Calc. Var. (2023) 62:113 https://doi.org/10.1007/s00526-022-02426-x Calculus of Variations Size of the zero set of solutions of elliptic PDEs near the boundary of Lipschitz domains with small Lipschitz constant Josep M. Gallegos1 Received: 31 March 2022 / Accepted: 29 December 2022 / Published online: 17 March 2023 © The Author(s) 2023 Abstract Let  ⊂ Rd be a C 1 domain or, more generally, a Lipschitz domain with small Lipschitz constant and A(x) be a d ×d uniformly elliptic, symmetric matrix with Lipschitz coefficients. Assume u is harmonic in , or with greater generality u solves div(A(x)∇u) = 0 in , and u vanishes on  = ∂ ∩ B for some ball B. We study the dimension of the singular set of u in , in particular we show that  there is a countable family of open balls (Bi )i such that u| Bi ∩ does not change sign and K \ i Bi has Minkowski dimension smaller than d − 1 −  for any compact K ⊂ . We also find upper bounds for the (d − 1)-dimensional Hausdorff measure of the zero set of u in balls intersecting  in terms of the frequency. As a consequence, we prove a new unique continuation principle at the boundary for this class of functions and show that the order of vanishing at all points of  is bounded except for a set of Hausdorff dimension at most d − 1 − . Mathematics Subject Classification 31B05 · 31B20 · 35J25 1 Introduction In this paper, we study the size of the zero set of solutions u of a certain class of elliptic PDEs (see Sect. 2.1) near the boundary of a Lipschitz domain. Assume  ⊂ Rd is a Lipschitz domain with small Lipschitz constant and  is an open set of the boundary ∂ where u vanishes. We investigate the dimension of the set S  (u) = {x ∈  | u −1 ({0})∩ B(x, r )∩  = ∅, ∀r > 0}, the set where u changes sign in every neighborhood. Communicated by X. Ros-Oton. Supported by the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme (Grant Agreement 101018680), by MICINN (Spain) under the Grant PID2020-114167GB-I00, and by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) under Germany’s Excellence Strategy - EXC-2047/1 - 390685813. B Josep M. Gallegos jgallegos@mat.uab.cat 1 Departament de Matemàtiques, Edifici C Facultat de Ciències, Universitat Autònoma de Barcelona, 08193 Bellaterra, Spain 123 113 Page 2 of 52 J. M. Gallegos In a more regular setting, for example in the case  is a C 1,Dini domain (see [2, 9, 19, 21, 22] for the definition) and u is harmonic, S  (u) coincides with the usual singular set at the boundary of u: S (u) = {x ∈  | |∇u(x)| = 0} (see Proposition 1.9). Note that all C 1,α domains (for any α > 0) are C 1,Dini and all C 1,Dini domains are C 1 , but the converse is not true. Nonetheless, in the case where  is a Lipschitz domain, ∇u(x) (or ∂ν u) only exists in  in a weaker sense (see “Appendix A”) as far as we know, which anticipates that we will not be able to find fine estimates of the size and dimension of S (u) (see Sect. 9). The situation  (u), for which we prove the following Minkowski dimension estimate: is different for S Theorem 1.1 Let  ⊂ Rd be a Lipschitz domain, and let A(x) be a uniformly elliptic symmetric matrix with Lipschitz coefficients defined on . Let B be a ball centered in ∂ and suppose that  = B ∩ ∂ is a Lipschitz graph with slope τ < τ0 , where τ0 is some positive constant depending only on d and the ellipticity of A(x). Let u  ≡ 0 be a solution of div(A(x)∇u(x)) = 0 in , continuous in  that vanishes in . Then there exists some small constant 1 (d) > 0 and a family of open balls (Bi )i , i ∈ N centered on  such that ∩ is either strictly positive or negative, for all i ∈ N, (1) u| Bi (2) K \ i Bi has Minkowski dimension at most d − 1 − 1 for any compact K ⊂ .  Moreover, in the planar setting (d = 2), the set K \ i Bi is finite for any compact K ⊂ . Recall that the upper Minkowski dimension of a set E ⊂ Rd−1 can be defined as log(#{dyadic cubes Q of side length 2− j that satisfy Q ∩ E  = ∅}) dimM E = lim sup . j→∞ j log 2 (1.1) The previous result gives the following corollary: Corollary 1.2 Assume  ⊂ Rd , , A(x), τ < τ0 , and u  ≡ 0 as in the statement of Theo- rem 1.1. Then, there exists a constant 1 (d) > 0 such that    dimM S (u) ∩ K ≤ d − 1 − 1 for any compact set K ⊂ . Remark 1.3 We remark that Theorem 1.1 and Corollary 1.2  (u) has not been well studied before (as • are new even in the harmonic case as the set S far as I know), • are valid for harmonic functions in Riemannian manifolds with Lipschitz boundary (with small Lipschitz constant depending on the metric), • include the case when  is a C 1 domain, situation where not too much is known either, • give Hausdorff dimension estimates for the set S  (u) by taking an exhaustion of  by compact sets. Note that Hausdorff dimension estimates are weaker than Minkowski dimension estimates but these were not known either. Remark 1.4 Some of the results of the present paper suggest that the set S  (u) might be a natural substitute of the usual singular set S (u) in the case of Lipschitz domains (and rougher). These results are the fact that S (u) = S  (u) in the case  is a C 1,Dini domain (Proposition 1.9), the existence of an example of a Lipschitz domain where dimH S (u) = d − 1 and no better (see Sect. 9), and Corollary 1.2 showing that better dimension estimates  (u). are true for S 123 Size of the zero set of solutions of elliptic PDEs near the… Page 3 of 52 113 Moreover, we show an upper bound estimate on the size of the zero set of u on balls centered at  in terms of the frequency function N (x, r ) (see Definition 3.3): Theorem 1.5 Assume , , A(x), τ < τ0 , and u  ≡ 0 as in the statement of Theorem 1.1. Let x ∈  and 0 < r < r0 with r0 depending on dist(x, ∂\), the Lipschitz constant L A and the ellipticity A of A(x), and d. There exists x̃ ∈  such that |x − x̃| ≈ A dist(x̃, ∂) ≈ A r and Hd−1 ({u = 0} ∩ B(x, r ) ∩ ) ≤ Cr d−1 (N (x̃, Sr ) + 1)α for some large S depending on L A , A and d, and some α ≥ 1 depending on d. The precise x̃ appearing in the statement of Theorem 1.5 is the center of a certain dyadic cube related to a Whitney cube decomposition of  but we have freedom in choosing it. For more details see Sect. 4.1 and Remark 6.5. This result is analogous to Theorem 2 in [28] for a more general class of functions but with a worse exponent. Further, we briefly discuss the application of this theorem to the study of the zero set of Dirichlet eigenfunctions of the operator div(A∇·) in  in Sect. 6.6. See [28] for more background on this result and its applications in the harmonic case. Let us give some historical background for the results of this paper. L. Bers asked the following question. Consider a harmonic function u in the upper half-plane Rd+ , C 1 up to the boundary such that there exists E ∈ ∂Rd+ = Rd−1 where u = |∇u| = 0 on E. Does measured−1 (E) > 0 imply u ≡ 0? This question has positive answer in the plane, thanks to the subharmonicity of log |∇u|. But in Rd+ , d ≥ 3, there are examples constructed by Bourgain and Wolff [6] which give a negative response in general. A related conjecture by Lin [23] which is still open is the following: Conjecture Let  ⊂ Rd be a Lipschitz domain and  = B ∩ ∂ for some ball B centered in ∂. Let u be a harmonic function in  and continuous up to the boundary that vanishes in . If the set where ∂ν u = 0 in  has positive surface measure, then u must be identically zero. This conjecture was proved in the C 1,1 case in [23], where it was also shown that S (u) is a (d − 2)-dimensional set (see also [7] for more quantitative estimates). V. Adolfsson, L. Escauriaza and C. Kenig also gave a positive answer to the conjecture in [3] in the case  is a convex Lipschitz domain. Their work was followed by Kukavica and Nyström [19], and Adolfsson and Escauriaza [2] where the conjecture is solved in the case  is C 1,Dini . Moreover, in [2] it is also proved that S (u) has Hausdorff dimension at most d − 2. In the interior of the domain, the singular and critical sets of u have been extensively studied too. In particular, the recent results from Naber and Valtorta [30] are remarkable: they find (d −2)-dimensional Minkowski content bounds and prove they are (d −2)-rectifiable with the introduction of a new quantitative stratification. Some of their techniques have been adapted to study the Minkowski content of the singular set at the boundary in the convex Lipschitz case by McCurdy [29] and in the C 1,Dini by Kenig and Zhao [21]. Unfortunately, the methods relying on pointwise monotonicity that have been commonly used to study this problem are no longer useful in general Lipschitz domains, and new ideas are needed in this case. The conjecture of F.-H. Lin saw no further progress until X. Tolsa proved the result for harmonic functions in Lipschitz domains with small Lipschitz constants in [33]. His proof uses the new powerful methods developed by A. Logunov and E. Malinnikova (see [24, 25, 27]) to study zero sets of Laplace eigenfunctions in compact Riemannian manifolds. These techniques are used on the boundary of the domain  with the trade-off of restricting its Lipschitz constant. This idea from [33] has also been successfully used by Logunov et al. [28] to study the zero 123 113 Page 4 of 52 J. M. Gallegos set of Dirichlet Laplace eigenfunctions in Lipschitz domains with small Lipschitz constant. In [28], the authors also develop novel methods to control the zero set near the boundary that will be relevant in the present paper. The main tool used in this paper is Almgren’s frequency function (see Definition 3.3), a quantity that controls the doubling properties of the L 2 averages of u in spheres. This function was also used in most of the works mentioned in this introduction. Our proof of Theorem 1.1 requires two technical lemmas. One is Lemma 4.1, an adaptation of Lemma 3.1 - Key lemma from [33] and it controls the behavior of the frequency function at points in  near the boundary (see also Lemma 7 from [28]). The other is Lemma 5.1, inspired by Lemma 8 - Second hyperplane lemma: cubes without zeros from [28] and it controls the size of balls centered at the boundary that contain no zeros of u. Both results were originally proved only for harmonic functions, hence we first extend them for solutions of second order linear elliptic PDEs in divergence form. Afterwards, we combine both lemmas in a new combinatorial argument that controls the size of the zero set of u near the boundary. We remark that the extension of Lemma 8 from [28] to the elliptic case presents many difficulties and it might be interesting on its own. Theorem 1.1 allows us to prove an analogous unique continuation at the boundary result to the one in [33] for more general elliptic PDEs in divergence form: Corollary 1.6 Assume , , A(x), τ < τ0 , and u  ≡ 0 as in the statement of Theorem 1.1. The set {x ∈  | ∂ν u(x) = 0} has (d − 1)-Hausdorff measure 0. Observe that the assumption that u vanishes continuously in  implies that ∇u exists σ -a.e. as a non-tangential limit in , and ∇u = (∂ν u)ν ∈ L 2loc (σ ). Here σ stands for the surface measure restricted to  and ν is the outer unit normal (see Remark 3.7 and “Appendix A”). The proof of Corollary 1.6 uses that the elliptic measure ω A associated to the elliptic operator div(A∇·) is an A∞ Muckenhoupt weight with respect to σ (see Definiton 7.1) and that, since |u| is locally comparable to a Green function near most points of  (thanks to not changing sign in a neighborhood), ∂ν u is comparable to dω A /dσ . Theorem 1.1 also has a second corollary that controls the vanishing order of the zeros in the set where u does not change sign nearby. Definition 1.7 The vanishing order of the zero at a point x ∈  is defined as the supremum of the α > 0 such that there exist Cα > 0 finite and r0 > 0 satisfying ˆ − |u|dy ≤ Cα r α , 0 < r ≤ r0 . B(x,r )∩ Corollary 1.8 Assume , , A(x), τ < τ0 , and u  ≡ 0 as in the statement of Theorem 1.1. There exists some small constant 2 > 0 depending on d, the Lipschitz constant τ of , and the ellipticity A of the matrix A(x) such that for all x ∈  outside a set of Hausdorff dimension d − 1 − 1 , the vanishing order of u at x is smaller than 1 + 2 . Moreover, for all x ∈ , the vanishing order of u at x is greater than 1 − 2 . This corollary is proved by comparing u locally (in the neighborhoods where it does not change sign) with the Green function of a certain cone with angular opening related to the Lipschitz constant τ of . In Sect. 9 we provide an example showing that Corollary 1.6 cannot be improved in the sense of Hausdorff dimension estimates. This contrasts with the higher regularity case (C 1,Dini ) where the set S (u) is known to be (d − 2)-rectifiable and, a fortiori, has Hausdorff dimension at most d − 2 (see [21]). Finally, in Sect. 10, we prove the following proposition relating S (u) and S  (u) in the smooth case. 123 Size of the zero set of solutions of elliptic PDEs near the… Page 5 of 52 113 Proposition 1.9 Let  ⊂ Rd be a C 1,Dini domain, B be a ball centered in ∂, and  = B ∩ ∂. Let u be a harmonic function defined in , continuous in  that vanishes in . Then  (u). S (u) coincides with S The proof of the proposition follows from a local expansion of u as the sum of a homogeneous harmonic polynomial and an error term of higher degree proved in [22]. 