 |∇u|2 sup G(X0 , X) j=1 B6δ (Ykj ) X∈B6δ (Ykj ) N1  ˆ 1/2 N1 n X X . δ 2 −2 |u|2 G(X0 , Aδ (Qkj )) . inf N2C1 δ (u)(P )ω X0 (∆δ (Qkj )) B10δ (Ykj ) P ∈∆kj j=1 j=1 N1 ˆ X ˆ . N2C1 δ (u)(P )dω X0 . N2C1 δ (u)dω X0 , j=1 ∆k j ∂Ω where Aδ (Qkj ) ∈ Γα (Qkj ) is the corkscrew point with dist(Aδ (Qkj ), ∂Ω) ≈ δ. I2 can be estimated in a similar manner. To estimate I3 , we write it as ˆ XN1 N1 ˆ X I3 = b∇( ηj ϕ) · ∇(uG) = b∇(ηj ϕ) · ∇(ξj uG), Ω j=1 j=1 Ω where ξj ∈ C0∞ (B7δ (Ykj )) with ξj ≡ 1 in B6δ (Ykj ) and |∇ξj | ≤ C/δ. We have used here the support property of ∇ϕ and (5.1.6). n By Sobolev inequality, uGξj ∈ W 1,p (Rn ) for some 1 < p ≤ n−1 . Then we can apply Proposition 2.2.1 and get N1 X |I3 | . Λ0 k∇(ηj ϕ)kLp0 k∇(uGξj )kLp j=1 N1 n n ˆ 1/p ˆ 1/p ˆ 1/p o −1 X . Λ0 δ p0 |u∇G|p + |G∇u|p + δ −1 |uG|p j=1 B7δ (Ykj ) B7δ (Ykj ) B7δ (Ykj ) (5.1.7) 85 Now we can estimate I3 as we did for I1 , namely, n ˆ 1/p n ˆ 1/2  ˆ 2p  2−p p0 −1 p p0 −1 2 2p δ |u∇G| .δ |∇G| |u| 2−p B7δ (Ykj ) B7δ (Ykj ) B7δ (Ykj ) . δ n−2 sup G(X0 , X) inf N2C1 δ (u)(P ) X∈B7δ (Ykj ) P ∈∆kj ˆ . N2C1 δ (u)dω X0 , ∆kj while the other two integrals in the right-hand side of (5.1.7) can be estimated similarly. Thus N1 ˆ X ˆ X0 |I3 | . Λ0 N2C1 δ (u)dω . Λ0 N2C1 δ (u)dω X0 . j=1 ∆kj ∂Ω where in the last inequality we have used that there are at most kˆ overlaps of {∆kj }N 1 j=1 . ´ Altogether we obtain |u(X0 )| ≤ C(n, ∂Ω)(1 + Λ0 ) ∂Ω N2C1 δ (u)dω X0 . Letting δ go to zero, we get u(X0 ) = 0. Remark 5.1.1. We can show that if f ∈ L1 (∂Ω, dω), then there exists a u with Lu = 0, N u ∈ L1,∞ (Ω, dω) which converges non-tangentially a.e. (dω) to f . But such u is not unique. 5.2 Dirichlet problem with Lp (dσ) data We now consider the Dirichlet problem with Lp (dσ) boundary data. Since we are con- sidering the surface measure σ, we should work on a domain where a surface measure is well-defined on its boundary. We are content with ourselves by assuming the domain Ω ⊂ Rn is Lipschitz. Definition 5.2.1. We say the Dirichlet problem with Lp (dσ) boundary data is solvable if 86 for any f ∈ C(∂Ω) ∩ Lp (∂Ω, dσ), the weak solution to the continuous Dirichlet problem   Lu = 0  in Ω,  u = f  on ∂Ω satisfies the estimate kN ukLp (dσ) . kf kLp (dσ) , (5.2.1) where the implicit constant does not depend on f . The following proposition shows that this is indeed an appropriate definition. Proposition 5.2.2. Suppose (5.2.1) holds for weak solution to the continuous Dirich- let problem for L = − div A∇ with A satisfying (1.0.1) and (1.0.3). Then given f ∈ Lp (∂Ω, dσ), there exists a unique u such that      Lu = 0 in Ω,   u → g ∈ Lp (∂Ω, dσ) non-tangentially a.e. dσ on ∂Ω    N u ∈ Lp (∂Ω, dσ).   Proof. The proof has a similar spirit as that of the proof of Theorem 5.1.2. Existence. Choose fn ∈ C(∂Ω) with fn → f ∈ Lp (∂Ω, dσ). Let un be the correspond- ing weak solution with boundary data fn . Let K ⊂ Ω be any compact set. Then as we argue in the proof of Theorem 5.1.2, there is some Y ∈ ∂K such that supX∈K |un (X) − um (X)| reaches its supremum at Y . Then there exists a set ∆Y ⊂ ∂Ω, such that for all P ∈ ∆Y , Y ∈ Γα (P ). And σ(∆Y ) ≈ dist(Y, ∂Ω)n−1 . So we have ˆ 1 sup |un (X) − um (X)| ≤ N (un − um )dσ X∈K ω(∆Y ) ∆Y  ˆ 1/p 1 p ≤ |N (un − um )| dσ ≤ CK kN (un − um )kLp (∂Ω,dσ) σ(∆Y ) ∆Y ≤ CK kfn − fm kLp (∂Ω,dσ) → 0 87 as m, n → ∞. This implies that un converges uniformly in K. Since K is arbitrary, un converges to some u ∈ C(Ω) in Ω pointwise. And un → u in L2 (K). By Caccioppoli’s 1,2 inequality, one can show un also converges in Wloc . Therefore, the limit u is a weak solution to L in Ω. We now show kN ukLp (∂Ω,dσ) ≤ C kf kLp . Observe that the uniform convergence of un on compact sets implies that for any Q ∈ ∂Ω, any δ > 0, N ∗δ (u)(Q) = limn→∞ N ∗δ (un )(Q). ∗δ ⇒ N u ≤ lim N ∗δ (un ) ≤ lim kN (un )kLp (∂Ω,dσ) Lp (∂Ω,dσ) n→∞ Lp (∂Ω,dσ) n→∞ ≤ C lim kfn kLp (∂Ω,dσ) = C kf kLp (∂Ω,dσ) . n→∞ ⇒ kN ukLp (∂Ω,dσ) ≤ lim N ∗δ u ≤ C kf kLp (∂Ω,dσ) . δ→0 Lp (∂Ω,dσ) Similarly, one has kN (u − un )kLp (∂Ω,dσ) ≤ C kf − fn kLp (∂Ω,dσ) , and thus there exists a set E ⊂ ∂Ω with σ(E) = 0 such that N (u − un )(Q) + |f (Q) − fn (Q)| → 0 ∀ Q ∈ ∂Ω \ E. Now let X ∈ Γ(Q) with Q \ E, we have |u(X) − f (Q)| ≤ |u(X) − un (X)| + |un (X) − fn (Q)| + |fn (Q) − f (Q)| → 0 as n → ∞. This proves that u converges to f non-tangentially a.e. (dσ) on ∂Ω. Uniqueness. To show uniqueness, we assume (5.2.1) holds for L, that N u ∈ Lp (∂Ω, dσ), and that u converges to 0 non-tangentially a.e. dσ. We want to show for any fixed X0 ∈ Ω, 88 u(X0 ) = 0. We proceed as in the proof of Theorem 5.1.2, and obtain that ˆ |u(X0 )| . N2C1 δ (u)dω X0 , ∂Ω where C1 and δ are as in the proof of Theorem 5.1.2. Note that by (5.2.1), Lemma X 0 5.1.1 and Lemma 2.5.1 item 7, we have k = dωdσ 0 ∈ Lp (dσ). Therefore, we can bound ´ X0 by (up to a constant) kN ∂Ω N2C1 δ (u)dω 2C1 δ ukLp (∂Ω,dσ) , which goes to 0 as δ → 0. Now, we can bring in the theory of weights due to Proposition 5.2.2, Lemma 5.1.1 and Lemma 2.5.1, and obtain the following corollary that characterizes the solvability of the Lp Dirichlet problem. Corollary 5.2.1. Let L = − div A∇ with A satisfying (1.0.1) and (1.0.3). Let Ω be a Lipschitz domain. Then the Dirichlet problem with Lp (σ) boundary data in Ω with 1 < p < ∞ is uniquely solvable if and only if ωL ∈ Bp0 (dσ). And the Dirichlet problem with Lp (σ) boundary data is uniquely solvable for some p ∈ (1, ∞) (sufficiently large) if and only if ωL ∈ A∞ (dσ). CHAPTER Six Characterization of the A∞ condition 6.1 Necessary condition for the A∞ in terms of square func- tion and non-tangential maximal function Definition 6.1.1. The square function with aperture determined by α is defined as ˆ 1/2 Sα u(Q) = |∇u(X)|2 δ(X)2−n dX , Γα (Q) where δ(X) = dist(X, ∂Ω). The truncated square function Sα,h u(Q) is defined similarly, integrating over the truncated cone Γhα (Q). The main lemma of this section is the following: ´ Lemma 6.1.1. Let Ω be an NTA domain. Let f ∈ L2 (∂Ω, dω), u(X) = ∂Ω f (Q)dω X (Q), where ω is the elliptic measure associated to the operator L = − div A∇ satisfying (1.0.1) 89 90 and (1.0.3). Let X0 ∈ Ω. Then ˆ ˆ 1 1 A(Y )∇u(Y ) · ∇u(Y )G(X0 , Y )dY = − u(X0 )2 + f (Q)2 dω X0 (Q). (6.1.1) Ω 2 2 ∂Ω This lemma has been proved for harmonic functions u in an NTA domain using a classical potential theoretic result of Riesz. See Theorem 5.14 in [JK82] for details. We present here a proof that does not use the Riesz theorem, and for solutions u to Lu = 0 in Ω. Once the lemma is proved, an argument in [DJK84] can be verified to work for operators L = − div A∇ with A satisfying (1.0.1) and (1.0.3). Thus we are able to obtain Theorem 6.1.2. Let Ω be a bounded NTA domain in Rn , n ≥ 3. Let µ be a positive measure satisfying A∞ with respect to ω X0 . Let u be the solution to Lu = 0 in Ω. Then for all 0 < p < ∞, ˆ 1/p ˆ 1/p p (Sα (u)) dµ ≤C (N u)p dµ ∂Ω ∂Ω where C depends on n,λ0 ,Λ0 ,p,α,Ω and X0 . If, in addition, u(X0 ) = 0, then ˆ 1/p ˆ 1/p p (N u) dµ ≤C (Sα (u))p dµ . ∂Ω ∂Ω Remark 6.1.1. Let Ω be a domain where the surface measure is well-defined on its bound- ary, say, Ω is a Lipschitz domain. Then this theorem implies that if u(x0 ) = 0 and ω x0 ∈ A∞ (dσ), then kSukLp (∂Ω,dσ) ≈ kN ukLp (∂Ω,dσ) . Before proving Lemma 6.1.1, we provide Green’s formula in the following form: Lemma 6.1.3 (Green’s formula). Let Ω be a domain with ∂Ω ∈ C 2 . Let A be a matrix with entries belong to C 0,1 (Rn ). Let L = − div A∇, and the adjoint operator L∗ = − div A∗ ∇. 0 Let G∗ be the Green function corresponding to L∗ . Then for all u ∈ W 1,p (Ω) with 1 < p < 91 n 1 1 n−1 and p0 + p = 1, ˆ ˆ ∗ ∗ u(X) = A (Y )∇G (Y, X)·∇u(Y )dY − A∗ (Q)∇G∗ (Q, X)· N ~ (Q)u(Q)dσ(Q) (6.1.2) Ω ∂Ω ∀ X ∈ Ω, where N (Q) is the outward unit normal at Q ∈ ∂Ω. Proof. For any  > 0, denote Ω = {X ∈ Ω : δ(X) > }. Choose η ∈ C0∞ (Ω) such that 0 ≤ η ≤ 1, and that   1  in Ω , η =  0  in Ω \ Ω 2 . Let  be sufficiently small, then ˆ ˆ   ˆ ∗ ∗ ∗ ∗ A (Y )∇G (Y, X) · ∇u(Y )dY = A ∇G · ∇ (1 − η )u dY + A∗ ∇G∗ · ∇(η u)dY Ω ˆ   Ω Ω = A∗ ∇G∗ · ∇ (1 − η )u dY + u(X) Ω\Ω2 ˆ ˆ ∗ ∗ =− div(A ∇G )(1 − η )udY + A∗ ∇G∗ · N ~ udσ + u(X) Ω\Ω2 ∂Ω The assumptions on the regularity of the boundary and the coefficients imply that G∗ ∈ W 2,2 (Ω \ Ω2 ), so that div(A∗ ∇G∗ ) is uniformly bounded in . Thus, as  → 0, ˆ − div(A∗ ∇G∗ )(1 − η )udY → 0. Ω\Ω2 Then the lemma follows. Remark 6.1.2. From the proof one can see that this lemma is also true for u ∈ W 1,2 (Ω) ∩ 1,p 0 0 Wloc (Ω). Also, note that if u ∈ W01,p (Ω), then the lemma is simply the definition of Green function. Corollary 6.1.1. Let A be a n × n matrix with entries belong to C k,1 (Rn ) with k ≥ 0 and 92 k> n 2 − 2. Let Ω, L , L∗ , G∗ be as in Lemma 6.1.3. Then dωLX (Q) = −A∗ (Q)∇G∗ (Q, X) · N ~ (Q)dσ. (6.1.3) Proof. For any ϕ ∈ Lip(∂Ω), let u ∈ W 1,2 (Ω) be the weak solution of   Lu = 0  in Ω,  u = ϕ  on ∂Ω. k+2,2 Since aij ∈ C k,1 (Rn ), u ∈ Wloc (Ω) ∩ C(Ω) (see eg. [GT01] Theorem 8.10). By the 1,p 0 n 1 1 Sobolev inequality, u ∈ Wloc (Ω) for some 1 < p < n−1 and p0 + p = 1. Then ˆ ˆ ∗ ∗ A (Y )∇G (Y, X) · ∇u(Y )dY = A(Y )∇u(Y ) · ∇G(X, Y )dY = 0. Ω Ω By Lemma 6.1.3, ˆ u(X) = − A∗ (Q)∇G∗ (Q, X) · N ~ (Q)ϕ(Q)dσ(Q), ∂Ω ´ ´ ⇒ X ∗ (Q)∇G∗ (Q, X) ~ (Q)ϕ(Q)dσ(Q). ∂Ω ϕ(Q)dωL (Q) = − ∂Ω A ·N Then a standard limiting argument completes the proof. We now prove Lemma 6.1.1. Proof. We first point out that the expression A∇u · ∇uG equals a∇u · ∇uG a.e. since b∇u · ∇u = 0 a.e.. In other words, the integral on the left-hand side of (6.1.1) is actually ´ Ω a(Y )∇u(Y ) · ∇u(Y )G(X0 , Y )dY . Let X0 ∈ Ω be any fixed point. We prove the lemma in 4 steps. Step 1. Assume ∂Ω ∈ C 2 , A = (aij ) with aij ∈ C ∞ (Rn ), (Aξ) · ξ ≥ λ0 |ξ|2 . Let 93 f ∈ C(∂Ω). Then (6.1.1) holds. To see this, we calculate, by Corollary 6.1.1 ˆ   ˆ − div u2 (Y )A∗ (Y )∇G∗ (Y, X0 ) dY = − f 2 A∗ (Q)∇G∗ (Q, X0 ) · N ~ (Q)dσ(Q) Ω ˆ ∂Ω = f 2 dω X0 . ∂Ω On the other hand, ˆ   ˆ ˆ − div u (Y )A (Y )∇G (Y, X0 ) dY = − div(A ∇G )u dY − 2 u∇u · (A∗ ∇G∗ )dY 2 ∗ ∗ ∗ ∗ 2 Ω ˆ   Ω Ω = u2 (X0 ) − 2 u(Y ) A(Y )∇u(Y ) · ∇G(X0 , Y )dY ˆΩ = u2 (X0 ) + 2 A(Y )∇u(Y ) · ∇u(Y )G(X0 , Y )dY, Ω where in the second equality we have used Lemma 6.1.3. Then (6.1.1) follows. Step 2. Let Ω a bounded NTA domain, A = (aij ) with aij ∈ C ∞ (Rn ), (Aξ) · ξ ≥ λ0 |ξ|2 , and let f ∈ Lip(∂Ω). There exists a sequence of domains {Ωt } with ∂Ωt ∈ C 2 , Ωt ⊂⊂ Ω such that Ωt increases to Ω as t → ∞. Let u be the solution to Lu = 0 in Ω with boundary data f . Then u ∈ C(Ω) ∩ C ∞ (Ω), and in particular, u ∈ C ∞ (∂Ωt ). So by the result we obtained in Step 1, we have ˆ ˆ 1 1 A(Y )∇u(Y ) · ∇u(Y )Gt (X0 , Y ) = − u2 (X0 ) + u2 dωtX0 , (6.1.4) Ωt 2 2 ∂Ωt where Gt is the Green function for domain Ωt , and ωt is the elliptic measure on domain Ωt . Extending G∗t to be zero in Ω \ Ωt . We still denote this extension as G∗t . 94 We now discuss the convergence of G∗t . By the estimate for Green function, we have kG∗t kW 1,k (Ω) ≤ Ck for all t, (6.1.5) where 1 < k < n ˜∗ So there exists a sequence tn → ∞ such that G∗tn converges to a G n−1 . weakly in W01,k , ⇒ there exists a subsequence, still denoted by tn such that G∗tn → G ˜ ∗ in Lp (Ω) for all 1 < p < nk ˜ ∗ a.e. in Ω. n−k . And then there is a subsequence converges to G For simplicity, we write ˜ ∗ weakly in W 1,k (Ω), strongly in Lp (Ω) ∀ 1 < p < nk , and a.e. in Ω (6.1.6) G∗t → G 0 n−k ˜ ∗. for some G ˜ ∗ is the Green function corresponding to L∗ on Ω. Claim 1. The limit G By a limiting process, it suffices to show ∀ ϕ ∈ C01 (Ω), ˆ A∗ (Y )∇G ˜ ∗ (Y, X) · ∇ϕ(Y )dY = ϕ(X), ∀ X ∈ Ω. (6.1.7) Ω For any fixed X ∈ Ω, choose t to be sufficiently large so that X ∈ Ωt and that supp ϕ ⊂ Ωt . Then ˆ ˆ A∗ (Y )∇G∗t (Y, X) · ∇ϕ(Y )dY = A∗ (Y )∇G∗t (Y, X) · ∇ϕ(Y )dY = ϕ(X). Ω Ωt By the weak convergence of G∗t (6.1.6), we obtain ˆ ˆ ∗ ˜∗ A (Y )∇G (Y, X) · ∇ϕ(Y )dY = lim A∗ (Y )∇G∗t (Y, X) · ∇ϕ(Y )dY = ϕ(X). Ω t→∞ Ω ˜ ∗ = G∗ is the Green function on Ω. By uniqueness of Green function, G ´ ´ Let vt (X) = ∂Ωt u2 dωtX . Then Lvt = 0 in Ωt , vt |∂Ωt = u2 . Let v(X) = ∂Ω f 2 dω X . 95 We have u, v ∈ C(Ω) and vt ∈ C(Ωt ). So for any  > 0, ∃ t1 such that whenever t ≥ t1 , sup∂Ωt |vt − v| < . Then by maximum principle, sup |vt − v| <  ∀ t ≥ t1 . Ωt Therefore, ˆ ˆ lim u 2 dωtX0 = f 2 dω X0 . t→∞ ∂Ω ∂Ω t Recall that we have ˆ ˆ 1 1 A(Y )∇u(Y ) · ∇u(Y )Gt (X0 , Y ) = − u2 (X0 ) + u2 dωtX0 , Ω 2 2 ∂Ωt so ˆ ˆ 1 1 lim A∇u · ∇uGt = − u2 (X0 ) + f 2 dω X0 < ∞. (6.1.8) t→∞ Ω 2 2 ∂Ω ´ By this, (6.1.6) and Claim 1, Fatou’s lemma implies that Ω A∇u·∇uG < ∞. Therefore, for any  > 0, ∃ r0 such that whenever r ≤ r0 , ˆ A∇u · ∇uG < . (6.1.9) Br (X0 ) Claim 2. ˆ ˆ lim A∇u · ∇uGt = A∇u · ∇uG. (6.1.10) t→∞ Ω Ω We write ˆ ˆ A∇u · ∇uGt − A∇u · ∇uG Ω Ω ˆ ˆ ˆ ≤ A∇u · ∇uG + A∇u · ∇uGt + A∇u · ∇u(Gt − G) Br (X0 ) Br (X0 ) Ω\Br (X0 ) . = I1 + I2 + I3 . 96 For any  > 0, choose r to be as in (6.1.9), so I1 < . By (4.1.6) and (4.1.7), we have 1 Gt (X0 , Y ) ≤ C(n, λ0 , Λ0 )G(X0 , Y ) ∀ Y such that |X0 − Y | ≤ δ(X0 ). 2 So I2 ≤ CI1 < C, where C = C(n, λ0 , Λ0 ). For I3 , since |Gt − G| is bounded in Ω \ Br (X0 ), we can use Fatou’s lemma to get ˆ lim A∇u · ∇u |Gt − G| = 0 t→∞ Ω\B (X ) r 0 Thus the claim is established, and we have shown that (6.1.1) holds for NTA domain Ω, L = − div A∇ with A smooth, and f ∈ Lip(∂Ω). Step 3. In this step, we show (6.1.1) holds for NTA domain Ω, A = a+b satisfying (1.0.1) (s) (s) (s) and (1.0.3), and f ∈ Lip(∂Ω). Let As = a(s) +b(s) = (aij )+(bij ), where aij (X) ∈ C ∞ (Rn ) (s) and bij (X) ∈ C ∞ (Rn ). a(s) and b(s) satisfy: for all s, (s) (a(s) ξ) · ξ ≥ λ0 |ξ|2 , (s) aij ≤ λ0 , b ≤ kbkBM O(Ω) . (6.1.11) L∞ BM O(Ω) And as s → ∞, a(s) → a in Lp (Ω), b(s) → b in Lp (Ω), ∀ p > 1, (6.1.12) a(s) → a a.e. in Ω, b(s) → b a.e. in Ω. (6.1.13) 97 Let Ls = − div As ∇. Let us be the solution to   Ls us = 0  in Ω,  us = f  on ∂Ω. Let Gs and ωs be the Green function and the elliptic measure associated to Ls , respectively. Then from Step 2., it follows that ˆ ˆ 1 1 As (Y )∇us (Y ) · ∇us (Y )Gs (X0 , Y )dY = − u2s (X0 ) + f 2 dωsX0 . (6.1.14) Ω 2 2 ∂Ω ´ Let vs (X) = ∂Ω f 2 dωsX . Then Ls vs = 0 in Ω, vs ∈ W 1,2 (Ω) ∩ C(Ω) and vs |∂Ω = f 2 . ´ Let v(X) = ∂Ω f 2 dω X . Using Remark 3.2.1, one can prove as in [KP93] 7.1–7.5, that kvs − vkW 1,2 (Ω) → 0 as s → ∞. (6.1.15) We postpone its proof to the end. Also, since vs ’s are equicontinuous (by Lemma 3.1.9, Lemma 3.2.5 and (6.1.11)) and uniformly bounded (by the maximum principle), vs → v uniformly in Ω. (6.1.16) For the same reason, kus − ukW 1,2 (Ω) → 0 and us → u uniformly in Ω. (6.1.17) Thus, ˆ ˆ  1 1  lim As ∇us · ∇us Gs = lim − u2s (X0 ) + f 2 dωsX0 s→∞ Ω s→∞ 2 2 ˆ ∂Ω 1 1 = − u2 (X0 ) + f 2 dω X0 < ∞. (6.1.18) 2 2 ∂Ω 98 Now consider the convergence of G∗s . By (4.1.5) and (6.1.11), kG∗s kW 1,k (Ω) ≤ Ck for all s, (6.1.19) n where Ck = C(n, λ0 , Λ0 , k) and 1 < k < n−1 . Then the same argument about the conver- gence of G∗t in Step 2 applies to G∗s . So we have nk G∗s → G∗ weakly in W01,k (Ω), strongly in Lp (Ω) ∀1 < p < , and a.e. in Ω n−k (6.1.20) for some G∗ . Claim 3. The limit G∗ is the Green function corresponding to L∗ in Ω. By a limiting process, it suffices to show ∀ ϕ ∈ C01 (Ω), ˆ A∗ (X)∇G∗ (X, Y ) · ∇ϕ(X)dX = ϕ(Y ). (6.1.21) Ω We have ˆ ˆ ˆ ˆ A∗s ∇G∗s · ∇ϕ − ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ A ∇G · ∇ϕ ≤ As ∇(Gs − ∇G ) · ∇ϕ + (As − A )∇G · ∇ϕ Ω Ω Ω Ω (6.1.22) The first integral on the right-hand side is bounded by (λ−1 ∗ ∗ 0 +Λ0 ) kGs − G kW 1,k (Ω) kϕkW 1,k0 (Ω) , while the second integral is bounded by ˆ 1/k0 kAs − AkLk00 (Ω) |∇G∗ · ∇ϕ|k0 Ω for some 1 < k0 < k. Thus by (6.1.12) and (6.1.20), ˆ ˆ lim A∗s ∇G∗s · ∇ϕ = A∗ ∇G∗ · ∇ϕ. (6.1.23) s→∞ Ω Ω 99 Since ˆ A∗s (X)∇G∗s (X, Y ) · ∇ϕ(X)dX = ϕ(Y ), Ω we have proved (6.1.21). Then by uniqueness of Green function, G∗ is the Green function corresponding to L∗ in Ω. Claim 4. ˆ ˆ lim As (Y )us (Y ) · ∇us (Y )Gs (X0 , Y )dY = A(Y )u(Y ) · ∇u(Y )G(X0 , Y )dY. (6.1.24) s→∞ Ω Ω Write ˆ ˆ As (Y )us (Y ) · ∇us (Y )Gs (X0 , Y )dY − A(Y )u(Y ) · ∇u(Y )G(X0 , Y )dY Ω Ω ˆ ˆ ≤ As ∇us · ∇us Gs + A∇u · ∇uG B δ (X0 ) B δ (X0 ) 2 2 ˆ ˆ + As ∇us · ∇us Gs − A∇u · ∇uG Ω\B δ (X0 ) Ω\B δ (X0 ) 2 2 . = I4 + I5 + I6 . For I4 , applying the identity (6.1.14) to the ball Bδ (X0 ) to obtain ˆ ˆ 1 1 X0 As ∇us · ∇us Gs,Bδ (X0 ) dY = − u2s (X0 ) + u2s dωs,B δ (X0 ) , Bδ (X0 ) 2 2 |X0 −Y |=δ where Gs,Bδ (X0 ) is the Green function for the ball Bδ (x0 ), and ωs,Bδ (X0 ) is the elliptic measure, associated to Ls and domain Bδ (X0 ). By (4.1.6), (4.1.7) and (6.1.11), we have Gs (X0 , Y ) ≤ C(n, λ0 , Λ0 )Gs,Bδ (X0 ) (X0 , Y ) ∀ Y ∈ B δ (X0 ). (6.1.25) 2 100 Also note that As ∇us · ∇us Gs = as ∇us · ∇us Gs ≥ 0, so ˆ ˆ A ∇u · ∇u G ≤C A ∇u · ∇u G dY s s s s s s s s,B (X ) δ 0 B δ (X0 ) Bδ (X0 ) 2 1 ˆ 1 X = C − u2s (X0 ) + us (Y )2 dωs,B0 δ (X0 ) (Y ) 2 2 |X0 −Y |=δ ˆ   X0 =C u2s (Y ) − u2s (X0 ) dωs,B |X0 −Y |=δ δ (X 0 ) 2 us (Y ) − u2s (X0 ) ≤ C sup |f | ≤C sup sup |us (Y ) − us (X0 )| |X0 −Y |=δ ∂Ω |X0 −Y |=δ ≤ C sup |f |2 δ α , (6.1.26) ∂Ω where C = C(n, λ0 , Λ0 ), 0 < α = α(n, λ0 , Λ0 ) < 1, and we used Lemma 3.1.9 to obtain the last inequality. For I5 , it follows from Fatou’s lemma, the convergence of G∗s and us , as well as (6.1.18) that ˆ ˆ ˆ 0≤ A∇u · ∇uG = lim inf As ∇us · ∇us Gs ≤ lim inf As ∇us · ∇us Gs s→∞ s→∞ Ω ˆ Ω Ω 1 1 = − u2 (X0 ) + f 2 dω X0 < ∞. (6.1.27) 2 2 ∂Ω ⇒ ∀  > 0, there exists a δ0 > 0, such that whenever δ < δ0 , ˆ A∇u · ∇uG < , i.e. I5 < . B δ (X0 ) 2 By (6.1.26), we can choose δ < δ0 to be so small that I4 < . Note that δ is independent of s. Let this δ be fixed. 101 We now estimate I6 . ˆ ˆ I6 ≤ As ∇us · ∇us (Gs − G) + (As ∇us · ∇us − A∇u∇u)G Ω\B δ (X0 ) Ω\B δ (X0 ) 2 2 . = Is + IIs . For Is , ˆ ˆ as ∇us · ∇us (Gs − G) ≤ λ−1 |∇us |2 |Gs − G| Is = 0 Ω\B δ (X0 ) Ω\B δ (X0 ) 2 2 Since Gs and G are bounded in Ω \ B δ (X0 ), |∇us |2 |Gs − G| is uniformly integrable. Then 2 Vitali convergence theorem gives that lims→∞ Is = 0. For IIs , we have ˆ IIs ≤ sup G(X0 , Y ) |as ∇us · ∇us − a∇u · ∇u| Ω\B δ (X0 ) Ω\B δ (X0 ) ˆ ˆ 2 2 ≤ λ−1 0 sup G(X0 , Y ) |∇us · ∇us − ∇u · ∇u| + |as − a| |∇u|2 . Ω\B δ (X0 ) 2 So lims→∞ IIs = 0. Note that δ has been fixed when estimating Is and IIs . Therefore, we have shown that for any  > 0, there exists δ > 0 such that ˆ ˆ lim As (Y )us (Y ) · ∇us (Y )Gs (X0 , Y )dY − A(Y )u(Y ) · ∇u(Y )G(X0 , Y )dY < , s→∞ Ω Ω which justifies Claim 4. Combining Claim 4 and (6.1.18), we obtain that (6.1.1) holds for NTA domain Ω, L = − div A∇ satisfying (1.0.1) and (1.0.2), and f ∈ Lip(∂Ω). Step 4. Finally, we prove (6.1.1) for Ω being NTA, L = − div A∇ satisfying (1.0.1) and ´ (1.0.3), and f ∈ L2 (∂Ω). Let u(X) = ∂Ω f dω X . Let fi → f in L2 (∂Ω, dω), with each 102 ´ X. fi ∈ Lip(∂Ω, dω). Define ui (X) = ∂Ω fi dω By Step 3, we have ˆ ˆ 1 1 A∇ui · ∇ui G = − ui (X0 )2 + fi2 dω X0 < ∞, Ω 2 2 ∂Ω ⇒ ˆ ˆ 1  2 1  lim A∇ui · ∇ui G = lim − ui (X0 ) + fi2 dω X0 i→∞ Ω i→∞ 2 2 ∂Ω ˆ 1 1 = − u(X0 )2 + f 2 dω X0 < ∞ (6.1.28) 2 2 ∂Ω By Caccioppoli’s inequality, ∇ui converges to ∇u in L2 (K) where K is any compact subset of Ω, and thus there exists a subsequence, still denoted by {∇ui }, converges to ∇u a.e. in K. Therefore, Fatou’s lemma implies that ˆ ˆ ˆ a∇u · ∇uG = A∇u · ∇uG ≤ sup lim A∇ui · ∇ui G Ω Ω K i→∞ K ˆ ˆ 1 1 ≤ lim A∇ui · ∇ui G = − u(X0 )2 + f 2 dω X0 < ∞. (6.1.29) i→∞ Ω 2 2 ∂Ω Similarly, for any fixed i, we have ˆ ˆ a∇(u − ui ) · ∇(u − ui )G = A∇(u − ui ) · ∇(u − ui )G Ω Ω ˆ ≤ sup lim A∇(uj − ui ) · ∇(uj − ui )G K j→∞ K 1 2 1 ˆ ≤ − u(X0 ) − ui (X0 ) + (f − fi )2 dω X0 . 