1.1 Further questions We present some open questions related to the previous results: • Is it true that dimM S  (u) ∩ K is at most d − 2 for any compact K ⊂  ( ≡ 1)? What 1 about the (slightly) easier case where  is a C 1 domain? • Is the set S (u) (d − 2)-rectifiable? • Do these estimates hold for general Lipschitz domains? For the moment, even the con- jecture of Lin is still open. • Can there exist points with vanishing order ∞ in  in the Lipschitz case? Thanks to the results of [2, 19, 21, 22] we know this cannot happen in the C 1,Dini case. 1.2 Outline of the paper In Sects. 2 and 3, we present some notation, tools and ideas that will be used throughout the paper (often without reference). The main aim of Sect. 4 is the proof of Lemma 4.1, although in Sect. 4.1 we construct a Whitney cube structure to  that will be used during the sections following after. Section 5 is devoted to Lemma 5.1. Both lemmas are then combined in a combinatorial argument in Sect. 6 to prove Theorems 1.1 and 1.5. In Sect. 6, we also briefly discuss the application of Theorem 1.5 to the study of the zero set of certain class of eigenfunctions. The rest of the paper is spent on the proofs of Corollaries 1.6, 1.8, the example of a Lipschitz domain and harmonic function u with “large" S (u), and the equality S (u) = S  (u) in the C 1,Dini case. This last part does not require the Whitney cube structure or Lemmas 4.1 and 5.1. In “Appendix A”, we discuss the existence of the non-tangential limit of ∇u in . 2 Lipschitz domains with small constant and some properties of elliptic PDEs Notation: the letters C, c, c , c̃ are used to denote positive constants that depend on the dimension d and whose values may change on different proofs. The constants c H and C N retain their values. The notation A  B is equivalent to A ≤ C B, and A ∼ B is equivalent to A  B  A. Sometimes, we will also use the notation A(x) = B(x) + O(x) to denote that |A(x) − B(x)|  |x|. In the whole paper, we assume that , , and u  ≡ 0 are as in Theorem 1.1. Moreover, we assume that  is a Lipschitz graph with Lipschitz constant τ with respect to the hyperplane H0 := {xd = 0} and that locally  lies above . Remark 2.1 Note that a C 1 domain is a Lipschitz domain with local Lipschitz constant as small as we need. In particular, Theorems 1.1 and 1.5, and its corollaries are valid for C 1 domains. 123 113 Page 6 of 52 J. M. Gallegos Remark 2.2 The Lipschitz constant of  is invariant by rescalings. If we consider a more general (anisotropic) scaling given by multiplication by a positive definite symmetric matrix with ellipticity ˜ , then the Lipschitz constant of  changes by a factor of at most ˜ 2 . 2.1 Divergence form elliptic PDEs with Lipschitz coefficients The function u we study solves div(A(x)∇u(x)) = 0 weakly in  where the matrix A(x) satisfies that • A(x) is symmetric. • There exists A > 1 such that −1 A |y| ≤ (A(x)y, y) ≤ 2 A |y| for all x ∈ , y ∈ R 2 d (uniform ellipticity). • There exists a Lipschitz constant L A > 0 such that |Ai j (x) − Ai j (y)| ≤ L A |x − y| (Lipschitz coefficients). By standard elliptic PDE theory (see [13, 16] for example), we know that u ∈ C 1,α in any ball with closure inside  for any 0 < α < 1. Also u ∈ W 1,2 () and u ∈ W 2,2 ( ) for any compactly embedded subdomain  ⊂⊂ . We extend the function u by 0 outside of  so that it is continuous through . We also extend the matrix A(x) in a way that it preserves the ellipticity A and Lipschitz L A constants up to a constant factor (the particular extension we choose will not matter). In particular, note that the absolute value of the extended function |u| is a subsolution in balls B such that B ∩ ∂ ⊂ . This means that ˆ (A∇|u|, ∇φ) dx ≤ 0, ∀φ ∈ Cc1 (B). B 2.2 Modifying the domain and A(x) Remark 2.3 Throughout this paper, we will require at different points the constants max( A − 1, 1 − −1 A ) and L A to be very small. We can obtain this by exploiting the fact that our considerations are local. Indeed, we can cover our initial domain  by a finite family of “small" domains and prove the results on the introduction on each one separately. By “zooming" on a small domain, we decrease the Lipschitz constant L A of the matrix. By rescaling the domain by multiplication with an adequate matrix, we can force A(x) = I at a particular point x. This, in addition to the small Lipschitz constant L A of the new matrix, implies small ellipticity A (we will show this below). Note, though, that this last operation changes the Lipschitz constant τ of . For this reason, in the statement of Theorem 1.1, τ0 depends on the ellipticity of A(x). Intuitively, we use the Lipschitz regularity of A(x) to exploit that, in small scales, A(x) is very close to a constant matrix. Let us show the effects of “zooming" and rescaling the domain when the domain is a ball. Suppose u solves weakly div(A∇u) = 0 in a ball B R for some R < 1, that is, for all φ ∈ Cc1 (B R ) we have ˆ (A(x)∇u(x), ∇φ(x)) dx = 0. BR If we “zoom" in on the origin by considering the function ũ(y) = u(Ry) defined on B1 , we can show it satisfies ˆ ( Ã(y)∇ ũ(y), ∇ φ̃(y)) dy = 0, ∀φ̃ ∈ Cc1 (B1 ), B1 123 Size of the zero set of solutions of elliptic PDEs near the… Page 7 of 52 113 where Ã(y) = A(Ry). Notice that Ã(y) has improved Lipschitz constant R · L A < L A . Now suppose that A(0)  = I and we want to change the function, the domain and the equation so that “A(0) = I " for the solution of the new equation. Consider ũ(x) = u( S̃x) defined in S̃ −1 B R where S̃ is the symmetric positive definite square root of A(0) and B R is a ball centered at the origin of radius R > 0. Then, for all φ̃(x) ∈ Cc∞ ( S̃ −1 B R ), we have ˆ ˆ ( S̃ −1 A( S̃x) S̃ −1 ∇ ũ(x), ∇ φ̃(x))dx = (A( S̃x) S̃ −1 ∇ ũ(x), S̃ −1 ∇ φ̃(x))dx S̃ −1 B R S̃ −1 B R ˆ = (A( S̃x)∇u( S̃x), ∇φ( S̃x))dx S̃ −1 B R ˆ = (det S̃)−1 (A(y)∇u(y), ∇φ(y))dy BR =0 where φ( S̃x) = φ̃(x). This implies that ũ is a weak solution of the equation div(A S̃ (x)∇ ũ) = 0 in S̃ −1 B R where A S̃ (x) := S̃ −1 A( S̃x) S̃ −1 . Properties of the new matrix A S̃ (x): • Observe that A S̃ (0) = S̃ −1 A(0) S̃ −1 = I . • The coefficients of A S̃ (x) are also Lipschitz with constant L A S̃ and the ratio between the Lipschitz constants of A S̃ and A can be bounded above and below by positive constants depending only on d and A . • A S̃ (x) is uniformly elliptic with ellipticity constant bounded by min( 2A , L A S̃ diam( S̃ −1 B R )d). Indeed, the largest and smallest eigenvalues λmax and λmin of A S̃ (z) at a point z ∈ S̃ −1 B R satisfy  max(λmax − 1, 1 − λmin ) = A S̃ (z) − I 2 ≤ A S̃ (z) − I  F = |a S̃,i, j (z) − δi, j |2 i, j −1 ≤ L A S̃ |z|d ≤ L A S̃ diam( S̃ B R )d where  ·  F is the Frobenius norm,  · 2 is the spectral norm, and δi, j is the Kronecker delta. 3 Frequency function for solutions of elliptic PDEs in divergence form Let x ∈  ∪  such that A(x) = I . For r > 0 such that Br (x) ∩ ∂ ⊂ , we denote μx (y) := (A(y)(y − x), y − x)/|y − x|2 and ˆ H (x, r ) := r 1−d μx (z)|u(z)|2 dσ (z) ∂ Br (x) where dσ is the surface measure of the ball. Note that the quantity H (x, r ) is nonnegative. To simplify the notation we will assume from now on that x is the origin and we will write H (r ) := H (x, r ) and μ(y) := μx (y). Remark 3.1 Observe that −1 A ≤ μ(y) ≤ A and, since A has Lipschitz coefficients, we have |μ(y) − 1|  L A |y|, A(y) − I 2  L A |y|, and |trA(y) − d|  L A |y|. 123 113 Page 8 of 52 J. M. Gallegos We also denote ˆ I (x, r ) := r 1−d (A∇u, ∇u)dy. Br (x)∩ Note that I (x, r ) is also nonnegative since A is positive definite. Moreover, I (x, r ) is finite thanks to Caccioppoli’s inequality supposing Br (0) ∩ ∂ ⊂ . As before, we will write I (r ) := I (x, r ). Proposition 3.2 Assume 0 ∈  ∪  and A(0) = I . For r > 0 such that Br ∩ ∂ ⊂ , the derivative of H (r ) is H  (r ) = 2I (r ) + O(L A H (r )). As a consequence there exists some constant c > 0 such that H  (r ) ≥ −cL A H (r ). For the rest of this paper, we denote c H := cL A . In particular, H (r )ec H r is a nondecreasing function of r . Note that in the harmonic case, c H = 0 and H is nondecreasing (which is also a corollary of the subharmonicity of |u|2 ). Note that the proof of Proposition 3.2 is quite simple in the case B(0, r ) ∩ ∂ = ∅. But in our setting we require some extra considerations to apply the divergence theorem in balls touching the boundary since u only belongs in W 1,2 (B(0, r ) ∩ ). To address this problem, we use “Appendix A” and follow the ideas of [33] in the harmonic case. Proof First, using that u ∈ W 1,2 (B(0, r )), we apply the divergence theorem to obtain ˆ −d H (r ) = r div(|u|2 A(x)x)dx. Br Using Remark 3.1, we expand the term inside the integral as  div(|u|2 A(x)x) = 2u(∇u, A(x)x) + |u|2 tr(A(x)) +|u|2 (∂i ai j )x j   d+O(L A |x|) i, j   O(L A |x|) =: 2u(∇u, A(x)x) + d|u|2 μ(x) + f (x) |u|2 .  O(L A |x|) We compute (formally) the derivative of H : ˆ   H  (r ) = −dr −1 H (r ) + r −d 2u(∇u, A(x)x) + d|u|2 μ(x) + |u|2 f (x) dσ (x) ∂ Br ˆ −d   =r 2u(∇u, A(x)x) + |u|2 f (x) dσ (x). ∂ Br ´ Note that the only problematic term is r −d ∂ Br 2u(∇u, A(x)x)dσ (x). Observe though, that ´ ´b´ ´ ∂ Br |∇u|dσ exists and is finite for almost every r since 0 ∂ Br |∇u| dσ dr = Br |∇u|dx is finite. Nonetheless, using the divergence theorem, we show that for all [a, b] ⊂ 123 Size of the zero set of solutions of elliptic PDEs near the… Page 9 of 52 113 [0, dist(0, ∂\)): ˆ b ˆ ˆ b   H  (s)ds = s −d 2u(∇u, A(x)x) + |u|2 f (x) dσ (x) ds a a ∂ Bs ˆ 1 1 = 2u(∇u, A(x)x) + d |u|2 f (x) dx A(0,a,b) |x| |x| d ˆ ˆ 1 d = div(|u| A(x)x)dx − 2 |u|2 μ(x) dx A(0,a,b) |x| d A(0,a,b) |x| d ˆ x = div |u|2 A(x) d dx A(0,a,b) |x| = H (b) − H (a). where A(0, a, b) is the annulus B(0, b)\B(0, a) for 0 < a < b. Thus, we have the identity for H  (r ) for almost every r . Now we want to prove that ˆ ˆ r −d 2u(∇u, A(x)x) dσ (x) = 2r 1−d (A∇u, ∇u)dx = 2I (r ). ∂ Br Br We denote  =  + ed and  =  + ed for small  > 0 where ed = (0, . . . , 0, 1) and we have ˆ ˆ −d −d r 2u(∇u, A(x)x)dσ (x) = lim r 2u(∇u, A(x)x)dσ (x) ∂ Br ∩ →0 ∂(Br ∩ ) from the fact that u vanishes continuously in  and ∇u converges as a non-tangential limit in L 2loc (), as shown in “Appendix A”. Now, since u ∈ W 2,2 (Br ∩  ), we may use divergence’s theorem and that u solves div(A∇u) = 0 to obtain ˆ ˆ r −d 2u(∇u, A(x)x)dσ (x) = r −d 2r div(u A(x)∇u)dx ∂(Br ∩ ) B ∩ ˆ r  = r −d 2r (A(x)∇u, ∇u)dx. Br ∩ Finally, the limit as  → 0 of the last term exists because u ∈ W 1,2 (Br ). Summing up, we have ˆ H  (r ) = 2I (r ) + r −d |u|2 f (x)dσ (x) ∂ Br with the second term being O(L A H (r )).   