2 2 ∂Ω So for any  > 0, there exists i sufficiently large, such that ˆ ˆ a∇(u − ui ) · ∇(u − ui )G = A∇(u − ui ) · ∇(u − ui )G < . (6.1.30) Ω Ω 103 We write ˆ ˆ ˆ ˆ a∇u · ∇uG − a∇ui · ∇ui G ≤ a∇(u − ui ) · ∇uG + a∇(u − ui ) · ∇ui G . Ω Ω Ω Ω (6.1.31) For the first integral on the right-hand side of (6.1.31), we have ˆ ˆ 1/2  ˆ a∇(u − ui ) · ∇uG ≤ 1 1/2 λ2 a∇(u − ui ) · ∇(u − ui )G a∇u · ∇uG Ω 0 ˆΩ Ω 1 1/2 < 2 a∇u · ∇uG . λ0 Ω The second integral can be estimated similarly, and is bounded by ˆ 1 1/2 a∇ui · ∇ui G  λ20 Ω So by (6.1.28) and (6.1.29), we have ˆ ˆ ˆ 1 1 A∇u · ∇uG = lim A∇ui · ∇ui G = − u(X0 )2 + f 2 dω X0 , Ω i→∞ Ω 2 2 ∂Ω which completes the proof. Proof of (6.1.15). Let g ∈ Lip(Ω) be the extension of f 2 ∈ Lip(∂Ω). Define ws = vs −g, w = v − g. Then it suffices to show ws converges to w in W 1,2 (Ω). Note that ws , w ∈ W01,2 (Ω) ∩ C(Ω), and   Ls ws = div As ∇g  in Ω,  ws = 0  on ∂Ω. By Remark 3.2.1, as well as (6.1.11),  1/p0 |∇ws |p0 ≤ C k∇gkL∞ (Ω) , (6.1.32) 104 for some p0 > 2, where p0 and C are independent of s. We now show that ws converges weakly to w in W 1,2 (Ω), equipped with inner product hA∇(·), ∇(·)i. Let ϕ ∈ W01,2 (Ω). Then ˆ ˆ ˆ ˆ ˆ A∇ws · ∇ϕ − A∇w · ∇ϕ ≤ (A − As )∇ws · ∇ϕ + As ∇ws · ∇ϕ − A∇w · ∇ϕ Ω Ω Ω Ω Ω ˆ 1/2  ˆ 1/2 ˆ |A − As |2 |∇ws |2 |∇ϕ|2 ≤ + (As − A)∇g · ∇ϕ Ω Ω Ω ˆ 2p0  p0 −2  ˆ 1 1/2 |∇ws |p0 |∇ϕ|2 2p0 p0 ≤ |A − As | p0 −2 ˆ Ω Ω 1/2  1/2 + k∇gkL∞ (Ω) |A − As |2 |∇ϕ|2 , Ω where p0 is as in (6.1.32). Then by (6.1.12), we have ˆ ˆ lim A∇ws · ∇ϕ − A∇w · ∇ϕ = 0 (6.1.33) s→∞ Ω Ω To see that ws converges to w in W 1,2 (Ω), it suffices to show k∇ws − ∇wkL2 (Ω) → 0 as s → ∞. We have ˆ ˆ 2 λ0 |∇(w − ws )| ≤ A∇(w − ws ) · ∇(w − ws ) ˆΩ ˆ Ω ˆ ˆ = A∇w · ∇w + A∇ws · ∇ws − A∇w · ∇ws − A∇ws · ∇w. (6.1.34) Ω Ω Ω Ω By (6.1.33), the last two integrals tend to zero as s → ∞. For the second integral in (6.1.34), we write ˆ ˆ ˆ A∇ws · ∇ws = (A − As )∇ws · ∇ws + As ∇ws · ∇ws . (6.1.35) Ω Ω Ω 105 The first integral on the right-hand side of (6.1.35) is bounded by ˆ  10  ˆ  1 ˆ  1 p00 p00 p0 0 0 p00 |A − As |p0 |∇ws |p0 p0 p0 |A − As | |∇ws | |∇ws | ≤ k∇gkL∞ (Ω) Ω Ω Ω ´ 0 0 Another use of H¨ older inequality on the integral Ω |A − As |p0 |∇ws |p0 shows that it goes to zero as s → ∞. For the second integral in (6.1.35), we have ˆ ˆ ˆ ˆ As ∇ws · ∇ws = (As − A)∇g · ∇ws + As ∇g · ∇ws = A∇g · ∇ws 1/2 ˆ Ω Ω Ω Ω  1/2  2 2 ≤ k∇gkL∞ (Ω) |A − As | |∇ws | + A∇g · ∇ws . (6.1.36) Ω As s → ∞, the first term in (6.1.36) tends to zero by (6.1.12), while the second term tends ´ ´ to Ω A∇g · ∇w = Ω A∇w · ∇w by (6.1.33). Combining these estimates and (6.1.34), we conclude that ˆ lim |∇(w − ws )|2 = 0, s→∞ Ω which completes the proof. 6.2 More characterizations of the A∞ condition We point out in this section that the −approximability result in [KKPT00] and the result in [KKPT16] also hold for operators L = − div A∇ with A satisfying (1.0.1) and (1.0.3), and domains being Lipschitz. Namely, we have the following Theorem 6.2.1. Let A satisfy (1.0.1) and (1.0.3) and let L = div A∇. Let Ω ⊂ Rn be a Lipschitz domain, X0 ∈ Ω. Then there exists an , depending on λ0 , Λ0 and the Lipschitz character of Ω such that if every solution u to Lu = 0, with kukL∞ ≤ 1, is −approxiable on Ω, then ωLX0 ∈ A∞ (dσ), where dσ is the surface measure on ∂Ω. 106 The proof proceeds exactly as in [KKPT00] since the solutions have all the properties to prove the theorem: the Harnack principle, H¨older continuity and boundary H¨ older continuity. Similarly, we obtain the following results in [KKPT16]: Theorem 6.2.2. Let L, Ω and X0 be as in Theorem 6.2.1. Assume that there exists some A < ∞ such that for all Borel sets H ⊂ ∂Ω, the solution to the Dirichlet problem   Lu = 0  in Ω  u = χH  on ∂Ω. satisfies the following Carleson bound ˆ 1 sup δ(X) |∇u(X)|2 dX ≤ A, ∆⊂∂Ω,diam(∆)≤diam(Ω) σ(∆) T (∆) where δ(X) = dist(X, ∂Ω). Then ωLX0 ∈ A∞ (dσ). Theorem 6.2.3. Let L, Ω and X0 be as in Theorem 6.2.1. Assume that for all Borel sets H ⊂ ∂Ω, the solution to the Dirichlet problem   Lu = 0  in Ω  u = χH  on ∂Ω. satisfies kSα (u)kLp (∂Ω,dσ) ≤ A kN ukLp (∂Ω,dσ) (6.2.1) 1 for some p, 1 + n−2 ≤ p < ∞ if n ≥ 3 or p0 ≤ p < ∞ if n = 2 with p0 depending on λ0 , Λ0 and the Lipschitz character of Ω. Then ωLX0 ∈ A∞ (dσ). CHAPTER Seven The Gaussian property 7.1 Sectorial operators and resolvent estimates We give a precise definition for the operator L. f −1,2 (Rn ) be the space of the bounded semilinear functionals on W 1,2 (Rn ), i.e. Let W f −1,2 , f ∈W hf, αu + βviW ¯ hf, uiW f −1,2 ,W 1,2 = α ¯ f −1,2 ,W 1,2 + βhf, viW f −1,2 ,W 1,2 , whenever α, β ∈ C and u, v ∈ W 1,2 (Rn ). Define L : W 1,2 (Rn ) → W f −1,2 (Rn ) as follows ˆ hL u, viW f −1,2 ,W 1,2 = A∇u · ∇¯ v Rn ˆ ˆ = As ∇u · ∇¯ v+ Aa ∇u · ∇¯ v, (7.1.1) Rn Rn 107 108 where A = (aij (x)) is n × n, real, As = 12 (A + A| ) = (asij (x)) is L∞ and elliptic, i.e. there exists 0 < λ0 ≤ 1 such that λ0 |ξ|2 ≤ asij (x)ξi ξj ∀ ξ ∈ Rn , kAs k∞ ≤ λ−1 0 , and Aa = 21 (A − A| ) = (aaij (x)) is in BMO(Rn ), with a aij . a aij − (aaij )Q dx ≤ Λ0 BM O = sup (7.1.2) Q⊂Rn Q for some Λ0 > 0. Then by Proposition 2.2.1, hL u, viW f −1,2 ,W 1,2 ≤ C k∇ukL2 k∇vkL2 , (7.1.3) with C depending on λ0 , Λ0 and dimension. Now define a sesquilinear form on L2 (Rn ) × L2 (Rn ): for any u, v ∈ W 1,2 (Rn ), let t[u, v] = hL u, viW f −1,2 ,W 1,2 . (7.1.4) The numerical range Θ(t) of t is defined as . Θ(t) = {t[u, u] : u ∈ D(t) with kukL2 = 1}. Proposition 7.1.1. t is a densely defined, closed, sectorial sesqulinear form in L2 , and π there exists 0 < θ0 < 2 such that for any ξ ∈ Θ(t), |arg ξ| ≤ θ0 . Proof. The domain D(t) of t is W 1,2 (Rn ), which is dense in L2 . So t is densely defined. To see that it is closed, let un ∈ D(t), un → u in L2 and t[un − um , un − um ] → 0. We want to show that u ∈ D(t) and t[un − u, un − u] → 0. Since t[un − um , un − um ] → 0, ˆ λ0 |∇(un − um )| ≤ < A∇(un − um ) · ∇(un − um ) → 0. Rn 109 So {un } is a Cauchy sequence in W 1,2 (Rn ), which implies that u ∈ W 1,2 = D(t) and |t[un − u, un − u]| ≤ Λ0 k|∇(un − u)|kL2 → 0. Now we show that t is sectorial, i.e., its numerical range Θ(t) is a subset of a sector of the form π |arg(ξ − γ)| ≤ θ, for some 0 ≤ θ < and γ ∈ R. (7.1.5) 2 For u ∈ D(t) with kukL2 = 1, write u = u1 + iu2 . Then ˆ ˆ 0, and (L + λI)−1 ≤ (< λ)−1 . This lemma implies that there is an unique L : D(L) ⊂ L2 (Rn ) → L2 (Rn ) with its do- main dense in W 1,2 corresponding to L : W 1,2 → W f −1,2 , and (Lu, v) = hL u, vi f −1,2 1,2 W ,W for any u ∈ D(L), v ∈ W 1,2 . We denote by Θ(L) the numerical range of L: Θ(L) = {(Lu, u) : u ∈ D(L) with kukL2 = 1.}. Let T = −L. Denote the resolvant set of T by ρ(T ). Note that since L is m-accretive, {λ ∈ C : < λ < 0} ⊂ ρ(L). So λ ∈ ρ(T ) whenever < λ > 0. Let Σ0 = C \ Θ(T ), and denote ˆ 0 the component of Σ0 that contains R+ . by Σ ˆ 0 ⊂ ρ(T ). And for any λ ∈ Σ Lemma 7.1.2. Σ ˆ 0, (λI − T )−1 = (λI + L)−1 ≤ 1 . (7.1.8) dist(λ; Θ(T )) Proof. For any fixed λ ∈ Σ0 , for any u ∈ D(T ) with kukL2 = 1, we have 0 < dist(λ, Θ(T )) ≤ |λ − (T u, u)| = |((λI − T )u, u)| ≤ k(λI − T )ukL2 . (7.1.9) ⇒ If λ ∈ ρ(T ), then (λI − T )−1 2 2 ≤ 1 L →L . (7.1.10) dist(λ; Θ(T )) ˆ 0 ⊂ ρ(T ). Consider ρ(T ) ∩ Σ Now we show Σ ˆ 0 . It is nonempty since R+ ⊂ ρ(T ) ∩ Σ ˆ 0. ˆ 0 is open in Σ The fact that ρ(T ) is open implies that ρ(T ) ∩ Σ ˆ 0 . But it is also closed in ˆ 0 since λn ∈ ρ(T ) ∩ Σ Σ ˆ 0 and λn → λ ∈ Σ ˆ 0 imply for n large enough, dist(λn , Θ(T )) > 1 2 dist(λ, Θ(T )), and consequently for n large enough |λn − λ| < dist(λn , Θ(T )). Write 111 λI − T = (λn I − T )(I + (λ − λn )(λn I − T )−1 ). From (7.1.10), (λ − λn )(λn I − T )−1 ≤ |λ − λn | (λn I − T )−1 ≤ |λ − λn | 1 < . dist(λn ; Θ(T )) 2 ⇒ (I + (λ − λn )(λn I − T )−1 )−1 is bounded in L2 , and consequently so is λI − T , i.e. ˆ 0 is closed in Σ λ ∈ ρ(T ). This implies that ρ(T ) ∩ Σ ˆ 0 . By the connectedness of Σ ˆ 0, ˆ0 = Σ ρ(T ) ∩ Σ ˆ 0 , or Σ ˆ 0 ⊂ ρ(T ). Fix a θ1 ∈ (θ0 , π2 ). Let Γπ−θ1 = {λ ∈ C : λ 6= 0, |arg λ| ≤ π − θ1 }. Then Γπ−θ1 ⊂ ρ(T ) and there exists a c0 = c0 (θ0 , θ1 ) ≥ 1 such that for any λ ∈ Γπ−θ1 , |λ| dist(λ; Θ(T )) ≥ . (7.1.11) c0 Corollary 7.1.1. There exists a C = C(θ0 , θ1 , λ0 ) > 0 such that for any λ ∈ Γπ−θ1 , (λI + L)−1 2 2 + 1 ∇(λI + L)−1 2 2 ≤ C . L →L L →L |λ|1/2 |λ| Proof. It follows immediately from Lemma 7.1.2 and (7.1.11) that (λI + L)−1 2 2 ≤ c0 . L →L (7.1.12) |λ| Since Γπ−θ1 ⊂ ρ(T ), for any λ ∈ Γπ−θ1 , the range of λI + L is L2 (Rn ). And for any u ∈ D(L), ((λI + L)u, u) = h(L + λ)u, uiW f −1,2 ,W 1,2 . ⇒ <((λI + L)u, u) ≤ k(λI + L)ukL2 kukL2 . On the other hand, ˆ ˆ <((λI + L)u, u) = <( A∇u · ∇¯ u+λ |u|2 ) ≥ λ0 k∇uk2L2 + <λ kuk2L2 . Rn Rn 112 Therefore, ˆ λ0 |∇u|2 ≤ |λ| kuk2L2 + k(λI + L)ukL2 kukL2 . Rn Since c0 kukL2 = (λI + L)−1 (λI + L)u L2 ≤ k(λI + L)ukL2 , |λ| one obtains k∇ukL2 ≤ C(c0 , λ0 ) |λ|−1/2 k(λI + L)ukL2 , and consequently ∇(λI + L)−1 ≤ C(c0 , λ0 ) |λ|−1/2 . (7.1.13) Let Ψ be a real-valued bounded Lipschitz function. Let eΨ denote the operator of pointwise multiplication by the function e−Ψ(x) . Define LΨ = e−Ψ L eΨ , i.e. for any u, v ∈ W 1,2 (Rn ) Ψ −Ψ hLΨ u, viW f −1,2 ,W 1,2 = hL (e u), e viW f −1,2 ,W 1,2 . (7.1.14) It follows from direct computation that LΨ = L − div(A∇Ψ) − ((∇Ψ)T A) · ∇ − A∇Ψ · ∇Ψ, (7.1.15) or ˆ ˆ ˆ hLΨ u, viW f −1,2 ,W 1,2 = A∇u · ∇v + A∇Ψ · ∇vu − A∇u · ∇Ψv Rn Rn Rn ˆ − A∇Ψ · ∇Ψuv (7.1.16) Rn ˆ ˆ ˆ ˆ s s = A∇u · ∇v + A ∇Ψ · ∇vu − A ∇u · ∇Ψv − Aa ∇(uv) · ∇Ψ Rn Rn Rn Rn ˆ − As ∇Ψ · ∇Ψuv. (7.1.17) Rn Since k∇ΨkL∞ is bounded, Proposition 2.2.3 implies that LΨ : W 1,2 → W f −1,2 is bounded. For any δ ∈ (0, λ0 ), define LΨ,δ = LΨ + δ. As before, we can define sesquilinear forms 113 on L2 tΨ and tΨ,δ associated to LΨ and LΨ,δ , respectively. That is, for u, v ∈ W 1,2 (Rn ), tΨ [u, v] = hLΨ u, viW f −1,2 ,W 1,2 , tΨ,δ [u, v] = hLΨ,δ u, viW f −1,2 ,W 1,2 . (7.1.18) δ Proposition 7.1.2. There exists M0 = M0 (n, λ0 , Λ0 ) ≥ 1, such that if k∇ΨkL∞ ≤ M0 , then for any ξ in the numerical range Θ(tΨ,δ ) of tΨ,δ , θ1 − θ 0 |arg ξ| ≤ θˆ0 = θ0 + . (7.1.19) 16 And tΨ,δ is a densely defined, closed, sectorial sesqulinear form in L2 . Proof. Let u ∈ W 1,2 (Rn ) = D(t) and write u = u1 + iu2 . Then by (7.1.17) and Proposition 2.2.3, 0 such |λ−δ| that for any λ − δ ∈ Γπ−θˆ1 , dist(λ − δ, −Θ(tΨ,δ )) ≥ c0 . Observe that there exists a M = M (θ0 , θ1 ) ≥ 1, such that for any λ ∈ Γπ−θ1 \ BM δ (0), λ − δ ∈ Γπ−θˆ1 . Therefore,   |λ − δ| 1 |λ| dist(λ − δ, −Θ(tΨ,δ )) ≥ ≥ 1− , ∀ λ ∈ Γπ−θ1 \ BM δ (0). (7.1.21) c0 M c0 Using this observation, one can follow the proof of Corollary 7.1.1 to obtain Corollary 7.1.2. Let δ ∈ (0, λ0 ) and Ψ be a bounded Lipschitz function satisfying the condition in Proposition 7.1.2. There exist C = C(θ0 , θ1 , λ0 ) ≥ 1 and M = M (θ0 , θ1 ) ≥ 1 such that for any λ ∈ Γπ−θ1 \ BM δ (0), (λI + LΨ )−1 2 2 + 1 ∇(λI + LΨ )−1 2 2 ≤ C . L →L L →L (7.1.22) |λ|1/2 |λ| 7.2 Proof of the Gaussian property We shall prove that the kernel of e−tL the Gaussian property (Theorem 7.2.6) following the idea of [Aus96]. We also obtain estimates for the derivatives of the heat kernel (Theorem 7.2.7), as well as the L2 estimates for ∂t e−tL and ∇e−tL (Theorem 7.3.1) using a contour integral expression of the semigroup. We start with reviewing the definition and some facts of Morrey-Campanato spaces. Definition 7.2.1. For 0 ≤ γ ≤ n, define the Morrey space L2,γ (Rn ) ⊂ L2loc (Rn ) by the 116 condition ˆ 1/2 .  kf kL2,γ = sup R−γ |f |2 < ∞. (7.2.1) x∈Rn ,0 0, ∇u ∈ L2,γ with γ = min {2 + α, 2 + β, n − 2 + 2µ0 − }. Proof. We first point out that due to Lemma 7.2.2, g ∈ L2,2+α , so the right-hand side of (7.2.5) makes sense. To prove (7.2.5), we can assume without loss of generality that x0 = 0 R and 0 < ρ ≤ 8. 118 Let v ∈ W 1,2 (BR/4 ) be a weak solution of div(A∇v) = 0 in BR/4 , v = u on ∂BR/4 . (7.2.6) Define   v − u  in BR/4 w= in BR/4 c .  0  Let η ∈ C0∞ (BR/2 ) with 0 ≤ η ≤ 1, η = 1 in BR/4 and |∇η| . 1 R. ˜ = u − (u)BR/2 . Set u We claim that ˆ ˆ |∇v|2 ≤ C |∇u|2 . (7.2.7) BR/4 BR/2 ´ In fact, take w as test function in (7.2.6), one obtains BR/4 A∇v · ∇w = 0. So ˆ ˆ ˆ 2 λ0 |∇w| ≤ < A∇w · ∇w = −< u · ∇w A∇˜ BR/4 BR/4 BR/4 nˆ ˆ o s = −< A ∇˜ u · ∇w + Aa ∇(η˜ u) · ∇w . (7.2.8) BR/4 BR/4 By Proposition 2.2.1, the right-hand side is bounded by ˆ C |∇u| |∇w| + C k∇(η˜ u)kL2 k∇wkL2 BR/4 ˆ ˆ ˆ λ0 C 2 2 ≤ |∇w| + 2 |˜ u| + C |∇u|2 8 BR/4 R BR/2 BR/2 ˆ ˆ λ0 2 ≤ |∇w| + C |∇u|2 . 8 BR/4 BR/2 So (7.2.7) follows. By Lemma 7.2.3, ˆ ˆ ρ 2 |∇v| ≤ C( )n−2+2µ0 |∇v|2 . (7.2.9) Bρ R BR/4 Combining this and (7.2.7), one obtains ˆ ˆ ˆ ρ 2 2 |∇u| ≤ C( )n−2+2µ0 |∇u| + |∇w|2 . (7.2.10) Bρ R BR/2 Bρ 119 ´ We now estimate |∇w|2 . Observe that w ∈ W01,2 (BR/4 ) is a weak solution of div(A∇w) = div(A∇Ψ)g + ((∇Ψ)T A) · ∇g + A∇Ψ · ∇Ψg − λg + f in BR/4 . (7.2.11) Take w as test function, then take the real part in both sides of the equation, one has ˆ ˆ 2 λ0 |∇w| ≤ < A∇w · ∇w BR/4 BR/4 nˆ ˆ ˆ =< A∇Ψ · ∇wg − A∇g · ∇Ψw − A∇Ψ · ∇Ψgw BR/4 BR/4 BR/4 ˆ ˆ o +λ gw − fw (7.2.12) BR/4 BR/4 We have ˆ ˆ ˆ As ∇Ψ · ∇wg + As ∇g · ∇Ψw + As ∇Ψ · ∇Ψgw BR/4 BR/4 BR/4 2+α α 2+α ≤ CR 2 k∇wkL2 kgkL2,2+α + CR 2 kwkL2 k∇gkL2,α + CR 2 kwkL2 kgkL2,2+α 2+α ≤ CR 2 k∇wkL2 (kgkL2,2+α + k∇gkL2,α ), (7.2.13) ´ ´ where in the last inequality we have used that ( B |w|2 )1/2 ≤ CR( BR/4 |∇w|2 )1/2 , and R/4 that R ≤ 1. By Proposition 2.2.3, we have ˆ ˆ a a A ∇Ψ · ∇(wg) = A ∇Ψ · ∇(wgη) BR/4 BR/4 2+α ≤ C(kwkL2 k∇(gη)kL2 + k∇wkL2 kgηkL2 ) ≤ R 2 k∇wkL2 (kgkL2,2+α + k∇gkL2,α ). (7.2.14) ˆ 2+α 2+α gw ≤ R 2 kwkL2 kgkL2,2+α . R1+ 2 k∇wkL2 kgkL2,2+α , (7.2.15) BR/4 and similarly, ˆ β β f w ≤ R 2 kwkL2 kf kL2,β . R1+ 2 k∇wkL2 kf kL2,β . (7.2.16) BR/4 120 Combining (7.2.12) – (7.2.16), one has ˆ !1/2 2+α 2+β 2 |∇w| ≤ CR 2 (max {1, |λ|} kgkL2,2+α + k∇gkL2,α ) + CR 2 kf kL2,β . BR/4 (7.2.17) This estimate together with (7.2.10) gives (7.2.5), or ˆ ˆ 2 ρ |∇u| ≤C( )n−2+2µ0 |∇u|2 Bρ R BR  n o  + CRmin{2+α,2+β} max 1, |λ|2 kgk2L2,2+α + k∇gk2L2,α + kf k2L2,β . (7.2.18) This implies, by Lemma 2.4.3, that ˆ ˆ ρ  n o  |∇u|2 ≤ C( )γ |∇u|2 + C max 1, |λ|2 kgk2L2,2+α + k∇gk2L2,α + kf k2L2,β ργ , Bρ R BR (7.2.19) where γ = min {2 + α, 2 + β, n − 2 + 2µ0 − }. Thus the conclusion follows from our as- sumption on u. Lemma 7.2.5. Let δ ∈ (0, λ0 ) and let Ψ satisfy the condition of Proposition 7.1.2. Then there exists µ0 = µ0 (λ0 , Λ0 , n) such that for all 0 ≤ γ < n, and all λ ∈ Γπ−θ1 \ BM δ (0), where M is as in (7.1.21), ∇(LΨ + λI)−1 : L2 ∩ L2,γ → L2 ∩ L2,γ1 , (7.2.20) (LΨ + λI)−1 : L2 ∩ L2,γ → L2 ∩ L2,γ 1 1 +2 , (7.2.21) where γ1 = min {γ + 2, n − 2 + 2µ0 − }, and  > 0 is arbitrary. Moreover, there exists C = C(λ0 , Λ0 , n) such that ∇(LΨ + λI)−1 h 2 2,γ ≤ C max(|λ|n/2 , 1 ) khk 2 2,γ , L ∩L 1 L ∩L (7.2.22) |λ| (LΨ + λI)−1 h 2 2,γ1 +2 ≤ C max(|λ|n/2 , 1 ) khk 2 2,γ . L ∩L1 L ∩L (7.2.23) |λ| 121 Proof. Let µ0 be as in Lemma 7.2.3 and 7.2.4. By Lemma 7.2.2, it suffices to prove (7.2.20) and (7.2.22). Let γ0 = n − 2 + 2µ0 . We begin with the case 0 ≤ γ < γ0 − 2, and we assume that γ0 − 2 > 0. Let k ≥ 1 be the greatest integer for which γ0 − 2k > 0 and let Ij = [γ0 − 2(j + 1), γ0 − 2j). We shall prove by induction that if γ ∈ Ij with j ≥ 2, then ∇(LΨ + λI)−1 : L2 ∩ L2,γ → L2 ∩ L2,2+γ with ∇(LΨ + λI)−1 h 2 2,γ+2 ≤ C max(|λ| 2 , 1 ) khk 2 2,γ , k−j+2 L ∩L L ∩L (7.2.24) |λ| Assume γ = 0 ∈ Ik and h ∈ L2 . Then by Corollary 7.1.2, there exists u ∈ W 1,2 such that (LΨ + λI)u = h. By (7.1.15), Lemma 7.2.4 applies, with g = u, f = h and α = β = 0. By Corollary 7.1.2, as well as Lemma 7.2.1 (3), we have k∇ukL2 . |λ|−1/2 khkL2 , (7.2.25) kukL2,2 . kukL2 + k∇ukL2 . (|λ|−1/2 + |λ|−1 ) khkL2 . (7.2.26) So (7.2.5) gives ˆ ˆ ρ n o 2 |∇u| ≤ C( )γ0 |∇u|2 + CR2 max |λ|2 , |λ|−2 khk2L2 , (7.2.27) Bρ (x0 ) R BR (x0 ) for any x0 ∈ Rn and 0 < ρ < R ≤ 1. Since we assume γ0 > 2, Lemma 2.4.3 yields ˆ ˆ ρ n o 2 |∇u| ≤ C( )2 |∇u|2 + Cρ2 max |λ|2 , |λ|−2 khk2L2 , (7.2.28) Bρ R BR which implies ∇u ∈ L2,2 with n o k∇ukL2,2 ≤ C max |λ| , |λ|−1 khkL2 . (7.2.29) Now let γ ∈ Ik with γ > 0. Let h ∈ L2,γ ∩ L2 and let u ∈ W 1,2 be a solution of (LΨ + λI)u = h. Applying Lemma 7.2.4 with g = u, f = h, α = 2 and β = γ, and using 122 (7.2.25) and (7.2.26), one obtains ∇u ∈ L2,2+γ with n o k∇ukL2,2+γ ≤ C max |λ| , |λ|−1 khkL2 ∩L2,γ . (7.2.30) So we have proved (7.2.24) for all γ ∈ Ik with γ ≥ 0. Now we prove that if j ≥ 2 and that (7.2.24) holds for γ ∈ Ij , then (7.2.24) holds for γ ∈ Ij−1 . If k = 1 there is nothing to prove. If k ≥ 2 and γ ∈ Ij−1 with j ≥ 2, then γ − 2 ∈ Ij . Let h ∈ L2 ∩ L2,γ and u ∈ W 1,2 be a solution of (LΨ + λI)u = h. Then by L2 ∩ L2,γ ⊂ L2 ∩ L2,γ−2 and the assumption, ∇u ∈ L2 ∩ L2,γ with k∇ukL2,γ ≤ n k−j+2 o C max |λ| 2 , |λ|−1 khkL2 ∩L2,γ−2 . By (7.2.5) with α = β = γ and its proof, as well as the estimate n k−j+2 o kukL2,2+γ . kukL2 + k∇ukL2,γ . max |λ| 2 , |λ|−1 khkL2 , we obtain ∇u ∈ L2 ∩ L2,γ+2 with k−j+3 k∇ukL2 ∩L2,γ+2 ≤ C max(|λ| 2 , |λ|−1 ) khkL2 ∩L2,γ . So we have proved (7.2.24) for all γ ∈ Ij with 2 ≤ j ≤ k. Finally, we turn to the case γ0 − 2 ≤ γ < n. If γ0 − 2 > 0, then choose  ∈ (0, 1) such that γ0 − 2 −  > 0. Then by previous result and L2,γ ⊂ L2,γ0 −2− , we have ∇(LΨ + λI)−1 : L2 ∩ L2,γ → L2 ∩ L2,γ0 − . If γ0 − 2 ≤ 0, then by an argument similar to the case γ = 0, one obtains the desired result. Corollary 7.2.1. Let m be the smallest integer for which 4m ≥ n + 2µ0 . Then for all 123 µ ∈ (0, µ0 ), (LΨ + λI)−m : L2 (Rn ) → L2 (Rn ) ∩ C µ (Rn ), (7.2.31) (LΨ + λI)−2m : L1 (Rn ) → L2 (Rn ) ∩ C µ (Rn ), (7.2.32) with  n o m (LΨ + λI)−m h µ ≤ C max |λ|n/2 , |λ|−1 C khkL2 (Rn ) , (7.2.33)  n o 2m (LΨ + λI)−2m h µ ≤ C max |λ|n/2 , |λ|−1 C khkL1 (Rn ) . (7.2.34) Proof. (7.2.31) and (7.2.33) follow from Lemma 7.2.1 and Lemma 7.2.5, and by a discussion of the case n + 2µ0 ≤ 4m < n + 4 and n + 4 ≤ 4m < n + 4 + 2µ0 . If n + 2µ0 ≤ 4m < n + 4, then 2,4(m−1) ∼ (LΨ + λI)−(m−1) : L2 → L2 ∩ L1 = L2 ∩ L2,4(m−1) . Applying Lemma 7.2.5 with γ = 4(m−1), then since 4m−2 ≥ n−2+2µ0 , γ1 = n−2+2µ0 −. Choose  < 2µ0 , then 0 − 2µ0 −  (LΨ + λI)−m : L2 → L2 ∩ L2,n+2µ 1 ⊂ C µ , with µ = . (7.2.35) 2 Now we turn to the case n + 4 ≤ 4m < n + 4 + 2µ0 . For all 0 > 0, 2,4(m−2) 0 (LΨ + λI)−(m−2) : L2 → L2 ∩ L1 ∼ = L2 ∩ L2,4(m−2) ⊂ L2 ∩ L2,4(m−2)− . Choose 0 > 4m − (n + 4), then 0 0 (LΨ + λI)−(m−1) : L2 → L2 ∩ L12,4m−4− ∼= L2 ∩ L2,4m−4− . Now if we choose 4 − 2µ0 ≥ 0 > 4m − (n + 4), then 4m − 2 − 0 ≥ n − 2 + 2µ0 . So again 124 we obtain 0 − 2µ0 −  (LΨ + λI)−m : L2 → L2 ∩ L2,n+2µ 1 ⊂ C µ , with µ = . (7.2.36) 2 To prove (7.2.32) and (7.2.34), consider the adjoint operator LΨ∗ = eΨ L ∗ e−Ψ : W 1,2 (Rn ) → f −1,2 (Rn ) and the induced operator L∗ : D(L∗ ) ⊂ L2 (Rn ) → L2 (Rn ). One can show W Ψ Ψ that Lemma 7.2.5 applies to (LΨ + λI)∗ . So for all 0 < µ < µ0 , (LΨ + λI)∗−m : L2 → L2 ∩ C µ ⊂ C µ . By duality,  ∗ (LΨ + λI)−m = (LΨ + λI)∗−m : C f2 = L2 . fµ → L fµ , this implies (LΨ + λI)−m : L1 → L2 . So (7.2.32) follows from (7.2.31). By Since L1 ⊂ C (7.2.33), to prove (7.2.34) it suffices to show n om (LΨ + λI)−m h 2 ≤ C max |λ|n/2 , |λ|−1 L khkL1 . (7.2.37) By duality, we have (LΨ + λI)−m h 2 ≤ sup khkL1 (LΨ + λI)∗−m g L∞ L kgkL2 =1 n om sup khkL1 (LΨ + λI)∗−m g C µ ≤ C max |λ|n/2 , |λ|−1 ≤ khkL1 . kgkL2 =1 Proposition 7.2.1. Let m and µ0 be as above, δ ∈ (0, λ0 ), and let λ ∈ Γπ−θ1 \ BM δ (0) 125 with M as in (7.1.21). Fix 0 < µ < µ0 . Then the kernel G2m −2m satisfies λ (x, y) of (L + λI) n o2m G (x, y) ≤ C max |λ|n/2 , |λ|−1 e−α1 |x−y| , 2m λ (7.2.38) 2m G (x, y) − G2m (x + h, y) + G2m (x, y) − G2m (x, y + h) λ λ λ λ n o2m ≤ C max |λ|n/2 , |λ|−1 e−α1 |x−y| |h|µ (7.2.39) when 2 |h| ≤ |x − y|, for some α1 = α1 (n, λ0 , Λ0 , δ) ∈ (0, 1). Proof. We shall fix a λ ∈ Γπ−θ1 \ BM δ (0) and simply write G2m (x, y) for the kernel of (L + λI)−2m . By Corollary 7.2.1, for all x ∈ Rn , there exists G2m (x, ·) ∈ L∞ (Rn ) such that for all ´ f ∈ L1 (Rn ), (L + λI)−2m (f )(x) = Rn G2m (x, y)f (y)dy. Furthermore, one can show by a limiting argument that there exists a measure zero set E0 such that for all x, x0 ∈ Rn , y ∈ Rn \ E0 , n o2m G (x, y) ≤ C max |λ|n/2 , |λ|−1 2m , n o2m G (x, y) − G2m (x0 , y) ≤ C max |λ|n/2 , |λ|−1 x − x0 µ . 