Definition 3.3 Assume x ∈  ∪  and A(x) = I . For r > 0 such that Br (x) ∩ ∂ ⊂  we define the Almgren’s frequency function as I (x, r ) N (x, r ) := r . H (x, r ) We will assume, as before, that x is the origin and we will write N (r ) := N (x, r ). The following geometric lemma is essential to ensure good behavior for N (x, r ) in balls intersecting . 123 113 Page 10 of 52 J. M. Gallegos Lemma 3.4 Suppose that x ∈  and A(x) = I . Assuming Br (x) ∩ ∂ ⊂ , r is small enough (depending on dist(x, ) and the Lipschitz constant L A of A(y)), and the Lipschitz constant τ of the domain is also small enough (depending on dist(x, )/r and L A r ), then (ν(y), A(y)(y − x)) ≥ 0 for Hd−1 -a.e. y ∈  ∩ Br (x) where ν is the outer unit normal to ∂. In particular, if x is the origin, we have ˆ μ(y)−1 (ν, Ay)(Aν, ν)|∂ν u|2 dσ (y) ≥ 0. Br ∩∂ Proof Without loss of generality, we assume x is the origin, denote T = dist(0, ) and suppose T ≤ r . Otherwise,  ∩ Br = ∅. First, note that A(y) − I 2 ≤ C L A r for all y ∈ Br ∩ ∂ by the Lipschitz continuity of A. Using this, we obtain (ν(y), A(y)y) = (ν(y), y) + (ν(y), (A − I )y) ≥ (ν(y), y) − |y|C L A r = |y| cos(α) − |y|C L A r where α is the angle between y and ν(y). If the domain were flat (τ = 0 and  were an open subset of a hyperplane) then all the points would have the same fixed normal vector ν̃ and we would have cos α ≥ Tr . Since the domain is not flat (τ  = 0), we have that the normal vector ν makes at most an angle of arctan(τ ) with ν̃. Using this we can bound from above α by arccos Tr + arctan τ . Now, assuming r is small enough so that T > C L Ar r and τ is also small enough so that T cos arccos + arctan τ ≥ C L Ar , r we obtain the desired inequality.   Definition 3.5 We will say that the origin 0 ∈  ∪  (assuming A(0) = I ) and a radius r are admissible if they satisfy the assumptions of the previous lemma. If A(0)  = I , we will 1/2 say that 0 ∈  ∪  and r are admissible if 0 and A r are admissible for the transformed −1 domain S̃  and the matrix A S̃ (see Remark 2.3). We can extend this definition to a point x  = 0 by translating the domain. Remark 3.6 (1) If dist(x, ) > r , we have that Br ∩ ∂ = ∅ and the integral is 0. In this case we also say that x and r are admissible. (2) If x and r  are admissible, then x and r are admissible for all 0 ≤ r < r  . (3) In the case A(x)  = I , we can ensure admissibility if we impose −1 dist(x, ) 1/2 cos arccos A + arctan( Aτ ) ≥ CLA A r. r The reason is that, in the transformed domain, the Lipschitz constant τ increases at most 1/2 by a factor A and the distance from the point to  decreases at most by a factor A . 123 Size of the zero set of solutions of elliptic PDEs near the… Page 11 of 52 113 Remark 3.7 Since the outer unit normal vector ν is only defined σ -a.e. the derivative ∂ν u may not exist everywhere. Its existence, the non-tangential convergence, and the fact that ∇u = (∇u, ν)ν in L 2 (σ ) are proven in “Appendix A”. We prove an almost monotonicity property for N (r ) which was first observed in [14] in the interior of the domain. Proposition 3.8 Assume 0 and r  are admissible, A(0) = I , and the ellipticity constant A of A(x) is smaller than 2. Then there exists C N > 0 depending on L A such that erC N N (r ) is nondecreasing in the interval (0, r  ). For the proof of this proposition we will only keep track of the relation between the constant C N and the Lipschitz constant L A . C N also depends on the ellipticity constant A but, by assuming (for example) A < 2, we can omit it. We may do so thanks to Remark 2.3. Notice also that C N ≡ 0 in the harmonic case. Our proof is an adaptation of the one in [26] but with the inclusion of the case Br ∩∂  = ∅. We need special care when Br ∩ ∂  = ∅ as with the proof of Proposition 3.2. Again, we use “Appendix A” and follow the ideas of [33] in the harmonic case to circumvent these problems. Proof Fix a compact interval I ⊂ (0, r  ). We will show that the derivative of N (r ) is positive a.e. r ∈ I . Since ˆ rˆ I (r ) = r 1−d (A∇u, ∇u) dσ ds 0 ∂ Bs and r is bounded away from 0, we have that I is absolutely continuous. Also note that we are only considering the case 0 ∈  (since we ask for admissibility). For this reason, H is of class C 1 and bounded away from 0 and N is also absolutely continuous. The derivative is (I (r ) + r I  (r ))H (r ) − r I (r )H  (r ) N  (r ) = . H (r )2 Let’s compute ˆ I  (r ) = (1 − d)r −1 I (r ) + r 1−d (A∇u, ∇u)dσ . ∂ Br ∩   A The previous identity is true for a.e. r ∈ I . Let w(y) := μ(y)−1 A(y)y be a vector field in . Observe that (w(y), y) = |y|2 and that y/r = ν, the normal vector in ∂ Br . We can rewrite A as ˆ 1 A = (A∇u, ∇u)(ν(y), w(y))dσ (y) r ∂ Br ∩ ˆ ˆ 1 1 = (A∇u, ∇u)(ν(y), w(y))dσ − (A∇u, ∇u)(ν(y), w(y))dσ . r ∂(Br ∩) r Br ∩∂     B C To study C we can use that Br ∩ ∂ ⊂  and, thus, ∇u = (∇u, ν)ν on  (in the sense of non-tangential limits, see “Appendix A”). 123 113 Page 12 of 52 J. M. Gallegos Then, we have that ˆ C = (Aν, ν)|∂ν u|2 (ν(y), w(y))dσ (y). Br ∩∂ This term is a bit problematic and we will treat it later with the help of Lemma 3.4. Let’s use the divergence theorem on B (we do this in Br ∩  and then let  → 0, as in the proof of Proposition 3.2), obtaining ˆ B = div(w(x)(A∇u, ∇u))dx. Br ∩ We have that div(w(x)(A∇u, ∇u)) = div(w)(A∇u, ∇u) + (w, ∇(A∇u, ∇u)) = div(w)(A∇u, ∇u) + 2(w, Hess(u)(A∇u)) + (A D,w ∇u, ∇u) (3.1) where A D,w = { k (∂k ai j )wk }i j . Furthermore, Hess(u) is symmetric and Hess(u)(w) = ∇(∇u, w) − (Dw)∇u. Let’s compute the integrals with all these terms in (3.1) one by one. First, we obtain ˆ ˆ (Hess(u)w, A∇u)dx = ((∇(∇u, w), A∇u) − ((Dw)∇u, A∇u)) dx. Br ∩ Br ∩ Using that u satisfies div(A∇u) = 0 in , we get ˆ ˆ (Hess(u)w, A∇u)dx = (div((∇u, w)A∇u) − ((Dw)∇u, A∇u)) dx Br ∩ Br ∩ ˆ ˆ = (∇u, w)(A∇u, ν)dσ − ((Dw)∇u, A∇u)dx . ∂(B ∩ ) B ∩  r    r   D E We can rewrite the previous equation using the divergence theorem: ˆ ˆ D = (∇u, w)(A∇u, ν)dσ + (∇u, w)(A∇u, ν)dσ ∂ Br ∩ Br ∩∂ Using that ∇u| converges to (∇u, ν)ν on  in as  → 0+ (see “Appendix A”), L 2loc (σ ) we get ˆ ˆ lim D = (∇u, w)(A∇u, ν)dσ + (ν, w)(Aν, ν)|∂ν u|2 dσ →0 ∂ Br ∩ Br ∩∂   C where the last term on the right hand side coincides with C defined previously. Remember that w(x) = μ(x)−1 A(x)x. Thus, we have the approximations Dw = I + O(L A |x|), and div(w) = d + O(L A |x|) for some a > 0. We can use these estimates to show that lim E = r d−1 I (r ) + O(L A r d I (r )). →0 123 Size of the zero set of solutions of elliptic PDEs near the… Page 13 of 52 113 One of the terms we missed in (3.1) behaves as ˆ div(w)(A∇u, ∇u)dx = dr d−1 I (r ) + O(L A r d I (r )) Br ∩ and the other as ˆ ˆ (A D,w ∇u, ∇u)dx  L A r |∇u|2 = O(L A r d I (r )). Br ∩ Br ∩ Summing everything up, ˆ I  (r ) = −r −1 I (r ) + O(L A I (r )) + 2r 1−d μ(y)−1 (A∇u, ν)2 dσ + r −d C ∂ Br ∩ and we obtain N  (r )N (r )−1 = (r I  (r ) + I (r ))(r I (r ))−1 − H  (r )(H (r ))−1 ˆ = 2r 1−d μ(y)−1 (A∇u, ν)2 dσ + r −d C I (r )−1 − 2r −1 N (r ) + O(L A ) ∂ Br ∩ ˆ = 2r 1−d H (r ) μ(y)−1 (A∇u, ν)2 dσ + r −d H (r ) C − 2I (r )2 (H (r )I (r ))−1 ∂ Br ∩ + O(L A ). Since C ≥ 0 thanks to the fact that 0 and r  are admissible, and Lemma 3.4, we can use Cauchy–Schwarz inequality to show that the whole first term is positive. Indeed, note that from the proof of Proposition 3.2, we have ˆ I (r ) = r 1−d u(∇u, A(x)ν) dσ (x). ∂ Br ∩ Thus, using Cauchy-Schwarz, we obtain ˆ ˆ I (r ) ≤ r 2 2−2d μ|u| dσ 2 μ(y)−1 (A∇u, ν)2 dσ ∂ Br ∩ ∂ Br ∩ ˆ = r 1−d H (r ) μ(y)−1 (A∇u, ν)2 dσ. ∂ Br ∩ Therefore, there exists C ≥ 0 such that N  (r ) ≥ −C L A N (r ) and we denote C N := C L A .   3.1 Frequency function centered at arbitrary points We have only considered H (x, r ) and N (x, r ) centered at points x ∈ ∪ where A(x) = I . We can treat general points by making a change of variables such as the one in Sect. 2.2. Assume A(0)  = I . Let S̃ be the symmetric positive definite square root of A(0), ũ(x) = u( S̃x), and A S̃ = S̃ −1 A( S̃x) S̃ −1 . For the transformed equation div(A S̃ ∇ ũ) = 0 with A S̃ (0) = I , we can compute ˆ H (r ) = r −1−d (A S̃ (x)x, x)|ũ(x)|2 dσ (x). ∂ Br 123 113 Page 14 of 52 J. M. Gallegos After some computations and a change of variables, we can check that it is equal (in the original domain) to ˆ −1 −d H (r ) = (det S̃) r |u(y)|2 |A(0)−1 y|(A(y)ν(y), ν(y))dσ (y). ∂( S̃ Br ) Remark 3.9 Note that det S̃, (A(y)ν(y), ν(y)) and r −1 |A(0)−1 y| can be upper and lower bounded by a constant depending only on the ellipticity constant A . In particular, by assuming A bounded, we have ˆ H (r ) ≈ r 1−d |u(y)|2 dσ (y). ∂( S̃ Br ) On the other hand, we have ˆ I (r ) = r 1−d − (A S̃ (x)∇ ũ, ∇ ũ)dx Br which, in the original domain, is equal to ˆ ˆ I (r ) = (det S̃)−1 r 1−d (A(y)∇u(y), ∇u(y))dy ≈ r 1−d |∇u|2 dy. S̃ Br S̃ Br This allows us to compute N (x, r ) for general points x. Beware that A S̃ may have different Lipschitz and ellipticity constants but this is not a problem since the change can be controlled as discussed in Sect. 2.2. 3.2 Auxiliar lemmas on the behavior of H(r) and N(r) First, we will present a lemma that controls the growth of H (r ) using N (r ). Lemma 3.10 Suppose 0 ∈  ∪ , A(0) = I , and α > 1. Then ˆ αρ ˆ αρ N (r ) H (αρ) N (r ) 2 − c H dr ≤ log ≤ 2 + c H dr . ρ r H (ρ) ρ r Moreover, if 0 and αρ are admissible, we have 2N (ρ)(log α)e−C N (α−1)ρ − c H (α − 1)ρ ≤ H (αρ) log ≤ 2N (αρ)(log α)eC N (α−1)ρ + c H (α − 1)ρ. H (ρ) Proof Using Propositions 3.2 and 3.8 on the interval [ρ, αρ], we can control H  (r ) ≤ 2I (r ) + c H H (r ) = (2r −1 N (r ) + c H )H (r ) ≤ (2r −1 N (αρ)eC N (αρ−r ) + c H )H (r ). Analogously, H  (r ) ≥ 2I (r ) − c H H (r ) ≥ (2r −1 N (ρ)eC N (ρ−r ) − c H )H (r ). Now we simply integrate H  /H in the interval [ρ, αρ].   The next lemma bounds L 2 norms in annuli by H (r ). We denote an annulus centered at x of outer radius r2 and inner radius r1 by A(x, r1 , r2 ) := B(x, r2 )\B(x, r1 ). 123 Size of the zero set of solutions of elliptic PDEs near the… Page 15 of 52 113 Lemma 3.11 Suppose 0 ∈  ∪ , r , δ ≥ 0, B(0, r + δ) ∩ ∂ ⊂  and A(0) = I . Then we have ˆ 1 1 ec H r H (r ) ≤− ec H |x| μ(x)|u(x)|2 dx ≤ ec H (r +δ) H (r + δ) . (3.2) d|B1 | A(0,r ,r +δ) d|B1 | Proof Using polar coordinates, write ˆ ˆ r +δ ec H |x| μ(x)|u(x)|2 dx = s d−1 ec H s H (s)ds, A(0,r ,r +δ) r and use that ec H r H (r ) is nondecreasing (Proposition 3.2), and |A(0, r , r + δ)| = |B1 |((r + δ)d − r d ).   In a similar fashion, if A(x0 )  = I and we assume A bounded we obtain the following result. 1/2 Lemma 3.12 Suppose x0 ∈  ∪  and B(x0 , A r ) ∩ ∂ ⊂ . Then we have ˆ 1/2 1/2 − |u|2 dx  ec H A r H (x0 , A r ). B(x0 ,r ) Proof Make a change of variables so that A(x0 ) = I . The ball B(x0 , r ) is sent to an ellipsoide 1/2 contained in the ball B(x0 , A r ). Proceed as in the previous lemma by using that μ ≈ 1.   The next lemma is a perturbation result for H (z, r ): it shows that we can bound H (0, r ) by C H (z, r  ) if 0 and z are close compared to r . Moreover, it does not assume that A(z) = I . Lemma 3.13 Assume that 0 ∈  ∪ , A(0) = I , A − 1 is small, z ∈  such that |z| ≤ γ r with γ ∈ (0, 1) and B100r (z) ∩ ∂ ⊂ . Then for any δ ∈ (0, 10), we have 1/2 H (0, r ) ≤ C(γ , δ, d, r , L A )H (z, A r (1 + γ + δ)) for some constant C > 0 depending on γ , δ, d, r , and L A . We omit the dependence of C on A in this lemma. Proof Let δ ∈ (0, 10), then using Lemma 3.11, we get ˆ d|B1 | e H (0, r ) ≤ cH r ec H |x| μ(x)|u(x)|2 dx. |A(0, r , r (1 + δ))| A(0,r ,r (1+δ)) √ Let S̃ := A(z), λmin be the minimum eigenvalue of S̃, λmax be the maximum eigenvalue −1 of S̃, A0 := A(0, r , r (1 + δ)), and Az = A(0, λ−1max (1 − γ )r , λmin (1 + δ + γ )r ). These two annuli are defined so that A0 ⊂ {z} + S̃ Az . Moreover, we have the following estimates −1/2 λmin ≥ max( A ,1 − O(L A γ r )) and 1/2 λmax ≤ min( A ,1 + O(L A γ r )) which will be useful in the proof of the next lemma. 123 113 Page 16 of 52 J. M. Gallegos Now we can bound | A0 | c H r e H (0, r ) d|B1 | ˆ ˆ c H |x| ≤ e μ(x)|u(x)| dx ≤ (1 + O(L A γ r ) + O(c H r (1 + δ)) 2 |u|2 dx A0 A0 ˆ ≤ (1 + O(L A r (1 + δ))) |u|2 dx {z}+ S̃ Az where we have used that A0 ⊂ {z} + S̃ Az and ec H |x| = 1 + O(L A |x|). We make the change of variables x = S̃ y, dx = (det S̃)dy = (1 + O(L A γ r )) dy and integrate in polar coordinates to get ˆ ˆ |u|2 dx = (1 + O(L A γ r )) |u( S̃ y)|2 dy {z}+ S̃ Az {z}+Az ˆ λ−1 min (1+γ +δ)r ≤ (1 + O(L A γ r )) s d−1 H (z, s)ds. λ−1 max (1−γ )r Finally, we use that ec H r H (r ) is increasing to see that ˆ λ−1 min (1+γ +δ)r s d−1 H (z, s)ds ≤(1 + O(L A r (1 + δ)))H (z, λ−1 min (1 + γ + δ)r ) λ−1 max (1−γ )r r d [((λ−1 −1 min (1 + δ)) − (λmax (1 − γ )) ] d d . d Summing up, we obtain H (0, r ) ≤ (1 + O(L A r (1 + δ))) [((λ−1 −1 min (1 + γ + δ)) − (λmax (1 − γ )) ] d d H (z, λ−1 min (1 + γ + δ)r ). (1 + δ)d − 1   Finally, we prove a perturbation result for the frequency function N . Lemma 3.14 Let r > 0 and z ∈  with |z| ≤ γ r for γ > 0 small enough. Assume L A r is small enough, 0 ∈ , A(0) = I , the ellipticity A of A(x) is small enough, and the point z and distance 4r are admissible. Then we have the following bound  √   √  N (0, r ) ≤ O( L A r ) + O( γ ) + N (z, 4r ) 1 + O( L A r ) + O( γ ) . If γ merely satisfies 0 < γ < ( A + 1)−1 , we obtain N (0, r ) ≤ C + C N (z, 4r ) for some constant C > 0. Proof Let δ ∈ (0, 1) to be chosen later. Using Lemma 3.11 we get the following upper and lower bounds ˆ d|B1 | ec H r H (0, 2r ) ≤ ec H |x| μ(x)|u(x)|2 dx |A(0, 2r , r (2 + δ))| A(0,2r ,r (2+δ)) 123 Size of the zero set of solutions of elliptic PDEs near the… Page 17 of 52 113 and ˆ d|B1 | ec H r H (0, r ) ≥ ec H |x| μ(x)|u(x)|2 dx. |A(0, r (1 − δ), r )| A(0,r (1−δ),r ) By Lemma 3.10, we have H (0, 2r ) 2N (0, r )(log 2)e−C N r − c H r ≤ log . H (0, r ) We aim to upper bound this quantity by something of the form log(H (z, r1 )/H (z, r2 )). To do this we will proceed as in Lemma 3.13. Since A(0) = I , we have λmax (A(z)) = 1 + O(L A γ r ) and λmin (A(z)) = 1 − O(L A γ r ) (maximum and minimum eigenvalues of A(z)). Let S̃z be the positive definite symmetric square root of A(z). From now on, we will denote λmax ( S̃z ) =: λmax and λmin ( S̃z ) =: λmin , both depending on z. We want to find an ellipsoidal annulus (with shape given by S̃z ) centered at z that contains A20 := A(0, 2r , r (2 + δ)) and another one that is contained in A10 := A(0, r (1 − δ), r ). If we take an annulus A(0, r1 , r2 ) and deform it by S̃z , we get an “ellipsoidal annulus" S̃z A(0, r1 , r2 ) such that A(0, λmax r1 , λmin r2 ) ⊂ S̃z A(0, r1 , r2 ) ⊂ A(0, λmin r1 , λmax r2 ). Using this, we can choose the following annuli −1 A20 ⊂ {z} + S̃z A(0, λ−1 max (2 − γ )r , λmin (2 + δ + γ )r ) =: {z} + S̃z Az 2 and this other one (recall |z| ≤ γ r ) A10 ⊃ {z} + S̃z A(0, λ−1 −1 min (1 − δ + γ )r , λmax (1 − γ )r ) =: {z} + S̃z Az . 1 For this last annulus to be well defined, we need −1 λmax λ−1 −1 min (1 − δ + γ ) < λmax (1 − γ ) ⇐⇒ δ > (1 + γ ) − (1 − γ ). λmin We also require 1 1 γ ≤ λmax < λmin +1 2 so that δ can satisfy δ ≤ 1. Now we can proceed in the exact same way as in Lemma 3.13 to get |A2z | H (0, 2r ) ≤ (1 + O(L A r (1 + δ))) H (z, λ−1 min (2 + γ + δ)r ). |A20 | In an analogous way, we can lower bound |A1z | H (0, r ) ≥ (1 + O(L A r (1 + δ))) H (z, λ−1 min (1 − δ + γ )r ). |A10 | 123 113 Page 18 of 52 J. M. Gallegos Putting both expressions together   H (0, 2r ) |A2z ||A10 | log ≤ (1 + O(L A r (1 + δ))) + log H (0, r ) |A1z ||A20 |   (3.3) H (z, λ−1 min (2 + γ + δ)r ) + log . H (z, λ−1 min (1 − δ + γ )r ) H (z,r1 ) Using Lemma 3.10 again, we can upper bound log H (z,r2 ) in terms of N as   H (z, λ−1 min (2 + γ + δ)r ) 2+γ +δ log −1 ≤ 2N (z, λ−1 min (2 + γ + δ)r ) log H (z, λmin (1 − δ + γ )r ) 1−δ+γ (1 + O(L A r (1 + δ))) + O(L A r (1 + δ)). (3.4) Now we need to choose δ. Remember that δ has to satisfy λmin 1 > δ > (1 + γ ) − (1 − γ ) = 2γ + O(L A r )(1 − γ ). (3.5) λmax We will choose δ equal to the geometric mean of the left hand side and the right hand side of inequality (3.5), that is  λmin δ = (1 + γ ) − (1 − γ ). λmax On the other hand, γ has to satisfy 1 1 1 0<γ < λmax = = − O(L A r ). λmin +1 2 + O(L A r ) 2 Notation: From now on, for this proof, we will write 1 −  := λλmax min . Thus  = O(L A r ). Notice that √   √  γ +  ≤ δ = 2γ +  − γ ≤ 2γ +  ≤ O( γ ) + O( L A r ) for  small enough (L A r small enough). Now let’s bound every term that has appeared before on Equations (3.3) and (3.4). First we bound √ √ 2+γ +δ 2 + (2 − ) γ +  + γ 2 + (2 − ) γ +  log = log √ ≤ log √ 1−δ+γ 1 − (2 − ) γ +  + γ 1 − (2 − ) γ +  √ 2 + 2γ +  √ √  ≤ log √ = log(2) + O( γ + ) ≤ log(2) + O( γ ) + O( L A r ). 1 − 2γ +  To bound     |A10 ||A2z | (1 − (1 − δ)d )((λ−1 −1 min (2 + δ + γ )) − (λmax (2 − γ )) ) d d log = log |A20 ||A1z | ((2 + δ)d − 2d )((λ−1 −1 max (1 − γ )) − (λmin (1 − δ + γ )) ) d d we separate it in two terms, 1 − (1 − δ)d log A = log ((1 − )(1 − γ ))d − (1 − δ + γ )d 123 Size of the zero set of solutions of elliptic PDEs near the… Page 19 of 52 113 and (2 + δ + γ )d − ((1 − )(2 − γ ))d log B = log . (2 + δ)d − 2d Let’s write the first order expansion of the terms in δ: • (1 − δ)d = 1 − dδ + O(δ 2 ) • (1 − )d = 1 − O(L A r ) • (1 − γ )d ≥ 1 − O(δ 2 ) • (1 − δ + γ )d = (1 − δ)d + O(γ ) = 1 − dδ + O(δ 2 ) Now we can use this in the first term dδ + O(δ 2 ) A = (1 − O(L A r ))(1 − O(γ )) − (1 − dδ + O(δ 2 )) dδ + O(δ 2 ) ≤ (1 − O(L A r ))(1 − O(δ 2 )) − (1 − dδ + O(δ 2 )) dδ + O(δ 2 ) = dδ − O(L A r ) − O(δ 2 ) = 1 + O(L A r /δ) + O(δ)  √ ≤ 1 + O( L A r ) + O( γ ). As for the other term, we proceed similarly (2 + δ)d + O(γ ) − (1 − O(L A r ))(2d − O(γ )) B = d2d−1 δ + O(δ 2 ) d2 δ + O(δ ) + O(γ ) + O(L A r ) d−1 2 = d2d−1 δ + O(δ 2 ) = 1 + O(δ) + O(γ /δ) + O(L A r /δ)  √ ≤ 1 + O( L A r ) + O( γ ). Using all the bounds obtained in the last paragraphs together with Eqs. (3.3) and (3.4), we get  √   √  N (0, r ) ≤ O( L A r ) + O( γ ) + Ñ (z, λ−1 min (2 + γ + δ)r ) 1 + O( L A r ) + O( γ ) for L A r and γ small enough. Finally, we can bound λ−1 min (2 + γ + δ)r ≤ 4r and use that eC N r N (r ) is increasing to get a simpler expression. Remark 3.15 To prove the second part of the lemma (γ not necessarily small), we just need to choose δ as the arithmetic mean of the left hand side and the right hand side of inequality (3.5). The rest of the proof is straightforward.   123 113 Page 20 of 52 J. M. Gallegos 4 Behavior of the frequency function on cubes near the boundary The aim of this section is to prove the first technical lemma concerning the behavior of N near the boundary. For an analogous proof in the harmonic case, see Section 3 of [33] or Sections 4.1 and 4.2 of [28]. 4.1 Whitney cube structure on Ä We will consider the same Whitney cube structure in  as [33]. Let H0 be the horizontal hyperplane through the origin, and B0 be a ball centered in  such that M B0 ∩ ∂ ⊂  for some very large M. We also assume that M B0 ∩ ∂ is a Lipschitz graph with slope τ small enough with respect to H0 . We consider the following Whitney decomposition of : a family W of dyadic cubes in Rd with disjoint interiors and constants W > 20 and D0 ≥ 1 such that  (1) Q∈W Q = , (2) 10Q ⊂ , ∀Q ∈ W , (3) W Q ∩ ∂  = ∅, ∀Q ∈ W , (4) there are at most D0 cubes Q  ∈ W such that 10Q ∩ 10Q   = ∅, ∀Q ∈ W . Further, for such cubes Q  we have 21 (Q  ) ≤ (Q) ≤ 2(Q  ). We will denote by (Q) the side length of Q and by x Q the center of the cube Q. From these properties it is clear that dist(Q, ∂) ≈ (Q). Also we consider the cubes small enough so that diam(Q) < 20 1 dist(Q, ∂). Now we will introduce a “tree" structure of parents, children and generations to this Whitney cube decomposition. Let  denote the orthogonal projection on H0 and choose R0 ∈ W such that R0 ⊂ M 2 B0 . It will be the root of the tree and we define DW0 (R ) = {R } (that is the set of cubes of 0 0 generation 0 of the rooted tree). To characterize the generations DW k (R ) for k ≥ 1, we 0 define first J (R0 ) = {(Q) : Q ∈ W such that (Q) ⊂ (R0 ) and Q is below R0 }. We have that J (R0 ) is a family of d −1 dimensional dyadic cubes in H0 , all of them contained in (R0 ). Let Jk (R0 ) ⊂ J (R0 ) be the subfamily of (d − 1)-dimensional dyadic cubes in H0 with side length equal to 2−k (R0 ). To each Q  ∈ Jk (R0 ) we assign some Q ∈ W such that (Q) = Q  and such that Q is below R0 (notice that there may be more than one choice for Q but the choice is irrelevant), see [33, Lemma B.2], and we write s(Q  ) = Q. Then we define k DW (R0 ) := {s(Q  ) : Q  ∈ Jk (R0 )} and  DW (R0 ) = k DW (R0 ). k≥0 Finally, for each R ∈ k DW (R0 ) and j ≥ 1, we denote   j k+ j DW (R) = Q ∈ DW (R0 ) : (Q) ⊂ (R) . By the properties of the Whitney cubes, we can observe that Q ∈ DW (R0 ) ⇒ dist(Q, ) = dist(Q, ∂) ≈ (Q). 123 Size of the zero set of solutions of elliptic PDEs near the… Page 21 of 52 113 Further, for any Q ∈ W , we denote its center by x Q , its associated cylinder by C (Q) := −1 ((Q)), and the (d − 1)-dimensional Lebesgue measure on the hyperplane H0 by m d−1 . In Appendix B of [33] one can find more details about the construction of this Whitney cube structure and its projections. 4.2 Lemma on the behavior of the frequency in the Whitney tree Now we can present the first main lemma required in the proofs of Theorems 1.1 and 1.5. This lemma controls probabilistically the behavior of the frequency function in the tree of Whitney cubes defined in the last section. See [33, Lemma 3.1] for a version of this lemma for harmonic functions. Our proof is very similar. The reader only needs to consider that the properties of the frequency function for elliptic PDEs are slightly worse than those of the frequency function for harmonic functions, and that A(x) is a perturbation of the identity matrix. Also this Lemma should be compared with the (interior) Hyperplane lemma of [24] and [28, Lemma 7]. Note that, in what follows, we refer to the frequency function N of a solution of div(A∇u) = 0 in  as in the statement of Theorem 1.1. Lemma 4.1 Let N0 > 1 be big enough. There exists some absolute constant δ0 > 0 such that for all S  1 big enough the following holds, assuming also the Lipschitz constant τ of  is small enough. Let R be a cube in DW (R0 ) with (R) small enough depending on S and L A that satisfies N (x R , S(R)) ≥ N0 . Then, there exists some positive integer K = K (S) big enough such that if we let   1 G K (R) = Q ∈ DW K (R) : N (x Q , S(Q)) ≤ N (x R , S(R)) 2 then:   (1) m d−1 Q∈G K (R) (Q) ≥ δ0 m d−1 ((R)), (2) for Q ∈ DWK (R), it holds N (x Q , S(Q)) ≤ (1 + C S −1/2 )N (x R , S(R)). Note that (2) does not require N (x R , S(R)) ≥ N0 . It is important that δ0 does not depend on S. Other constants such as M, K , and the upper bound on τ do depend on S. Finally, N0 only depends on the dimension d. Remark 4.2 Fix S, T > 0 and R ∈ DW (R0 ) with (R) small enough depending on L A and T . If x ∈  satisfies dist(x, R) ≤ T (R) and dist(x, ∂) ≥ T −1 (R), then x and S(R) are admissible, assuming M  T , M  S, and that τ is small enough. The proof is analogous to the proof of [33, Remark 3.3]. An important tool for the proof of the Lemma 4.1 is the following quantitative Cauchy uniqueness theorem. Theorem 4.3 (Quantitative Cauchy uniqueness) Let u be a solution of div(A(x)∇u(x)) = 0 in the half ball     2  B+ = y = y  , y  ∈ Rd−1 × R |  y   + |y  |2 < 1, y  > 0 123 113 Page 22 of 52 J. M. Gallegos and