2m Now that for fixed x, G2m (x, y) is measurable in y and for fixed y ∈ Rn \ E0 , H¨ older continuous in x, one can show G2m (x, y) is measurable. Since Corollary 7.2.1 also holds for the adjoint operator (L + λI)∗−2m , it follows from a similar argument that there exist a measure zero set E00 , such that for all y, y 0 ∈ Rn , x ∈ Rn \ E00 , n o2m G (x, y) ≤ C max |λ|n/2 , |λ|−1 2m , n o2m G (x, y) − G2m (x, y 0 ) ≤ C max |λ|n/2 , |λ|−1 y − y 0 µ . 2m 126 Combining these estimates one obtains ∀ x, x0 , y, y 0 ∈ Rn , n o2m G (x, y) ≤ C max |λ|n/2 , |λ|−1 2m , (7.2.40) n o2m G (x, y) − G2m (x0 , y) ≤ C max |λ|n/2 , |λ|−1 x − x0 µ , 2m (7.2.41) n o2m G (x, y) − G2m (x, y 0 ) ≤ C max |λ|n/2 , |λ|−1 y − y 0 µ . 2m (7.2.42) Fix any x0 , y0 ∈ Rn with x0 6= y0 . For any  > 0, define α·x Ψ (x) = − , 1 +  |x|2 α(x0 −y0 ) where α = |x0 −y0 | , with α = α(n, λ0 , Λ0 , δ) ∈ (0, 1) so small that Ψ satisfies the condition in Proposition 7.1.2. Then the kernel G2m −2m satisfies estimates (7.2.40) Ψ (x, y) of (LΨ +λI) – (7.2.42) as well. Also, ∀ f ∈ S (Rn ), (LΨ + λI)−2m (f )(x) = e−Ψ (x) (L + λI)−2m (eΨ f )(x) ˆ = e−Ψ (x) G2m (x, y)eΨ (y) f (y)dy, Rn ⇒ G2m (x, y) = G2m Ψ (x, y)e Ψ (x)−Ψ (y) . It follows from estimates (7.2.40) – (7.2.42) for G2m Ψ (x, y) that n o2m G (x0 , y0 ) ≤ C max |λ|n/2 , |λ|−1 2m eΨ (x0 )−Ψ (y0 ) , which gives (7.2.38) by letting  go to 0+ . 127 We also have G2m (x0 , y0 ) − G2m (x0 + h, y0 ) = G2m Ψ (x0 , y0 )e Ψ (x0 )−Ψ (y0 ) − G2m Ψ (x0 + h, y0 )e Ψ (x0 +h)−Ψ (y0 ) = G2m 2m  Ψ (x0 +h)−Ψ (y0 ) Ψ (x0 , y0 ) − GΨ (x0 + h, y0 ) e + G2m Ψ (x0 + h, y0 )e Ψ (x0 )−Ψ (y0 ) (1 − eΨ (x0 +h)−Ψ (x0 ) ), Letting  go to 0+ one obtains n o2m α G (x0 , y0 ) − G2m (x0 + h, y0 ) ≤ C max |λ|n/2 , |λ|−1 e− 4 |x0 −y0 | |h|µ 2m for 2 |h| ≤ |x0 − y0 |. G2m (x0 , y0 ) − G2m (x0 , y0 + h) can be estimated similarly. This proves (7.2.39). Since L is m-accretive, there exists an analytic contraction semigroup e−tL on L2 gen- erated by −L. Moreover, the semigroup can be expressed as follows (see e.g. [Paz83] Theorem 7.7) ˆ −tL 1 e = etλ (L + λI)−1 dλ, 2πi Γ where the path Γ consists of two half-rays Γ± = λ = re±i(π−θ1 ) , r ≥ R and of the arc  Γ0 = λ = Reiθ , |θ| ≤ π − θ1 , for any fixed R > 0. After integration by parts 2m − 1  times, it becomes ˆ −tL (2m − 1)! e = etλ (L + λI)−2m dλ. (7.2.43) 2πit2m−1 Γ We shall use this expression of etL , as well as Proposition 7.2.1, to obtain the following bounds for the heat kernel. Note that since the expression holds for any fixed R > 0, we shall always choose R > M δ, where δ ∈ (0, λ0 ) and M is as in (7.1.21). Therefore, for any λ on the path Γ, λ ∈ Γπ−θ1 \ BM δ (0). 128 Theorem 7.2.6. The kernel Kt (x, y) of e−tL satisfies the following bounds: for all t > 0, n β|x−y|2 |Kt (x, y)| ≤ Ct− 2 e− t , (7.2.44)   n |h| β|x−y|2 |Kt (x, y) − Kt (x + h, y)| + |Kt (x, y) − Kt (x, y + h)| ≤ Ct− 2 e− t t1/2 + |x − y| (7.2.45) when 2 |h| ≤ t1/2 + |x − y|. Here, 0 < µ < µ0 and µ0 is defined in Lemma 7.2.3, C = C(λ0 , Λ0 , n), β = β(λ0 , Λ0 , n) ∈ (0, 1). Proof. Let G2m −2m . Then by (7.2.43), λ be the kernel of (L + λI) ˆ (2m − 1)! Kt (x, y) = etλ G2m λ (x, y)dλ 2πit2m−1 Γ ˆ ˆ ˆ cm cm cm = 2m−1 etλ G2m λ (x, y)dλ + 2m−1 e tλ 2m Gλ (x, y)dλ + 2m−1 etλ G2m λ (x, y)dλ t Γ0 t Γ+ t Γ− . = I0 + I+ + I− . (7.2.46) We shall derive the following bounds of G2m λ (x, y) and then substitute them in (7.2.46) to obtain (7.2.44) and (7.2.45): n G (x, y) ≤ C |λ| 2 −2m e−α1 |λ|1/2 |x−y| , 2m λ (7.2.47) 2m G (x, y) − G2m (x + h, y) + G2m (x, y) − G2m (x, y + h) λ λ λ λ n 1/2 −2m ≤ C |λ| 2 e−α1 |λ| |x−y| |h|µ (7.2.48) when 2 |h| ≤ |x − y|. Here C and µ are the same as in Proposition 7.2.1, α1 ∈ (0, 1) only depends on λ0 , Λ0 and n. Fix any f ∈ S (Rn ), λ = reiθ ∈ Γπ−θ1 , and define u(j) = (L + λI)−j f . Then (L + λI)u(1) (x) = (− div A∇+reiθ I)u(1) (x) = f (x). For any function g, define ge(x) = g(r−1/2 x). 129 Then a change of variable gives (− div A∇ (1) (x) = r −1 fe(x). e + eiθ I)ug (7.2.49) e = − div A∇, Define L e and define G e iθ I)−2m . Then G e 2m to be the kernel of (L+e e 2m satisfies r,θ r,θ the estimates in Proposition 7.2.1, with λ = eiθ . And since |λ| = 1, we can fix δ ∈ (0, λ0 ) such that M δ < 1/2 in Proposition 7.2.1. So we have Gr,θ (x, y) ≤ Ce−α1 |x−y| , e 2m (7.2.50) e 2m e 2m (x + h, y) + G e 2m (x, y) − G e 2m (x, y + h) Gr,θ (x, y) − G r,θ r,θ r,θ ≤ Ce−α1 |x−y| |h|µ (7.2.51) when 2 |h| ≤ |x − y|, where C and µ are as in Proposition 7.2.1, α1 = α1 (n, λ0 , Λ0 ). By (7.2.49) and u(j+1) = (L + λI)−1 u(j) , we have ˆ (2m) (x) = r −2m (L u] e + eiθ I)−2m (fe)(x) = r−2m e 2m (x, y)fe(y)dy G r,θ Rn ˆ n −2m e 2m (x, r1/2 y)f (y)dy. =r 2 G r,θ Rn Therefore, we have n −2m e 2m 1/2 G2m λ (x, y) = r 2 Gr,θ (r x, r1/2 y), which combined with (7.2.50) and (7.2.51) implies (7.2.47) and (7.2.48), respectively. Now we estimate (7.2.46). Using (7.2.47), we have ˆ ∞ −n −α1 R1/2 |x−y| n |I± | ≤ cm t 2 e es cos(π−θ1 ) s 2 −2m ds, tR 130 and n n |I0 | ≤ cm t− 2 (tR) 2 −2m+1 exp{tR − α1 R1/2 |x − y|} n ≤ cm t− 2 (tR) exp{tR − α1 R1/2 |x − y|}, α21 |x−y|2 1 n o where the last inequality follows from 4m ≥ n + 2µ0 . Choosing R = max 4t2 ,t we obtain ( ) −n α2 |x − y|2 |I± | ≤ cm t 2 exp − 1 , 2t and ( ) −n α2 |x − y|2 |I0 | ≤ cm t 2 exp − 1 ∀  > 0, 4(1 + )t α21 which proves (7.2.44) with β = 4(1+) for arbitrary  > 0. The argument for the bound (7.2.45) is similar. Theorem 7.2.7. For l = 0, 1, 2, ..., ∂tl Kt (x, y) satisfies the following estimates: β|x−y|2 n ∂t Kt (x, y) ≤ Cl t− 2 −l e− t , l (7.2.52) l l l l ∂t Kt (x, y) − ∂t Kt (x + h, y) + ∂t Kt (x, y) − ∂t Kt (x, y + h)   −n −l |h| − β|x−y|2 ≤ Ct 2 e t t1/2 + |x − y| when 2 |h| ≤ t1/2 + |x − y|. Here, 0 < µ < µ0 and µ0 is defined in Lemma 7.2.3, C = C(λ0 , Λ0 , n), β = β(λ0 , Λ0 , n) ∈ (0, 1). Proof. We have l X ˆ −2m+1−k ∂tl Kt (x, y) = cm t λl−k eλt G2m λ (x, y)dλ, k=0 Γ where Γ and G2m λ (x, y) are the same as in Theorem 7.2.6. Then the theorem follows immediately from bounds (7.2.47) (7.2.48) of G2m λ (x, y) and a similar argument as in the proof of Theorem 7.2.6. 131 7.3 L2 estimates for ∂tl e−tL and ∇e−tL Theorem 7.3.1. For any f ∈ L2 (Rn ), t > 0, x ∈ Rn , denote h(x, t) = e−tL (f )(x). Then h ∈ C (0, ∞), W 1,2 (Rn ) , ∂tl h ∈ C (0, ∞), L2 (Rn ) for any l ∈ N. And for all t > 0,   ˆ 1 ∂tl h(x, t)= λl eλt (λI + L)−1 (f )(x)dλ, (7.3.1) 2πi Γ ˆ 1 ∂j h(x, t) = eλt ∂j (λI + L)−1 (f )(x)dλ, ∀ j = 1, 2, . . . , n. (7.3.2) 2πi Γ Moreover, we have for all t > 0, l −tL ∂t e ≤ cl t−l for all l ∈ N, (7.3.3) L2 →L2 ∇e 2 2 ≤ Ct− 21 . −tL L →L (7.3.4) Proof. Recall that ˆ −tL 1 e = etλ (L + λI)−1 dλ, (7.3.5) 2πi Γ where Γ consists of two half-rays Γ± = λ = re±i(π−θ1 ) , r ≥ R and of the arc Γ0 =  λ = Reiθ , |θ| ≤ π − θ1 , for any fixed R > 0.  Assuming (7.3.1), then (7.3.3) follows from (7.1.12). In fact, we split the right-hand side of (7.3.1) into three integrals and use (7.1.12) to estimate each of them directly. We have ˆ ˆ π−θ1 l λt −1 Rl eRt cos θ dθ kf kL2 . Rl eRt kf kL2 , λ e (λI + L) (f )dλ . Γ0 L2 −π+θ1 and ˆ ˆ ∞ −1 l λt rl−1 e−rt cos θ1 dr kf kL2 . t−l (tR)−1 . λ e (λI + L) (f )dλ . Γ± L2 R 132 1 Choosing R = t we obtain (7.3.3). Assuming (7.3.2), (7.3.4) follows from estimate (7.1.13) and a similar argument as above. It remains to prove (7.3.1) and (7.3.2). Proof of (7.3.1) c0 Since (λI + L)−1 f L2 ≤ kf kL2 , for any τ0 > 0, there exists F (x, t) ∈ L∞ [τ0 , ∞), L2 (Rn )  |λ| τ0 such that for all 0 < |τ | < 4, x ∈ Rn , t ≥ τ0 , |∆τ h(x, t)| ≤ F (x, t), h(x,t+τ )−h(x,t) where ∆τ h(x, t) = τ . Therefore, one can apply Dominated Convergence The- orem to get ˆ λ(t+τ ) 1 e − eλt ∂t h(x, t) = lim (λI + L)−1 (f )(x)dλ 2πi τ →0 Γ τ ˆ 1 = λeλt (λI + L)−1 (f )(x)dλ, t ≥ τ0 . 2πi Γ Similarly, we have ˆ 1 ∂tl h(x, t) = λl eλt (λI + L)−1 (f )(x)dλ, t ≥ τ0 . 2πi Γ This gives (7.3.1) since τ0 > 0 is arbitrary. Proof of (7.3.2) We need the following lemma. For its proof see [Mik98] Theorem 4.2.10. Lemma 7.3.2. Suppose X and Y are Banach spaces and A : D(A) ⊂ X → Y is a closed linear operator. Let I be an interval in R. If F : I → D(A) a.e., F ∈ L1 (I, X) and ´ ´ ´ AF ∈ L1 (I, Y ), then I F (t)dt ∈ D(A) and A I F (t)dt = I AF (t)dt. 133 In application, we let X = Y = L2 (Rn ), A = ∂j , and F (t) = eλt (L + λI)−1 f . In fact, ´ we should first write the contour integral Γ eλt (L + λI)−1 f dλ into real integrals, and let F (t) be the corresponding integrands. We omit this process. We want to show 1. The weak differentiation operator ∂j is a closed operator from L2 (Rn ) to L2 (Rn ); ´ 2. Γe λt (L + λI)−1 f (·)dλ ∈ L2 ; ´ 3. Γ ∂j e λt (L + λI)−1 f (·)dλ ∈ L2 . Then (7.3.4) follows from the lemma. (1) is a well-known fact. (2) follows immediately from the expression (7.3.5) and e−tL : L2 → L2 . (3) has been proved; since we have shown the right-hand side of (7.3.2) is L2 bounded. 1 ´ tλ −1 Finally, it follows from the definition of h(x, t) = 2πi Γ e (L + λI) (f )(x)dλ, (7.3.1) and (7.3.2) that h ∈ C (0, ∞), W 1,2 (Rn ) and ∂tl h ∈ C (0, ∞), L2 (Rn ) for any l ∈ N.   Remark 7.3.1. Alternatively, we can obtain (7.3.3) using the kernel estimates (7.2.52). Actually, it gives Lp estimates for ∂tl e−tL for all 1 ≤ p < ∞. One benefit of proving (7.3.1) is that we can see immediately from (7.3.2) and the proof of (7.3.1) that ˆ 1 ∂tl ∂j h(x, t) = λl eλt ∂j (λI + L)−1 (f )(x)dλ, ∀ j = 1, 2, . . . , n, l ∈ N. 2πi Γ Therefore, ∂tl ∂j h ∈ L2loc (0, ∞), L2 (Rn ) ∩ C (0, ∞), L2 (Rn ) . Moreover, (7.1.13) implies   1 l −tL ∂t ∇e ≤ cl t−l− 2 , for any l ∈ N. L2 →L2 134 This can not be proved by using the kernel estimates. CHAPTER Eight The Lp theory for the semigroup and square root 8.1 L2 off-diagonal estimates for the semigroup Definition 8.1.1. Let T = (Tt )t>0 be a family of operators. We say that T satisfies L2 off-diagonal estimates if for some constants C ≥ 0 and α > 0 for all closed sets E and F , all h ∈ L2 with support in E and all t > 0 we have αd(E,F )2 kTt hkL2 (F ) ≤ Ce− t khkL2 . (8.1.1) Here and subsequntly, d(E, F ) is the semi-distance induced on sets by the Euclidean dis- tance. π If T = (Tz )z∈Σµ is a family defined on a complex sector Σµ with 0 ≤ µ < 2, then we adopt the same definition and replace t by |z| in the right hand side of (8.1.1). In this case, the constants C and α may depend on the angle µ. Proposition 8.1.1. There exists ω0 = ω0 (n, λ0 , Λ0 ) ∈ (0, π2 ), such that for all µ ∈ (0, π2 − √ ω0 ), the families (e−zL )z∈Σµ , (zLe−zL )z∈Σµ and ( z∇e−zL )z∈Σµ satisfy L2 off-diagonal 135 136 estimates. Proof. We begin with the case of real times t > 0. Let ϕ be a bounded Lipschitz function with Lipschitz constant 1 and ρ > 0. Define Lρ = eρϕ L e−ρϕ as (7.1.14), and define Lρ0 = Lρ + c0 ρ2 , with c0 to be determined. Using Proposition 2.2.3, we estimate 0, such that 0 =hLρ u, uiW f −1,2 ,W 1,2 ≤ C0 , 0. Then by L0ρ = Lρ + c0 ρ2 I and ∂t e−tLρ = −L0ρ e−tLρ , a direct computation shows −tL √ 2 e ρ (f ) 2 + t(Lρ + c0 ρ2 I)e−tLρ (f ) 2 + −tLρ ≤ Cec0 ρ t kf kL2 . t∇e (f ) L L L2 This implies −tL √ 2 e ρ (f ) 2 + t∂t e−tLρ (f ) 2 + −tLρ ≤ Ce2c0 ρ t kf kL2 t∇e (f ) ∀ t > 0. (8.1.3) L L L2 Let E and F be two closed sets and f ∈ L2 , with compact support contained in E. For d(x,E) any ε > 0, choose ϕ(x) = ϕε (x) = 1+εd(x,E) . Then e−tL f = e−ρϕε e−tLρ (eρϕε f ) = e−ρϕε e−tLρ f, (8.1.4) where the last equality follows from the facts that supp f ⊂ E and that ϕε (x) = 0 if x ∈ E. Observe that there exists R0 > 1 sufficiently large, such that for any R ≥ R0 , d(E, F ) ≤ d(E, F ∩ BR (0)) ≤ 2d(E, F ). For any fixed R ≥ R0 , there exists ε0 = ε0 (n, R, E) > 0, such that for any 0 < ε ≤ ε0 , 1 εd(x, E) ≤ 2 for all x ∈ F ∩ BR (0), and thus d(x, E) 2 ϕε (x) = ≥ d(x, E), ∀ x ∈ F ∩ BR (0). (8.1.5) 1 + εd(x, E) 3 Now fix 0 < ε ≤ ε0 , then by (8.1.3) and (8.1.5), for all t > 0 and ρ > 0 −tL 2 2 e f 2 L (F ∩B ≤ Ce− 3 ρd(E,F ) e2c0 ρ t kf kL2 . R (0)) 138 Letting R → ∞ and optimizing with respect to ρ > 0 yield d(E,F )2 −tL − c t e f 2 L (F ) ≤ Ce 1 kf kL2 , (8.1.6) where c1 = 18c0 . Differentiating (8.1.4), one has ∂t e−tL f = e−ρϕε ∂t e−tLρ f . Then by a same argument one obtains d(E,F )2 t∂t e−tL f 2 − c t L (F ) ≤ Ce 1 kf kL2 . (8.1.7) Applying the gradient operator to (8.1.4) gives ∇e−tL f = −ρ(e−ρϕε e−tLρ f )(∇ϕε )+e−ρϕε ∇e−tLρ f . So −tL 2 2 1 2 2 ∇e f 2 L (F ∩B ≤ Cρe− 3 ρd(E,F ) e2c0 ρ t kf kL2 + Ct− 2 e− 3 ρd(E,F ) e2c0 ρ t kf kL2 . R (0)) d(E,F ) Letting R → ∞ and choosing ρ = 6c0 t yield √   d(E, F ) − d(E,F )2 d(E,F )2 −tL − βt t∇e f ≤C 1+ √ e c t kf kL2 ≤ Ce kf kL2 (8.1.8) 1 L2 (F ) t for some β > c1 . Now we extend (8.1.6), (8.1.7) and (8.1.8) to complex times. Consider the operator eiα L , which has coefficients eiα A(x). Note that if ξ is in the numerical range of eiα L , then |arg ξ| ≤ θ0 + |α|, where the angle θ0 is as in (7.1.1). Therefore, in light of (8.1.2), the argument above applies to eiα L as long as |α| < π 2 − ω0 . Observe that when z = teiα , iα L) e−zL = e−t(e . From this the desired estimates follow. Remark 8.1.1. The same argument applies to the adjoint operator L∗ . So for all µ ∈ ∗ ∗ √ ∗ (0, π2 − ω0 ), the families (e−zL )z∈Σµ , (zL∗ e−zL )z∈Σµ and ( z∇e−zL )z∈Σµ satisfy L2 off- diagonal estimates. Using the L2 off-diagonal estimates, we can define the action of the semigroup on L∞ 139 and on Lipschitz function in the L2loc sense. Lemma 8.1.1 (Definition). Let f ∈ L∞ (Rn ). Then for any x0 ∈ Rn , limR→∞ e−tL (f 1BR (x0 ) ) exists in L2loc (Rn ) and the limit does not depend on x0 . We define the limit to be e−tL f . Proof. We first fix any x0 ∈ Rn . Fix any R0 > 1, and let R2 > R1 > 8R0 . Then there exists l ∈ N such that 2l R1 < R2 ≤ 2l+1 R1 . We write   e (f 1BR2 (x0 ) ) − e−tL (f 1BR1 (x0 ) ) = e−tL f 1BR2 (x0 ) − f 1BR1 (x0 ) −tL   X l   ≤ e−tL f 1BR2 (x0 ) − 1B2l R 1B2k R (x0 ) − 1B2k−1 R −tL + e f (x0 ) . (x0 ) 1 1 1 k=1 Observe that d(BR0 (x0 ), B2k R1 (x0 ) \ B2k−1 R1 (x0 )) & 2k−1 R1 for k = 1, 2, . . . , l. Then by L2 off-diagonal estimates for e−tL t>0 , we have  e (f 1BR2 (x0 ) ) − e−tL (f 1BR1 (x0 ) ) −tL L2 (BR0 (x0 )) l c(2l R1 )2 X c(2k−1 R1 )2 . e− t kf kL2 (BR (x0 )\B2l R (x0 )) + e− t kf kL2 (B (x0 )) 2 1 2 k R1 k=1 l+1 c(2k−1 R1 )2 −n/2 X . e− t (2k R1 )n/2 kf kL∞ . tn/2 R1 kf kL∞ , k=1 where the implicit constant depends only on λ0 , Λ0 and n. This shows that e−tL (f 1BR (x0 ) ) is a Cauchy sequence in L2loc (Rn ), when f ∈ L∞ . We now show the limit is independent of the choice of x0 . Let x1 ∈ Rn be a different point than x0 . Then for R sufficiently large, the symmetric difference BR (x0 )∆BR (x1 ) is contained in B2R (0) \ B R (0). So by the L2 off-diagonal estimates for e−tL , and that 2 d(BR (x0 )∆BR (x1 ), B R (0)) & R, we have 4 cR2 e f 1B (x ) − e−tL f 1B (x ) 2 −tL . e− R 0 R 1 L (B t kf kL2 (BR (x0 )∆BR (x1 )) R (0)) 4 n cR2 . R 2 e− t kf kL∞ . 140 This implies that limR→∞ e−tL f 1BR (x0 ) = limR→∞ e−tL f 1BR (x1 ) in the L2loc sense. Remark 8.1.2. The limit limR→∞ e−tL (f 1BR (x0 ) ) actually exists in Wloc 1,2 (Rn ). One can √ show this by using the L2 off-diagonal estimates for t∇e−tL f t>0 instead of the L2  off-diagonal estimates for e−tL t>0 in the argument above.  Similarly, we can define e−tL f for f Lipschitz. Lemma 8.1.2 (Definition). Let f be a Lipschitz function. Then for any x0 ∈ Rn , the limit limR→∞ e−tL (f 1BR (x0 ) ) exists in L2loc (Rn ) and not depend on x0 . We define the limit to be e−tL f . Proof. Fix any x0 ∈ Rn , any R0 > 1, and let R2 > R1 > 8R0 . Then there exists l ∈ N such that 2l R1 < R2 ≤ 2l+1 R1 . We write e (f 1BR2 (x0 ) ) − e−tL (f 1BR1 (x0 ) ) −tL     ≤ e−tL (f − f (x0 ))(1BR2 (x0 ) − 1BR1 (x0 ) ) + e−tL f (x0 )(1BR2 (x0 ) − 1BR1 (x0 ) ) . = I1 + I2 Since f (x0 ) is a bounded constant function, the proof of Lemma 8.1.1 applies to I2 , and we have −n/2 kI2 kL2 (BR (x0 )) . tn/2 R1 |f (x0 )| . (8.1.9) 0 For I1 , we have (2l R1 )2 kI1 kL2 (BR . e−c (f − f (x0 ))1BR2 (x0 )\B2l R t (x0 )) (x0 ) 0 1 L2 l (2k−1 R1 )2 (f − f (x0 ))1B2k R X + e−c t (x0 )\B2k−1 R (x0 ) 1 1 L2 k=1 − n+1 l+1 ∞  k X −c (2k−1 R1 )2 k 1+ n X (2 R1 )2 2 n . e t (2 R1 ) 2 k∇f kL∞ (Rn ) . (2k R1 )1+ 2 k∇f kL∞ t k=1 k=1 n+1 −n .t 2 R1 2 k∇f kL∞ . 141 This and (8.1.9) show that e−tL (f 1BR (x0 ) ) is a Cauchy sequence in L2loc (Rn ), when f is Lipschitz. By a similar argument as in the proof of Lemma 8.1.1, one can show the limit is independent of x0 . √ Remark 8.1.3. By applying the L2 off-diagonal estimates for t∇e−tL f t>0 instead of  the L2 off-diagonal estimates for e−tL t>0 in the argument above, one can show that as  R → ∞, e−tL (f 1BR (x0 ) ) converges in Wloc 1,2 (Rn ), when f is Lipschitz. Proposition 8.1.2. The conservation property e−tL 1 = 1, for any t > 0, holds in the sense of L2loc . Proof. Let Φ be an L2 (Rn ) function with compact support, and suppose the support of Φ is contained in a cube Q with l(Q) = r0 . We first show e−tL Φ ∈ L1 . Let Q0 = 2Q, and decompose Rn into a union of nonoverlapping cubes with same size. That is, Rn = ∞ F k=0 Qk , with |Qk | = |Q0 |. Let xQ denote the center of Q and Q0 , and let xk denote the center of Qk , for k = 1, 2, . . . . Define Fl = {Qk : 2lr0 ≤ |xk − xQ | < 2(l + 1)r0 } , for l = 1, 2, . . . . By the L2 boundedness of the semigroup, we have ˆ e Φ ≤ |Q0 |1/2 e−tL Φ 2 . |Q0 |1/2 kΦk 2 . −tL L L (8.1.10) Q0 For any Qk ∈ Fl , we use the L2 off-diagonal estimates for (e−tL )t>0 to obtain ˆ c(lr0 )2 e Φ ≤ |Qk |1/2 e−tL Φ 2 . |Q0 |1/2 e− −tL L (Q t kΦkL2 . k) Qk 142 Since |Fl | ≈ ln , we have X ˆ c(lr )2 n e Φ . ln |Q0 |1/2 e− t0 kΦk 2 . l−2 r− 2 −2 t n+2 −tL 2 kΦk 2 . L 0 L Qk ∈Fl Qk Summing in l yields ˆ ∞ −tL X − n −2 n+2 − n −2 n+2 e Φ . l−2 r0 2 t 2 kΦkL2 . r0 2 t 2 kΦkL2 . (8.1.11) Rn \Q0 l=1 ∗ Then e−tL Φ ∈ L1 follows from (8.1.10) and (8.1.11). The argument also applies to e−tL ∗ and so e−tL Φ ∈ L1 . Therefore, by Lemma 8.1.1, we have ˆ ˆ ˆ ˆ e−tL 1Φ = lim e−tL (1BR )Φ = lim 1BR e−tL∗ Φ = e−tL∗ Φ. (8.1.12) Rn R→∞ Rn R→∞ Rn Rn Here and subsequently, BR is the ball centered at the origin with radius R. We shall first ´ show Rn e−tL 1Φ does not depend on t > 0 by proving ˆ d e−tL 1Φ = 0, (8.1.13) dt Rn and then show ˆ ˆ −tL e 1Φ = Φ. (8.1.14) Rn Rn This implies that e−tL 1 = 1 in the sense of L2loc . Observe that we can define ∂t e−tL f = limR→∞ ∂t e−tL (f 1BR (x0 ) ) in L2loc , for f ∈ L∞ , for any x0 ∈ Rn . This is because the argument in the proof of Lemma 8.1.1 applies to ∂t e−tL if one uses the L2 off-diagonal estimates for tLe−tL t>0 instead of that for e−tL t>0 . In   particular, ∂t e−tL 1 = limR→∞ ∂t e−tL 1BR . Also, using the L2 off-diagonal estimates for ∗ ∗ tL∗ e−tL t>0 , one can show ∂t e−tL Φ ∈ L1 . Therefore, ˆ ˆ ˆ ˆ ∂t e −tL 1Φ = lim ∂t e −tL 1BR Φ = lim 1BR ∂t e−tL∗ Φ = ∂t e−tL∗ Φ. Rn R→∞ R→∞ 143 Let η ∈ C0∞ (Rn ) with η = 1 in B1 , and supp η ⊂ B2 . Let ηR (x) = η x  R for R > 0. Then ˆ ˆ ˆ ˆ d e−tL 1Φ = ∂t e−tL 1Φ = ηR ∂t e−tL∗ Φ + (1 − ηR )∂t e−tL∗ Φ. (8.1.15) dt Rn Rn Rn Rn ∗ Since ∂t e−tL Φ ∈ L1 , the last term goes to 0 as R → ∞. We write ˆ ˆ ˆ ηR ∂t e−tL∗ Φ =− ηR = L∗ e−tL∗ Φ A∇ηR · ∇e−tL∗ Φ Rn n n ˆ R R ˆ s = ∗ A ∇ηR · ∇e−tL Φ + Aa ∇ηR · ∇ (ζR e−tL∗ Φ). Rn Rn x , where ζ ∈ C0∞ (Rn ) with ζ = 1 in B2 \ B1 and supp ζ ⊂ B5/2 \ B1/2 .  Here, ζR (x) = ζ R Note that ζR = 1 in the support of ∇ηR . Choose R to be sufficiently large so that Q ⊂ B R . We estimate 8 ˆ s 1 −tL∗ n −1 − 21 − cRt 2 ∇e−tL∗ Φ A ∇ηR · ≤ λ0 k∇η R kL2 ∇e Φ . R 2 t e kΦkL2 , Rn L2 (B2R \BR ) √ ∗ where the last inequality follows from the L2 off-diagonal estimates for t∇e−tL  t>0 . By Proposition 2.2.1, we have ˆ   a −tL∗ e−tL∗ Φ) A ∇ηR · ∇ (ζR ≤ CΛ k∇η k ∇ ζ e Φ 2. 