suppose u is C 1 smooth up to the boundary and A(x) is as discussed in Sect. 2.1. Let  := {x ∈ Rd | xd = 0, |x| ≤ 3/4} Suppose that u L 2 (B + ) ≤ 1 and uW 1,∞ () ≤  for some  ∈ (0, 1). Then  sup |u| ≤ C α and sup |u| ≤ C   α B((0,...,0,1/2),1/4) 1 4 B+ where C, C  , α, and α  are positive constants depending only on the ellipticity and the Lipschitz constant of the matrix A(x) and the dimension d. This result is proved in great generality in [5, Theorem 1.7]. It will also be useful for the proof of Lemma 5.1. Before starting the proof of Lemma 4.1, we note that we will require both A − 1 and L A very small in what follows (see again Remark 2.3 to see why we can do so). Proof of Lemma 4.1 Let S  1 and then choose R ∈ DW (R0 ) with (R) small enough depending on S and L A . For some j  1 independent of S that will be fixed below, consider the hyperplane L parallel to H0 (and above H0 ) such that dist(L,  ∩ C (R)) = 2− j (R). From now on, we will denote by J the family of cubes from W that intersect L ∩ C ( 21 R). By our construction of the Whitney cubes, we have (Q) ≈ 2− j (R) and (Q) ⊂ (R) for all Q ∈ J . Notice that if τ is small enough (depeding on j), then dist(x,  ∩ C (10R)) ≈ 2− j (R) for x ∈ L ∩ C (10R). Denote by Adm(2W Q) the set of points x ∈  ∩ 2W Q such that the interval (0, diam(25W Q)) is admissible for x. We assume that the Lipschitz constant of the domain τ is small enough so that 3Q ⊂ Adm(2W Q) (using Remark 4.2). Then by Lemma 3.14, for Q small enough sup N (x, diam(5W Q)) ≤ C0 N (x Q , diam(20W Q)) + C0 (4.1) x∈Adm(2W Q) where C0 is an absolute constant. Note that |x − x Q | ≤ diam(W Q) < 41 diam(5W Q) since x ∈ 2W Q, hence it satisfies the conditions required by Lemma 3.14 (we use that the ellipticity constant is small enough). Claim There exists some Q ∈ J such that N (x R , S(R)) N (x Q , diam(20W Q)) ≤ (4.2) 4C0 if j is big enough (but independent of S) and we assume that τ0 is small enough depending on j, and also N0 is big enough. Proof of the claim From now on, we denote N = N (x R , S(R)). Our aim is to prove the claim using Theorem 4.3 in a small half-ball centered at z R , the projection of x R onto the hyperplane L. Set B+ := {x ∈ B(z R , (R)/4) | xd > (z R )d } ⊂ . Also, let z̃ R := z R + (0, . . . , 0, (R)/8) ∈ B+ . Note that, rescaling B+ , z̃ R corresponds to the point (0, . . . , 0, 1/2) in the statement of Theorem 4.3. 123 Size of the zero set of solutions of elliptic PDEs near the… Page 23 of 52 113 We aim for a contradiction, so we assume that N (x Q , diam(20W Q)) > N /(4C0 ) for all Q ∈ J , where N = N (x R , S(R)). For each Q ∈ J , we have ˆ sup |u|2 − |u|2 dx 2Q B(x Q ,diam(3Q)) by standard properties of solutions of elliptic PDEs. Then, by Lemma 3.12, we obtain ˆ − |u|2 dx  H (x Q , diam(20W Q))ec H diam(20W Q) B(x Q ,diam(3Q)) where c H is the constant from Lemma 3.10. Note that ec H diam(20W Q) = 1 + O(L A (Q)), omiting the dependence on A (which we may assume is very close to 1). Using Lemma 3.10 we obtain the following bound using the frequency function: H (x Q , diam(20W Q)) 2N (x Q ,diam(20W Q))(1+O(L A (Q))) diam(20W Q) ≤ H (x Q , (R)) (1 + O(L A (Q))). (R) Note that for the previous step we need τ small enough depending on j. At other points of the proof we will require τ small enough but without further reference. Now we estimate H (x Q , (R)) as follows ˆ H (x Q , (R)) ≈ (R)1−d |u(x Q + A(x Q )1/2 y)|2 dσ (y) ∂ B((R)) ˆ (1 + O(L A (R)))  |u(x Q + A(x Q )1/2 y)|2 ec H |y| dy |A(x Q , (R), 2(R))| A(0,(R),2(R)) ˆ (1 + O(L A (R)))  |u(z̃ R + A(z̃ R )1/2 y)|2 ec H |y| dy |B(z̃ R , C1 (R))| B(0,C1 (R)) where we have used Remark 3.9, and that for some fixed C1 > 0 we have that A(x Q )1/2 A(x Q , (R), 2(R)) ⊂ A(z̃ R )1/2 B(z̃ R , C1 (R)). Finally, using Lemma 3.11 we can bound H (x Q , (R))  (1 + O(L A (R)))H (z̃ R , C1 (R)). Moreover, using again Lemma 3.10, we can further bound as follows H (z̃ R , C1 (R)) ≤ H (z̃ R , (R)/16)(16C1 )2N (z̃ R ,C1 (R))(1+O(L A (R))) (1 + O(L A (R))) ˆ (1 + O(L A (R)))  ec H |y| |u(z̃ R + A(z̃ R )1/2 y)|2 dy |B(z̃ R , (R)/8)| B(0,(R)/8) (16C1 )2N (z̃ R ,C1 (R))(1+O(L A (R))) and recalling that B(z̃ R , (R)/8) ⊂ B+ even after considering the reescaling by A(z̃ R )1/2 , we obtain ˆ (1 + O(L A (R))) H (z̃ R , C1 (R))  |u|2 dy · (16C1 )2 Ñ (z̃ R ,C1 (R))(1+O(L A (R))) . |B+ | B+ Now, using that L A (R) is very small, we can bound all the terms O(L A (R)) by 1, for example. Thus, summing up all the computations we have done, we can write ˆ diam(20W Q) N (x Q ,diam(20W Q)) sup |u|2  (16C1 )4N (z̃ R ,C1 (R)) − |u|2 dy. (4.3) 2Q (R) B+ 123 113 Page 24 of 52 J. M. Gallegos By Lemma 3.14, we have N (z̃ R , C1 (R)) ≤ C N (x R , S(R)) + C ≤ C  N (4.4) for a suitable constant C > 2C1 if S is large enough, and N0 too. Recalling that we assumed for all Q ∈ J that N (x Q , diam(20W Q)) > N 4C0 , we get N /4C0 ˆ Ndiam(20W Q) sup |u|2  (16C1 )C − |u|2 dy 2Q (R) B+ ˆ   = 2− jC N +C N − |u|2 dy B+ for some positive constants C  and C  . We have used that diam(20W Q) < (R) by choosing j large enough. Using interior estimates for solutions of elliptic PDEs ˆ sup2Q |u|2 22 j − jC  N +C  N sup |∇u|2   2 − |u|2 dy. 3 Q (Q)2 (R)2 B+ 2 From the last two estimates we deduce that if j is big enough (depending on the absolute constants C  and C  ) and N0 (and thus also N ) is big enough too, then there exists some c > 0 such that ˆ  2  − jc N sup |u| + (R) |∇u|  2 2 2 − |u|2 dy. 3 Q B+ 2 Since the cubes 23 Q with Q ∈ J cover the flat part of the boundary of B+ , we can apply a rescaled version of Theorem 4.3 to B+ to get ˆ  sup |u|2  2− jc N α − |u|2 dy (4.5) B(z̃ R ,(R)/16) B+ for some α > 0. Observe that we can lower bound sup |u|2  H (z̃ R , (R)/16) B(z̃ R ,(R)/16) and upper bound ˆ   2− jc N α − |u|2 dy  2− jc N α H (z̃ R , (R)). B+ Using these bounds in (4.5) we obtain  H (z̃ R , (R)/16)  2− jc N α H (z̃ R , (R)). By Lemma 3.10, this implies H (z̃ R , (R)) 2N (z̃ R , (R))(log 16)(1 + O(L A (R))) + O(L A (R)) ≥ log  c j N α H (z̃ R , (R)/16) for some fixed c > 0. But for j big enough this contradicts the fact that N (z̃ R , (R))  N by (4.4). Observe though that j “big enough" does not depend on the election of S.   123 Size of the zero set of solutions of elliptic PDEs near the… Page 25 of 52 113 Now we may introduce the set G K (R). Fix Q 0 ∈ J such that (4.2) holds for Q 0 . Notice that, by (4.1) 2 N 10 sup N (x, diam(5W Q 0 )) ≤ C0 N (x Q 0 , diam(20W Q 0 )) + C0 ≤ · x∈Adm(2W Q 0 ) 2 11 (4.6)  10 2 since N ≥ N0 and we assume N0 big enough. The precise value of the constant 11 is not important. Finally, we can define j+k G K (R) = {Q ∈ DW (R) : (Q) ⊂ (Q 0 )} with k = log2 S. Thus, we have G K (R) ⊂ DW K (R) with K = j + k and it holds (Q) = −k 2 (Q 0 ) for every Q ∈ G K (R). The property (1) follows from (4.6). Indeed if P ∈ G K (R), then taking into account that x P ∈ Adm(2W Q 0 ) for τ small enough (depending on S) and using Proposition 3.8 we get 2 11 11 N N (x P , S(P)) ≤ N (x P , (Q 0 )) ≤ N (x P , W (Q 0 )) ≤ 10 10 2 where we have bounded the terms 1 + O(L A (Q 0 )) by 11/10. Notice also that ⎛ ⎞   d−1 m d−1 ⎝ (Q)⎠ = l (Q 0 )d−1 ≈ 2− j (R) Q∈G K (R) and recall that j is independent of S. So (1) holds with δ0 ≈ 2− j(d−1) . The property (2) is a consequence of Lemma 3.14. Indeed for any P ∈ DWK (R), since −1 |x P − x R |  (R), taking γ ≈ S in the Lemma 3.14, we deduce N (x P , S(P)) ≤ (1 + O(L A S(R)))N (x P , S(R)/3)     ≤ (1 + O(L A S(R))) O( L A S(R)) + O( S −1/2 )     + (1 + O(L A S(R)))N (x R , S(R)) 1 + O( L A S(R)) + O( S −1/2 )) . Assuming that (R)  S −2 and N0 large enough, we obtain N (x P , S(P)) ≤ (1 + C S −1/2 )N (x R , S(R)) for certain constant C.   5 Balls without zeros near the boundary In this section we will prove the second main lemma concerning the behavior of N near the boundary. This lemma shows that if we have a ball near the boundary with bounded frequency, then we can find a smaller ball centered at the boundary where u does not change sign. The following lemma should be compared with [28, Lemma 8] that treats the harmonic case. Note that, in what follows, we refer to the frequency function N (x, r ) of a solution of div(A∇u) = 0 in  as in the statement of Theorem 1.1. Moreover, we consider  with the Whitney structure defined in Sect. 4.1. 123 113 Page 26 of 52 J. M. Gallegos Lemma 5.1 For any N > 0 and S  1 large enough there exist positive constants τ0 (N , S) and ρ(N , S) such that the following statement holds. Suppose the Lipschitz constant τ of  is smaller than τ0 and Q is a cube in DW (R) such that N (x Q , S(Q)) ≤ N . Then there exists a ball B centered in  ∩ C (Q) with radius ρ(Q) such that u does not vanish in B ∩ . First, we will prove a “toy" version of this lemma on the half ball B+ for harmonic func- tions. The following lemma is essentially [28, Lemma 9] but formulated using the frequency function instead of the doubling index (a closely related quantity used in [28]). Lemma 5.2 Let B be the unit ball in Rd and let B+ be the half ball,     B+ = y = y  , y  ∈ Rd−1 × R | |y  |2 + |y  |2 < 1, y  > 0 .  Let u be a function harmonic in B+ such that u ∈ C(B+ ), u = 0 on  := ∂ B+ ∩ y  = 0 , and sup |u| = 1. 1 4 B+ For any N > 0 and 0 < r0 < 1/16, there exist ρ = ρ(N , r0) > 0 and c0 = c0 (N , r0 ) > 0 such that if N (0, 1/2) ≤ N , then there is x  ∈ Rd−1 with x   < r0 such that |u(y)| ≥ c0 y  , for any y = (y  , y  ) ∈ B((x  , 0), ρ) ∩ B+ .    In particular, u does not vanish in B x  , 0 , ρ ∩ B+ . The notation n1 B+ used in the previous statement stands for {y ∈ Rd | ny ∈ B+ }. Proof Let B− be the reflection of the half-ball B+ with respect to  = {y  = 0}∩∂ B+ . Since u vanishes on , u can be extended to a harmonic function in B by the Schwarz reflection principle. We also denote this extension by u. Using Cauchy estimates we can uniformly bound every partial derivative of u inside B(0, 1/8) obtaining sup |∇u(x)|  sup |u| = 1. x∈B(0,1/8) B(0,1/4) Let δ := maxx  ∈Rd−1 , |∇u(x  , 0)|. Then, Theorem 4.3 applied to r0 B+ implies that |x  |≤r0 sup |u| ≤ Cδ γ B(0,r0 /4) ´ for some positive C and γ ∈ (0, 1). Then −∂ B(0,r0 /4) u 2 dσ ≤ Cδ 2γ and by subharmonicity ´ ∂ B(0,1/2) u dσ ≥ c sup B(0,1/4) u = c. − 2 2 By the monotonicity of N in the harmonic case, we also have ´ 2 − H (0, 1/2) ∂ B(0,1/2) u dσ log2/r0 ´ = log2/r0 ≤ N (0, 1/2) ≤ N . − 2 ∂ B(0,r0 /4) u dσ H (0, r0 /4) We can conclude that N − 2γ c 2 log2/r0 ≤ N ⇒ δ ≥ c . Cδ 2γ r0 Now, choose x∗ ∈ Rd−1 , |x∗ | ≤ r0 such that |∇u(x∗ , 0)| = δ. Clearly, at this point, |∇u(x∗ , 0)| = |∂d u(x∗ , 0)| (the derivative in the direction normal to ). Without loss of 123 Size of the zero set of solutions of elliptic PDEs near the… Page 27 of 52 113 generality, assume that ∂d u(x∗ , 0) = δ. Observe that the second derivatives of u are uniformly bounded in B(0, 1/8) using Cauchy estimates (as we have done before). Thus, we have ∂d u(y) > δ/2 when dist(y, (x∗ , 0)) < ρ = min{c0 δ, r0 } where c0 only depends on the bound on the second derivatives, and thus is an absolute constant. Using this, we finally get N − 2γ δ  2 u(y) ≥ y ≥ c y  , 2 r0      for y = y  , y  ∈ B x∗ , 0 , ρ ∩ B+ .   Now we will prove an elliptic extension of the previous lemma. Unfortunately, the proof presents some complications since we do not have an adequate substitute to Schwarz’s reflection principle. To overcome this, we assume that our function u is a solution of div(A(x)∇u) = 0 where A(x) is a small (Lipschitz) perturbation of the identity matrix, and we show that there is a harmonic function v very close to u in C 1 norm for which the previous lemma holds. Then, we obtain that there is a smaller ball where u does not vanish either.  