0 R 2 R L Rn L Using the support property of ζR and ∇ζR , we have  ∗  ∗ ∗ ∇ ζR e−tL Φ . ∇(e−tL Φ) + R−1 e−tL L2 L2 (B 5R \B R ) L2 (B 5R \B R ) 2 2 2 2 2 2 − 12 − cRt − cRt .t e kΦkL2 + R−1 e kΦkL2 , √ ∗ where the last inequality follows from the L2 off-diagonal estimates for t∇e−tL t>0 and ∗ ´ e−tL t>0 . Combining these estimates, we obtain Rn ηR ∂t e−tL∗ Φ → 0 as R → ∞. So from (8.1.15), the desired result (8.1.13) follows. 144 To prove (8.1.14), we fix R > 0 sufficiently large so that 2Q ⊂ BR . Write ˆ ˆ ˆ −tL e 1Φ = ηR e−tL∗ Φ + (1 − ηR )e−tL∗ Φ. Rn Rn Rn ∗ Since e−tL is strongly continuous in L2 at t = 0, ˆ ˆ ˆ lim ηR e−tL∗ Φ = ηR Φ = Φ. t→0 Rn Rn ´ ´ ∗ We have Rn (1 − ηR )e−tL∗ Φ ≤ Rn \Q0 e−tL Φ , where Q0 is constructed in the begin- ∗ ning. Since we can also obtain (8.1.11) for e−tL , we then have ˆ −tL∗ − n −2 n+2 (1 − ηR )e Φ . r0 2 t 2 kΦkL2 , Rn ´ which goes to 0 as t → 0. Therefore, since we have shown that Rn e−tL 1Φ is independent of t, we obtain (8.1.14). 8.2 Lp theory for the semigroup We now study the uniform boundedness of the semigroup (e−tL )t>0 and of the family √ ( t∇e−tL )t>0 on Lp spaces. We begin with a few definitions. Let T = (Tt )t>0 be a family of uniformly bounded operators on L2 . Definition 8.2.1. We say that T is Lp − Lq bounded for some p, q ∈ [1, ∞] with p ≤ q if for some constant C, for all t > 0 and all h ∈ Lp ∩ L2 γpq kTt hkLq ≤ Ct− 2 khkLp , 145 where γpq = nq − np . We shall use γp to denote γp2 = n2 − np . Definition 8.2.2. We say that T satisfies Lp − Lq off-diagonal estimates for some p, q ∈ [1, ∞] with p ≤ q if for some constants C, c > 0, for all closed sets E and F , all h ∈ Lp ∩L2 with support in E and all t > 0 we have γpq cd(E,F )2 kTt hkLq (F ) ≤ Ct− 2 e− t khkLp . Note that the uniform Lp boundedness of (e−tL )t>0 and of (t∂t e−tL )t>0 follows from the kernel estimates (7.2.52). Lemma 8.2.1. Let p ≥ 1. There is some constant C = C(n, λ0 , Λ0 , p) such that for all t > 0 and all f ∈ Lp , −tL e f p ≤ C kf k p , (8.2.1) L L t∂t e−tL f p ≤ C kf k p . L L (8.2.2) Proof. By (7.2.7), we have ˆ ˆ |Kt (x, y)| dx ≤ C |Kt (x, y)| dy ≤ C, (8.2.3) Rn Rn ˆ ˆ −1 |∂t Kt (x, y)| dx ≤ Ct |∂t Kt (x, y)| dy ≤ Ct−1 . (8.2.4) Rn Rn Then ˆ ˆ  10 ˆ 1 p p p −tL e f (x) = Kt (x, y)f (y)dy ≤ |Kt (x, y)| dy |Kt (x, y)| |f (y)| dy . R n Rn Rn So (8.2.1) follows from (8.2.3). And (8.2.2) follows from (8.2.4) by the same argument. Proposition 8.2.1. 1. (e−tL )t>0 is Lp − L2 bounded for any 1 ≤ p < 2. 146 2. (e−tL )t>0 satisfies the Lp − L2 off-diagonal estimates for any 1 < p < 2. 3. (e−tL )t>0 is L2 − Lp bounded and satisfies the L2 − Lp off-diagonal estimates for p > 2. Proof. (1). Let p ∈ [1, 2). Then by the Gagliardo-Nirenberg inequality, −tL 2 e f 2 ≤ C ∇e−tL f 2α2 e−tL f 2βp L L L (8.2.5) for all t > 0 and f ∈ L2 ∩ Lp , where α + β = 1 and (1 + γp )α = γp . By ellipticity and Lemma 3.3.1, one has ˆ ∇e f 2 ≤ 1 < 1 −tL 2 A∇e−tL f · ∇e−tL f = < Le−tL f, e−tL f  L λ0 λ0 Rn 1 1 d e−tL f 2 2 . = − 0 : ϕ(t) = 0}. Then ϕ(t) > 0 when 0 < t < t0 , and thus ϕ(t) ≤ Ct−γp for 0 < t < t0 . When t ≥ t0 , ϕ(t) = 0. So we have ϕ(t) ≤ Ct−γp for all t > 0. 147 (2). Fix two closed sets E and F , and fix any p ∈ (1, 2). Let f ∈ L1 (Rn ) ∩ L2 (Rn ) with supp f ⊂ E. By the L2 off-diagonal estimates for the semigroup, we have −tL cd(E,F )2 e f 2 L (F ) ≤ Ce− t kf kL2 (E) . By (1), (e−tL )t>0 is L1 − L2 bounded. So we have −tL γ1 e f 2 L (F ) ≤ Ct− 2 kf kL1 (E) . 1 αp There exists an αp ∈ (0, 1) such that p = 2 + 1 − αp . Interpolating by the Riesz-Thorin theorem one obtains −tL γ1 cd(E,F )2 γp c0 d(E,F )2 e f 2 L (F ) ≤ Cp t−(1−αp ) 2 e−αp t kf kLp (E) = Cp t− 2 e− t kf kLp (E) . Then (2) follows from a standard limiting argument. ∗ (3). Since (1) and (2) also holds for e−tL = (e−tL )∗ , (3) follows from duality. The following results for (t∂t e−tL )t>0 are in the same spirit of Propostion 8.2.1. Proposition 8.2.2. 1. (t∂t e−tL )t>0 is Lp − L2 bounded for any 1 ≤ p < 2. 2. (t∂t e−tL )t>0 satisfies the Lp − L2 off-diagonal estimates for any 1 < p < 2. 3. (t∂t e−tL )t>0 is L2 − Lp bounded and satisfies the L2 − Lp off-diagonal estimates for p > 2. Proof. (1). Let p ∈ [1, 2). Since ∂t e−tL f ∈ W 1,2 (Rn ) (see Remark 7.3.1), we can apply the 148 Gagliardo-Nirenberg inequality and get, −tL 2 ∂t e f 2 ≤ C ∇∂t e−tL f 2α2 ∂t e−tL f 2βp L L L (8.2.6) for all t > 0 and f ∈ L2 ∩ Lp , where α + β = 1 and (1 + γp )α = γp . Using ∂t2 e−tL f ∈ L2 and Lemma 3.3.1, we have ˆ A∇∂t e−tL f · ∇∂t e−tL f = < L∂t e−tL f, ∂t e−tL f = −0 , ∂t e−tL f Lp ≤ Cp t−1 . Then by (8.2.6) and (8.2.7), one obtains 2β 1 t α ϕ(t) α ≤ −Cϕ0 (t). ˆ ˆ 2t 2β 2t ϕ0 (t) 2−α ⇒ t α dt ≤ −C dt, ⇒ ϕ(t) ≤ Ct− 1−α . t t ϕ(t)1/α Here we assumed that ϕ(t) 6= 0. Otherwise, considering ϕ(t) + ε and then letting ε → 0 2 α would give the same result. Thus t∂t e−tL f L2 ≤ Ct− 1−α = Ct−γp , which proves (1). (2). As in the proof of Proposition 8.2.1, it follows from the L2 off-diagonal estimates of (t∂t e−tL )t>0 , the L1 − L2 boundedness of (t∂t e−tL )t>0 , and the Riesz-Thorin interpolation theorem. (3). Since e−tL f ∈ D(L) for any f ∈ L2 , t∂t e−tL f = −tLe−tL f = −te−tL Lf . So we have ∗ ∗ (t∂t e−tL )∗ = −tL∗ e−tL = t∂t e−tL . ∗ Since (2) and (3) also hold for t∂t e−tL , (3) follows from duality and a limiting argument. 149 √ For ( t∇e−tL )t>0 , when p < 2, we immediately obtain the Lp − L2 boundedness and the Lp − L2 off-diagonal estimates. Proposition 8.2.3. √ 1. ( t∇e−tL )t>0 is Lp − L2 bounded for any 1 ≤ p < 2. √ 2. ( t∇e−tL )t>0 satisfies Lp − L2 off-diagonal estimates for any 1 < p < 2. √ √ tL tL Proof. Let p ∈ [1, 2) and f ∈ L2 ∩ Lp . Write t∇e−tL f = t∇e− 2 e− 2 f . Then by the √ L2 boundedness of ( t∇e−tL )t>0 and the Lp − L2 boundedness of (e−tL )t>0 , one has √ tL γp −tL t∇e f ≤ C e− 2 f ≤ Cp t − kf kLp , 2 L2 L2 which proves (1). √ (2) follows from the L2 off-diagonal estimates of ( t∇e−tL )t>0 , the L1 −L2 boundedness √ of ( t∇e−tL )t>0 , and the Riesz-Thorin interpolation theorem. When p > 2, a duality argument would not give us the desired results as in Proposition 8.2.1 (3) and Proposition 8.2.2 (3). However, we are able to derive a reverse H¨older type inequality for ∇e−tL f , and then use the L2 − Lp boundedness of (t∂t e−tL )t>0 to obtain the √ L2 − Lp boundedness of ( t∇e−tL )t>0 . Proposition 8.2.4. √ 1. ( t∇e−tL )t>0 is L2 −Lp bounded for any 2 ≤ p ≤ 2+1 for some 1 = 1 (λ0 , Λ0 , n) > 0. √ 2. ( t∇e−tL )t>0 satisfies the L2 − Lp off-diagonal estimates for any 2 ≤ p < 2 + 1 , where 1 is as in (1). 150 Proof. Let f ∈ S (Rn ), and let u(x, t) = e−tL f (x). Then u satisfies the equation ∂t u+Lu = 0 in L2 . That is, for any w ∈ W 1,2 (Rn ), ˆ ˆ A(x)∇u(x, t) · ∇w(x)dx = − ∂t u(x, t)w(x)dx, ∀ t > 0. Rn Rn Fix t > 0 and fix a cube Q ⊂ Rn with l(Q) = ρ0 , where ρ0 is to be determined. Let x0 ∈ 3Q and let 0 < ρ < min 21 dist(x0 , ∂(3Q)), ρ0 . Let Qs (x) denote the cube centered at x with  side length s. Choose ϕ ∈ C02 (Rn ), with 0 ≤ ϕ ≤ 1, ϕ = 1 in Qρ (x0 ), supp ϕ ⊂ Q 3 ρ (x0 ), 2 1 and |∇ϕ| . ρ. ffl Let w(x) = (u(x, t) − c) ϕ2 (x), where c = Q2ρ (x0 ) u(x, t)dx. Then we have ˆ ˆ 2 ∂t u(x, t) (u(x, t) − c) ϕ2 (x)dx.  A(x)∇u(x, t) · ∇ (u(x, t) − c)ϕ (x) dx = − Rn Rn We have ˆ ˆ ˆ s 2 s 2 As ∇u · ∇ϕ ((u − c)ϕ) dx  A ∇u · ∇ (u − c)ϕ dx = A ∇u · ∇uϕ dx + 2 Rn Rn Rn ˆ ˆ ˆ ˆ λ0 λ0 ≥ |∇u|2 ϕ2 − C |u − c|2 |∇ϕ|2 ≥ |∇u|2 − Cρ−2 |u − c|2 . 2 Rn R n 2 Qρ (x0 ) Q 3 (x0 ) 2ρ ´ Aa ∇u·∇ (u − c)ϕ2 dx, we introduce another bump function η ∈ C02 (Rn )  To deal with Rn with 0 ≤ η ≤ 1, η = 1 on Q 3 ρ (x0 ), supp η ⊂ Q2ρ (x0 ), and |∇η| . ρ1 . Then we have 2 ˆ ˆ a 2 Aa ∇u · ∇(ϕ2 )(u − c)dx  A ∇u · ∇ (u − c)ϕ dx = Rn Rn ˆ 1 Aa ∇ (u − c)2 η 2 · ∇(ϕ2 ).  = 2 Rn 151 By Proposition 2.2.3, one has ˆ a 2  A ∇u · ∇ (u − c)ϕ dx ≤ C k∇ϕkL∞ k(u − c)ηkL2 k∇ ((u − c)η)kL2 Rn ˆ ˆ ˆ λ0 2 2 2 2 2 ≤ |∇u| η dx + C k∇ϕkL∞ |u − c| η dx + C k∇ϕkL∞ |u − c|2 |∇η|2 dx 4 Rn n n ˆ ˆ R R λ0 2 −2 2 ≤ |∇u| dx + Cρ |u − c| dx. 4 Q2ρ (x0 ) Q2ρ (x0 ) ´ For Rn ∂t u(x, t) (u(x, t) − c) ϕ2 (x)dx, we use Cauchy-Schwarz inequality to get ˆ ˆ ˆ 1 1 ∂t u(x, t) (u(x, t) − c) ϕ (x)dx ≤ ρ2 2 |∂t u|2 ϕ2 dx + ρ−2 |u − c|2 ϕ2 dx n 2 n 2 n R ˆR ˆ R 1 2 2 1 ≤ ρ0 |∂t u| dx + ρ −2 |u − c|2 dx 2 Q2ρ (x0 ) 2 Q2ρ (x0 ) Combining these estimates, we obtain ˆ ˆ ˆ ˆ λ0 C2 2 λ0 2 ρ2 |∇u| dx ≤ 2 |u − c| dx + |∇u| dx + 0 |∂t u|2 dx. 2 Qρ (x0 ) ρ Q2ρ (x0 ) 4 Q2ρ (x0 ) 2 Q2ρ (x0 ) The Sobolev-Poincar´e inequality gives ! n+2 n 2 2n 1 |∇u| dx ≤ C |∇u| n+2 + |∇u|2 dx + Cρ20 |∂t u|2 dx. Qρ (x0 ) Q2ρ (x0 ) 2 Q2ρ (x0 ) Q2ρ (x0 ) Then by Lemma 2.4.1, there is some 1 = 1 (λ0 , Λ0 , n) > 0, such that for all p ∈ [2, 2 + 1 ],  1/p n 1/2  1/p o |∇u|p dx ≤C |∇u|2 + |ρ0 ∂t u|p Q 2Q 2Q ˆ ˆ p/2 ˆ p −pγ 2 ⇒ |∇u| dx ≤ Cρ0 p |∇u| + |ρ0 ∂t u|p dx. (8.2.8) Q 2Q 2Q Decompose Rn into a union of disjoint cubes Rn = t∞ j=1 Qj with each Qj having side length ρ0 . For each Qj , applying (8.2.8) and then summing in j, one has −γp k∇ukLp (Rn ) ≤ Cρ0 k∇ukL2 (Rn ) + C kρ0 ∂t ukLp (Rn ) , ∀ p ∈ [2, 2 + 1 ]. 152 √ Choosing ρ0 = t gives √ γp √ t∇u ≤ Ct− t∇u + C kt∂t ukLp , ∀ p ∈ [2, 2 + 1 ]. 2 Lp L2 √ Then by the L2 boundedness of ( t∇e−tL )t>0 and the L2 −Lp boundedness of (t∂t e−tL )t>0 , we obtain √ γp t∇e−tL f ≤ Ct− kf kL2 , ∀ p ∈ [2, 2 + 1 ], f ∈ S (Rn ). 2 Lp Thus (1) follows from a standard limiting argument. (2) can be proved using the L2 off-diagonal estimates and the L2 − L2+1 boundedness √ of ( t∇e−tL )t>0 , and the Riesz-Thorin interpolation theorem. 8.3 Lp Theory for the square root Since L is an m-accretive operator, there is a unique m-accretive square root L1/2 such that L1/2 L1/2 = L in D(L). (8.3.1) Also, L1/2 is m-sectorial with the numerical range contained in the sector |arg ξ| ≤ π4 . And D(L) is a core of L1/2 , i.e. (u, L1/2 u) : u ∈ D(L) is dense in the graph (u, L1/2 u) : u ∈ D(L1/2 )   (see [Kat76] p.281 for a proof for these facts). Our goal in this section is to prove the Lp bounds for the square root. And we shall only focus on Lp bound with p ≥ 2, as this is all we need to derive estimates for the square functions in section 6. 153 Many formulas can be used to compute L1/2 . The one we are going to use is ˆ ∞ −1/2 dt L 1/2 f =π e−tL Lf √ . (8.3.2) 0 t Observe that the integral converges in L2 when f ∈ D(L). Since for f ∈ D(L), Lf ∈ L2 , ´1 then by the L2 boundedness of the semigroup, 0 e−tL Lf √ dt t converges. And the L2 bound ´∞ of (t∂t e−tL )t>0 implies that 1 e−tL Lf √ dt t converges in L2 . Moreover, we have the following estimate Lemma 8.3.1. ˆ ∞ −tL dt 1/2 e Lf √ . L f 2 n ∀ f ∈ D(L). 0 t L2 (Rn ) L (R ) Proof. It suffices to show ˆ N −tL dt 1/2 e Lf √ . L f 2 n (8.3.3)  t L2 (Rn ) L (R ) uniformly in ε, N , for all f ∈ D(L). To this end, we write, for any g ∈ D(L1/2 ), ˆ N ˆ ˆ ˆ dxdt N −tL −tL 1/2 1/2 dxdt e Lf (x)g(x) √ = e L L f (x)g(x) √  Rn t  Rn t ˆ N ˆ dxdt e−tL/2 e−tL/2 L1/4 L1/4 L1/2 f (x)g(x) √ =  Rn t ˆ N ˆ −tL/2 1/4  1/2  ∗ 1/4 −tL∗ /2 dt = e L L f (L ) e √ gdx  Rn t ˆ ∞ ˆ  2 dxdt 1/2 ˆ ∞ ˆ 1/2 −tL∗ /2 ∗ 1/4 2 dxdt  −tL/2 1/4 1/2 ≤ e (tL) L f e (tL ) g 0 Rn t 0 Rn t . L1/2 f 2 n kgkL2 , L (R ) where in the last step we have used the McIntosh-Yagi theorem ([MY90] Theorem 1) to obtain the quadratic estimates. Since L1/2 is m-accretive, the domain of L1/2 is dense in L2 (see eg. [Kat76] p.279). Therefore, taking supremum over all g ∈ D(L1/2 ) with kgkL2 ≤ 1, (8.3.3) follows from the estimate above. 154 The determination of the domain of the square root of L has become known as the Kato square root problem. It has been shown by Auscher, Hofmann, Lacey, McIntosh, and Tchamitchian [AHL+ 02] that for a uniformly complex elliptic operator L = − div(A∇) with bounded measurable coefficients, one has in all dimensions 1/2 L f ≈ k∇f kL2 , (8.3.4) L2 and the domain of L1/2 is W 1,2 , which was known as the Kato’s conjecture. Recently, Escauriaza and Hofmann ([EH18]) extended the result to the same kind of operators that we are interested in, that is, operators with a BMO anti-symmetric part. Note that although they only showed one side of (8.3.4), that is 1/2 L f . k∇f kL2 , (8.3.5) L2 the other direction follows from a duality argument. In fact, the same argument applies to L∗ so one has ∗ 1/2 (L ) f . k∇f kL2 . (8.3.6) L2 It turns out that (8.3.5) and (8.3.6) are enough: Lemma 8.3.2. If (8.3.5) holds for all f ∈ D(L), and (8.3.6) holds for all f ∈ D(L∗ ). Then k∇f kL2 . L1/2 f , ∀ f ∈ D(L). (8.3.7) L2 And the domain of L1/2 is W 1,2 (Rn ). Proof. We first show that W 1,2 ⊂ D(L1/2 ). Since D(L) is dense in W 1,2 , for any u ∈ W 1,2 , there are {uk } ⊂ D(L) such that uk → u in W 1,2 . Then by (8.3.5), L1/2 (uk − uj ) L2 . k∇(uk − uj )kL2 . This shows that L1/2 uk is Cauchy in L2 . Suppose L1/2 uk → v ∈ L2 .  Since L1/2 is closed, we have L1/2 u = v, and u ∈ D(L1/2 ). 155 Now we show (8.3.7) holds. Let f ∈ S (Rn ). Let g ∈ S (Rn ) with kgkL2 ≤ 1. For any . δ > 0, define hδ = (L∗ + δI)−1 (− div g) ∈ D(L∗ ). That is, ˆ ˆ ˆ δ hδ w dx + A∗ ∇hδ · ∇w dx = − div g w dx ∀ w ∈ W 1,2 (Rn ). (8.3.8) Rn Rn Rn Letting w = hδ and taking real parts of (8.3.8), then ellipticity and Young’s inequality give ˆ ˆ ˆ 2 λ0 2 δ |hδ | dx + |∇hδ | dx ≤ C |g|2 dx ≤ C. (8.3.9) Rn 2 Rn Rn We write (∇f, g) = − (f, div g) = (f, (L∗ + δI)hδ ) = (f, (L∗ )1/2 (L∗ )1/2 hδ ) + δ(f, hδ ) = (L1/2 f, (L∗ )1/2 hδ ) + δ(f, hδ ). 1/2 ∗ 1/2 ⇒ |(∇f, g)| ≤ L f 2 (L ) hδ 2 + δ kf kL2 khδ kL2 L L . L1/2 f 2 k∇hδ kL2 + δ kf kL2 khδ kL2 L . L1/2 f 2 + δ 1/2 kf kL2 , L where we have used (8.3.6) in the second inequality, and (8.3.9) in the last inequality. Therefore, k∇f kL2 = sup |(∇f, g)| . L1/2 f + δ 1/2 kf kL2 . g∈S (Rn ) L2 kgkL2 ≤1 Letting δ → 0, we obtain (8.3.7) holds for all f ∈ S (Rn ). Since S (Rn ) is dense in W 1,2 (Rn ), (8.3.7) holds for all f ∈ W 1,2 (Rn ), which contains the domain of L. Finally, we show D(L1/2 ) ⊂ W 1,2 , and thus proves D(L1/2 ) = W 1,2 . To this end, let u ∈ D(L1/2 ). Since D(L) is a core of L1/2 , there exist {un } ⊂ D(L) such that un → u in 156 L2 , and L1/2 (un ) → L1/2 u in L2 . Since k∇(un − um )kL2 . L1/2 (un − um ) , L2 {un } is a Cauchy sequence in W 1,2 . This implies u ∈ W 1,2 . Remark 8.3.1. Based on the proof, one can see that (8.3.7) actually holds for all f ∈ D(L1/2 ). From the Kato’s estimate (8.3.5) one can see that L1/2 can be extended to the homo- ˙ 1,2 . In particular, L1/2 extends to an isomorphism from W geneous Sobolev space W ˙ 1,2 to L2 and g = L1/2 L−1/2 g ∀ g ∈ L2 . (8.3.10) In fact, by (8.3.7), L1/2 is one-to-one. So it suffices to justify that the range of L1/2 is the whole L2 . To this end, we first show that the range of L is dense in L2 . Lemma 8.3.3. The range of L is dense in L2 . Proof. Let g ∈ L2 . For any δ > 0, let gδ ∈ S (Rn ) such that kgδ − gkL2 < δ. Define . f(δ) = (L + I)−1 gδ ∈ D(L). (δ) We claim that Lf − g < Cδ when  is sufficiently small. And this would complete L2 the proof of the lemma. We write Lf(δ) − g = L(L + I)−1 gδ − g = gδ − (L + I)−1 gδ − g 157 ≤ kgδ − gkL2 +  (L + I)−1 gδ < δ +  (L + I)−1 gδ . (8.3.11) (δ) ⇒ Lf − g L2 We have ˆ ∞ −1 (L + I) gδ = e−t(L+I) (gδ )dt 0 ˆ ∞ (L + I)−1 gδ 2 ≤ e−t e−tL gδ L2 dt. ⇒ L 0 Fix any 1 < p < 2, then by the Lp − L2 bound of the semigroup, we obtain ˆ ∞ γp (L + I)−1 gδ 2 . e−t t− 2 kgδ kLp dt L 0 ˆ ∞ γp γp γp −1 . 2 e−τ τ − 2 dτ kgδ kLp .  2 −1 . 0 (δ) Therefore, by choosing  sufficiently small, this and (8.3.11) imply that Lf − g < 2δ. L2 Since δ > 0 is arbitrary, it proves that the range of L is dense in L2 . Remark 8.3.2. By a similar argument and interpolation, one can show that the range of L is dense in Lp , for any 1 < p < ∞. ˙ 1,2 is L2 . Corollary 8.3.1. The range of L1/2 acting on W Proof. Since L1/2 L1/2 = L in D(L), L1/2 maps D(L) into D(L1/2 ) = W 1,2 , and L1/2 (W 1,2 ) contains the range of L. So the range of L1/2 acting on W 1,2 is dense in L2 . Extending L1/2 to W˙ 1,2 , we claim that L1/2 has closed range in L2 . To see this, suppose L1/2 fn  is a Cauchy sequence in L2 with limn→∞ L1/2 fn = y ∈ L2 . By (8.3.7), k∇(fn − fm )kL2 . ˙ 1,2 . So fn → f ∈ W ˙ 1,2 . Then 1/2 L (fn − fm ) 2 , which implies that {fn } is Cauchy in W L 158 we have y − L1/2 f ≤ y − L1/2 fn + L1/2 (fn − fm ) L2 L2 <  + k∇(fn − fm )kL2 < 2 for n sufficiently large. This implies that y = L1/2 f , and thus the range of L1/2 is closed. ˙ 1,2 (Rn ) is the following representation formula A consequence of D(L1/2 ) = W ˙ 1,2 then Lemma 8.3.4. If f , h ∈ W   ˆ ∗ 1/2 1/2 (L ) f, L h = A∇f · ∇h. Rn ˙ 1,2 , L1/2 h and (L∗ )1/2 f belong to L2 . So both sides of the equality are Proof. For f , h ∈ W well-defined (we use Proposition 2.2.1 for the right-hand side). Since the domain of L is ˙ 1,2 , it suffices to show the equality holds for h ∈ D(L) dense in W 1,2 and thus dense in W ˙ 1,2 . By (8.3.1), and f ∈ W     (L∗ )1/2 f, L1/2 h = f, L1/2 L1/2 h = (f, Lh) , ´ which equals to Rn ∇f · A∇h by construction of L. Another consequence of (8.3.7) and (8.3.10) is the L2 boundedness of ∇L−1/2 , the Riesz ˙ 1,2 to L2 , letting transform associated to L. In fact, since L1/2 is an isomorphism from W . ˙ 1,2 in (8.3.7) one obtains f = L−1/2 g ∈ W ∇L−1/2 g . kgkL2 ∀ g ∈ L2 . (8.3.12) L2 159 Note that by the formula (8.3.2), we immediately get the following formula for L−1/2 ˆ ∞ −1/2 −1/2 dt L g=π e−tL g √ , ∀ g ∈ R(L). 0 t Fix any 1 < p < 2, this is also true for all g ∈ L2 ∩ Lp : Lemma 8.3.5. ˆ ∞ −1/2 −1/2 dt L g=π e−tL g √ (8.3.13) 0 t ∗ is valid and converges in Lp + L2 for all g ∈ L2 ∩ Lp , where 2∗ = 2n n−2 . Proof. Let g ∈ L2 ∩ Lp , and write ˆ ∞ ˆ 1 ˆ ∞ −tL dt −tL dt dt . e g√ = e g√ + e−tL g √ = I + II. 0 t 0 t 1 t Then by the Lp boundedness of the semigroup, I converges in Lp norm. For II, note that we have −tL 1+γp e f 2∗ . t− 2 kf k p , (8.3.14) L L √ which is a consequence of the Lp − L2 bound of ( t∇e−tL )t>0 and Sobolev embedding ∗ (and if n = 2, one should substitute the BMO norm). So II converges in L2 norm (or in BMO norm if n = 2). It remains to show that (8.3.13) holds for any g ∈ L2 ∩ Lp . By Lemma 8.3.3 and the remark after it, R(L) contains a dense subset of L2 ∩ Lp . Therefore, there exists {gn } ⊂ R(L) ∩ Lp such that gn → g ∈ L2 ∩ Lp . Then ˆ ∞ −1/2 −1/2 dt L gn = π e−tL gn √ , 0 t and thus from the convergence argument above, one can see that L−1/2 gn is a Cauchy  ∗ ∗ sequence in Lp + L2 . Suppose L−1/2 gn → f ∈ Lp + L2 . Since L1/2 is an isomorphism 160 ˙ 1,2 to L2 , L−1/2 g is well-defined. We compute from W −1/2 L g − f ≤ L−1/2 g − L−1/2 gn p 2∗ + L−1/2 gn − f p 2∗ Lp +L2∗ L +L L +L −1/2 ≤ L g − L−1/2 gn 2∗ + L−1/2 gn − f p 2∗ . L L +L And by Sobolev embedding and (8.3.12), −1/2 −1/2 −1/2 L g−L gn . ∇L (gn − g) . kgn − gkL2 . L2∗ L2 Thus, we have proved that L−1/2 g = f , which implies that (8.3.13) holds for all g ∈ L2 ∩Lp , for all 1 < p < 2. Corollary 8.3.2. We have the following representation for the Riesz transform: ˆ ˆ ∞  −1/2  −1/2 dt ∇L f, v = π lim ∇e−tL f √ v dx →0 Rn  t for all f ∈ L2 ∩ Lp , and for all Cn - valued v ∈ C0∞ . Proof. Since we have observed that the improper integral defining L−1/2 f converges in ∗ Lp + L2 (or in Lp + BM O if n = 2), and since div v is compactly supported and belongs to every Lp , (and to the Hardy space H1 ), we can write   ˆ ˆ −1/2 −1/2 π 1/2 ∇L f, v = π 1/2 ∇L f v dx = −π 1/2 L−1/2 f div v dx Rn Rn ˆ ˆ 1 ˆ 1ˆ  dt  dt = − lim e−tL f √ div v dx = lim ∇e−tL f v dx √ →0 Rn  t →0  n t ˆ 1ˆ ˆ ∞ˆ R dt dt = lim ∇e−tL f v dx √ + ∇e−tL f v dx √ (8.3.15) →0  n t 1 n t ˆ ˆ 1 ˆ ˆ ∞ R R dt dt = lim ∇e−tL f √ v dx + ∇e−tL f √ v dx →0 Rn  t Rn 1 t ˆ ˆ ∞ dt = lim ∇e−tL f √ v dx →0 Rn  t √ where in (8.3.15), we have used the Lp −L2 bound of ( t∇e−tL )t>0 , so the second improper integral converges, and we can then exchange the order of integration. 161 ˙ 1,p , with p > 1, it suffices to Lemma 8.3.6. To show that (L∗ )1/2 f Lp . k∇f kLp , f ∈ W show 0 −1/2 ∇L g . kgkLp0 ∀ g ∈ L2 ∩ Lp . (8.3.16) Lp0 Proof. Let f ∈ S (Rn ).     ∗ 1/2 (L ) f = sup (L∗ )1/2 f, g = sup (L∗ )1/2 f, L1/2 L−1/2 g Lp 0 0 g∈L2 ∩Lp g∈L2 ∩Lp kgk p0 ≤1 kgk p0 ≤1 L L   = sup (L∗ )1/2 f, L1/2 h 0 g∈L2 ∩Lp kgk p0 ≤1 L where h = L−1/2 g. Then by Lemma 8.3.4 and Proposition 2.2.1   ˆ ∗ 1/2 1/2 ∇f · A∇h . k∇f kLp ∇L−1/2 g p0 . (L ) f, L h = R n L Therefore, (8.3.16) gives ∗ 1/2 (L ) f . sup k∇f kLp kgkLp0 . k∇f kLp ∀ f ∈ S (Rn ). (8.3.17) Lp 0 g∈L2 ∩Lp kgk 0 ≤1 Lp Since S (Rn ) is dense in W ˙ 1,p , (L∗ )1/2 can be extended to W ˙ 1,p and (8.3.17) holds for all ˙ 1,p . f ∈W Therefore, to prove that (L∗ )1/2 f Lp . k∇f kLp for p > 2, it suffices to show ∇L−1/2 g . kgkLp0 ∀ g ∈ L2 ∩ Lp0 , (8.3.18) Lp0 0 where p0 ∈ [1, p0 ). This is because L2 ∩ Lp0 is dense in L2 ∩ Lp . In order to prove (8.3.18), 162 we show that the Riesz transform is of weak type (p0 , p0 ), and then the strong type (p, p) bound follows from interpolation with the strong type (2,2) bound (8.3.12). Proposition 8.3.1. Let p0 ∈ [1, 2). Then ∇L−1/2 g . kgkLp0 ∀ g ∈ Lp0 ∩ L2 . (8.3.19) L∞,p0 As a consequence, for any 1 < p < 2, ∇L−1/2 g . kgkLp ∀ g ∈ Lp ∩ L2 . (8.3.20) Lp We shall use the following lemma Lemma 8.3.7 ([Aus07] Theorem 2.1). Let p0 ∈ [1, 2). Suppose that T is a sublinear operator of strong type (2, 2), and let Ar , r > 0, be a family of linear operators acting on L2 . For a ball B, let C1 (B) = 4B, Cj (B) = 2j+1 B \ 2j B if j ≥ 2. Assume for j ≥ 2 ˆ !1/2  ˆ 1/p0 1 T (I − Ar(B) )f 2 1 p0 ≤ g(j) |f | (8.3.21) |2j+1 B| Cj (B) |B| B and for j ≥ 1 ˆ !1/2  ˆ 1/p0 1 Ar(B) f 2 1 |f |p0 ≤ g(j) (8.3.22) |2j+1 B| Cj (B) |B| B for all ball B with radius r(B) and all f supported in B. If Σ = Σj g(j)2nj < ∞, then T is of weak type (p0 , p0 ), with a bound depending only on the strong type (2, 2) bound of T , p0 and Σ. Proof of Proposition 8.3.1. Let T = ∇L−1/2 . Then T is of strong type (2, 2) ((8.3.12)). 163 Let m 2 2 X Ar = I − (I − e−r L )m = ck e−kr L , k=1 where m ≥ 1 to be determined. Let B be any ball, with radius r. Let f ∈ L2 ∩ Lp0 with supp f ⊂ B. Then by Lemma 8.3.7, it suffices to show (8.3.21) and (8.3.22). Proof of (8.3.22). By the Lp0 − L2 off-diagonal estimates for the semigroup, we have γp0 c4j c4j −kr2 L e f . (kr2 )− 2 e− k kf kLp0 ≤ C(m)r−γp0 e− m kf kLp0 . L2 (Cj (B)) ˆ !1/2  ˆ 1/p0 1 2 − c4 j − jn 1 p0 ⇒ |Ar (f )| ≤ C(m)e m 2 2 |f | , |2j+1 B| Cj (B) |B| B c4j jn which proves (8.3.22) with g(j) = C(m)e− m 2− 2 . 2 Proof of (8.3.21). Since T (I − e−r L )m (f ) ∈ L2 for any f ∈ Lp0 ∩ L2 with supp f ⊂ B, 2  2  T (I − e−r L )m (f ) = sup T (I − e−r L )m (f ), v . (8.3.23) L2 (Cj (B)) v∈C0∞ (Cj (B)) kvkL2 ≤1 2 Let h = (I − e−r L )m (f ). Then by Corollary 8.3.2, ˆ (T h, v) = lim π −1/2 F (h) v dx, →0 Rn . ´∞ where F (h) =  ∇e−tL h √ dt t . Therefore, |(T h, v)| ≤ π −1/2 lim kF (h)kL2 (Cj (B)) kvkL2 (Cj (B)) (8.3.24) →0 164 We now estimate lim→0 kF (h)kL2 (Cj (B)) . We write m ˆ ∞ m ˆ ∞ X 2 )L dt X dt F (h) = ak ∇e−(t+kr (f ) √ = ak ∇e−tL (f ) √ k=0  t k=0 +kr2 t − kr2 ˆ ˆ Xm ∞ 1{t>+kr2 } −tL ∞ = ak √ ∇e (f )dt = gr() (t)∇e−tL (f )dt, k=0 0 t − kr2 0 () . P 1{t>+kr2 } where gr (t) = m , ak = (−1)k m  k=0 ak . √ t−kr2 k √ By the Lp0 − L2 off-diagonal estimates for ( t∇e−tL )t>0 , ˆ ∞ () − γp20 − 12 − c4jtr2 kF (h)kL2 (Cj (B)) . gr (t) t e dt kf kLp0 (B) . 0 ˆ ∞ γp0 c4j r 2 − 12 ⇒ lim kF (h)kL2 (Cj (B)) . |gr (t)| t− 2 e− t dt kf kLp0 (B) , →0 0 . P 1{t>kr2 } where gr (t) = mk=0 ak t−kr2 . √ Now we can adopt the method using in [Aus07]. Namely, we use the following estimates for gr : C |gr (t)| ≤ √ if kr2 < t ≤ (k + 1)r2 ≤ (m + 1)r2 t − kr2 and 1 |gr (t)| ≤ Cr2m t−m− 2 if t > (m + 1)r2 . For the proof see [Aus07] p.44. This yields the estimate ˆ ∞ γp0 c4j r 2 − 12 |gr (t)| t− 2 e− t dt ≤ Cr−γp0 2−j(2m+γp0 ) . 0 Then by (8.3.23) and (8.3.24), direct computation gives ˆ !1/2  ˆ 1/p0 1 T (I − Ar(B) )f 2 −(2m+ pn )j 1 p0 .2 0 |f | . |2j+1 B| Cj (B) |B| B 165   1 Therefore, choosing 2m > 1 − p0 n, one obtains (8.3.21) with Σg(j)2nj < ∞. Now it follows from Lemma 8.3.6, that (L∗ )1/2 is Lp bounded, for p > 2. And since the same argument applies to L, the same result holds for L1/2 as well. We summarize the results in the following proposition. ˙ 1,p . Then Proposition 8.3.2. Let p ≥ 2, let f ∈ W ∗ 1/2 1/2 (L ) f . k∇f kLp , L f . k∇f kLp . Lp Lp CHAPTER Nine Estimates for Square functions and non-tangential maximal functions involving the heat semigroup 9.1 Lp estimates for square functions Proposition 9.1.1. For any f ∈ L2 , ˆ ˆ ∞ 2 1/2 −tL 2 (L e f )(x) dtdx . kf kL2 (Rn ) , (9.1.1) Rn 0 where the implicit constant depends on λ0 , Λ0 and n. By a change of variable, we have ˆ ˆ ∞ 2 dtdx 1/2 −t2 L tL e f (x) . kf k2L2 (Rn ) . (9.1.2) Rn 0 t Proof. We first prove ˆ ∞ˆ (∇e−tL f )(x) 2 dtdx . kf k2 2 L (Rn ) . (9.1.3) 0 Rn 166 167 Actually, the converse of (9.1.3) is also true. We have ˆ ∞ d e−tL f 2 2 dt. kf k2L2 =− L (9.1.4) 0 dt We postpone its proof to the end. Using Lemma 3.3.1, one has ˆ A∇e−tL f · ∇e−tL f = < Le−tL f, e−tL f = − 0, choose ϕε ∈ C0∞ (Rn ) such that and e−tL (f − ϕε ) L2 < ε, kϕε kL2 ≤ kf kL2 + 1 ∀ t > 0. And suppose supp ϕε ⊂ Qε . Let p ∈ (1, 2). Then the Lp boundedness of (e−tL )t>0 gives −tL 2−p e ϕε p ≤ C kϕε k p ≤ C |Qε | 2p kϕε k 2 . L L L 0 And the L2 − Lp boundedness of (e−tL )t>0 gives −tL γp0 e ϕε p0 ≤ Ct− 2 kϕε k 2 . L L 168 Therefore, we have 2−p γp0 e ϕε 2 ≤ e−tL ϕε 1/2 −tL −tL 1/2 e ϕε p0 ≤ C |Qε | 4p t− 4 (kf k 2 + 1) . L Lp L L Then there is a t0 = t0 (ε) > 0, such that for any t > t0 , e−tL ϕε L2 < ε, and thus −tL e f 2 < 2ε. This proves (9.1.5). L Proposition 9.1.2.  ˆ ∞ 2 1/2 −t2 L dt tLe F p n ≤ Cp k∇F kLp (Rn ) (9.1.6) 0 t L (R ) for all p ≥ 2, and F ∈ W 1,2 ∩ W 1,p . Equivalently,  ˆ ∞  −t2 L 2 dt 1/2 ∂t e F p n ≤ Cp k∇F kLp (Rn ) . (9.1.7) 0 t L (R ) Proof. We can assume that F ∈ S (Rn ). When p = 2, applying (9.1.2) to L1/2 F , then 2 (9.1.6) follows from commutativity of e−t L and L1/2 , L1/2 L1/2 = L and L1/2 F L2 . ´ k∇F kL2 . For fixed (y, t) ∈ Rn × R+ , Rn 1Γα (x) (y, t)dx ≈ tn . So by Fubini’s theorem, we have ˆ ˆ 2 dydt ˆ 2 dtdx −t2 L 2 tLe F (y) n+1 dx ≈ tLe−t L F (x) . k∇F k2L2 (Rn ) . R n |x−y|<αt t n+1 R+ t (9.1.8) 2 2 2 2 We now show tLe−t L F (x) dxdt t is a Carleson measure. Since ∂t e −t L = −2tLe−t L , it is 2 2 equivalent to saying ∂t e−t L F (x) dxdtt is a Carleson measure. We claim that ˆ ˆ l(Q) 1 −t2 L 2 dtdx sup ∂t e F (x) ≤ C k∇F k2L∞ . (9.1.9) Q⊂Rn |Q| Q 0 t 2L Suppose the kernel of ∂t e−t is Vt (x, y), then by (7.2.52), C |x−y|2 |Vt (x, y)| ≤ n+1 e− Ct2 . (9.1.10) t 169 Fix a cube Q ⊂ Rn . Write . F = F + (F − F )14Q + (F − F )1Rn \4Q = F1 + F2 + F3 . (9.1.11) 4Q 4Q 4Q It suffices to show ˆ ˆ l(Q) 2 dtdx −t2 L ∂t e Fi (x) ≤ C k∇F k2L∞ |Q| , for i = 1, 2, 3. (9.1.12) Q 0 t 2 Since F1 is a constant, ∂t e−t L F1 ≡ 0 in the sense of L2loc . Using (9.1.7) for p = 2, we have ˆ 2 dxdt ˆ −t2 L ∂t e F2 (x) . |∇F |2 dx . k∇F k2L∞ |Q| . (9.1.13) n+1 R+ t 4Q ´ ´ l(Q) −t2 L 2 dtdx To estimate ∂ e F (x) t , we use the Gaussian decay (9.1.10) of the kernel Q 0 t 3 Vt (x, y) as well as Poincar´e inequality. To be precise, for x ∈ Q, we have ˆ 2 −t L ∂t e F3 (x) = Vt (x, y)(F (y) − (F )4Q )dy Rn \4Q ∞ ˆ ( ) X 1 |x − y|2 . n+1 exp − |F (y) − (F )4Q | dy 2 k+1 Q\2k Q t Ct2 k=2 ∞ ˆ  2k−2 l(Q)2  1 X 2 . n+1 exp − |F (y) − (F )4Q | dy (9.1.14) t 2k+1 Q Ct2 k=2 Since |F − (F )Q | ≤ |F − (F )2Q | + |(F )Q − (F )2Q | ≤ |F − (F )2Q | + 2n |F − (F )2Q | , 2Q Poincar´e inequality gives F − (F )2k+1 Q ≤ Cn 2k+1 l(Q) F − (F )2k Q ≤ Cn |∇F | . 2k+1 Q 2k+1 Q 2k+1 Q 170 ⇒ ˆ |F − (F )4Q | 2k+1 Q ˆ   ≤ F − (F )2k Q + (F )2k Q − (F )2k−1 Q + · · · + |(F )8Q − (F )4Q | 2k+1 Q   k+1 k+1 k ≤ Cn 2 Q 2 l(Q) |∇F | + 2 l(Q) |∇F | + · · · + 8l(Q) |∇F | 2k+1 Q 2k Q 8Q ≤ Cn k∇F kL∞ (2k+2 l(Q))n+1 . (9.1.15) n 2k o l(Q)2 2 n+2 Since exp − 2 Ct2 ≤ ( 22kCt l(Q)2 ) 2 , (9.1.14) and (9.1.15) imply −t2 L t ∂t e F3 (x) ≤ Cn k∇F kL∞ ∀ x ∈ Q. (9.1.16) l(Q) Thus ˆ ˆ l(Q) 2 dtdx −t2 L ∂t e F3 (x) ≤ C k∇F k2L∞ |Q| , Q 0 t which proves the claim. Recall that in [CMS85], the tent space T p is defined as ˆ dydt 1/2 p T = {f : A(f ) ∈ L (R )} , p n 1 ≤ p < ∞, where A(f )(x) = |f (y, t)|2 , Γ(x) tn+1 and  ˆ 1/2 ∞ ∞ n 1 dydt2 T = {f : C(f ) ∈ L (R )} , where C(f )(x) = sup |f (y, t)| , x∈B |B| ˆ B t ˆ = {(y, t) : dist(y, B c ) ≥ t}. These spaces are equipped with norm kf k p = kA(f )k p and B T L and kf kT ∞ = kC(f )kL∞ , respectively. Also, one has the tent space interpolation result: 1 1−θ θ [T p0 , T p1 ]θ = T p , for 1 ≤ p0 < p < p1 ≤ ∞, with = + . (9.1.17) p p0 p1 2 By definition, (9.1.8) implies that ∂t e−t L F . k∇F kL2 , and (9.1.9) implies that T2 171 −t2 L ∂t e F . k∇F kL∞ . Thus by tent space interpolation, we obtain T∞ ˆ ˆ 2 dydt p/2 −t2 L ∂t e F (y) n+1 dx ≤ Cp k∇F kpLp (Rn ) , ∀ p ≥ 2. (9.1.18) Rn Γα (x) t Finally, we show (9.1.18) implies (9.1.7). It suffices to prove ˆ ∞ 2 dt ˆ 2 dydt −t2 L −t2 L ∂t e F (x) . ∂t e F (y) n+1 ∀ x ∈ Rn . (9.1.19) 0 t Γα (x) t We have ˆ 2 dydt ˆ ∞ ˆ −t2 L −t2 L 2 dt ∂t e F (y) n+1 = ∂t e F (y) dy n+1 Γα (x) t 0 |y−x|<αt t ˆ ∞ 2 dt −t2 L ≈ ∂t e F (y) dy . (9.1.20) 0 Bαt (x) t On the other hand, ˆ ∞ 2 dt X ˆ 2k+1 2 dt −t2 L −t2 L ∂t e F (x) = ∂t e F (x) 0 t 2k t k∈Z 2k+1 2 2 X −t2 L X −t2 L ≤ ∂ e F (x) dt ≤ sup ∂ e F (y) t t k 2k k k+1 ] k B2k−2 α (x)×[2 ,2 Xˆ 2k+1 −t2 L 2 dydt . ∂t e F (y) 2k−1 B2k−1 α (x) t k Xˆ 2k+1 −t2 L 2 dydt ˆ ∞ −t2 L 2 dydt ≤ ∂t e F (y) . ∂t e F (y) . (9.1.21) 2k−1 Bαt (x) t 0 Bαt (x) t k We have used Proposition 3.3.2 in the third inequality. So we proved (9.1.19). Lemma 9.1.1 ([Aus07] Theorem 2.2). Let p0 ∈ (2, ∞]. Suppose that T is sublinear operator acting on L2 , and let Ar , r > 0, be a family of linear operators acting on L2 . Assume  ˆ 1/2 1 T (I − Ar(B) )f 2  1/2 ≤ C M (|f |2 ) (y), (9.1.22) |B| B 172 and  ˆ 1/p0 1 T Ar(B) f p0  1/2 ≤ C M (|T f |2 ) (y), (9.1.23) |B| B for all f ∈ L2 , all ball B with radius r(B) and all y ∈ B. If 2 < p < p0 and T f ∈ Lp when f ∈ Lp ∩ L2 , then T is strong type (p, p). More precisely, for all f ∈ Lp ∩ L2 , kT f kLp ≤ c kf kLp where c depends only on n, p, p0 and C. Remark 9.1.1. Note that the lemma requires T f ∈ Lp when f ∈ Lp ∩ L2 and the purpose of the statement is to bound the Lp norm of T f . In practice, we would apply this lemma to suitable approximation of T and obtain uniform Lp bounds. The uniformity of the bounds allows a limiting argument to deduce Lp boundedness of T . We now use this lemma to obtain the following Proposition 9.1.3.  ˆ ∞ 2 1/2 2 −t2 L dt t ∇Le F p n ≤ Cp k∇F kLp (Rn ) (9.1.24) 0 t L (R ) for 2 ≤ p ≤ 2+0 , with 0 = 0 (λ0 , Λ0 , n) > 0, and for all F ∈ W 1,2 ∩W 1,p . Or equivalently,  ˆ ∞ 2 1/2 −t2 L dt t∇∂ e F p n ≤ Cp k∇F kLp (Rn ) t t 0 L (R ) for 2 ≤ p ≤ 2 + 0 , F ∈ W 1,2 ∩ W 1,p . Proof. We shall establish the following  ˆ ∞ 2 1/2 2 1/2 −t2 L dt t ∇L e f p n ≤ Cp kf kLp (Rn ) (9.1.25) 0 t L (R ) for 2 ≤ p ≤ 2 + 0 and f ∈ L2 ∩ Lp . Once this is proved, setting f = L1/2 F ∈ L2 ∩ Lp and 173 then using 1/2 L F . k∇F kLp Lp one obtains (9.1.24). 2L 2 L/2 2 L/2 We first show (9.1.25) holds for p = 2. We write e−t = e−t e−t , and use the 2 L/2 fact that t∇e−t is bounded on L2 (Rn ), uniformly in t (by (7.3.4)), to obtain ˆ ˆ ∞ ˆ ∞ ˆ 1/2 −t2 L dt 2 2 2 −t2 L/2 1/2 −t2 L/2 t ∇L e f dx = t t∇e (L e f ) dxdt t Rn 0 Rn ˆ ∞ ˆ 0 ˆ ˆ ∞ 1/2 −t2 L/2 2 2 2 ≤C t L e f dxdt = C t L1/2 e−t L/2 f dtdx. (9.1.26) 0 Rn Rn 0 By Proposition 9.1.1, ˆ ˆ ∞ 1/2 −t2 L/2 2 t L e f dtdx . kf k2L2 (Rn ) , Rn 0 which finishes the proof of L2 boundedness. We now prove (9.1.25) for 2 < p ≤ 2 + 0 for some 0 using Lemma 9.1.1. Define ˆ ∞ 2 dt 1/2 . 2 t ∇L1/2 e−t L f 2 G(f ) = , 0 t and define Gε to be the approximation of G ˆ ∞ 2 1/2 . 2 1/2 −t2 L dt Gε (f ) = t ∇L e f . ε t We shall first show Gε (f ) ∈ Lp for f ∈ Lp ∩ L2 , and then derive the uniform estimates for Gε (f ). Namely,  ˆ 2 1/2 1 −r2 L m Gε (I − e ) f . (M (|f |2 ))1/2 (y) ∀y ∈ B (9.1.27) |B| B and  ˆ p 1/p0 1 −r2 L 0 Gε (e f ) . (M (|Gε f |2 ))1/2 (y) ∀y ∈ B (9.1.28) |B| B 174 uniformly in ε, for any ball B with radius r, and for some integer m large enough. We shall prove (9.1.28) with p0 = 2 + 1 , where 1 is as in Proposition 8.2.4. Then by Lemma 9.1.1, kGε (f )kLp . kf kLp for all 2 < p < p0 , uniformly in ε. Letting ε → 0, one obtains kGf kLp . kf kLp for all 2 < p < p0 . This proves the Proposition, and 0 can be any positive number smaller than 1 . Proof of Gε (f ) ∈ Lp for f ∈ Lp ∩ L2 . We rewrite Gε (f ) to be ˆ ∞ 2 1/2 1/2 −tL dt Gε (f ) = t∇L e f . ε2 t By Minkowski’s inequality, nˆ ˆ !2/p ∞  2 p/2 o1/2 1/2 −tL kGε (f )kLp ≤ t ∇L e f dx dt ε2 Rn nˆ ∞ ˆ √ p 2/p o1/2 1/2 −tL = t∇L e f dx dt ε2 Rn nˆ ∞  ˆ √ p 2/p o1/2 −tL/2 1/2 −tL/2 = t∇e L e f dx dt . ε2 Rn √ We first use the L2 − Lp bounds for ( t∇e−tL )t>0 , then (8.3.5), and finally the L2 bounds √ for ( t∇e−tL )t>0 to obtain ˆ ∞ ˆ 1/2 1/2 −tL/2 2 −γp kGε (f )kLp . t L e f dxdt ε2 Rn ˆ ∞ ˆ 1/2 ˆ ∞ 1/2 −tL/2 2 . t−γp ∇e f dxdt . t−γp −1 dt kf kL2 ≤ Cε,p kf kL2 . ε2 Rn ε2 Proof of (9.1.28) Since the domain of e−tL is L2 , the operators commute for f ∈ L2 : 2 2 2 2 ∇L1/2 e−t L (e−r L f ) = ∇e−r L L1/2 e−t L f. (9.1.29) 175 By Minkowski inequality, n 1 ˆ ˆ ∞ 2  p0 o 2 2 1/2 −t2 L −r2 L dt 2 p t ∇L e (e f ) dx 0 |B| B ε t ˆ ∞ ˆ  2 dt 1 2 1/2 −t2 L −r2 L p0 p0 ≤ t ∇L e (e f ) dx |B| t ε B ˆ ∞ ˆ 2 1 2 −r2 L 1/2 −t2 L p0  p dt = t ∇e (L e f ) dx 0 (9.1.30) ε |B| B t √ Using the L2 − Lp0 off-diagonal estimates for ( r∇e−rL )r>0 , as well as Poincar´e inequality, one can show  1 ˆ  1 ˆ −r2 L p0 1/p0 X  1/2 2 ∇e f ≤ g(j) |∇f | |B| B |2j+1 B| 2j+1 B j≥1 P with j≥1 g(j) < ∞. By this and H¨older inequality we can bound (9.1.30) by ˆ ∞nX ˆ  1 2 1/2 −t2 L 2 1/2 o2 dt g(j) t ∇L e f dx |2j+1 B| 2j+1 B t ε j≥1 ˆ ∞X ˆ 1 2 2 dt t ∇L1/2 e−t L f dx 2 ≤C g(j) j+1 ε |2 B| 2j+1 B t j≥1 ˆ ˆ ∞ 1 2 2 dt t ∇L1/2 e−t L f 2 ≤ C sup j+1 dx j≥1 |2 B| 2j+1 B ε t ≤ C(M |Gε f |2 )1/2 (y) ∀ y ∈ B. (9.1.31) Proof of (9.1.27) P Now write f = j≥1 fj , where  fj = (f − (f )4B )12j+1 B\2j B   j ≥ 2, f1 = (f − (f )4B )14B .   Then  ˆ 2 1/2 X  1 ˆ 2 1/2 1 −r2 L m −r2 L m G (I − e ) f ≤ G (I − e ) f . ε ε j |B| |B| B B j≥1 176 2 For f1 , the L2 bound of G and that of (I − e−r L )m (the latter is a consequence of the holomorphic functional calculus on L2 ) imply  ˆ 2 1/2 ˆ 1 −r2 L m 1 Gε (I − e ) f1 ) ≤ C( |f1 |2 )1/2 |B| B |B| Rn  ˆ 1/2 1 2 ≤C |f | ≤ C(M |f |2 )1/2 (y) ∀ y ∈ B. (9.1.32) |B| 4B 2 2 For fj with j ≥ 2, let ϕ(z) = tz 1/2 e−t z (1 − e−r z )m . Then (see e.g. [Aus07] section 3.2) ˆ ˆ 2 2 tL1/2 e−t L (1 − e−r L )m = ϕ(L) = e−zL η+ (z)dz + e−zL η− (z)dz, (9.1.33) Γ+ Γ− π where Γ± is the half-ray R+ e±i( 2 −θ) , ˆ 1 η± (z) = eζz ϕ(ζ)dζ, z ∈ Γ± , (9.1.34) 2πi γ± with γ± being the half-ray R+ e±iν , and 0 < ω0 < θ < ν < π2 , where ω0 is as in Proposition (8.1.1). One can show Ct r2m |η± (z)| ≤ inf(1, ), z ∈ Γ± , (9.1.35) (|z| + t2 )3/2 (|z| + t2 )m whose proof is postponed to the end. Then, ˆ 1 −r2 L m 2 Gε (I − e ) fj dx |B| B ˆ ∞ ˆ ˆ ˆ 2 t −zL −zL = ∇e η+ (z)dzfj + ∇e η − (z)dzfj dxdt ε |B| B Γ+ Γ− ˆ ∞ ˆ ˆ 2 t ∇e−zL η+ (z)dzfj dxdt ≤C 0 |B| B Γ+ ˆ ∞ ˆ ˆ 2 t −zL + ∇e η− (z)dzfj dxdt 0 |B| B Γ− . = I+ + I− . (9.1.36) 177 By Minkowski inequality and (9.1.35), ˆ ∞ ˆ ˆ t n −zL 2 1/2 o2 I+ ≤ ∇e η+ (z)fj dx |dz| dt 0 |B| Γ+ B ˆ ∞ ˆ ˆ √ 2m 2 t n  t r 1/2 o2 . z∇e−zL 1/2 )f dx |dz| dt. 2 m j |B| Γ+ |z| (|z| + t2 )3/2 (|z| + t ) 0 B π Since z ∈ Γ+ = R+ ei( 2 −θ) and θ < ω0 , we can apply the L2 − L2 off-diagonal estimates for √ z∇e−zL z∈Σ π , and bound the expression above by  2 −θ ˆ ∞ ˆ t  j 2 − c4|z|r t r2m 2 e 2 m ) |dz| kf k j L2 dt (9.1.37) 0 |B| Γ+ |z|1/2 (|z| + t2 )3/2 (|z| + t ) We use the following lemma to estimate (9.1.37) Lemma 9.1.2 ([Aus07] Lemma 5.5). Let γ, α ≥ 0, m > 0 be fixed parameters, and c a positive constant. For some C independent of j ∈ N, r, t > 0, the integral ˆ ∞ c4j r 2 1 tα r2m I= e− s ds 0 sγ/2 (s + t)1+α (s + t)m satisfies the estimate C  t j 2 α 4 r m  I≤ inf ( ) , ( ) . (9.1.38) 4jm (2j r)γ 4j r 2 t Letting γ = 1, α = 12 , and t replaced by t2 in the lemma, we obtain ˆ ∞  t2 ˆ t 4j r2 2m  1 (9.1.37) . jm (2j r))2 inf ( j 2 ), ( 2 ) dt |f |2 dx 0 (4 4 r t |B| 2j+1 B ˆ ∞  t2 j r2 ˆ t 4  1 jn −2jm .2 4 inf j 2 , ( 2 ) 2m dt j+1 |f |2 dx 0 (2j r)2 4 r t |2 B| 2j+1 B . 2j(n−4m) M |f |2 (y) ∀ y ∈ B. I− can be estimated similarly. Choose 4m > n, we get ˆ 1 2 2 2 Gε (I − e−r L )m f . M |f | (y) ∀ y ∈ B, |B| B 178 which proves (9.1.27). Proof of (9.1.35) We only show the estimate for η+ . The proof for η− is similar. i( π2 −θ) −ρ|z| −t2 ζ 2 = e−t ρ cos ν . ζz Write ζ = ρeiν , and z = |z| e . Then e =e sin (ν−θ) , e 2 2 Since 0 < θ < ν < π2 , eζz e−t ζ ≤ e−cρ(|z|+t ) for some 0 < c < 1. So ˆ ∞ 2 |η+ (z)| . t ρ1/2 e−cρ(|z|+t ) H(ρ)m dρ, 0 where H(ρ) = 1 − exp(−r2 ρeiν ) . Observe that H(0) = 0, H(ρ) ≤ 2, and that H is a Lipschitz function with [H]C 0,1 ≤ r2 . So we have H(ρ) ≤ C inf(1, r2 ρ). Using this estimate of H, we can bound |η+ (z)| by ˆ ˆ ∞ 1 2 Ct ∞ 1 CtΓ( 32 ) Ct ρ 2 e−cρ(|z|+t ) dρ = s 2 e−s ds = , (9.1.39) 0 (|z| + t2 )3/2 0 (|z| + t2 )3/2 and by ˆ ∞ 1 2 CtΓ(m + 32 )r2m Ctr 2m ρ 2 +m e−cρ(|z|+t ) dρ = . (9.1.40) 0 (|z| + t2 )m+3/2 Combining the two bounds we obtain Ct r2m |η+ (z)| ≤ inf(1, ). (|z| + t2 )3/2 (|z| + t2 )m Remark 9.1.2. As seen in the proof, the 0 we chose here might be different from the 0 in Proposition 10.2.1. For convenience, we shall take the smaller one to be 0 and fix the notation from now on. 179 We also have a similar estimate when the derivative falls on t. But in this case, the Lp estimates hold for any p ≥ 2: Proposition 9.1.4.  ˆ ∞ 2 1/2 2 −t2 L dt t ∂t Le F p n ≤ Cp k∇F kLp (Rn ) (9.1.41) 0 t L (R ) for all 2 ≤ p < ∞, and all F ∈ W 1,2 ∩ W 1,p . Proof. Let ˆ ∞ 1/2 −t2 L dt 2 1/2 2 G(f ) = t t∂ L e f . t 0 Then by the Lp bounds of the square root of L, it suffices to show kGf kLp ≤ Cp kf kLp . (9.1.42) When p = 2, it can be copied almost verbatim from the proof in Proposition 9.1.3. The 2 L/2 only difference is that we would use the uniform L2 boundedness of (t∂t e−t )t>0 , rather 2 L/2 than that of (t∇e−t )t>0 . When p > 2, we shall apply Lemma 9.1.1 again. And as we saw in the proof of Proposition 9.1.3, we should derive the analog of (9.1.27) and (9.1.28) for the approximation operator Gε . We do not bother to go through the limiting process again, but derive the analogous estimates for G instead. Let p0 > 2. We wish to prove  ˆ p 1/p0 1 −r2 L 0 G(e f ) . (M (|Gf |2 ))1/2 (y) ∀ y ∈ B. |B| B To this end, we first claim that 2 2 2 2 t2 ∂t L1/2 e−t L (e−r L f ) = −2t3 e−r L L1/2 Le−t L f, ∀ f ∈ L2 . (9.1.43) 180 Then by this and Minkowski inequality, n 1 ˆ ˆ ∞ 2  p0 o 2 2 1/2 −t2 L −r2 L dt 2 p t ∂t L e (e f ) dx 0 |B| B 0 t ˆ ∞ ˆ  2 dt 1 2 1/2 −t2 L −r2 L p 0 p0 ≤ t t ∂ L e (e f ) dx |B| t 0 B ˆ ∞ ˆ 2 1 3 −r2 L 1/2 −t2 L p0  p dt =4 t e L Le f dx 0 (9.1.44) 0 |B| B t 2 Using the L2 − Lp0 off-diagonal estimates for (e−r L )r>0 , one can show  1 ˆ 2 p0 2/p0 X C ˆ j |f |2 −r L e f ≤ (9.1.45) |B| B |2j+1 B| 2j+1 B j≥1 j with Cj = Ce−c4 . So n 1 ˆ ˆ ∞ 2 2 2 dt  p0 o 2 t ∂t L1/2 e−t L (e−r L f ) 2 2 p dx 0 |B| B t ˆ ∞X 0 ˆ Cj 3 1/2 −t2 L 2 dt ≤4 j+1 B| t L Le f dx 0 j≥1 |2 2j+1 B t ˆ ˆ ∞ 1 2 2 dt t ∂t L1/2 e−t L f 2 . sup j+1 dx j |2 B| 2j+1 B 0 t . M (|G(f )|2 )(y) ∀ y ∈ B. (9.1.46) Now we prove  ˆ 2 1/2 1 −r2 L m G(I − e ) f . (M (|f |2 ))1/2 (y) ∀ y ∈ B. (9.1.47) |B| B Define {fj }j≥1 as in the proof of Proposition 9.1.3. The estimates for f1 again follows 2 from the L2 bound of G and that of (I − e−r L )m . For fj with j ≥ 2, we let ϕ(z) = 2 2 t3 z 3/2 e−t z (1 − e−r z )m . Then 1 2 2 2 2 − t2 ∂t L1/2 e−t L (I − e−r L )m = t3 L1/2 Le−t L (I − e−r L )m 2 ˆ ˆ = ϕ(L) = e−zL η+ (z)dz + e−zL η− (z)dz, Γ+ Γ− where η± and Γ± are defined as in the proof of Proposition 9.1.3. By a similar argument 181 as in the proof of (9.1.35), one can show ˆ ∞ 2 Ct3 r2m |η± (z)| . t3 ρ3/2 e−cρ(|z|+t ) H(ρ)m dρ ≤ inf(1, ). (9.1.48) 0 (|z| + t2 )5/2 (|z| + t2 )m Then, ˆ 1 −r2 L m 2 G(I − e ) fj |B| ˆ B ∞ ˆ 1 3 1/2 −t2 L 2 2 dt =4 t L Le (I − e−r L )m fj dx (9.1.49) 0 |B| B t ˆ ∞ ˆ ˆ ˆ 2 1 −zL −zL dt =4 e η+ (z)dzfj + e η− (z)dzfj dx 0 |B| B Γ+ Γ− t ˆ ∞ ˆ ˆ 2 ˆ ∞ ˆ ˆ 2 1 −zL dt 1 −zL dt . e η+ (z)dzfj dx + e η− (z)dzfj dx 0 |B| B Γ+ t 0 |B| B Γ− t . = I+ + I− . (9.1.50) By Minkowski inequality and (9.1.48), ˆ ∞ ˆ ˆ 2 1 n 3 −zL r2m 1/2 o2 dt I+ . t e fj dx |dz| . 