u be a solution of div(A∇u) = 0 in B+ such that u ∈ C(B+ ), u = 0 on Lemma 5.3 Let  := ∂ B+ ∩ y  = 0 , and sup |u| = 1 1 4 B+   where 41 B+ = 14 B + . For any N > 0 and 0 < r0 < 1/32, there exist ρ = ρ(N , r0 ) > 0   and c0 = c0 (N , r0 ) > 0 such that if N (0, 1/2) ≤ N , then there is x  ∈ Rd−1 with x   < r0 such that      |u(y)| ≥ c0 y  , for any y = y  , y  ∈ B x  , 0 , ρ ∩ B+ assuming that L A and A − 1 are small enough depending on N and r0 (where L A and A are the Lipschitzand ellipticity   constants of A(x), respectively). In particular, u does not have zeros in B x  , 0 , ρ ∩ B+ . Proof Let v be the harmonic extension of u|∂( 1 B+ ) defined in 21 B+ . We intend to use 2 Lemma 5.2 to find a ball B such that |v(y  , y  )|  y  for (y  , y  ) ∈ B ∩ . Afterwards, we will see that if L A and A − 1 are small enough, then the difference v − u is arbitrarily small in W 1,∞ (B ∩ ) which will prove the lemma (for a smaller concentric ball). First, we bound the frequency (associated to ) of v as follows ´ ´ 1 21 B+ |∇v| dx B |∇u| dx 2 2 1 v N (0, 1/2) = ´  ´2 + 2 ∂ 1 B+ v 2 dσ ∂ 1 B+ u dσ 2 2 2 using that v is the minimizer of the Dirichlet energy for the boundary condition u|∂(B+ /2) . We also have that ´ ´ 1 21 B+ |∇u| dx 1 21 B+ (A(x)∇u, ∇u)dx 2 ´ ≈ ´ = N u (0, 1/2), 2 ∂ 1 B+ u 2 dσ 2 1 ∂ B+ μu 2 dσ 2 2 123 113 Page 28 of 52 J. M. Gallegos obtaining an upper bound for N v (0, 1/2). Analogously, we may also obtain a lower bound for N v (0, 1/2) in terms of N u (0, 1/2) using that u minimizes a weighted Dirichlet energy. Consider the function h = v − u defined in 21 B+ . Note that h is a solution of ! div(A∇h) = div((A − I )∇v), in 21 B+ , h = 0, on ∂ 21 B+ . We aim to bound h(x) and ∇h(x) in 16 1 B+ in terms of N , L A , and A . Using the Green’s function G A (x, y) for the elliptic operator div(A(x)∇·) in 21 B+ , we can represent h(x) as ˆ h(x) = − (∇ y G A (x, y), (A − I )(y)∇v(y))dy 1 2 B+ for x ∈ B+ /16. We split this integral in two parts and take absolute values: ˆ |h(x)| ≤ |∇ y G A (x, y)||(A − I )(y)||∇v(y)|dy 1 B+ ˆ 8 + |∇ y G A (x, y)||(A − I )(y)||∇v(y)|dy. 1 2 B+ \ 18 B+ In both integrals we bound |A − I | by A − 1. Also, in the first integral, we are going to bound |∇v(y)|  2cN for some c > 0 using Cauchy estimates. To this end, we consider v extended to 21 B using Schwarz reflection principle, and then, for y ∈ 18 B, we obtain ˆ |∇v(y)|2  v 2 dσ ≈ H v (0, 1/4) ≤ H v (0, 1/2) ≈ H u (0, 1/2)  H u (0, 1/4) 2cN  1 · 2cN ∂ 41 B where we have also used H u (0, 1/4)  sup 1 B+ |u|2 = 1. Remember that H v (x, r ) = ´ 4 r 1−d ∂ Br (x) |u(z)|2 dσ (z), as v is harmonic. d Then, using that ∇ y G A (x, ·) has weak L d−1 norm bounded by a constant depending only on A and d (see [17, estimate (1.6)] together with the symmetry of A), we get ˆ |∇ y G A (x, y)|dy ≤ C(d, A ). 1 8 B+ Thus, we can bound the first integral by ˆ |∇ y G A (x, y)||(A − I )(y)||∇v(y)|dy  C  (d, A )2 cN . 1 8 B+ For the other integral, we use Cauchy–Schwarz to obtain ˆ ˆ 1/2 ˆ 1/2 |∇ y G A (x, y)||∇v(y)|dy ≤ |∇ y G A (x, y)| dy 2 |∇v(y)| dy 2 . 1 2 B+ \ 18 B+ 1 2 B+ \ 18 B+ 1 2 B+     A B Invoking [17, Theorem 3.3], we have that |∇ y G A (x, y)| ≤ C|x − y|1−d  1 for x ∈ 1 16 B+ and y ∈ 21 B+ \ 18 B+ , which allows us to bound A . We estimate B as follows: ˆ ˆ |∇v(y)|2 dy = 2N v (0, 1/2) v 2 dσ  2N 2cN , 1 2 B+ ∂ 21 B+ 123 Size of the zero set of solutions of elliptic PDEs near the… Page 29 of 52 113 where we have used that H v (0, 1/2)  2cN , as shown before. Summing up all the previous estimates, we have obtained |h(x)|  C(N , A )( A − 1) for x ∈ 16 1 B+ . Now, [16, Corollary 8.36] gives us hC 1,α ( 1 B+ )  h L ∞ ( 1 B+ ) + (A − I )∇vC 0,α ( 1 B+ ) 32 16 16 for some α ∈ (0, 1). Using the product rule for derivatives, we get (A − I )∇vC 0,α ( 1 B+ )  (A − I )∇vC 0,1 ( 1 B+ )  L A ∇v L ∞ ( 1 B+ ) +( A − 1)D 2 v L ∞ ( 1 B+ ) . 16 16 16 16 Finally, using interior Cauchy estimates to estimate D 2 v L ∞ ( 1 B+ ) and ∇v L ∞ ( 1 B+ ) in 16 16 terms of N , we get 1 |∇h(x)| ≤ C(N ) (L A + ( B+ . A − 1)) , x ∈ 32 To end the proof, we apply Lemma 5.2 to v to get a ball where      |v(y)| ≥ c0 y  , for any y = y  , y  ∈ B x  , 0 , ρ ∩ B+ and we make L A and A −1 small so that ∂d h(x) ≤ c0 /2 in 32 1 B+ . This implies |h(y  , y  )| ≤ c0  2 y and, since |u| ≥ |v| − |h|, it finishes the proof.   We need to state a last lemma before proving Lemma 5.1. Lemma 5.4 Let  ⊂ Rd be a Lipschitz graph with respect to the hyperplane H0 with Lipschitz constant τ < 1/2, assume that 0 ∈ , and let u be a solution of div(A∇u) = 0 in the domain  = {x + r ed | x ∈  ∩ B(0, 1), r ∈ (0, 1)}. Then, there exist positive constants M and δ depending on the Lipschitz constant L A and the ellipticity constant A of A(x) such that the following statement holds. If u ≥ −1 in  ∩ B(0, 1), u ≡ 0 on  and u ≥ M on {x + r ed | x ∈  ∩ B(0, 1), r ∈ (δ, 1)} ∩ B(0, 1), then u ≥ 0 in  ∩ B(0, 1/2). The proof of this lemma is Step 1 of the proof of [11, Theorem 1.1]. For a simpler proof in the harmonic case (in Lipschitz domains with small Lipschitz constant), see Appendix B in [13]. In the following proof, we will assume again that L A and A − 1 are very small (see Remark 2.3). Proof of Lemma 5.1 Let S  1 and Q be a cube of our Whitney cube structure such that x Q (the center of Q) and S(Q) are admissible and N (x Q , S(Q)) ≤ N . Note that, to attain this, we need the Lipschitz constant of the domain τ small enough depending on S. Further, we assume that S(Q) = 8 by rescaling the domain and the Whitney cube structure. This rescaling changes the Lipschitz constant L A of the matrix A(x) corresponding to the elliptic operator. But if the cube Q is small enough, the rescaling improves it, that is, makes L A smaller. Let x̃ be the projection (in the direction ed ) of x on . Then, if S is big enough, we have H (x̃, S(Q)/2k ) H (x Q , S(Q)/2k−1 ) log ≤ log c + log H (x̃, S(Q)/2k+1 ) H (x Q , S(Q)/2k+2 ) ≤ log c + N c , ∀k ∈ {1, . . . , 5}. (5.1) by Lemmas 3.10 and 3.13 (since we do not assume A(x Q ) = I or A(x̃) = I , we use first Remark 3.9). We need S large enough so that B(x̃, A S(Q)/2k ) ⊂ B(x Q , S(Q)/2k−1 ) 123 113 Page 30 of 52 J. M. Gallegos Fig. 1 The domain , the Lipschitz graph , and the half ball B2,+ xQ B2,+ ∩ Ω Σ x̃ x1 B2,+ ∩ Ωc Γ0 and B(x̃, S(Q)/2k+1 ) ⊃ B(x Q , A S(Q)/2k+2 ) (we are using that A is very close to 1) for k = 1, . . . , 5. From now on, S is fixed. We also fix the following normalization for u: sup |u| = 1 (5.2) B(x̃,3)∩ (remember that S(Q) = 8). Let x1 = x̃ − 3τ ed . Observe that 0 = {x = (x  , x  ) ∈ B(x1 , 2) : x  = x1 } doesn’t intersect . Also let B1 := B(x1 , 1), B2 := B(x1 , 2) and Bk,+ the upper half of Bk , k = 1, 2 (the half of Bk that intersects ), see Fig. 1. Let g0 be the solution of div(A∇g0 ) = 0 on B2,+ such that g0 ≡ 1 on ∂ B2,+ \0 and g0 ≡ 0 on 0 . By the maximum principle g0 ≥ 0 on B2,+ and g0 ≥ |u| on  ∩ B2 ⊂ B2,+ because of the normalization (5.2) of u. Notice that, moreover, we have the bound g0 (x) ≤ C1 (x  − x1 ) for x ∈ B1,+ which gives us a bound for |u| in  ∩ B1 . In the case A ≡ I , this follows from reflection and interior Cauchy estimates for ∇u. In the general case, we may use that g0 (x) is comparable to the Green function G A (x, y) of the domain B2,+ (with pole y = (0, . . . , 0, 1.5) for example) by the boundary Harnack inequality inside B1,+ . By [17, Theorem 3.3], since B2,+ satisfies an exterior sphere condition, we have that G(x, y)  dist(x, ∂ B2,+ )|x − y|1−d ≈ dist(x, ∂ B2,+ ). Further, for x ∈ B1,+ we have that dist(x, ∂ B2,+ ) = x  − x1 which gives us the desired bound. Let g be the solution of div(A∇g) = 0 in B1,+ with Dirichlet boundary conditions g ≡ u on ∂ B1,+ ∩  and g ≡ 0 on ∂ B1,+ \. Since |u| ≤ g0 in B2,+ ∩ , we have |g| ≤ g0 ≤ C1 (x  − x1 ). Also, ! 0, on  ∩ ∂ B1,+ , |g − u| = (5.3) |g| ≤ 4C1 τ, on ∂ ∩ B1,+ , because of the bound of |g| ≤ g0 ≤ C1 (x  − x1 ). 123 Size of the zero set of solutions of elliptic PDEs near the… Page 31 of 52 113 By (5.1), Remark 3.9, Lemma 3.12, Lemma 3.11, Proposition 3.2, and (5.2), we have ˆ ˆ ˆ    |u|2 dσ  e−c N |u|2 dσ ≈ e−c N H (x̃, 4)  e−c N |u|2 dy ∂ B(x̃, 81 )∩ ∂ B(x̃,4)∩ B(x̃,4) ˆ −c N −c N −c N ≥e |u|2 dy  e |u(xsup )|2 = e B(xsup ,1) where xsup is a point in ∂ B(x̃, 3) where |u(xsup )| = 1. Assuming that τ is small enough, we have B(x̃, 18 ) ∩  ⊂ 14 B1,+ . Using (5.3) and Lemma 3.13, we get ˆ 1/2 ˆ 1/2 2 g dσ ≥ 2 u dσ − C2 τ ∂ 41 B1,+ ∂ 14 B1,+  ˆ 1/2 ≥ c̃  u 2 dσ − C2 τ. ∂ B x̃, 18 If we assume that τ (depending on N ) is small enough, we conclude that  g    ≥ c1 e−c N /2 . (5.4) L 2 ∂ 41 B1,+ Now let’s estimate the doubling of H g (x1 , r ) for g by using (5.2) and (5.4): ´ 2 ∂ 1 B1,+ g dσ σ (∂ 21 B1 ) log ´ 2 ≤ log  = log c + c N /2 ≤ C(N + 1). 2 ∂ 1 B1,+ g dσ c1 e−c N /2 4 This gives us an upper bound for N g (x1 , 1/4), the frequency for g at x1 . Also note that (5.4) implies  sup |g| ≥ ce−c N /2 . 1 4 B1,+ Then, by Lemma 5.3, there exists x∗ ∈ 0 ∩ B1/S , c2 = c2 (C(N + 1), S) > 0, and ρ = ρ(C(N + 1), S) such that     |g(x)| ≥ c2 x  − x1 for x = x  , x  ∈ B (x∗ , ρ) ∩ B1,+ . We may assume that g > 0 in B (x∗ , ρ) ∩ B1,+ , otherwise we consider −u and −g in place of u and g. From (5.3), we obtain   u(x) ≥ g(x) − 4C1 τ ≥ c2 x  − x1 − 4C1 τ in B (x∗ , ρ) ∩ . (5.5)   Note that ρ does not depend on τ and for τ small enough we have B x∗ , ρ4 ∩ ∂  = ∅. Our goal is to show that u > 0 on B(x ∗ , ρ2 ) ∩  and we will use Lemma 5.4 to this end (from now on, constants δ and M come from the statement of Lemma 5.4). Restrict τ to be small enough so that u > 0 in B(x∗ , ρ) ∩  ∩ {(x  , x  )|x  > x∗ + δρ}, which we can do thanks to (5.5). Now, we multiply the function u by a very large constant K so that K u ≥ M in the same set B(x∗ , ρ) ∩  ∩ {(x  , x  )|x  > x∗ + δρ}. Finally, we make τ even smaller so that K u ≥ −1 in B(x∗ , ρ) ∩ , thanks again to (5.5). Now, by a rescaled version of Lemma 5.4 to B(x∗ , ρ) ∩ , the function K u restricted to B(x∗ , ρ/2) ∩  is positive, which implies that u is positive too and ends the proof.   123 113 Page 32 of 52 J. M. Gallegos 6 Proof of Theorems 1.1 and 1.5 We will combine Lemmas 4.1 and 5.1 to prove Theorems 1.1 and 1.5. Note that, in the present section, we refer to the frequency function N of a solution of div(A∇u) = 0 in  as in the statement of Theorem 1.1. Without further mention, we will use that the constants L A and A − 1 from the matrix A(x) are very small thanks to Remark 2.3. Moreover, we consider  with the Whitney structure defined in Sect. 4.1. First, we give a corollary to Lemma 5.1 in a language closer to that of Lemma 4.1. Corollary 6.1 For any N > 0 and S  1 large enough there exist positive constants τ0 (N , S) and K (N , S) such that the following statement holds. Suppose  ⊂ Rd has Lipschitz constant τ < τ0 and Q is a cube in DW (R) such that N (x Q , S(Q)) ≤ N . Then, for all K̃ ≥ K , there exist cubes Q 1 , . . . , Q (d−1)( K̃ −K ) such that, for all j, 2 (1) the center of Q j lies in , (2) u| Q j ∩ does not have zeros, (3) there exists Q j ∈ DW K̃ (Q) such that Q  is a vertical translation of Q  . j j In particular, there exists δ0 (N , S) > 0 such that     m d−1 (Q i ) ≥ δ0 m d−1 ((Q)). i These cubes Q j are the cubes that are contained in the ball B given by Lemma 5.1 and are K̃ (Q). From now on, given a cube Q ∈ D (R), we denote vertical translation of cubes in DW W by t(Q) the unique cube Q such that its center lies on  and Q  is a vertical translation of  Q. Next, we present a modified frequency function for which we prove good behavior as a consequence of Lemmas 4.1 and 5.1. 