5 0 |B| Γ+ B (|z| + t2 ) 2 +m t We use the L2 off-diagonal estimate for (ezL )z∈Σ π −θ to bound the above expression by 2 ˆ ∞ ˆ 1  − c4|z|r j 2 t3 r2m 2 dt e |dz| kfj kL2 . 0 |B| Γ+ (|z| + t2 )5/2+m t Applying Lemma 9.1.2 and letting α = 32 , γ = 0 and t = t2 there, we obtain ˆ ∞  t2 j 2 1 3 4 r 2m  1 dt I+ . jm )2 ( inf ( j r2 ) , ( 2 ) kf k2L2 (2j+1 B) 0 4 4 t |B| t ˆ j  2 r t5 ˆ ∞ 2jn−4jm (4j r2 )2m  . j+1 kf k2L2 (2j+1 B) dt + dt |2 B| 0 (4j r2 )3 2j r t 4m+1 . 2jn−4jm M (|f |2 )(y) ∀ y ∈ B. 182 I− can be estimated similarly. Choose 4m > n, we get ˆ 1 2 2 2 G(I − e−r L )m f . M |f | (y) ∀ y ∈ B. |B| B Therefore, (9.1.42) holds for all 2 ≤ p < p0 . And since p0 > 2 is arbitrary, (9.1.42) holds for all 2 ≤ p < ∞. Finally, we justify (9.1.43). Proof of (9.1.43). 2 2 We know that for any g ∈ L2 , e−t L g ∈ D(L). And ∂t e−t L g ∈ W 1,2 = D(L1/2 ). The latter follows from analyticity of the semigroup ˆ 2 1 2 ∂j e−t L g = eλt ∂j (λI + L)−1 (g)dλ 2πi Γ ⇒ ˆ −t2 L 2t 2 ∂t ∂j e g= eλt λ∂j (λI + L)−1 (g)dλ ∈ L2 . 2πi Γ We then use the following lemma Lemma 9.1.3 ([Kat76] Chapter V, Theorem 3.35). For any bounded linear operator B on L2 , if BL = LB in D(L), then L1/2 B = BL1/2 in D(L1/2 ). 2 2L Since e−r L f ∈ D(L) ⊂ D(L1/2 ) and that e−t is a bounded, linear operator on L2 , the lemma implies 2 2 2 2 t2 ∂t L1/2 e−t L (e−r L f ) = t2 ∂t e−t L L1/2 (e−r L f ). (9.1.51) So 2 2 2 2 t2 ∂t L1/2 e−t L (e−r L f ) = −2t3 Le−t L (L1/2 e−r L f ) 2 2 2 2 = −2t3 L1/2 L1/2 e−t L (L1/2 e−r L f ) = −2t3 L1/2 Le−t L e−r L f (9.1.52) 183 where the last equality follows from Lemma 9.1.3 and L1/2 L1/2 = L. 2 2 2 2 − 2t3 L1/2 Le−t L e−r L f = −2t3 L1/2 Le−r L e−t L f 2 2 2 2 = −2t3 L1/2 e−r L Le−t L f = t2 L1/2 e−r L ∂t e−t L f 2 2 = −2t3 e−r L L1/2 Le−t L f, (9.1.53) 2 2 where in the last step we have used −2tLe−t L f = ∂t e−t L f ∈ W 1,2 = D(L1/2 ) and Lemma 9.1.3. Combining (9.1.51)-(9.1.53), we have proved (9.1.43). 9.2 Lp estimates for non-tangential maximal functions Definition 9.2.1. The non-tangential maximal function is defined as . N α (u)(x) = sup sup |u(y, t)| . (9.2.1) t>0 (y,t):|x−y|<αt The integrated non-tangential maximal function is defined as !1/2 eα . 2 N (u)(x) = sup sup |u(z, t)| dz . (9.2.2) t>0 (y,t):|x−y|<αt |y−z|<αt 2 2 We shall consider functions such as N α (∂t e−t L f ), where we think of ∂t e−t L f (x) as a function of x and t. Proposition 9.2.1. Let η > 0, α > 0. Then −1 ηα −(ηt)2 L η N (∂t e f ) ≤ Cα,p k∇f kLp Lp for all p > 1, and f ∈ W 1,p . The constant Cα,p also depends on λ0 , Λ0 and n, but not on η. 184 Proof. Fix any x ∈ Rn . Let (y, t) ∈ Γηα (x), i.e. |x − y| < ηαt. Assume f ∈ S (Rn ). Claim: −1 −(ηt)2 L η ∂t e f (y) ≤ Cα M (∇f )(x). (9.2.3) 2 Let Vt (x, y) be the kernel associated to ∂t e−t L . Then by (7.2.52), we have |x−y|2 |x−y|2 −n−1 − η Vηt (x, y) . (ηt)−n−1 e− c(ηt)2 −1 |Vt (x, y)| . t e ct2 , where the implicit constant depends on λ0 , Λ0 and n. We write ! −1 −(ηt)2 L −1 −(ηt)2 L η ∂ e f (y) = η ∂ e f − f (y) t t B2ηαt (x) ˆ ! −1 = η Vηt (y, z) f − f (z)dz Rn B2ηαt (x) ˆ |y−z| 2 −1 −(ηt)2 L 1 − c(ηt)2 ⇒ η ∂t e f (y) . e f (z) − f dz B2ηαt (x) (ηt)n+1 B2ηαt (x) ∞ ˆ |y−z|2 X 1 − c(ηt)2 f (z) − + e f dz (ηt)n+1 k=1 2k+1 Bηαt (x)\2k Bηαt (x) B2ηαt (x) . = I1 + I2 . |y−z|2 − For I1 , trivially bound e c(ηt)2 by 1, then the Poincar´e inequality gives αn I1 . f (z) − f dz . αn+1 |∇f | . αn+1 M (∇f )(x), ηt B2ηαt (x) B2ηαt (x) B2ηαt (x) where the implicit constants depend only on n. For I2 , we have ∞ ˆ (2k − 1)2 α2  X 1 I2 . exp − f (z) − f dz (ηt)n+1 c k=1 2k+1 Bηαt (x) B2ηαt (x) ∞ 2n(k+1) αn (2k − 1)2 α2 X   . exp − f (z) − f dz. ηt c 2k+1 Bηαt (x) B2ηαt (x) k=1 185 By a calculation similar to (9.1.15), we obtain ∞ k+1 4k 2 α2 X   X I2 . exp − 2n(k+1) αn+1 2l M (∇f )(x) .α M (∇f )(x), c k=1 l=2 and thus (9.2.3) follows. By the choice of (y, t), this implies 2 η −1 N ηα (∂t e−(ηt) L f )(x) ≤ Cα M (∇f )(x), −1 ηα −(ηt)2 L ⇒ η N (∂t e f ) ≤ Cα,p k∇f kLp ∀ p > 1, f ∈ S (Rn ). Lp Then the proposition follows from a standard limiting argument. We also have Lp estimates for the integrated non-tangential maximal function: Proposition 9.2.2. Let η > 0. Then for any p > 2, f ∈ W 1,p (Rn ), 2 N (∇e−(ηt) L f ) eη ≤ Cp k∇f kLp , Lp where the constant depends on p, λ0 , Λ0 and n, but not on η. Proof. Let f ∈ S (Rn ). Define u(x, t) = e−tL f (x). Then u satisfies the equation ∂t u − div(A∇u) = 0 in L2 . Now fix x ∈ Rn , and fix (y, t) ∈ Γη (x). Define Bs = B(y, (1 + s)ηt), the ball centered at y with radius (1 + s)ηt. For 0 ≤ s < s0 < 21 , choose 1 Ψ ∈ C0∞ (B s+s0 ), with Ψ = 1 on Bs , |∇Ψ| . , 2 (s0 − s)ηt and e ∈ C ∞ (Bs0 ), 1 Ψ with Ψ e = 1 on B s+s0 , ∇Ψ . e 0 2 (s0 − s)ηt 186 ffl Let u ¯= B(y, 32 ηt) u(x, 0)dx. ¯)Ψ2 as test function to obtain Taking (u − u ˆ ˆ 2 ¯)Ψ2 dx  A(x)∇u(x, τ ) · ∇ (u(x, τ ) − u ¯)Ψ dx = − ∂τ u(x, τ )(u(x, τ ) − u ∀ τ > 0. Rn Rn Then by an argument similar to the proof of Proposition 3.3.3, one obtains ˆ ¯)Ψ2 dx  A(x)∇u(x, τ ) · ∇ (u(x, τ ) − u Rn ˆ ˆ ˆ λ0 2 Cθ 2 ≤ |∇u(x, τ )| dx − 0 |u(x, τ ) − u ¯| dx − θ |∇u(x, τ )|2 dx, 2 Bs (s − s)2 (ηt)2 Bs0 Bs0 and ˆ ∂τ u(x, τ )(u(x, τ ) − u ¯) Rn ˆ ˆ 1 0 2 . (s − s) (ηt) 2 |∂τ u(x, τ )|2 dx + 0 ¯|2 . |u(x, τ ) − u B s+s0 (s − s)2 (ηt)2 B s+s0 2 2 Combining, we have ˆ |∇u(x, τ )|2 dx Bs ˆ ˆ ˆ C 2 ≤ 0 2 2 (u(x, τ ) − u 2 ¯) dx + C(ηt)2 |∂τ u(x, τ )| dx + Cθ |∇u(x, τ )|2 dx. (s − s) (ηt) Bs0 Bs0 Bs0 Choosing θ sufficiently small, then Lemma 2.4.2 gives 1 |∇u(x, τ )|2 dx . ¯|2 dx + (ηt)2 |u(x, τ ) − u |∂τ u(x, τ )|2 dx, B(y,ηt) (ηt)2 B(y, 32 ηt) B(y, 32 ηt) (9.2.4) for any τ > 0. Let w(z, t) = u(z, η 2 t2 ). Then it suffices to show eη N (∇x w) ≤ Cp k∇f kLp ∀ p > 2. Lp 187 To this end, let τ = η 2 t2 in (9.2.4). Noticing that ∂t w(z, t) = 2η 2 t∂t u(z, η 2 t2 ), we have 1 |∇w(z, t)|2 dz . ¯ 2 dz + η −2 |w(z, t) − w| |∂t w(z, t)|2 dz B(y,ηt) (ηt)2 B(y, 32 ηt) B(y, 32 ηt) (9.2.5) ffl where w ¯= B(y, 23 ηt) w(z, 0)dz. ffl 2 can be controlled by M N η (∂t w)2 (x). To  The second term B(y, 23 ηt) |∂t w(z, t)| dz estimate the first term on the right-hand side of (9.2.5), we write |w(z, t) − w| ¯ ≤ |w(z, t) − w(z, 0)| + |w(z, 0) − w| ¯ . Using Poinca´re inequality, we have 1 ¯ 2 dz . |w(z, 0) − w| |∇w(z, 0)|2 dz (ηt)2 B(y, 32 ηt) B(y, 32 ηt)     . M |∇w(·, 0)|2 (x) = M |∇f |2 (x). Since ˆ t |w(z, t) − w(z, 0)| ≤ |∂τ w(z, τ )| dτ ≤ t sup |∂τ w(z, τ )| , 0 0≤τ ≤t |w(z, t) − w(z, 0)|2 dz ≤ t2 sup ∂τ w(z, t)2 dz ≤ t2 M (N η (∂t w)2 )(x). B(y, 23 ηt) B(y, 32 ηt) 0≤τ ≤t Therefore, we have obtained   |∇w(z, t)|2 dz . M |∇f |2 (x) + η −2 M N η (∂t w)2 (x)  B(y,ηt) for any (y, t) ∈ Γη (x). This implies, by the definition of the integrated non-tangential e η , that maximal function N    1/2 1/2 e η (∇z w)(x) . M |∇f |2 (x) + η −1 M N η (∂t w)2 (x) ∀ x ∈ Rn .  N 188 Thus for any p > 2,  2 2   2 2  N ∇e−η t L f . k∇f kLp + η −1 N η ∂t e−η t L f eη ≤ Cp k∇f kLp (Rn ) . Lp (Rn ) Lp (Rn ) In the last inequality we have used Proposition 9.2.1 with α = 1. CHAPTER Ten The A∞ property for operators with coefficients independent of the transverse variable In this chapter, we prove the A∞ property for elliptic operators with t-independent, BMO coefficients in the upper half space Rn+1 + = {(x, t) ∈ Rn × (0, ∞)} with n ≥ 3. To be precise, let A = A(x) be an (n + 1) × (n + 1) matrix of real, t-independent coefficients such that   1. The symmetric part As = 21 (A + A| ) = Asij (x) is L∞ (Rn ), and satisfies the ellip- ticity condition n+1 X 2 s λ0 |ξ| ≤ hA (x)ξ, ξi = Asij (x)ξi ξj ∀ ξ ∈ Rn+1 , kAs k∞ ≤ λ−1 0 (10.0.1) i,j=1 for some 0 < λ0 < 1. 2. The anti-symmetric part Aa = 21 (A − A| ) = (Aaij (x)) is in the space BMO(Rn ), with a a Aij − (Aaij )Q dx ≤ Λ0 Aij = sup (10.0.2) BM O Q⊂Rn Q 189 190 1 ´ for some Λ0 > 0. Here (f )Q denotes the average |Q| Q f (x)dx. We define in Rn+1 a second order divergence form operator L = − divx,t A(x)∇x,t , (10.0.3) Recall that Definition 10.0.1 (A∞ (dx)). We say that the elliptic measure ω associated to L belongs to the A∞ class with respect to the Lebesgue measure dx, if there are positive constants C and θ such that for every cube Q ⊂ Rn , and every Borel set E ⊂ Q, we have  θ |E| ω(E) ≤ C ω(Q). |Q| Definition 10.0.2. We say the Dirichlet problem with Lp (dx) data in Rn+1 + is solvable for L if the weak solution u for continuous boundary data f satisfies the estimate kN α (u)kLp (Rn ,dx) . kf kLp (Rn ,dx) , where the implicit constant does not depend on f . Here, N α (u) denotes the non-tangential maximal function of u: . N α (u)(x) = sup |u(y, t)| , (10.0.4) (y,t):|x−y|<αt where α > 0 is the aperture. Our main result is the following Theorem 10.0.1. Assume that A is t-independent, satisfies (10.0.1) and (10.0.2). Let L be an elliptic operator as above, defined in Rn+1 + with n > 2. Then the elliptic measure ω associated to L belongs to A∞ (dx) with constants depending only on the ellipticity constant 191 and the BMO semi-norm. In particular, there is some p ∈ (1, ∞) sufficiently large, such that the Lp Dirichlet problem is solvable. 10.1 Outline of the proof of Theorem 10.0.1 We shall prove Theorem 10.0.1 by establishing the Carleson measure estimate (1.0.9). It suffices to show that for any bounded weak solution u to L in Rn+1 + with kukL∞ ≤ 1, it satisfies ˆ l(Q) ˆ |∇u(x, t)|2 t dx . |Q| , (10.1.1) 0 Q for any cube Q ⊂ Rn , and the implicit constant depends only on the ellipticity constants and the BMO semi-norm. Note that this formulation allows us to assume that A is smooth as long as the bounds do not depend on the regularity of the coefficients. It turns out that we do not need to show (10.1.1) holds for integral over all of the cube Q, but only on a subset F of Q that has a big portion of the measure of Q. To be precise, we have the following lemma. Lemma 10.1.1. Let u be a weak solution to L in Rn+1 + . Then to show the Carleson measure estimate (10.1.1) for u, it suffices to show the following holds: there is a uniform constant c, and for each cube Q ⊂ Rn , a Borel set F ⊂ Q, with |F | ≥ c |Q|, for which ˆ l(Q) ˆ |∂t u(x, t)|2 t dx . |Q| (10.1.2) 0 F holds. And the implicit constant should not depend on Q and F , but on c, kukL∞ , the ellipticity constants and the BMO semi-norm. The proof of the Lemma 10.1.1 requires two steps of reduction. First, one can show by 192 integration by parts and the Caccioppoli inequality on Whitney cubes, that ˆ l(Q) ˆ ˆ 2l(Q) ˆ 2 |∇u(x, t)| t dxdt . |∂t u(x, t)|2 t dxdt + |Q| . (10.1.3) 0 Q 0 2Q The details can be found in [HKMP15]. Second, since the coefficients are independent of t, ∂t u is also a weak solution of L (see Appendix 3.3, Remark 3.3.1), and thus ∂t u satisfies Harnack Principle and interior H¨older estimates (see section 3.1). It allows us to apply a well-known result of Carleson measures, which was first wrote down explicitly in [AHLT01] Lemma 2.14, to conclude that (10.1.2) implies ˆ l(Q) ˆ |∂t u(x, t)|2 t dx . |Q| . 0 Q Combining this with (10.1.3), Lemma 10.1.1 follows. This lemma gives us the freedom to choose the set F . We now take a closer look at the structure of the matrix A by writing it as   A|| b A= , c d where A|| denotes the n × n submatrix of A with entries (A|| )i,j , 1 ≤ j ≤ n, b denotes the column vector (Ai,n+1 )1≤i≤n , c denotes the row vector (An+1,j )1≤j≤n , and d = An+1,n+1 . We observe that if divx c = 0, then the Carleson measure estimates follow easily. We define an n-dimensional divergence form operator . . L|| = div A|| ∇, and its adjoint L∗|| = − div A∗|| ∇. The construction of the set F is presented in section 10.3.1. Basically, we will construct F such that on the set F , the non-tangential maximal function involving . 2 . −t2 L∗|| Pt = e−t L|| and Pt∗ = e , 193 as well as some other maximal functions are small (see (10.3.4)). We will exploit this property of the set F in the proof of the Carleson measure estimates. Namely, as long as a term can be bounded by maximal functions showing up in the definition of F , then there is hope to control that term with desired bounds. We shall also take the advantages of the Lp bounds for the square functions involving 2L −t2 L∗|| Pt = e−t || and Pt∗ = e to obtain the Carleson measure estimates. These bounds are obtained in section 9.1. Using these estimates, we are ready to prove the Carleson measure estimate (10.1.2). However, having coefficients in BMO forces us not to work directly with the weak solution to the equation L = − div A∇. Instead, we shall work with weak solutions to the operator L0 = − div A0 ∇, where A0 is defined in (10.4.3). We observe in Lemma 10.4.2 that a weak solution of L is also a weak solution of L0 . This observation enables us to work with the equation L0 u = 0 in Rn+1 + , for which we have good control of the BMO coefficients. 10.2 Hodge decomposition Recall that we write the matrix A = A(x) as follows   A|| b A= , c d where A|| is the n × n submatrix of A, b is a n × 1 vector, c is a 1 × n vector, d is a scalar function. We consider the symmetric part As and anti-symmetric part Aa of A: " # " # . As|| bs Aa|| ba A = As + Aa = + . cs d ca 0 194 We assume that As is L∞ and elliptic, with the ellipticity constant λ0 and kAs k∞ ≤ λ−1 0 . And assume Aa is in BMO(Rn ), with the BMO semi-norm a aij . a aij − (aaij )Q dx ≤ Λ0 . BM O = sup Q⊂Rn Q e ∈ W01,2 (5Q) that solve Proposition 10.2.1. For any cube Q ⊂ Rn , there exist ϕ, ϕ − divx (A∗|| ∇x ϕ) = divx (c15Q − (ca )2Q ), (10.2.1) e = divx (b15Q − (ba )2Q ), divx (A|| ∇x ϕ) (10.2.2) respectively. Moreover, there exists some 0 = 0 (n, λ0 , Λ0 ) > 0 and C = C(n, λ0 , Λ0 ) > 0 such that for all p ∈ [2, 2 + 0 ], |∇ϕ(x)|p dx ≤ C, |∇ϕ(x)| e p dx ≤ C. (10.2.3) 5Q 5Q ffl p Proof. We only prove 5Q |∇ϕ| e ≤ C, as the estimate for ∇ϕ can be derived similarly. We will identify ϕ e with its zero extension outside of 5Q. Let QR0 be a cube in Rn with QR0 ∩ 5Q 6= ∅. For any x ∈ QR0 and 0 < R < 1 2 dist(x, ∂QR0 ), we have three possibilities: (i) Q 3 R (x) ∩ 5Q = ∅ 2 (ii) Q 3 R (x) ∩ (QR0 \ 5Q) = ∅ 2 (iii) Q 3 R (x) ∩ 5Q 6= ∅ and Q 3 R (x) ∩ (QR0 \ 5Q) 6= ∅. 2 2 In case (ii), Q 3 R (x) ⊂ 5Q, then by the interior Caccioppoli inequality and Poincar´e- 2 195 Sobolev inequality, we ˆ ˆ 2 2 −2 |∇ϕ| e dy ≤ CR ϕ e Q3/2R (x) dy + CRn e − (ϕ) (10.2.4) QR (x) Q 3 R (x) 2 ˆ  2/r ≤C e r |∇ϕ| + CRn , Q 3 R (x) 2 2n where r = n+2 . In case (iii), we also have ˆ ˆ  2/r 2 |∇ϕ| e dy ≤ C  e r |∇ϕ| + CRn , QR (x) Q 3 R (x) 2 which follows from the boundary Caccioppoli inequality, ˆ ˆ e 2 dy ≤ CR−2 |∇ϕ| e 2 dy + CRn , |ϕ| (10.2.5) QR (x) Q 3 R (x)∩5Q 2 and a Sobolev-Poincar´e theorem. The proof for (10.2.5) is postponed to the end. Now we can apply Lemma 2.4.1 to get !p/2 p 2 |∇ϕ| e ≤C |∇ϕ| e +C 1. Q R0 ∩5Q QR0 ∩5Q Q R0 2 Choose Q R0 ⊇ 5Q then 2  p/2 p 2 |∇ϕ| e ≤C |∇ϕ| e + Cn . 5Q 5Q We claim that e 2 ≤ C(n, λ0 , Λ0 ), |∇ϕ| (10.2.6) 5Q ffl which would imply the desired bound for e p. 5Q |∇ϕ| e ∈ W01,2 (5Q) as test In fact, taking ϕ 196 function, then (10.2.2) and ellipticity of As imply ˆ ˆ ˆ ˆ 2 λ0 |∇ϕ| e ≤ As|| ∇ϕ e· ∇ϕ e= A|| ∇ϕ e · ∇ϕ e= b · ∇ϕ. e 5Q 5Q 5Q 5Q We have ˆ ˆ ˆ ˆ ˆ λ0 2 2 bsj baj (baj )5Q  a a b · ∇ϕ e = + − e ≤ ∂j ϕ |∇ϕ| e +C bj − (bj )5Q + C 1. 5Q 5Q 2 5Q 5Q 5Q Then (10.2.6) follows from the John-Nirenberg inequality. It remains to prove (10.2.4) and (10.2.5). Proof of (10.2.5) 3 For any R ≤ t < s ≤ 2 R, define ξ ∈ C02 (Q t+s (x)) and η ∈ C02 (Qs (x)) such that 2 1 0 ≤ ξ, η ≤ 1, ξ = 1 in Qt (x), η = 1 in Q t+s (x), and |∇ξ| , |∇η| . s−t . 2 e 2 ∈ W01,2 (5Q) as test function, then (10.2.2) gives Choose ϕξ ˆ ˆ A|| ∇ϕ e 2) = e · ∇(ϕξ b15Q · ∇(ϕξ e 2 ). (10.2.7) Q t+s (x) Q t+s (x) 2 2 For the left-hand side of (10.2.7), we have ˆ ˆ ˆ λ0 2C As|| ∇ϕ e· ∇(ϕξ e )≥ 2 |ξ∇ϕ| e − e2 . ϕ Q t+s (x) 2 Q t+s (x) (s − t)2 Q t+s (x) 2 2 2 197 And ˆ ˆ a 2 1 a 2 2  A|| ∇ϕe · ∇(ϕξ e ) = aij ∂j ϕ∂ e ) − ∂i ϕ∂ e i (ϕξ e j (ϕξ e ) 2 Q t+s (x) Q t+s (x) 2 2 ˆ 1 a 2 2 2 2  = a ∂j (ϕ e )∂i (ξ ) − ∂i (ϕ e )∂j (ξ ) 4 Q t+s (x) ij 2 ˆ 1 a 2 2 2 2  = e ∂i (ξ ) − ∂i (ϕη) a ∂j (ϕη) e ∂j (ξ ) . 4 Q t+s (x) ij 2 By Proposition 2.2.3, this is bounded by ˆ ˆ C Cθ 2 kϕηk e L2 k∇(ϕη)k e L2 ≤ |ϕ| e +θ e2 |∇ϕ| s−t (s − t)2 Qs (x) Qs (x) for any 0 < θ < 1. For the right-hand side of (10.2.7), we have ˆ ˆ ˆ λ0 C b 15Q · ∇(ϕξ s 2 2 e 2 + Csn . e ) ≤ |ξ∇ϕ| e + 2 |ϕ| Q t+s (x) 8 Q t+s (x) (s − t) Q t+s (x) 2 2 2 And by Proposition 2.2.2, ˆ ˆ  1/2 C  ba 15Q · ∇(ϕξ e 2 ) ≤ e 2  sn/2 + kξkL2 k∇(ϕξ)k |ϕξ| e L2 s−t Q t+s (x) Q t+s (x) ˆ2 ˆ 2 λ0 C ≤ e2+ |ξ∇ϕ| e 2 + Csn . |ϕ| 8 Q t+s (x) (s − t)2 Q t+s (x) 2 2 Combining these estimates with (10.2.7), we fix 0 < θ < 1 to be sufficiently small and obtain ˆ ˆ ˆ ˆ 2 Cθ 2 2 |∇ϕ| e ≤ |ξ∇ϕ| e ≤ 2 |ϕ| e + Cθ e 2 + Csn |∇ϕ| Qt (x) Q t+s (x) (s − t) Qs (x) Qs (x) 2 ˆ ˆ Cθ 1 ≤ 2 e2+ |ϕ| |∇ϕ|e 2 + CRn . (s − t) Q 3 (x) 2 Qs (x) 2R Then (10.2.5) follows from Lemma 2.4.2. 198 The interior Caccioppli (10.2.4) can be shown in the same manner if one chooses   ϕ e Q 3 (x) ξ 2 as test function in the beginning. e − (ϕ) 2R Remark 10.2.1. Note that one can replace (ca )2Q and (ba )2Q in the right-hand side of (10.2.1) and (10.2.2), respectively, by any constant vector C without changing the result. ´ This follows from the simple fact that 5Q C · ∇v = 0 for any test function v ∈ W01,2 (5Q). Moser-type interior estimates for the weak solution to the homogeneous equation − divx A|| ∇x u = ˇ 0 have been shown in section 3.1, or [SSSZ12] for the parabolic equation. We show that similar estimates hold for weak solutions to the nonhomogeneous equations. e be as in Proposition 10.2.1. Let B2R = B2R (x0 ) ⊂ 5Q. Proposition 10.2.2. Let ϕ and ϕ Then for any p > 1,  1/p p sup |ϕ e − c0 | ≤ C |ϕ e − c0 | + CR(kbs kL∞ + kba kBMO ), (10.2.8) BR B2R where c0 is any constant, and C = C(n, λ0 , Λ0 , p). And a similar estimate holds for ϕ:  1/p sup |ϕ − c0 | ≤ C |ϕ − c0 |p + CR(kcs kL∞ + kca kBMO ). (10.2.9) BR B2R Proof. The proof is similar to that of Lemma 3.1.4. Fix any p > 1 and 1 2 < k0 < p2 . Let 1 2s0 2n n 2 < k1 < min{1, k0 } and k ≥ k0 . Define q = 2−s0 ∈ (2, n−2 ), where s0 ∈ (1, n−1 ). Define as in Lemma 3.1.4, for any δ > 0, N >> 1 and β ≥ k0 ,   tβ ,  t ∈ [δ, N ], Hδ,N (t) =  Nβ + β β−k1 (tk1 k1 N − N k1 ), t > N.   βtβ−1 , t ∈ (δ, N ),   0 ⇒ Hδ,N (t) =  βN β−k1 tk1 −1 ,  t > N. 199 Define ˆ w 0 Gδ,N (w) = |Hδ,N (t)|2 dt, w ≥ δ. δ Then for w ≥ δ, we have (3.1.8), (3.1.9) and (3.1.10). In the sequel we omit the subscripts in Gδ,N and Hδ,N . Let δ = R(kbs kL∞ + kba kBMO ), and define Ψ = |ϕ e − c0 | + δ, where c0 is an arbitrary e = divx (b15Q − (ba )2Q ). constant. Then Ψ is a subsolution to the equation divx (A|| ∇x ϕ) Also, since Ψ ≥ δ, one can define H(Ψ), G(Ψ) etc.. For any R ≤ r0 < r ≤ 2R, let η ∈ C02 (Br ) with η = 1 in Br0 and |∇η| . (r − r0 )−1 . Choose v = G(Ψ)η 2 > 0 as test function. Then since Ψ is a subsolution, one has ˆ ˆ A|| ∇Ψ · ∇v ≤ b · ∇v. (10.2.10) Br Br For the left-hand side of (10.2.10), we have (see the proof of Lemma 3.1.4) ˆ ˆ 2/q C(n, λ0 , k0 ) β 2 rn  λ0 2 As|| ∇Ψ · ∇v ≥ |∇H(Ψ)| η − 2 Ψ βq (10.2.11) Br 2Br (2k0 − 1)2 (r − r0 )2 Br ˆ ˆ 2/q C(n, λ0 , Λ0 , q, k0 ) β 2 rn  a λ0 2 2 βq A|| ∇Ψ · ∇v ≤ |∇H(Ψ)| η + Ψ . Br 8 Br (2k0 − 1)2 (r − r0 )2 Br (10.2.12) The right-hand side of (10.2.10) equals ˆ ˆ s 2 ba · ∇ G(Ψ)η 2   b · ∇ G(Ψ)η + B Br ˆ r ˆ ˆ 0 s 2 s (ba − (ba )Br ) · ∇H(Ψ) H 0 (Ψ) η 2 = b · ∇H(Ψ) H (Ψ) η + 2 b · ∇ηG(Ψ)η + Br Br Br ˆ +2 (ba − (ba )Br ) · ∇ηG(Ψ)η Br . = I1 + I2 + I3 + I4 . 