6.1 Modified frequency function Let R be a cube in W such that it satisfies the conditions of Lemma 4.1 with S = S1  1 −1/2 large enough so that C S1 (also from the statement of Lemma 4.1) is small enough (we will specify the precise relation later). The use of this lemma gives us constants K 1 := K (S1 ) and δ1 := δ0 (that does not depend on S1 ). Consider also Corollary 6.1 with fixed N = 2N0 + 1 (where N0 is the constant of Lemma 4.1) and S = S2 large enough but smaller than S1 (note that both constants are independent). The use of this corollary gives us constants K 2 := K (N , S2 ) and δ2 := −1/2 δ0 (N , S2 ). In particular, we may assume that K 2 is smaller or equal than K 1 and that C S1 is small enough depending on min(δ1 , δ2 ). We remark that the use of Lemma 4.1 and Corollary 6.1 with constants S1 and S2 respec- tively requires that the domain has small enough Lipschitz constant τ . For the rest of this −1/2 section, we will denote  := C S1 , K := K 1 , and δ0 = min(δ1 , δ2 ) (in particular, δ0 and S1 are independent constants). Define N (Q) := N (x Q , S1 (Q)) 123 Size of the zero set of solutions of elliptic PDEs near the… Page 33 of 52 113 jK for any Q ∈ DW (R) for j ≥ 0. Notice that 1 + 2N (Q) ≥ N (x Q , S2 (Q)) thanks to Proposition 3.8 (we assume x Q and S1 (Q) are admissible for all children Q of R in the Whitney tree structure, thanks to τ being small enough). We define the modified frequency function N  (Q) for Q ∈ DW (R), j ≥ 0, inductively. jK For j = 0, we define N (R) = max(N (R), N0 /2). Assume we have N  (P) defined for all  cubes P ∈ DW " ∈ D( j−1)K (R) and consider its vertical translation i K (R) for 0 ≤ i < j. Fix Q W " t( Q) centered on . Then: " (a) if u restricted to t( Q)∩ " for δ0 ·card{D K ( Q)} has no zeros, define N  (Q) = N  ( Q)/2 " W of the cubes Q in DW ( Q) K " and N (Q) = (1 + )N ( Q)   " for the rest of cubes in D K ( Q) " W (the particular choice is irrelevant), " ∩  has zeros, choose Q ∈ D K ( Q), (b) if u restricted to t( Q) " and W 1. if its vertical translation t(Q) satisfies that u restricted to t(Q) ∩  has no zeros, " define N  (Q) = N  ( Q)/2, 2. else define N  (Q) = max(N (Q), N0 /2). Note that if a cube Q satisfies that u restricted to t(Q) ∩  has no zeros, then all its descen- dants in the Whitney cube structure will satisfy the same property and (a) applies to them. Alternatively, if a cube Q satisfies that u restricted to t(Q) ∩  changes sign, then all its predecessors in the Whitney cube structure will satisfy the same and (b2) applies to them. Now, a combination of Lemma 4.1 and Corollary 6.1 yields the following behavior for N  (Q) for Q ∈ DW (R), j ≥ 0. Consider a cube Q jK " and its vertical translation t( Q). " Then: • If u restricted to t( Q)" ∩  has zeros and N ( Q) " ≥ N0 , then Lemma 4.1 tells us that at least δ0 · card{DW " cubes in D K ( Q) K ( Q)} " satisfy N (Q) ≤ N ( Q)/2. " Moreover, in W this case, N  (Q) = max(N (Q), N0 /2) ≤ N ( Q)/2 " " = N  ( Q)/2 where we have used " > N0 and that (b) applies. For the rest of the cubes Q in D K ( Q), that N ( Q) " we have W  N (Q) = max(N (Q), N0 /2) ≤ (1 + )N ( Q) " = (1 + )N  ( Q)" where we have used again that N ( Q)" > N0 and (b) applies. " • If u restricted to t( Q)∩ " < N0 , then Corollary 6.1 enters in play and has zeros and N ( Q) " it tells us that at least δ0 · card{DW ( Q)} cubes Q in DW K " satisfy that t(Q) ∩  does K ( Q)   " not contain zeros of u. For these cubes, N (Q) = N ( Q)/2 = max(N ( Q), " N0 /2)/2 < N0 /2. For the rest of the cubes Q in DW " we have N  (Q) = max(N (Q), N0 /2) ≤ K ( Q), max((1 + )N ( Q), " N0 /2) ≤ (1 + )N  ( Q)." • If u restricted to t( Q)" ∩  has no zeros, then we have defined N  (Q) = N  ( Q)/2 " for δ0 · card{DW " cubes Q in D K ( Q). K ( Q)} " We have defined N  (Q) = (1 + )N  ( Q) " for W the rest of cubes Q in DW " . K ( Q) Summing up, for Q" ∈ DW (R) and random Q ∈ D K ( Q), " we have W ! " N  ( Q)/2, with probability at least δ0 , N  (Q) ≤  " + ), with probability at most 1 − δ0 . N ( Q)(1 This is similar to the behavior of the frequency function N given by Lemma 4.1 but without " ≥ N0 . the restriction N  ( Q) Let’s summarize the dependence of the constants that have appeared in this section (omit- ting its dependence on the dimension d and on L A and A by assuming we are in a setting like the one described in Remark 2.3). On one hand, the constants K 2 and δ2 given by Corol- lary 6.1 are absolute since they depend on S2 and N = 2N0 + 1 which also are absolute constants. We do require τ small enough to use Corollary 6.1. On the other hand, we have 123 113 Page 34 of 52 J. M. Gallegos −1/2 K 1 depending on S1 ,  = C S1 , and an absolute constant δ1 given by Lemma 4.1. To use Lemma 4.1, we require τ small enough depending on S1 . For the arguments that follow, we need  small enough depending on δ0 = min(δ1 , δ2 ). Though, since both constants ( and δ0 ) are independent, we can choose S1 large enough and, thus, we require τ small enough (depending on S1 ) to use Lemma 4.1. 6.2 Proof of Theorem 1.1 The idea behind the proof is that Lemma 4.1 allows us to use that most cubes in any generation of the Whitney tree satisfy N (x Q , S(Q)) ≤ N0 . Then, we can apply Lemma 5.1 to these cubes, thus covering most of  by balls where u does not change sign. Proof of Theorem 1.1 We will prove the result for the projection of a single cube (R). Afterwards, we can cover any compact in  by a finite union of such cubes which leaves stable the Minkowski dimension estimate. jK For x ∈ (R), we denote by Q j (x) the unique cube Q j ∈ DW (R) such that x ∈ (Q j ) for some integer K large enough that will be fixed later. We will say that Q j (x) is a good cube if N  (Q j ) ≤ 21 N  (Q j−1 ) and that it is bad otherwise. Remark 6.2 Note that, with the previous definitions, N  (Q) < N0 /2 implies that for all x ∈ t(Q)∩ there is a neighborhood where u does not vanish in . Thus we only need to study the Minkowski dimension of the set of points x ∈ (R) such that they are not in (Q) for some Q ∈ DW (R) with N  (Q) < N0 /2. Also, notice that the map  :  ∩ −1 ((R))  → (R) is biLipschitz and thus it preserves Minkowski dimensions. We define the goodness frequency of a point x ∈ (R) as 1 F j (x) = #{good cubes in Q 1 (x), . . . , Q j (x)} j for j ∈ N. We define F(x) = lim supn→∞ Fn (x). Let α(δ0 ) > 0 be such that δ0 1 − α =3 1 − δ0 α (in particular, α < δ0 ) and 0 (α) > 0 such that log(1 + 0 ) α= . log(1 + 0 ) + log 2 Note that for any 0 <  < 0 , we have α 1 (1 + )1−α < 1. (6.1) 2 For all j > 1, define 1 2N  (R) μj = log2 ≥ 0. j N0 Claim For all j > 1, the following holds N0 F j (x) ≥ α + μ j ⇒ N  (Q j (x)) < . 2 123 Size of the zero set of solutions of elliptic PDEs near the… Page 35 of 52 113 Proof of claim We have that j(α+μ j ) jμ j 1 1 N0 N  (Q j (x)) ≤ (1 + ) j(1−α−μ j ) N  (R) < (1 + )− jμ j N  (R) < 2 2 2 by (6.1), and the definition of F j (x) and μ j .   Now, thanks to the previous claim and Remark 6.2, we can reduce the problem to studying the Minkowski dimension of the set of points E = {x ∈ (R) | F j (x) < α + μ j , ∀ j ∈ N}. # a random sequence of cubes (Q j ) j with Q 0 = R and Q j ∈ DW (Q j−1 ), If we consider K and let x ∈ j≥0 (Q j ), the probability that F j (x) ≤ β j for j ∈ N and β j ∈ (0, 1) is bounded above by  jβ j  j i δ (1 − δ0 ) j−i . i 0 i=0 Note that choosing randomly such a sequence is equivalent to choosing a random x ∈ (R) δ0 1−β j uniformly. In what follows we will assume that β j satisfy 2 < 1−δ 0 βj < 4 for all j > 0, in particular β < δ0 . Let’s find an upper bound for the previous quantity for very large j:  jβ j  jβ j  j  j δ0 i A := (1 − δ0 ) j−i δ0i = (1 − δ0 ) j . j i i 1 − δ0 i=0 i=0 Observe that for β j < 1/2 we have j βj j < , for 0 < k ≤ β j j . k−1 1 − βj k This is because  j  k−1 k!( j − k)! k βj  j = = < . (k − 1)!( j − k + 1)! j −k+1 1 − βj k Iterating this inequality, we obtain  jβ j −i j βj j < , for i <  jβ j . i 1 − βj  jβ j Using this observation we can bound A j by  jβ j  jβ j  j δ0 i βj  jβ j j  δ0 1 − β j i (1 − δ0 ) j ≤ (1 − δ0 ) j . i 1 − δ0 1 − βj  jβ j 1 − δ0 β j i=0 i=0 We use Stirling’s formula to approximate √  j j 2π j ej ≈       jβ j  βj j βj j (1−β j ) j (1−β j ) j 2πβ j j e 2π(1 − β j ) j e ⎛ ⎞j 1 ⎝ 1 ⎠ =  . 2πβ j (1 − β j ) j β β j (1 − β j )1−β j j 123 113 Page 36 of 52 J. M. Gallegos We also estimate  jβ j βj  jβ j βj jβ j  δ0 1 − β j i δ0 1 − β j jβ j ≈ and ≈ . 1 − βj 1 − βj 1 − δ0 β j 1 − δ0 β j i=0 Observe that the comparability constants in the previous approximations depend on the upper and lower bounds of β j but not on j for j large enough depending on δ0 . Summing everything up, we obtain ⎛ ⎞j 1−β j δ β j A   2 ⎝ (1 − δ0 ) 0 ⎠ , j 2πβ j (1 − β j ) j βj β j (1 − β j ) j 1−β   z(β j ) j for j large enough. Observe that z(β j ) < 1 for β j < δ0 . jK Now, we will choose a suitable covering of the set E by projections of cubes in DW (R). First, pick  < 0 (equivalently S1 large enough, recall the discussion in Sect. 6.1). Observe that by choosing  we also fix K . For j ≥ 1, set E j := {x ∈ (R) | F j (x) ≤ α + μ j }, # so that E = j E j . Let’s upper bound the (K -adic) Minkowski dimension of E by finding a certain cover of E j by projections of cubes in DW (R) (note that there are M = 2(d−1)K jK cubes in DWK (R)). Using the previous asymptotics (setting β j = α + μ j ), we can cover E j (for j large enough) with 2 CM j  z(α + μ j ) j 2π(α + μ j )(1 − α − μ j ) j projections of cubes in DW (R) and each of those cubes has side length M − j/(d−1) . jK Now we are ready to upper bound the Minkowski dimension of the set E. We will use the following definition of upper Minkowski dimension log #({K − adic cubes Q of side length K − j that satisfy Q ∩ E  = ∅}) dimM E = lim sup . j→∞ j log K which is equivalent to the dyadic one in (1.1). By covering a single set E j ⊃ E and making j → ∞, we obtain the following upper Minkowski dimension estimate j(ln M + ln z(α + μ j )) + ln √ 2 2π(α+μ j )(1−α−μ j ) j dimM (E) ≤ lim j→∞ (d − 1)−1 j ln M ln M + ln z(α) = (d − 1) 0 and where x Q is the center of Q. Remark 6.4 If we assume N (x Q , 16 diam(Q)) > N0 for some N0 positive, we can rewrite the previous theorem as Hd−1 ({u = 0} ∩ Q) ≤ C(Q)d−1 (N (x Q , 16 diam(Q)))α where now C depends also on N0 and α. 6.5 Proof of Theorem 1.5 To prove Theorem 1.5 we will follow the ideas of [28] but exchanging the use of the Donnelly– Fefferman estimate for the size of nodal sets (see [10]) by Logunov’s estimate (Theorem 6.3). This gives rise to a worse estimate (polynomial in the frequency) than the one of [28] (which is linear in the frequency but only valid for harmonic functions). 123 113 Page 38 of 52 J. M. Gallegos Proof of Theorem 1.5 Let Q be a small enough cube of the Whitney structure (so that we can use the modified frequency function defined in Sect. 6.1). Again, we will use the notation t(Q) for the unique cube Q  with center on  such that Q  is a vertical translation of Q. Claim The following equation holds Hd−1 ({u = 0} ∩ t(Q) ∩ K ) ≤ C0 N  (Q)α (Q)d−1 (6.2) for any K compact inside  with C0 and α independent of K . Proof of claim Note that Eq. (6.2) holds for all cubes Q small enough, since t(Q) ∩ K = ∅ if Q is small enough. This is because dist(K , ) > 0. We will proceed to prove this estimate by induction going from small cubes to large cubes. Assume it holds for all small cubes with (Q) < s . Now choose a larger cube P in the Whitney cube structure with (P) < 2 K (Q) (with K given by Lemma 4.1 as discussed in Sect. 6.1). Given such a cube P, we can cover t(P) ∩  with small cubes t(Q i ) (intersecting the boundary) where Q i ∈ DW K (P) and with small cubes Q  far from the boundary (small i enough so that we can apply Theorem 6.3 on them). Using that the Lipschitz constant of ∂ is small, we can bound the number of small cubes Q i necessary. Moreover, using Lemma 3.14, we can bound N (x Q  , 16 diam(Q  )) < 2N (P) + 1. Note that N (P) ≤ N  (P) inthe case that t(P) ∩  ∩ {u = 0}  = ∅. This allows us to bound the size of the nodal set on Q i ∩  using Theorem 6.3 by  Hd−1 ({u = 0} ∩ Q i ) ≤ C1 N  (P)(P)d−1 . Q i We still need to bound the size of the nodal set in the boundary cubes t(Q i ), which satisfy N  (Q i ) ≤ (1 + )N  (P). Moreover, we know that at least for δ0 card{DW K (P)} cubes Q i  N  (P) we have N (Q i ) ≤ 2 . Now we can use the induction hypothesis   Hd−1 ({u = 0} ∩ K ∩ t(Q i )) ≤ Hd−1 ({u = 0} ∩ K ∩ t(Q i )) Qi N  (Q i )>N  (P)/2  + Hd−1 ({u = 0} ∩ K ∩ t(Q i )) N  (Q i )≤N  (P)/2  ≤ C0 N  (Q i )α (Q i )d−1 N  (Q i )>N  (P)/2  + C0 N  (Q i )α (Q i )d−1 N  (Q i )≤N  (P)/2 δ0 ≤ C0 N  (P)α (P)d−1 (1 + )α (1 − δ0 ) + . 