200 For I1 , using Cauchy-Schwartz, (3.1.9), as well as Young’s inequality, we obtain ˆ ˆ λ0 2 |I1 | ≤ |∇H(Ψ)| η + 2 C(n, λ0 ) kbs k2L∞ β 2 Ψ2β−2 η 2 . 8 2n Recalling that Ψ ≥ δ = R(kbs kL∞ + kba kBMO ) and 2 < q < n−2 , |I1 | is bounded by ˆ ˆ λ0 2 2 2 −2 |∇H(Ψ)| η + C(n, λ0 )β R Ψ2β η 2 8 ˆ  2/q λ0 2 2 2 −2 n qβ ≤ |∇H(Ψ)| η + C(n, λ0 )β R r Ψ . (10.2.13) 8 Br For I2 , we use (3.1.10) and obtain ˆ ˆ kbs kL∞ β 2 C(n, k0 )β 2 |I2 | ≤ Ψ2β−1 |η| ≤ Ψ2β r − r0 2k1 − 1 (r − r0 )(2k0 − 1)R Br 2/q )β 2 rn  C(n, k0 ≤ Ψqβ . (10.2.14) (r − r0 )(2k0 − 1)R Br For I3 , we estimate ˆ  10 ˆ 1/2 ˆ 1/q s0 s00 2 H (Ψ) q η q a a 2 0 |I3 | ≤ |b − (b )Br | |∇H(Ψ)| η Br n ˆ 1/2 ˆ 1/q a s00 2 2 qβ−q q ≤ C(n, q) kb kBM O r |∇H(Ψ)| η β Ψ η ˆ 2/q C(n, λ0 , q)β 2 rn  λ0 2 2 qβ ≤ |∇H(Ψ)| η + Ψ , (10.2.15) 8 R2 Br s0 where s00 = s0 −1 . For I4 , we have ˆ  q−2 ˆ 2/q q q a a q/2 q/2 |I4 | ≤ |b − (b )Br | q−2 |G(Ψ)| |∇η| Br ˆ 2/q a 1 β2 (q−2)n qβ− 2q ≤ C(n, q) kb kBM O r Ψq 2k1 − 1 r − r0 Br ˆ 2/q β 2 rn qβ ≤ C(n, q, k0 ) Ψ . (10.2.16) (2k0 − 1)(r − r0 )R Br 201 Combining (10.2.10)–(10.2.16), we get ˆ 2/q λ0 2 2 0 −2 0 −1 −1 −2  qβ |∇H(Ψ)| ≤ Cβ (r − r ) + (r − r ) R +R Ψ . 8 Br 0 Br By Sobolev’s inequality, (3.1.8), and letting N go to infinity, we obtain ! n−2 n ˆ 2/q 2n β n−2 2 02 0 −2 0 −1 −1 −2  qβ Ψ ≤ Cβ r (r − r ) + (r − r ) R +R +1 Ψ . Br0 Br Now let l = 2n (n−2)q > 1, β = βi = kli , r = ri = R + R 2i and r0 = ri+1 for i = 0, 1, 2, . . . , one finds ! 1 ! 1 kli+1 q 2(i+1) kli q 1 kli+1 q kli q Ψ ≤ (Ckli ) kli 2 kli Ψ ≤ ... Bri+1 Bri  1 Pi 1 Pi j 2 Pi j+1 kq kq ≤ (Ck) j=0 klj l j=0 klj 2 k j=0 lj Ψ . B2R ffl 1 kq kq Letting i → ∞, we have supBR Ψ ≤ C(n, λ0 , Λ0 , q, k0 ) B2R Ψ , and thus  1 kq kq sup |ϕ e − c0 | ≤ C |ϕ e − c0 | + Cδ, BR B2R where C = C(n, λ0 , Λ0 , q, k0 ). Then by choosing k and q to be such that kq = p, we obtain (10.2.8). The proof of (10.2.9) is similar and thus omitted. 202 10.3 Construction of F and sawtooth domains associated with F 10.3.1 The set F We define the following maximal differential operator  p !1/p . |f (x) − f (y)| Dp f (x) = sup dy . (10.3.1) r>0 |x−y| 0 and let 0 < s ≤ r. Let ω ∈ ∂B(0, 1). Then ˆ s ˆ s |f (x + sω) − f (x)| = Df (x + tω) · ωdt ≤ |Df (x + tω)| dt. 0 0 203 Integrating on the unit sphere gives ˆ ˆ sˆ |f (x + sω) − f (x)| dσ(ω) ≤ |Df (x + tω)| dσ(ω)dt ∂B(0,1) 0 ∂B(0,1) ˆ sˆ ˆ ˆ dσ(y) |Df (y)| |Df (y)| = |Df (y)| n−1 dt = dy ≤ dy 0 ∂B(x,t) t B(x,s) |x − y|n−1 B(x,r) |x − y|n−1 On the other hand, ˆ ˆ 1 |f (x + sω) − f (x)| dσ(ω) = |f (y) − f (x)| dσ(y). ∂B(0,1) sn−1 ∂B(x,s) ˆ ˆ n−1 |Df (y)| ⇒ |f (y) − f (x)| dσ(y) ≤ s dy. ∂B(x,s) B(x,r) |x − y|n−1 Integrating with respect to s from 0 to r gives ˆ ˆ rn |Df (y)| |f (y) − f (x)| dy ≤ dy. (10.3.3) B(x,r) n B(x,r) |x − y|n−1 Let W = B(x, r) ∩ B(y, r), and let r = |x − y|, then we have |f (x) − f (y)| ≤ |f (x) − f (z)| dz + |f (y) − f (z)| dz. W W Applying (10.3.3), one obtains ! |Df (z)| |Df (z)| |f (x) − f (y)| ≤ Cn rn n−1 dz + n−1 dz B(x,r) |x − z| B(y,r) |y − z| ˆ ˆ ! dz dz ≤ Cn M (∇f )(x) n−1 + M (∇f )(y) n−1 B(x,r) |x − z| B(y,r) |y − z| ≤ Cn |x − y| (M (∇f )(x) + M (∇f )(y)) , which proves (10.3.2). 2L −t2 L∗|| We introduce a few notations. Recall that we use Pt to denote e−t || , and Pt∗ = e . 204 Define 1/2 .  Λ1 = η −1 N η (∂t Pηt ∗ ϕ) + N (∂t Pt∗ ϕ) + N ∗ e η (∇x Pηt ϕ) + M (|∇ϕ|2 ) , 1/2 .  Λ2 = η −1 N η (∂t Pηt ϕ) e + N (∂t Pt ϕ) e η (∇x Pηt ϕ) e +N e 2) e + M (|∇ϕ| , where ϕ and ϕ e are as in Proposition 10.2.1, and the non-tangential maximal operator N in the second terms on the two right hand sides in defined with respect to cones of aperture 1. Let Q ⊂ Rn and κ0  1 be given. Fix p1 ∈ (1, 2) and define the set F as follows . F = {x ∈ Q : Λ1 (x) + Λ2 (x) + Dp1 ϕ(x) + Dp1 ϕ(x) e ≤ κ0 } . (10.3.4) Lemma 10.3.2. Let 0 be as in Proposition 10.2.1. Then |Q \ F | . κ0−2−0 |Q| (10.3.5) uniformly in η. Proof. By Tchebychev’s inequality, ˆ κ2+ 0 0 |Q \ F | ≤ e 2+0 dx (Λ1 + Λ2 + Dp1 ϕ + Dp1 ϕ) Q∩{Λ1 +Λ2 +Dp1 ϕ+Dp1 ϕ>κ e 0} We apply Proposition 9.2.1, Propostion 9.2.2, and Proposition 10.3.1, and their analogs for the adjoint operators, with p = 2 + 0 , to obtain that the right-hand side is bounded by ˆ   2+0   2+0 |∇ϕ|2+0 + M |∇ϕ|2 e2 e 2+0 + M |∇ϕ| 2 2 C + |∇ϕ| dx, Rn 205 which is bounded by   C |5Q| e 2+0 . |∇ϕ|2+0 + |∇ϕ| Q Then the lemma follows from (10.2.3). We can now choose κ0 , depending only on λ0 , Λ0 and n, such that 1 |Q \ F | ≤ |Q| . (10.3.6) 1000 This completes the construction of F and from now on κ0 is fixed. 10.3.2 Sawtooth domains and related estimates Define Ω0 to be the sawtooth domain . [ Ω0 = Γη (x). (10.3.7) x∈F Define . . θt = ϕ − Pt∗ ϕ, e − Pt ϕ. θet = ϕ e (10.3.8) We observe that ˆ ηt ˆ ηt θηt (x) = − ∂s Ps∗ ϕ(x), and θeηt (x) = − ∂s Ps ϕ(x). e 0 0 So by the definition of the set F , |θηt (x)| ≤ ηtκ0 , θηt (x) ≤ ηtκ0 ∀ (x, t) ∈ F × (0, ∞). (10.3.9) e We show that such estimates also holds in the truncated sawtooth domain. Note that we shall eventually choose η > 0 to be small, so we can assume in the sequel that η < 1/2. 206 Lemma 10.3.3. |θηt (x)| . ηtκ0 and θηt (x) . ηtκ0 ∀ (x, t) ∈ Ω0 ∩ (2Q × (0, 4l(Q))) . e Proof. We only show the estimate for θηt , for the proof for θeηt is similar. Let (x, t) ∈ Ω0 ∩ (2Q × (0, 4l(Q))). Then there exists x0 ∈ F such that |x − x0 | ≤ ηt. Since t < 4l(Q), and η < 21 , we have 2B(x0 , ηt) ⊂ 5Q. We write   ∗ |θηt (x)| ≤ |ϕ(x) − ϕ(x0 )| + |θηt (x0 )| + Pηt ϕ − (ϕ)2Bηt (x0 ) (x0 )   ∗ + Pηt ϕ − (ϕ)2Bηt (x0 ) (x) (10.3.10) ffl where (ϕ)2Bηt (x0 ) = 2Bηt (x0 ) ϕ. Note that we have used the conservation property, so that ∗ (ϕ) Pηt Bηt (x0 ) is a constant. By Proposition 10.2.2, the first term on right-hand side of (10.3.10) is bounded by !1/p1 C |ϕ − ϕ(x0 )|p1 + Cηt(kcs kL∞ + kca kBMO ). 2Bηt (x0 ) By the definition of Dp1 and the set F , this is bounded by Cηt (Dp1 ϕ(x0 ) + λ0 + Λ0 ) ≤ Cηt(κ0 + λ0 + Λ0 ) ≤ Cηtκ0 , with C = C(λ0 , Λ0 , n, p1 ). By (10.3.9), the second term on the right-hand side of (10.3.10) is also bounded by Cηtκ0 . Now we take care of the last two terms in (10.3.10). We claim that for any (y, s) ∈ Γη (x0 ),   ∗ Pηs ϕ − (ϕ)2Bηs (x0 ) (y) . ηsM (∇ϕ)(x0 ) . ηsκ0 . And this would complete the proof of the lemma. 207 ∗ ∗ Consider the kernel K(ηs)2 associated to Pηs . Then by (7.2.44), 2 ∗ 1 − c|y−z| K(ηs)2 (y, z) . e (ηs)2 . (ηs)n Let (y, s) ∈ Γη (x0 ), then we have   ˆ 1 − c|y−z| 2 ∗ (ηs)2 ϕ(z) − (ϕ) Pηs ϕ − (ϕ)2Bηs (x0 ) (y) . e 2Bηs (x0 ) dz n (ηt) n ˆ R 1 . n ϕ(z) − (ϕ) 2B ηs (x 0 ) dz 2Bηs (x0 ) (ηs) ∞ ˆ 2 X 1 − c|y−z| (ηs)2 ϕ(z) − (ϕ) + n e 2Bηs (x0 ) dz 2k+1 k Bηs (x0 )\2 Bηs (x0 ) (ηs) k=1 Since |y − x0 | ≤ ηs, |y − z| ≥ (2k − 1)ηs for z ∈ 2k+1 Bηs (x0 ) \ 2k Bηs (x0 ). Therefore,   ∗ Pηs ϕ − (ϕ)2Bηs (x0 ) (y) . (ηs) |∇ϕ(z)| dz 2Bηs (x0 ) ∞ ˆ X 1 −c(2k −1)2 + n e ϕ(z) − (ϕ)2Bηs (x0 ) dz 2k+1 Bηs (x0 )\2k Bηs (x0 ) (ηs) k=1 ∞ k −1)2 X . (ηs)M (∇ϕ)(x0 ) + 2k(n+1) e−c(2 ηsM (∇ϕ)(x0 ) k=1    1/2 . ηsM (∇ϕ)(x0 ) . ηs M |∇ϕ|2 (x0 ) . ηsκ0 . Lemma 10.3.4. ¨ ¨ dxdt 2 2 dxdt |θηt (x)| . η 2 |Q| θηt (x) . η 2 |Q| e n+1 t 3 n+1 t 3 R+ R+ where the implicit constants only depend on λ0 , Λ0 and n. Proof. We only prove the estimate for θeηt , for the proof for θηt is similar. We have the 208 following weighted Hardy’s inequality: ˆ ∞ ˆ t p ˆ ∞ 1 dt dt |f (s)| ds ≤ |f (t)|p , ∀ 1 < p < ∞. (10.3.11) 0 t 0 t 0 t A short and direct proof of (10.3.11) is provided at the end. Recall that ˆ ηt ˆ ηt θηt = ∂s Ps ϕds e ≤ |∂s Ps ϕ| e ds, e 0 0 so ˆ ˆ ∞  ˆ ηt 2 ˆ ∞ ˆ t 2 ∞ 1 e 2 dt  1 dt 2 1 dt θ ηt ≤ |∂ P s s ϕ| e ds = η |∂ P s s ϕ| e ds 0 t t 0 t 0 t 0 t 0 t ´∞ By (10.3.11), the last term is bounded by η 2 0 e2 |∂t Pt ϕ| dt t . Then Proposition 9.1.2 gives ˆ ˆ ∞ p/2 ˆ ˆ ∞ p/2 ˆ e 2 dt 2 dt θηt 3 dx ≤ η 2 |∂t Pt ϕ| e dx . η 2 e p dx |∇ϕ| Rn 0 t Rn 0 t Rn for any p ≥ 2. Let p = 2 and we obtain ¨ 2 dxdt ˆ θ (x) . η 2 e 2 dx . η 2 |∇ϕ| e 2 dx |Q| . η 2 |Q| . |∇ϕ| e ηt 3 n+1 R+ t Rn 5Q Proof of (10.3.11). By H¨ older’s inequality, ˆ t ˆ t 1/p p 1− 1 |f (s)| ds ≤ |f (s)| ds t p. 0 0  ˆ t p ˆ 1 1 t ⇒ |f (s)| ds ≤ |f (s)|p ds. t 0 t 0 ˆ ∞ ˆ t p ˆ ∞ˆ t ˆ ∞ ˆ ∞ 1 dt p dt p 1 ⇒ |f (s)| ds ≤ |f (s)| ds 2 = |f (s)| dtds 0 t 0 t 0 0 t 0 s t2 ˆ ∞ ds = |f (s)|p . 0 s 209 10.3.3 The cut-off function In this subsection, we define the cut-off function adapted to a thinner sawtooth domain. Define [ Ω1 = Γ η (x). (10.3.12) 8 x∈F Let Φ ∈ C ∞ (R) with 0 ≤ Φ ≤ 1, Φ(r) = 1 if r ≤ 1 16 , and Φ(r) = 0 if r > 18 . Define       . . δ(x) t t Ψ(x, t) = Ψη, = Φ Φ 1−Φ , (10.3.13) ηt 32l(Q) 16 . where δ(x) = dist(x, F ). Then Ψ has following properties: S Property 1. Ψ = 1 on x∈F Γ η ∩ {2 < t ≤ 2l(Q)}; 16 Property 2. supp Ψ ⊂ Ω1 ∩ { < t < 4l(Q)}; Property 3. supp ∇Ψ ⊂ E1 ∪ E2 ∪ E3 , where   ηt ηt E1 = (x, t) ∈ 2Q × (0, 4l(Q)) : ≤ δ(x) ≤ 16 8   ηt E2 = (x, t) ∈ 2Q × (2l(Q), 4l(Q)) : δ(x) ≤ 8   ηt E3 = (x, t) ∈ 2Q × (, 2) : δ(x) ≤ . 8 Direct computation shows 1 1 1 |∇Ψ(x, t)| . 1E1 + 1E2 + 1E3 . (10.3.14) ηt l(Q)  A consequence of (10.3.14) is the following 210 Lemma 10.3.5. ˆ ˆ ∞ p α α−1 |∇Ψ| t dt dx ≤ C(η, α, p, n) |Q| (10.3.15) Rn 0 for any α > 0, p > 0. And we have ¨ dxdt ≤ Cn |Q| . (10.3.16) t supp ∇Ψ Proof. Using (10.3.14), we compute ˆ ˆ ∞ p ˆ ˆ 16δ(x)  α !p α α−1 η 1 α−1 |∇Ψ| t dt dx .n t dt dx Rn 0 2Q 8δ(x) ηt η ˆ ˆ 4l(Q)  α !p ˆ ˆ 2  α p 1 α−1 1 α−1 + t dt dx + t dt dx 2Q 2l(Q) l(Q) 2Q   ˆ ˆ 16δ(x) !p 1 1 η .n dt dx + Cα,p |2Q| η αp 2Q t 8δ(x) η   1 ≤ C(α, p, n) 1 + αp |Q| ≤ C(η, α, p, n) |Q| . η And (10.3.16) can be derieved similarly: ¨ ¨ ¨ ¨ dxdt dxdt dxdt dxdt ≤ + + t E1 t E2 t E3 t supp ∇Ψ ˆ ˆ 16δ(x) ˆ ˆ 4l(Q) ˆ ˆ 2 η dt dt dt ≤ dx + dx + dx 2Q 8δ(x) t 2Q 2l(Q) t 2Q  t η ≤ Cn |Q| . 211 10.4 Proof of the Carleson measure estimate Throughout this section, let Q ⊂ Rn be fixed, and construct F ⊂ Q and the cut-off function Ψ as in section 10.3. Recall that κ0 is fixed to ensure that (10.3.6) holds. Recall that we have the matrix A = A(x) whose entries are functions on Rn , or,   A|| b independent of t, and we write A = . A precise definition of the operator L = c d − div A∇ was given in section 7.1. Write the n × 1 vector b = b1 + b2 , with divx b2 = 0. We define a new matrix A1 as follows: " # A|| b1 A1 = c + b|2 d and define L1 = − div A1 ∇. Then L1 and L are actually the same operator. To be precise, we have the following 1,2 Lemma 10.4.1. For any u ∈ W 1,2 (Rn+1 n+1 + ), v ∈ W0 (R+ ), ¨ ¨ A(x)∇u(x, t) · ∇v(x, t)dxdt = A1 (x)∇u(x, t) · ∇v(x, t)dxdt. (10.4.1) Rn+1 + Rn+1 + In particular, a weak solution to Lu = 0 in Rn+1 + is also a weak solution to L1 u = 0 in Rn+1 + . Proof. We first show (10.4.1) for u ∈ W 1,2 (Rn+1 2 n+1 + ) and v ∈ C0 (R+ ). We write ¨ A(x)∇u(x, t) · ∇v(x, t)dxdt Rn+1 ¨+ = A|| ∇x u · ∇x v + (b|1 + b|2 ) · ∇x v ∂t u + c · ∇x u ∂t v + d ∂t u ∂t v dxdt (10.4.2) 212 We have ¨ ¨ b|2 · ∇x v ∂t u dxdt = − ∂t (b|2 · ∇x v) u dxdt Rn+1 Rn+1 + ¨ ¨+ ¨ =− b|2 · ∇x (∂t v) u dxdt = − b|2 · ∇x (∂t v u)dxdt + b|2 · ∇x u ∂t v dxdt ¨ = b|2 · ∇x u ∂t v dxdt, where in the second equality we have used that b2 is t-independent and that v ∈ C 2 , and in the last equality we have used divx b2 = 0. Then (10.4.2) equals to ¨ A|| ∇x u · ∇x v + b|1 · ∇x v ∂t u + (c + b|2 ) · ∇x u ∂t v + d ∂t u ∂t v dxdt Rn+1 ¨+ = A1 (x)∇u(x, t) · ∇v(x, t)dxdt. Rn+1 + 1,2 Now since C02 (Rn+1 n+1 + ) is dense in W0 (R+ ), a limiting argument shows that (10.4.1) holds 1,2 for all u ∈ W 1,2 (Rn+1 n+1 + ), v ∈ W0 (R+ ). Define b − (ba )2Q " # A|| A0 = , (10.4.3) c − (ca )2Q d ffl . where (ba )2Q = 2Q b a, and let L0 = − div A0 ∇. Note that (ba )|2Q = − (ca )2Q by definition of ba and ca . Also, (ba )2Q is a constant vector so we of course have divx (ba )2Q = 0. So we can apply the lemma with b2 = (ba )2Q . Moreover, observe that b − (ba )2Q Aa|| ba − (ba )2Q " # " # " # A|| As|| bs A0 = = + , c − (ca )2Q d cs d ca − (ca )2Q 0 " # As|| bs where is the symmetric part of A0 , which is the same as the symmetric part of cs d Aa|| ba − (ba )2Q " # A, and is anti-symmetric, BMO, and having the same BMO ca − (ca )2Q 0 semi-norm as Aa . We summarize these observations in the following lemma. 213 Lemma 10.4.2. A weak solution to Lu = 0 in Rn+1 + is also a weak solution to L0 u = 0 in Rn+1 + .And the operator L0 has the same ellipticity constant and BMO semi-norm as L. Let u be a bounded weak solution to Lu = 0 in Rn+1 + with kukL∞ ≤ 1. Then u is also a bounded weak solution to L0 u = 0 in Rn+1 + . Define ¨ . Jη, = A0 ∇u · ∇u Ψ2 t dxdt (10.4.4) Rn+1 + Then by ellipticity of A0 and support property of Ψ, we have ˆ l(Q) ˆ Jη, ≥ λ0 |∇u(x, t)|2 t dxdt. 2 F The goal of this section is to prove the following estimate of Jη, . Lemma 10.4.3 (Main Lemma). Let σ, η ∈ (0, 1). Then there exists some finite constant c = c(λ0 , Λ0 , n) > 0, and some finite constant c˜ = c˜(σ, η, λ0 , Λ0 , n) > 0, such that Jη, ≤ (σ + cη)Jη, + c˜ |Q| . Once it is proved, we choose σ and η to be sufficiently small, so that Jη, ≤ 2˜ c |Q| . Now that η is fixed, and c˜ is independent of , we let  → 0 and thus obtain ˆ l(Q) ˆ |∇u(x, t)|2 t dxdt ≤ 2˜ c. 0 F We can assume that A is smooth (and thus A0 is smooth), as long as all bounds depend e Pt ϕ, on A only through its ellipticity constant and BMO semi-norm. In that case, ϕ, ϕ, e Pt∗ ϕ e and u are smooth by interior regularity of elliptic equations. 214 In the sequel, we shall simply write J for Jη, . We shall not specify a column vector and a row vector, namely, we shall not write transpose of a vector in equations, and its meaning should be clear from context. And we denote by c some constant depending only on λ0 , Λ0 and n, and denote by c˜ some constant depending additionally on σ and η. Since u is a weak solution to L0 u = 0 in Rn+1 + , ¨ A0 ∇u · ∇(uΨ2 t)dxdt = 0, Rn+1 + where we have chosen uΨ2 t to be the test function. Therefore, ¨ ¨ ¨ 2 2 J= A0 ∇u · ∇u Ψ t dxdt = − A0 ∇u · ∇(Ψ )ut dxdt − A0 ∇u · ∇t uΨ2 dxdt Rn+1 + . = J1 + J2 ¨ ¨ 2 J1 = − A|| ∇x u · ∇x (Ψ )ut dxdt − (b − (ba )2Q ) · ∇x (Ψ2 )∂t u ut dxdt ¨ ¨ − (c − (ca )2Q ) · ∇x u ∂t (Ψ2 )ut dxdt − d ∂t u ∂t (Ψ2 )ut dxdt . = J11 + J12 + J13 + J14 . For J11 , we claim that ¨   J11 = − A|| − (Aa|| )2Q ∇x u · ∇x (Ψ2 )ut dxdt. (10.4.5) This is because ¨ ¨ 1 (Aa|| )2Q ∇x u 2 · ∇x (Ψ )ut dxdt = (Aa|| )2Q ∇x (u2 ) · ∇x (Ψ2 t)dxdt, 2 and the last integral is 0 because (Aa|| )2Q is a constant anti-symmetric matrix, and Ψ2 t is 215 C 2 . Therefore, we write ¨ ¨   J11 = − As|| ∇x u 2 · ∇x (Ψ )ut dxdt − Aa|| − (Aa|| )2Q ∇x u · ∇x (Ψ2 )ut dxdt . = J111 + J112 . ¨ ¨ s 2 |J111 | = 2 A|| ∇x u Ψ u∇x Ψ t dxdt ≤ |∇x u| |Ψ| t1/2 |∇x Ψ| t1/2 dxdt Rn+1 λ 0 Rn+1 ¨ ¨ + + c ≤σ |∇u|2 Ψ2 t dxdt + |∇Ψ|2 t dxdt n+1 R+ σ n+1 R+ ≤ σJ + c˜ |Q| , where in the first inequality we have used As|| ≤ λ−1 0 and kukL∞ ≤ 1, and in the last L∞ step we have used Lemma 10.3.5. ˆ ˆ 4l(Q)   a a |J112 | = 2 A|| − (A|| )2Q ∇x u · ∇x Ψ(uΨt)dtdx 2Q 0 ˆ s0 1/s0 ˆ ˆ 4l(Q) !s !1/s a a |∇u| Ψt |∇Ψ| t1/2 dt 1/2 ≤2 A|| − (A|| )2Q dx dx 2Q 2Q 0 nˆ ˆ ˆ !s/2 !s/2 4l(Q) 4l(Q) o1/s 1/s0 a 2 2 2 ≤ cn |Q| A|| |∇u| Ψ t dt |∇Ψ| t dt dx , BM O 2Q 0 0 (10.4.6) 2 where s is any number between 1 and 2. Now we use H¨older inequality with p = s to bound (10.4.6) by  2−s ˆ ˆ 4l(Q) !1/2 ˆ ˆ 4l(Q) ! s 2−s 2s 1/s0 2 2 2 c |Q| |∇u| Ψ t dtdx  |∇Ψ| t dt dx , 2Q 0 2Q 0 which by Lemma 10.3.5 can be bounded by 1 2−s c˜J 1/2 |Q| s0 + 2s = c˜J 1/2 |Q|1/2 . 216 Then Young’s inequality gives |J112 | ≤ σJ + c˜ |Q| . Note that J12 and J13 can be estimated similar as (10.4.5). So both of then are bounded by σJ + c˜ |Q|. Since kdkL∞ ≤ λ−1 ˜ |Q| using Young’s 0 , J14 can be also bounded by σJ + c inequality and Lemma 10.3.5. For J2 , we compute ¨ J2 = − A0 ∇u · en+1 uΨ2 dxdt Rn+1 ¨ + ¨ a 2 =− (c − (c )2Q ) · ∇x u(uΨ )dxdt − d ∂t u(uΨ2 )dxdt Rn+1 + Rn+1 + . = J21 + J22 . For J22 , since d is t-independent, integration by parts gives ¨ ¨ 1 J22 = − d ∂t (u2 )Ψ2 dxdt = d u2 Ψ∂t Ψ dxdt. 2 Rn+1 + Rn+1 + Thus |J22 | ≤ c˜ |Q| again by Lemma 10.3.5. For J21 , we write ¨ ¨ u2 Ψ2   a J21 = − (c − (c )2Q ) · ∇x dxdt + (c − (ca )2Q ) · ∇x Ψ(u2 Ψ)dxdt Rn+1 + 2 Rn+1 + . = J211 + J212 . Write ¨ ¨ s 2 . J212 = c · ∇x Ψ(u Ψ)dxdt + (ca − (ca )2Q ) · ∇x Ψ(u2 Ψ)dxdt = J2121 + J2122 . Rn+1 + Rn+1 + 217 Then again by Lemma 10.3.5, |J2121 | ≤ c˜ |Q|. For J2122 , ˆ ˆ 4l(Q) ! |ca − (ca )2Q | ∇x Ψ(u2 Ψ) dt dx |J2122 | ≤ 2Q 0 ˆ 1/s0 ˆ ˆ 4l(Q) !s !1/s s0 |ca − (ca )2Q | dx |∇Ψ| u2 Ψ dt ≤ 2Q 2Q 0 ˆ ˆ 4l(Q) !s !1/s 1/s0 ≤ c |Q| |∇Ψ| dt dx ≤ c˜ |Q| . 2Q 0 For J211 , we use (10.2.1) to get ¨ u2 Ψ2   J211 = A∗|| ∇x ϕ · ∇x dxdt. Rn+1 + 2 ∗ ϕ in Section 10.3.2. We compute Recall we defined θηt = ϕ − Pηt ¨ ¨ u2 Ψ2 u2 Ψ2     J211 = A∗|| ∇x θηt · ∇x dxdt + A∗|| ∇x Pηt ∗ ϕ · ∇x dxdt Rn+1 + 2 Rn+1 + 2 ¨ ¨ u2 Ψ2 u2 Ψ2       = A∗|| − (A∗a || )2Q ∇x θηt · ∇x dxdt + A∗|| ∇x Pηt ∗ ϕ · ∇x dxdt Rn+1 + 2 Rn+1 + 2 . = J2111 + J2112 , where in the second equality we have used the assumption that the coefficients are smooth, which implies u2 is smooth, and thus (A∗a || )2Q being a constant anti-symmetric matrix gives ¨ u2 Ψ2   (A∗a || )2Q ∇x θηt · ∇x dxdt = 0. n+1 R+ 2 For J2112 , integration by parts with respect to t gives ¨   2 2  u Ψ J2112 = − ∂t A∗|| ∇x Pηt∗ ϕ · ∇x t dxdt n+1 R+ 2 ¨  2 2 ¨  2 2 ∗ ∗ u Ψ ∗ ∗ u Ψ =− A|| ∇x ∂t Pηt ϕ · ∇x t dxdt − A|| ∇x Pηt ϕ · ∇x ∂t t dxdt n+1 R+ 2 R+n+1 2 . = I1 + I2 . 218 By the same reasoning as for (10.4.5), we have ¨ u2 Ψ2     I1 = − A∗|| − (A∗a || )2Q ∗ ∇x ∂t Pηt ϕ · ∇x t dxdt Rn+1 + 2 ¨ u2 Ψ2   =− A∗s ∗ || ∇x ∂t Pηt ϕ · ∇x t dxdt Rn+1 + 2 ¨ u2 Ψ2     − A∗a || − (A∗a || )2Q ∗ ∇x ∂t Pηt ϕ · ∇x t dxdt Rn+1 + 2 . = I11 + I12 . Applying Proposition 9.1.3 to the operator L∗|| = − div A∗|| ∇, with p = 2, one has ¨ !1/2 ∗ 2 ∇x ∂t Pηt ϕ t dxdt ≤ c k∇ϕkL2 (Rn ) . Rn+1 + So by Cauchy-Schwartz and then by (10.2.3), we have ¨ !1/2 ¨ !1/2 ∗ 2 ∇x (u2 Ψ2 ) 2 t dxdt |I11 | ≤ c ∇x ∂t Pηt ϕ t dxdt Rn+1 + Rn+1 + ¨ ¨ !1/2 ≤ c |Q|1/2 |∇x u|2 Ψ2 t dxdt + |∇x Ψ|2 t dxdt . n+1 R+ Rn+1 + Then Lemma 10.3.5 and Young’s inequality give |I11 | ≤ c |Q|1/2 (J + c˜ |Q|)1/2 ≤ σJ + c˜ |Q| . For I12 , we use H¨ older inequality to get ˆ s0 1/s0 n ˆ ˆ ∞ s o 1 ∗a ∗a ∗ 2 2  1/s |I12 | ≤ A|| − (A|| )2Q dx ∇x ∂t Pηt ϕ ∇x u Ψ t dt dx 2 2Q Rn 0 ˆ ˆ ∞  s 2−s ! 2−s ˆ 2s ˆ ∞ 1/2 1/s0 ∗ 2 2 2 2 ≤ c |Q| ∇x ∂t Pηt ϕ t dt dx ∇x (u Ψ ) t dtdx . Rn 0 Rn 0 219 s 2+0 Letting 2−s = 2 , then by Proposition 9.1.3, (10.2.3) and Lemma 10.3.5, one gets 0 2−s |I12 | ≤ c |Q|1/s |Q| 2s (J + c˜ |Q|)1/2 ≤ σJ + c˜ |Q| . For I2 , by the definition of L∗|| , we can write I2 as ˆ ∞ˆ 1 I2 = − L∗|| Pηt ∗ ϕ ∂t (u2 Ψ2 )dx t dt 2 0 Rn ˆ ˆ ∞ 1/2 ˆ ∞ 1/2 ∗ ∗ 2 ∂t (u2 Ψ2 ) 2 t dt ⇒ |I2 | ≤ c L|| Pηt ϕ t dt dx Rn 0 0 ¨ !1/2 ¨ !1/2 ∗ ∗ 2 2 ∂t (u2 Ψ2 ) t dtdx ≤c L|| Pηt ϕ t dtdx . Rn+1 + Rn+1 + By Proposition 9.1.2, ¨ !1/2 ∗ ∗ 2 L|| Pηt ϕ t dtdx ≤ c k∇ϕkL2 (Rn ) . Rn+1 + So we have ¨ !1/2 2 2 2 |I2 | ≤ c k∇ϕkL2 (Rn ) ∂t (u Ψ ) t dtdx Rn+1 + ≤ c |Q|1/2 (J + c˜ |Q|)1/2 ≤ σJ + c˜ |Q| . We now return to J2111 . Write ¨   J2111 = A∗|| − (A∗a 2 || )2Q ∇x θηt · ∇x u(uΨ )dxdt Rn+1 + ¨ 1   + A∗|| − (A∗a || )2Q ∇x θηt · ∇x (Ψ2 )u2 dxdt 2 Rn+1 + . = II1 + II2 For II2 , we split it up into the integral involving A∗s ∗a || and the integral involving A|| − 220 (A∗a ∗a ∗a a || )2Q as before. We only treat the integral involving A|| − (A|| )2Q (denoted by II2 ) as the estimate for the former is similar and easier. By Cauchy-Schwarz and (10.3.14), we write ˆ 2 1/2 ˆ ˆ  !2 1/2 4l(Q) ∗a |II2a | ≤ c A|| − (A∗a || )2Q dx |∇x θηt | |∇x Ψ| dt dx  2Q 2Q 0 ˆ ˆ  !2 1/2 4l(Q) n dt ≤ c |Q|1/2  |∇x θηt | 1E1 dx 2Q 0 t ˆ ˆ ˆ ˆ  !2 1/2  !2 1/2 4l(Q) 4l(Q) dt dt o + |∇x θηt | 1E2 dx + |∇x θηt | 1E3 dx 2Q 0 l(Q) 2Q 0  .   = c |Q|1/2 (II21 a 1/2 a 1/2 ) + (II22 a 1/2 ) + (II23 ) (10.4.7) ´ 4l(Q) ´ 16δ(x) By Cauchy-Schwarz, and note that 0 1 dt E1 (x, t) t = η 8δ(x) dt t = ln 2, we have η ˆ ˆ 4l(Q) ! ˆ 4l(Q) ! dt dt a II21 ≤ |∇x θηt | 1E1 2 1E1 dx 2Q 0 t 0 t X X ˆ ˆ 2−k+1 dt ≤c |∇x θηt |2 1E1 dx, 0 η Q0 2−k t k Q ∈Dk where Dkη denotes the grid of dyadic cubes such that 1 −k 1 η2 ≤ l(Q0 ) < η2−k , Q0 ∈ Dkη . (10.4.8) 64 32 Consider for any fixed k and Q0 ∈ Dkη , for which Q0 × [2−k , 2−k+1 ] ∩ E 6= ∅. One can show that for such Q0 , there exists some x0 ∈ F such that 2Q0 ⊂ B(x0 , η2−k ). (10.4.9) 221 This implies that for any t ∈ [2−k , 2−k+1 ], 2 |∇x θηt |2 dx .n ∗ |∇ϕ(x)|2 dx ∇x Pηt ϕ(x) dx + Q0 B(x0 ,η2−k ) B(x0 ,η2−k ) 2 ∗ |∇ϕ(x)|2 dx .n ∇x Pηt ϕ(x) dx + B(x0 ,ηt) B(x0 ,ηt)  2   .n N ∗ e η (∇x Pηt ϕ) (x0 ) + M |∇x ϕ|2 (x0 ) . κ20 (10.4.10) by definition of the integrated non-tangential maximal function (9.2.2) and the definition of the set F . By (10.4.8) and the definition of E1 , one can show there exists some uniform constant C > 1 such that . n ηs o Q0 × [2−k , 2−k+1 ] ⊂ E e1 = (y, s) ∈ 2Q × (0, 4l(Q)) : ≤ δ(y) ≤ Cηs , C which implies ˆ ˆ 2−k+1 ds 1Ee1 (y, s) 0 Q . dy. (10.4.11) Q0 2−k s a Using (10.4.10) and (10.4.11), we estimate II21 X X ˆ 2−k+1 dt a II21 ≤c |∇x θηt |2 dx Q0 2−k Q0 t k Q0 ∈Dkη ˆ 2−k+1 ! ˆ ˆ −k+1 2 dt ds 1Ee1 (y, s) X X ≤ cκ20 dy 2−k t Q0 2−k s k Q0 ∈Dkη ¨ ˆ ˆ Cδ(y) dsdy η ds ≤c 1Ee1 (y, s) ≤c dy ≤ c |Q| . Rn+1 + s 2Q δ(y) s Cη 222 a , we compute For II22 ˆ ˆ 4l(Q) !2 ˆ 4l(Q) ˆ dt 1 a II22 = |∇x θηt | dx ≤ |∇x θηt |2 dxdt 2Q 2l(Q) l(Q) l(Q) 2l(Q) 2Q ˆ 4l(Q) ηn ∗ ∇x Pηt 2 .n ϕ(x) dx |Q| dt l(Q) 2l(Q) B(x0 ,2ηl(Q)) ˆ 4l(Q) 1 + |∇ϕ(x)|2 dx |Q| dt l(Q) 2l(Q) B(x0 ,2l(Q)) e η and the set F , we where x0 is any point in the set F . Therefore, by the definition of N have ˆ 4l(Q) ˆ 4l(Q) cn η n e η (∇x P ∗ ϕ)2 (x0 ) |Q| dt + cn   a II22 ≤ N ηt M |∇ϕ|2 (x0 ) |Q| dt l(Q) 2l(Q) l(Q) 2l(Q) ≤ c˜κ20 |Q| ≤ c˜ |Q| . a ≤ c By a similar argument, one can show II23 ˜ |Q| as well. Combining these results with (10.4.7), we have shown |II2a | ≤ c˜ |Q|, and thus |II2 | ≤ c˜ |Q|. We now deal with II1 . Write ¨   II1 = A∗|| − (A∗a 2 || 2Q ∇x (θηt uΨ ) · ∇x udxdt ) Rn+1 ¨+   − A∗|| − (A∗a 2 || 2Q ∇x u · ∇x u (θηt Ψ )dxdt ) Rn+1 ¨ +   − A∗|| − (A∗a 2 || 2Q ∇x (Ψ ) · ∇x u (uθηt )dxdt ) Rn+1 + . = II11 + II12 + II13 . We use Lemma 10.3.3 to bound II12 and II13 . We rewrite Lemma 10.3.3 in the following way |θηt (x)| . κ0 ηt for (x, t) ∈ supp Ψ. (10.4.12) 223 Note that by anti-symmetry, ¨ II12 = − A∗s 2 || ∇x u · ∇x u (θηt Ψ )dxdt. Rn+1 + ¨ ⇒ |II12 | ≤ cη |∇u|2 Ψ2 t dxdt ≤ cηJ. Rn+1 + For II13 , we have ¨ ∗ A|| − (A∗a ∇x (Ψ2 ) |∇x u| t dxdt, |II13 | ≤ cκ0 η ) || 2Q Rn+1 + which is bounded by σJ + c˜ |Q| by the same reasoning for the term J11 . For II11 , observe first that ¨ ¨ II11 = A∗|| ∇x (θηt uΨ2 ) · ∇x udxdt = A|| ∇x u · ∇x (θηt uΨ2 )dxdt. Rn+1 + Rn+1 + Taking θηt uΨ2 as test function (this is valid due to the smoothness assumption) in the equation L0 u = 0 in Rn+1 + , one gets ¨ 0= A0 ∇u · ∇(θηt uΨ2 )dxdt Rn+1 ¨ + ¨ 2 = A|| ∇x u · ∇x (θηt uΨ ) + (b − (ba )2Q ) · ∇x (θηt uΨ2 )∂t u Rn+1 Rn+1 ¨+ + ¨ + (c − (ca )2Q ) · ∇x u ∂t (θηt uΨ2 ) + d ∂t u ∂t (θηt uΨ2 ). Rn+1 + Rn+1 + ¨ ⇒ II11 = − (b − (ba )2Q ) · ∇x (θηt uΨ2 )∂t u Rn+1 ¨ + ¨ a 2 − (c − (c )2Q ) · ∇x u ∂t (θηt uΨ ) − d ∂t u ∂t (θηt uΨ2 ) Rn+1 + Rn+1 + . = II111 + II112 + II113 . 224 We treat II113 first. Write ¨ ¨ ¨ 2 2 II113 = − d ∂t u ∂t θηt (uΨ ) − d ∂t u ∂t u(θηt Ψ ) − d ∂t u ∂t (Ψ2 )θηt u Rn+1 + Rn+1 + Rn+1 + . = II1131 + II1132 + II1133 . ∗ ϕ. So II ˜ ∗ ϕ(uΨ2 ). We first use Cauchy- Note that ∂t θηt = −∂t Pηt 1131 = Rn+1 d ∂t u ∂t Pηt + Schwartz and then Proposition 9.1.2 to get ¨ !1/2 ¨ !1/2 ∗ 2 dt |∂t u|2 Ψ2 t dxdt |II1131 | ≤ c ∂t Pηt ϕ dx Rn+1 + Rn+1 + t ≤ cJ 1/2 k∇ϕkL2 (Rn ) ≤ σJ + c˜ |Q| . For II1132 , we use (10.4.12) to get ¨ |II1132 | ≤ cκ0 η |∇u|2 Ψ2 t dxdt ≤ cηJ. Rn+1 + By (10.4.12), Young’s inequality and Lemma 10.3.5, ¨ |II1133 | ≤ cκ0 η |∂t u| |∂t Ψ| Ψtdxdt ≤ σJ + c˜ |Q| . Rn+1 + We now treat II112 . Write ¨ II112 = − (c − (ca )2Q ) · ∇x u ∂t u (θηt Ψ2 )dxdt Rn+1 ¨ + ∗ + (c − (ca )2Q ) · ∇x u ∂t Pηt ϕ(uΨ2 )dxdt Rn+1 ¨+ −2 (c − (ca )2Q ) · ∇x u ∂t Ψ(θηt uΨ)dxdt Rn+1 + . = II1121 + II1122 + II1123 . For II1122 , we only focus on the anti-symmetric part, namely, the integral involving ca − 225 (ca )2Q (denoted by I1122 a ), for the integral involving cs is easier to estimate. We have ˆ 1/s0 ˆ ˆ 4l(Q) !s !1/s a s0 ∗ 2 |c − (ca )2Q | |II1122 |≤ |∇x u| ∂t Pηt ϕ Ψ dt dx 2Q 2Q 0 ˆ ˆ ∞ s/2 ˆ ∞ s/2 !1/s 1/s0 2 ∗ 2 dt ≤ c |Q| |∇u| Ψ t dt ∂t Pηt ϕ dx Rn 0 0 t ˆ ˆ ∞  s ! 2−s 2s ∗ 2 dt 2−s 1/s0 1/2 ≤ |Q| J ∂t Pηt ϕ dx . Rn 0 t s 2+0 2s Choosing s so that 2−s = 2 , then applying Proposition 9.1.2 with p = 2−s = 2 + 0 , and (10.2.3) to get 0 a |II1122 | ≤ cη |Q|1/s J 1/2 k∇ϕk 2s ≤ cηJ 1/2 |Q|1/2 ≤ σJ + c˜ |Q| . L 2−s (Rn ) Using the bound (10.4.12), II1123 can be estimated like II13 , and hence bounded by σJ + c˜ |Q|. For II1121 , we write ¨ ¨ s 2 II1121 = − c · ∇x u ∂t u θηt Ψ − (ca − (ca )2Q ) · ∇x u ∂t u θηt Ψ2 Rn+1 Rn+1 ¨ + ¨+ s 2 =− c · ∇x u ∂t u θηt Ψ + (ba − (ba )2Q ) · ∇x u ∂t u θηt Ψ2 . (10.4.13) Rn+1 + Rn+1 + The first term in (10.4.13) can be estimated as II1132 . We leave the second term aside for now. 226 We write II111 as follows ¨ II111 = − (b − (ba )2Q ) · ∇x θηt (uΨ2 ∂t u)dxdt Rn+1 ¨ + − (b − (ba )2Q ) · ∇x u (θηt Ψ2 ∂t u)dxdt Rn+1 ¨+ −2 (b − (ba )2Q ) · ∇x Ψ(θηt uΨ∂t u)dxdt Rn+1 + . = II1111 + II1112 + II1113 . |II1113 | can be estimated like II1123 , and hence bounded by σJ + c˜ |Q|. For II1112 , we write ¨ ¨ s 2 II1112 = − b · ∇x u (θηt Ψ ∂t u) − (ba − (ba )2Q ) · ∇x u (θηt Ψ2 ∂t u) (10.4.14) Rn+1 + Rn+1 + The first term can be estimated as the first term in (10.4.13). And the second term in (10.4.14) cancels the second term in (10.4.13). It remains to estimate II1111 . Integration by parts in t gives ¨ ¨ a 2 2 2II1111 = (b − (b )2Q ) · ∂t (∇x θηt )u Ψ + (b − (ba )2Q ) · ∇x θηt ∂t (Ψ2 )u2 Rn+1 Rn+1 ¨ + ¨ + a ∗ ∗ = (b − (b )2Q ) · ∇x (∂t Pηt ϕΨ2 u2 ) − (b − (ba )2Q ) · ∇x (Ψ2 u2 )∂t Pηt ϕ Rn+1 Rn+1 ¨+ + + (b − (ba )2Q ) · ∇x θηt ∂t (Ψ2 )u2 Rn+1 + . = III1 + III2 + III3 . III2 can be estimated as II1122 , and thus |III2 | ≤ σJ + c˜ |Q|. III3 can be estimated as II2 , and thus |III3 | ≤ c˜ |Q|. ∗ ϕ. It turns out For III1 , note that it is similar to J211 except that it has an extra ∂t Pηt that this term will do our favor. We proceed like J211 by recalling that divx (b − (ba )2Q ) = 227 divx A|| ∇x ϕ e = −L|| ϕ e (see (10.2.2)). So we have ¨ ∗ III1 = A|| ∇x ϕ e · ∇x (∂t Pηt ϕΨ2 u2 ). Rn+1 + e = θeηt + Pηt ϕ, Writing ϕ e we get ¨ ¨ ∗ ∗ ϕ(Ψ2 u2 ) + ϕ(Ψ2 u2 )   III1 = A|| ∇x θeηt · ∇x ∂t Pηt A|| ∇x Pηt ϕ e · ∇x ∂t Pηt Rn+1 Rn+1 ¨  +  + ¨ ∗ ∗ (Aa|| )2Q ϕ(Ψ2 u2 ) + ϕ(Ψ2 u2 )  = A|| − ∇x θeηt · ∇x ∂t Pηt L|| Pηt ϕ e ∂t Pηt Rn+1 + Rn+1 + . = III11 + III12 , where in the second equality we have used the smoothness assumption to obtain ¨ ∗ (Aa|| )2Q ∇x θeηt · ∇x ∂t Pηt ϕ(Ψ2 u2 ) = 0.  Rn+1 + For III12 , Cauchy-Schwartz gives ¨ !1/2 ¨ !1/2 ∗ 2 dxdt 2 |III12 | ≤ c t L|| Pηt ϕ e dxdt ∂t Pηt ϕ . Rn+1 + Rn+1 + t So by Proposition 9.1.2, |III12 | ≤ c˜ |Q|. For III11 , we write ¨   ∗ III11 = A|| − (Aa|| )2Q ∇x θeηt · ∇x (u2 )∂t Pηt ϕ Ψ2 Rn+1 ¨+   ∗ + A|| − (Aa|| )2Q ∇x θeηt · ∇x (Ψ2 )∂t Pηt ϕ u2 Rn+1 ¨ +   ∗ + A|| − (Aa|| )2Q ∇x θeηt · ∇x ∂t Pηt ϕ(Ψ2 u2 ) Rn+1 + . = III111 + III112 + III113 . ∗ ϕ (x) ≤ cκ η for any x ∈ F by the construction of F , ∂ P ∗ ϕ ≤ cκ η Since N η ∂t Pηt  0 t ηt 0 228 on the support of Ψ. Therefore, III112 can be estimated like the term II2 and thus |III112 | ≤ c˜ |Q|. For III113 , note that Proposition 9.1.4 implies ¨ ∗ dxdt 2 2 ∗ t L|| ∂t Pηt ϕ ≤ cη −2 |Q| . (10.4.15) Rn+1 + t We write ¨ III113 = ∇x (θeηt u2 Ψ2 ) · A∗|| ∇x ∂t Pηt ∗ ϕ Rn+1 ¨+   − θeηt ∇x (u2 Ψ2 ) · A∗|| − (Aa∗ || )2Q ∗ ∇x ∂t Pηt ϕ Rn+1 ¨ + ¨   = θeηt u2 Ψ2 L∗|| ∇x ∂t Pηt ∗ ϕ− θeηt ∇x (u2 Ψ2 ) · A∗|| − (Aa∗ || ) 2Q ∗ ∇x ∂t Pηt ϕ Rn+1 + Rn+1 + . = III1131 + III1132 . By Cauchy-Schwartz, Lemma 10.3.4 and (10.4.15), ¨ 2 dxdt !1/2 ¨ !1/2 ∗ dxdt 2 2 ∗ |III1131 | ≤ c θηt t L|| ∂t Pηt ϕ e n+1 R+ t3 Rn+1 + t ≤ c |Q| . By (10.4.12), |III1132 | is bounded by ¨ ∗ ∗ A|| − (Aa∗ ∇x (u2 Ψ2 ) ∇x ∂t Pηt cκ0 η ) ϕ t dxdt, || 2Q Rn+1 + which is bounded by σJ + c˜ |Q| using the same method of estimating I12 . Now it remains to estimate III111 . Note that the integral is on the support of Ψ instead of support of ∇Ψ, so we cannot use the same method as estimating II2 . Like before, we only deal with the term involving Aa|| − (Aa|| )2Q , as the term with the symmetric matrix As|| is easier to estimate. 229 We have ¨   a ∗ |III11 |= Aa|| − (Aa|| )2Q ∇x θeηt · ∇x (u2 )∂t Pηt ϕ Ψ2 n+1 R+ ˆ s0 1/s0 ˆ ˆ 4l(Q) !s !1/s a a ∗ ∇x θηt |∇x u| Ψ2 ∂t Pηt ≤ A|| − (A|| )2Q dx ϕ dt dx e 2Q 2Q 0 ˆ ˆ ∞ ˆ ∞  s !1/s e 2 ∗ 2 2 dt 2−s 1/s0 2 2 ≤ c |Q| σ |∇u| Ψ t dt + cσ ∇x θηt ∂t Pηt ϕ Ψ dx 2Q 0 0 t ˆ ˆ ∞  s !1/s e 2 dt 2−s ∇x θηt ∂t Pηt ϕ 1supp Ψ 0 ∗ 2 ≤ c |Q|1/s σJ + cσ dx (10.4.16) Rn 0 t We write ˆ ˆ ∞  s e 2 dt 2−s ∇x θηt ∂t Pηt ϕ 1supp Ψ ∗ 2 dx Rn 0 t ¨ s e 2 dxdt 2−s ∇x θηt ∂t Pηt ϕ ξ(x)1supp Ψ ∗ 2 = sup . t ξ∈S (Rn ) Rn+1 + s kξk 2s−2 ≤1 As before, let Dkη be the grid of dyadic cubes such that (10.4.8) holds. Then ¨ e 2 dxdt ϕ ξ(x)1supp Ψ ∗ 2 ∇x θηt ∂t Pηt n+1 R+ t X X ˆ ˆ 2−k+1 2 dtdx ϕ ξ(x)1supp Ψ ∗ 2 = x ηt ∂t Pηt ∇ θ . (10.4.17) e 0 η Q 0 2−k t k Q ∈Dk We observe that we have the following estimates: ˆ ˆ 2−k+1 ∗ ∂t Pηt 2 η ∗ 2 dt sup f (x) ≤ C(n, λ0 , Λ0 ) 0 ∂t Pηt f (x) dx (10.4.18) Q0 ×(2−k ,2−k+1 ] |Q | 3 0 Q 2−k− 2 1 t 2 for all f ∈ L2 (Rn ). This can be obtained using Proposition 3.3.2. 230 By (10.4.18), we bound (10.4.17) by ˆ ˆ 2−k+1 !ˆ 2−k+1 ˆ ∗ 2 dydt 1 e 2 dtdx ∇x θηt |ξ(x)| 1supp Ψ X X cη ∂t Pηt ϕ 2Q0 2−k−1 t 2−k |Q0 | Q0 t k Q0 ∈Dkη (10.4.19) ´ 2−k+1 ´ e 2 x ηt |ξ(x)| 1supp Ψ t . 1 dtdx We now estimate 2−k |Q0 | Q0 ∇ θ Note that as a consequence of Proposition 3.3.3, for any t ∈ (2−k , 2−k+1 ),  1/p  1/2  1/p p 2 −1 p |∇x Pηt ϕ(x)| e dx ≤C |∇x Pηt ϕ(x)| e dx +η |∂t Pηt ϕ(x)| e dx Q0 2Q0 2Q0 (10.4.20) for all p ∈ [2, 2 + ), where C = C(n, λ0 , Λ0 ) and  = (n, λ0 , Λ0 ) are positive constants. Let r = 1 +  with  > 0 sufficiently small. We use H¨older’s inequality, then definition of θeηt , and (10.4.20) as well as the reverse H¨older estimates for ∇ϕ, e to get ˆ 2−k+1 ˆ 1 e 2 dtdx 0 ∇ x θ ηt |ξ(x)| 1supp Ψ |Q | Q0 t 2−k ˆ 2−k+1  2r0 1/r0  1/r dt ≤ ∇x θηt dx e r |ξ(x)| dx 1supp Ψ 2−k Q0 Q0 t  1/r ≤ |ξ(x)|r dx · Q0 ˆ 2−k+1 n 1/r0  1/r0 o dt |∇x Pηt ϕ| e 2r0 dx + |∇ϕ| e 2r0 dx 1supp Ψ 2−k Q0 Q0 t  1/r ˆ 2−k+1 n  1/r0 r 2 −2 2r0 ≤c |ξ(x)| dx |∇x Pηt ϕ| e dx + η |∂t Pηt ϕ| e Q0 2−k 2Q0 2Q0 o dt + e2 |∇ϕ| 1supp Ψ 2Q0 t  1/r ˆ 2−k+1 n ≤c |ξ(x)|r dx e 2 dx |∇x Pηt ϕ| Q0 2−k B(x0 ,η2−k ) !1/r0 o dt +η −2 |∂t Pηt ϕ| e 2r0 dx + e 2 dx |∇ϕ| 1supp Ψ , B(x0 ,η2−k ) B(x0 ,η2−k ) t where in the last inequality we have used (10.4.9), with x0 ∈ F . Therefore, we can bound 231 this by  1/r ˆ 2−k+1 r   dt c |ξ(x)| dx N e 0 )2 + η −2 N η (∂t Pηt ϕ)(x e η (∇x Pηt ϕ)(x e 2 (x0 ) e 0 )2 + M |∇ϕ| Q0 2−k t  1/r ≤ cκ20 |ξ(x)|r dx . Q0 So (10.4.19) is bounded by ˆ ˆ 2−k+1 ! 1/r ∗ 2 dydt X X r cη ∂t Pηt ϕ |ξ(x)| dx 2Q0 2−k−1 t Q0 k Q0 ∈Dkη X X ˆ ˆ 2−k+1 2 dt (M (|ξ|r ))1/r (y) ∗ ≤ cη ∂t Pηt ϕ(y) dy 2Q0 2−k−1 t k Q0 ∈Dkη ˆ ˆ ∞ r ∗1/r 2 dt ≤ cη M (|ξ| )(y) ∂t Pηt ϕ(y) dy Rn 0 t ˆ 1/q ˆ ˆ ∞ q0 !1/q0 2 dt M (|ξ|r )(y)q/r dy ∗ ≤ cη ∂t Pηt ϕ(y) Rn Rn 0 t s Choosing q = 2s−2 , the above is bounded by ˆ  2s−2 ˆ ˆ ∞  s ! 2−s s s s ∗ 2 dt 2−s cη |ξ| 2s−2 ∂t Pηt ϕ(y) dy Rn Rn 0 t 2−s ≤ c˜ kξk s |Q| s , L 2s−2 where in the last step we have used Proposition 9.1.2. Combining these estimates with (10.4.16), we obtain 0 σ a |III111 | ≤ c |Q|1/s (σJ + c˜ |Q|)1/s ≤ J + c˜ |Q| . 2 This finishes the proof of Lemma 10.4.3. BIBLIOGRAPHY [AEN16] Pascal Auscher, Moritz Egert, and Kaj Nystr¨om, The Dirichlet problem for ´ second order parabolic operators in divergence form, Journal de l’Ecole poly- techniqueMath´ematiques (2016). [AHL+ 02] Pascal Auscher, Steve Hofmann, Michael Lacey, Alan McIntosh, and Philippe Tchamitchian, The solution of the Kato square root problem for second order elliptic operators on Rn , Annals of mathematics 156 (2002), no. 2, 633–654. [AHLT01] Pascal Auscher, Steve Hofmann, John L Lewis, and Philippe Tchamitchian, Extrapolation of Carleson measures and the analyticity of Kato’s square-root operators, Acta mathematica 187 (2001), no. 2, 161–190. [Aus96] Pascal Auscher, Regularity theorems and heat kernel for elliptic operators, Jour- nal of the London Mathematical Society 54 (1996), no. 2, 284–296. [Aus07] , On necessary and sufficient conditions for Lp -estimates of riesz trans- forms associated to elliptic operators on Rn and related estimates, American Mathematical Soc., 2007. [Cal76] AP Calder´ on, Inequalities for the maximal function relative to a metric, Studia Mathematica 57 (1976), no. 3, 297–306. [CF74] Ronald Coifman and Charles Fefferman, Weighted norm inequalities for max- imal functions and singular integrals, Studia Mathematica 51 (1974), no. 3, 241–250. [CFK81] Luis A Caffarelli, Eugene B Fabes, and Carlos E Kenig, Completely singu- lar elliptic-harmonic measures, Indiana University Mathematics Journal 30 (1981), no. 6, 917–924. [CFMS81] Lo Caffarelli, E Fabes, S Mortola, and So Salsa, Boundary behavior of non- negative solutions of elliptic operators in divergence form, Indiana University Mathematics Journal 30 (1981), no. 4, 621–640. [CLMS93] R Coifman, PL Lions, Y Meyer, and S Semmes, Compensated compactness and Hardy spaces, Journal de math´ematiques pures et appliqu´ees 72 (1993), no. 3, 247–286. [CMS85] RR Coifman, Yves Meyer, and EM Stein, Some new function spaces and their applications to harmonic analysis, Journal of functional Analysis 62 (1985), no. 2, 304–335. 232 233 [DJK84] Bj¨ orn EJ Dahlberg, David S Jerison, and Carlos E Kenig, Area integral esti- mates for elliptic differential operators with non-smooth coefficients, Arkiv f¨ or Matematik 22 (1984), no. 1, 97–108. [DK18] Hongjie Dong and Seick Kim, Fundamental solutions for second-order parabolic systems with drift terms, Proceedings of the American Mathematical Society 146 (2018), no. 7, 3019–3029. [EH18] Luis Escauriaza and Steve Hofmann, Kato square root problem with unbounded leading coefficients, Proceedings of the American Mathematical Society (2018). [Eva98] Lawrence C. Evans, Partial differential equations, vol. 19, American Mathe- matical Society, 1998. [Fef71] Charles Fefferman, Characterizations of bounded mean oscillation, Bulletin of the American Mathematical Society 77 (1971), no. 4, 587–588. [Gia83] Mariano Giaquinta, Multiple integrals in the calculus of variations and nonlin- ear elliptic systems.(am-105), vol. 105, Princeton University Press, 1983. [GT01] David Gilbarg and Neil S Trudinger, Elliptic partial differential equations of second order, springer, 2001. [GW82] Michael Gr¨ uter and Kjell-Ove Widman, The green function for uniformly el- liptic equations, Manuscripta Mathematica 37 (1982), no. 3, 303–342. [HKMP15] Steve Hofmann, Carlos Kenig, Svitlana Mayboroda, and Jill Pipher, Square function/non-tangential maximal function estimates and the Dirichlet problem for non-symmetric elliptic operators, Journal of the American Mathematical Society 28 (2015), no. 2, 483–529. [HLM02] Steve Hofmann, Michael Lacey, and Alan McIntosh, The solution of the Kato problem for divergence form elliptic operators with gaussian heat kernel bounds, Annals of Mathematics 156 (2002), 623–631. [HW68] Richard A Hunt and Richard L Wheeden, On the boundary values of harmonic functions, Transactions of the American Mathematical Society 132 (1968), no. 2, 307–322. [JK81] David S Jerison and Carlos E Kenig, The Dirichlet problem in non-smooth domains, Annals of mathematics 113 (1981), no. 2, 367–382. [JK82] , Boundary behavior of harmonic functions in non-tangentially accessi- ble domains, Advances in Mathematics 46 (1982), no. 1, 80–147. [JN61] Fritz John and Louis Nirenberg, On functions of bounded mean oscillation, Communications on pure and applied Mathematics 14 (1961), no. 3, 415–426. [Jon80] Peter W Jones, Extension theorems for BMO, Indiana Univ. Math. J 29 (1980), no. 1, 41–66. [Jon81] , Quasiconformal mappings and extendability of functions in Sobolev spaces, Acta Mathematica 147 (1981), no. 1, 71–88. [Kat76] Tosio Kato, Perturbation theory for linear operators, second ed., vol. 132, Springer, 1976. 234 [Ken94] Carlos E Kenig, Harmonic analysis techniques for second order elliptic bound- ary value problems, vol. 83, American Mathematical Soc., 1994. [KKPT00] Carlos Kenig, Herbert Koch, Jill Pipher, and Tatiana Toro, A new approach to absolute continuity of elliptic measure, with applications to non-symmetric equations, Advances in Mathematics 153 (2000), no. 2, 231–298. [KKPT16] C Kenig, B Kirchheim, J Pipher, and T Toro, Square functions and the A∞ property of elliptic measures, The Journal of Geometric Analysis 26 (2016), no. 3, 2383–2410. [KP93] Carlos E Kenig and Jill Pipher, The Neumann problem for elliptic equations with non-smooth coefficients, Inventiones mathematicae 113 (1993), no. 1, 447– 509. [KS00] David Kinderlehrer and Guido Stampacchia, An introduction to variational inequalities and their applications, vol. 31, SIAM, 2000. [KT01] Herbert Koch and Daniel Tataru, Well-posedness for the Navier-Stokes equa- tions, Advances in Mathematics 157 (2001), no. 1, 22–35. [Lie96] Gary M Lieberman, Second order parabolic differential equations, World scien- tific, 1996. [LP19] Linhan Li and Jill Pipher, Boundary behavior of solutions of elliptic opera- tors in divergence form with a BMO anti-symmetric part, Communications in Partial Differential Equations (2019), 1–49. [LSW63] Walter Littman, Guido Stampacchia, and Hans F Weinberger, Regular points for elliptic equations with discontinuous coefficients, Annali della Scuola Nor- male Superiore di Pisa-Classe di Scienze 17 (1963), no. 1-2, 43–77. [LZ04] Vitali Liskevich and Qi S Zhang, Extra regularity for parabolic equations with drift terms, manuscripta mathematica 113 (2004), no. 2, 191–209. [Mik98] Milan Miklavˇciˇc, Applied functional analysis and partial differential equations, World Scientific, 1998. [Muc72] Benjamin Muckenhoupt, Weighted norm inequalities for the hardy maximal function, Transactions of the American Mathematical Society (1972), 207–226. [Muc74] , The equivalence of two conditions for weight functions, Studia Math 49 (1974), no. 1, 1–1. [MY90] Alan McIntosh and Atsushi Yagi, Operators of type ω without a bounded H∞ functional calculus, Miniconference on operators in analysis, Centre for Math- ematics and its Applications, Mathematical Sciences Institute , 1990, pp. 159– 179. [MyV06] Vladimir G Maz ya and Igor E Verbitsky, Form boundedness of the general second-order differential operator, Communications on Pure and Applied Math- ematics: A Journal Issued by the Courant Institute of Mathematical Sciences 59 (2006), no. 9, 1286–1329. [Osa87] Hirofumi Osada, Diffusion processes with generators of generalized divergence form, Journal of Mathematics of Kyoto University 27 (1987), no. 4, 597–619. 235 [Paz83] Amnon Pazy, Semigroups of linear operators and applications to partial differ- ential equations, vol. 44, Springer Applied Mathematical Sciences, 1983. [QX18] Zhongmin Qian and Guangyu Xi, Parabolic equations with singular divergence- free drift vector fields, Journal of the London Mathematical Society (2018). ˇ [SSSZ12] Gregory Seregin, Luis Silvestre, Vladim´ır Sver´ˇ ak, and Andrej Zlatoˇs, On divergence-free drifts, Journal of Differential Equations 252 (2012), no. 1, 505– 540. [Ste70] Elias M Stein, Singular integrals and differentiability properties of functions, vol. 30, Princeton university press, 1970. [Zha04] Qi S Zhang, A strong regularity result for parabolic equations, Communications in mathematical physics 244 (2004), no. 2, 245–260.