2α We choose  small enough (by increasing S1 ) so that δ0 (1 + )α (1 − δ0 ) + <1 2α noting that δ0 does not depend on S1 or . Finally, we choose C0 large enough so that it absorbs all terms, that is, δ0 C0 (1 + )α (1 − δ0 ) + + C1 < C0 . 2α 123 Size of the zero set of solutions of elliptic PDEs near the… Page 39 of 52 113 Note that the previous estimates do not depend on the compact K chosen and we prove the claim.   To treat a general (small) ball B centered at , we first cover it by a comparable cube Q centered at . Now, we choose a translate and dilation of the dyadic cube structure of Rd such that Q = t(P) for some P in the Whitney cube structure and we apply the previous claim.   Remark 6.5 In the statement of Theorem 1.5, the point x̃ that appears is the center of a particular Whitney cube appearing in the proof. Nonetheless, Lemma 3.14 (and that the Lipschitz constant τ of the boundary is small to preserve admissibility) gives us a lot of freedom to choose x̃. 6.6 (d − 1)-dimensional Hausdorff measure of Dirichlet eigenfunctions Theorem 1.5 allows us to study the zero set of solutions of the Dirichlet eigenvalue problem ! div(A∇u λ ) = −λu λ , in , u λ = 0, on ∂. In fact, one can show that α Hd−1 ({u λ = 0}) ≤ C(, A , L A )λ 2 (6.3) for bounded domains  with local Lipschitz constant small enough for some α = α(d) > 1. For a detailed account of the proof in the harmonic case (and a sharper result), see Section 6 in [28]. We will only briefly sketch the main ideas behind its proof. Also note that this problem is intimately related to Yau’s conjecture on nodal sets of Laplace eigenfunctions in manifolds (see [10, 24, 25, 27]). The first step consists in passing from √ eigenfunctions to solutions of div(A∇u) = 0. Consider the function u(x, t) = u λ (x)e λt in the cylinder domain  × R ⊂ Rd+1 . Let A(x) 0 Ã(x) = , 0 1 then we have that u solves √ √ λt λt div( Ã∇u) = div(A∇u λ ) e + λu λ (x)e = 0.   −λu λ (x) Clearly, we have {u(x, t) = 0} = {u λ (x) = 0} × R. Thus, we can restrict us to the study of the nodal set of u(x, t). The next necessary step is the Donnelly–Fefferman frequency estimate [10]. For small balls √ B contained in the domain, it is shown in [10, 26] that N u (x B , r (B)) ≤ C(, A , L A ) λ. The previous result is also true for balls intersecting the boundary (see Lemma 10 in [28] for a proof in the harmonic case). Finally, using the previous estimate together with Theorems 1.5 and 6.3, one obtains the result in (6.3). 123 113 Page 40 of 52 J. M. Gallegos 7 Proof of Corollary 1.6 Theorem 1.1 tells us that we can decompose  in its intersection with a countable family of balls (Bi )i and a set of Hausdorff dimension smaller than d − 1 by taking an exhaustion of  by compacts. Thanks to countable additivity, we only need to prove that Hd−1 ({x ∈  | ∂ν u(x) = 0} ∩ B) = 0 for any ball B ∈ (Bi )i given by the decomposition of Theorem 1.1. Before starting the proof, we define the concept of A∞ weight. Definition 7.1 We say that a measure ω ∈ A∞ (σ ) if there exist 0 < γ1 , γ2 < 1 such that for all balls B and subsets E ⊂ B, σ (E) ≤ γ1 σ (B) implies ω(E) ≤ γ2 ω(B). Proof of Corollary 1.6 Consider B centered on  such that u| B∩ does not change sign. With- out loss of generality, we assume that u is positive. By Dahlberg’s theorem [8], harmonic measure for the domain B ∩  is an A∞ weight with respect to surface measure. By [12], since the matrix A(x) is uniformly elliptic and has Lipschitz coefficients, its associated elliptic measure ω A is another A∞ weight. In particular, it is well known that this implies that the density dω A dσ can only vanish in a set of zero surface measure. On the other hand, the density of elliptic measure is comparable with (A∇g, ν) at the boundary (where g is the Green function with pole outside 2B). By the boundary Harnack inequality (see [11] for example), since u is positive in B ∩ , we have that A∇u on  ∩ B is comparable to A∇g. This finishes the proof.   Remark 7.2 There is also a different approach to proving Corollary 1.6, which is the one adopted in [33] for harmonic functions. The ingredients are Lemma 4.1, [2, Lemma 0.2] (which is also valid for the type of PDEs we consider, see the paragraph below the proof of [2, Lemma 2.2], also [33, Lemma 4.3]), and a modification of [33, Lemma 4.1]. In particular, the tools of Sects. 5 and 6 are not indispensable for this result. 8 Proof of Corollary 1.8 Definition 8.1 We define a non-truncated cone C̃τ with aperture τ ∈ R and vertex at 0 as C̃τ (0) = {(y  , y  ) ∈ Rd−1 × R | y  > τ |y  |}. Notice that when τ = 0, then C̃0 is a half space and when τ < 0, C̃τ is a non-convex cone. A truncated cone Cτ,s is a non-truncated cone C̃τ intersected with the ball B(0, s). First we will prove the result for harmonic functions. Proof of Corollary 1.8 in the harmonic case Let x ∈ B ∩  for some ball B where u does not vanish given by Theorem 1.1. Assume without loss of generality that u| B∩ > 0. Since  is Lipschitz with Lipschitz constant τ , we can find a small truncated cone Cτ,s of aperture τ with vertex at x and contained in . Let g be the Green function (for the Laplacian) with pole at infinity of the non-truncated cone C̃τ . Since u|Cτ,s is nonnegative up to the boundary, we can lower bound it by some adequate multiple of g (that is u  g in Cτ,s ) by boundary Harnack inequality. Notice that g is of the form g(x) = gr (|x|)gθ ( |x| x ) where gθ is the first Dirichlet eigenfunction 123 Size of the zero set of solutions of elliptic PDEs near the… Page 41 of 52 113 of the Laplace-Beltrami √ operator  in the domain C̃τ ∩ ∂ B(0, 1) with eigenvalue λτ and d (d−2)2 −4λτ gr (|x|) = |x|− 2 +1+ 2 (see [1, Theorem 1.1]). Thus gr gives an upper bound on the order of vanishing at the origin. Also notice that λ0 = 1 − d (when C̃0 is a half space) and thus gr (|x|) = |x|. Since the Dirichlet eigenvalues λτ of C̃τ ∩ ∂ B(0, 1) for the Laplace- Beltrami operator vary continuously with the aperture τ of C̃τ , we can ensure that the order of vanishing of g is close enough to 1 by making τ small enough. The continuity of the variation of the eigenvalues with the aperture can be easily shown using the Rayleigh quotient. For the lower bound on the order of the vanishing at the origin, consider a cone C̃−τ of aperture −τ (concave cone) and its Green function g with pole at the infinity. Now g|∩C̃  u and we can follow the same argument.   Next, we deal with the elliptic case, but first, we need several lemmas. Lemma 8.2 Let u be a solution of div(A∇u) = 0 in  and x 0 ∈  ∪ . Then the vanishing order α of u at x0 satisfies √ H (x0 , 2r ) α(x0 ) = lim log2 √ r ↓0 H (x0 , r ) when the limit exists. Proof The proof has two parts. First, we see that ˆ  − |u| dx ≤ Cα r α ⇐⇒ H (x0 , r ) ≤ Cα r α . B(x0 ,r ) Notice that for r small enough, we have ˆ ˆ 1/2  − |u|dy ≤ − |u(y)| dy 2  H (x0 , r ) −1 −1 B(x0 , A r) B(x0 , A r) thanks to Cauchy–Schwarz and Lemma 3.12. On the other hand, if we let x∗ ∈ B(x0 , r ) be such that |u(x∗ )| = sup B(x0 ,r ) |u|, then  ˆ ˆ H (x0 , r )  sup |u|  − |u|dy  − |u|dy −1 −1 B(x0 ,r ) B(x∗ , A r) B(x0 ,2 A r) using standard estimates for solutions of elliptic PDEs. The second part consists in showing that √ H (x0 , 2r ) lim log2 √ =α r ↓0 H (x0 , r ) implies √  H (x0 , r ) H (x0 , r ) ≤ Cγ r γ , 0 < γ < α, and lim = ∞, γ > α. r →0 rγ Fix γ < α. Then there exists some kγ > 0 such that for all k > kγ , we have  H (x0 , 2−k+1 ) log2  > γ. H (x0 , 2−k ) 123 113 Page 42 of 52 J. M. Gallegos Then, for all n large enough, we have    $ n H (x0 , 2− j )  γ H (x0 , 2−n ) = H (x0 , 1)  < C2−γ (n−kγ ) = C  2−n . j=1 H (x0 , 2− j+1 ) Now, fix γ > α. As before,there exists some kγ > 0 such that for all k > kγ , we have  H (x0 , 2−k+1 ) log2  < γ − δ. H (x0 , 2−k ) for some small δ > 0. Finally, for all n large enough, we obtain    $n H (x0 , 2− j )  γ −n H (x0 , 2 ) = H (x0 , 1)  > C2−(γ +δ)(n−kγ ) > C  2−n 2δn . j=1 H (x0 , 2− j+1 ) Since limn→∞ 2δn = ∞, this finishes the proof.   Lemma 8.3 Let u be a positive solution to div(A∇u) = 0 in a cone Cτ,s such that u = 0 on ∂Cτ,s ∩ B(0, s). Assume also that A(0) = I . Then, its frequency N (0, r ) is bounded above for 0 < r < 2s . Proof Remember that ´ ´ B(0,r ) (A∇u, ∇u)dy B(0,r ) |∇u|2 dy N (0, r ) = r ´ ≈r ´ . ∂ B(0,r ) μu dσ 2 2 dσ ∂ B(0,r ) u Using that u extended by 0 outside of Cτ,s is a subsolution, by Caccioppoli’s inequality we have ˆ ˆ 1 |∇u| dx  2 2 u 2 dx. B(0,r ) r B(0,2r ) By Remark 3 (page 953) in [2], we obtain the doubling property ˆ ˆ u 2 dx  u 2 dx. −1 B(0,4r ) B(0, A r) uniformly for all r small enough. The remark is stated for convex domains, but the condition needed is that the domain is star-shaped with respect to 0. In the case of cones, this is trivially true. Using Lemma 3.12, we obtain that ˆ ˆ − u 2 dx  − u 2 dσ. −1 B(0, A r) ∂ B(0,r ) Summing up, we get ´ B(0,r ) |∇u|2 dx N (0, r ) ≈ r ´ 2 dx 1 ∂ B(0,r ) u for all r small enough.   The following lemma shows that the blow-up of positive solutions in cones converges to the Green function for the Laplacian in the domain C̃τ with pole at ∞ (see the proof of Corollary 1.8 in the harmonic case). 123 Size of the zero set of solutions of elliptic PDEs near the… Page 43 of 52 113 Lemma 8.4 Let u be a positive solution of div(A∇u) = 0 in a truncated cone Cτ,s that vanishes on ∂Cτ,s ∩ B(0, s) and assume that A(0) = I . Consider any sequence of radii rk ↓ 0 (such that r1 < | p|). Let u(rk x) u k (x) := ´ 1/2 . ∂ B1 |u(r k y)| 2 dσ (y)   1,2 1,α Then u k converges in Wloc (C̃τ ) and in Cloc C̃τ \{0} to a multiple of the Green function g with pole at ∞ for the Laplacian in C̃τ for some exponent 0 < α < 1. In particular, u and g have the same vanishing order at 0. Proof Without loss of generality assume that s > 2 and r1 = 1. Clearly, u ∈ W 1,2 (Cτ,1 ) and u k ∈ W 1,2 (Cτ,1/rk ) for all k > 0. Also notice that N u (0, rk ) = N u k (0, 1) where N u k is the frequency for the new PDE. Moreover, each u k satisfies 1 ≈ H u k (0, 1) and ∇u k 2L 2 (C ) ≈ N u k (0, 1). Observe that we can also bound ∇u k  L 2 (Cτ,1/r ) and H u k (0, r j ) τ,1 j for any j ≤ k (with bound depending on r j ). Thanks to Lemma 8.3, we have that the sequence (u k )k is bounded in Sobolev norm in any bounded set. Moreover, by boundary Schauder estimates (see Lemma 6.18 in [16]), u k ∈ C 1,α (K ) for any compact K ⊂ C̃τ \{0} (for k large enough depending on K ). By Arzelà-Ascoli, there is a subsequence (u k j ) j that converges in norm C 1 to ũ in Cτ,1 \B1/8 . 1,2 Moreover, we can also assume that ũ is a weak limit in Wloc (C̃τ ). Thus, ũ is harmonic in C̃τ . To see this, fix any compact K ⊂⊂ C̃τ . Then, for all φ ∈ CC1 (K ), we have ˆ (A(rk x)∇u k , ∇φ)dx = 0, ∀k > 0. K On the other hand, since ũ is the weak limit of u k , we have ˆ ˆ (∇ ũ, ∇φ)dx = lim (∇u k , ∇φ)dx K k→∞ K ˆ ˆ = lim (A(rk x)∇u k , ∇φ)dx − ((A(rk x) − I )∇u k , ∇φ)dx . k→∞ K   K  A B The term A is zero by the definition of weak solution and we can estimate B as ˆ ((A(rk x) − I )∇u k , ∇φ)dx  L A rk (dist(0, K ) K + diam(K ))∇u k  L 2 (K ) ∇φ L 2 (K ) →k→∞ 0. Summing up, we get that ũ is harmonic in K . Claim The only functions which are positive and harmonic in C̃τ are the multiples of the Green function g for the Laplacian in C̃τ with pole at ∞. For the proof of the claim see [18, Lemma 3.7]. ´ Since we have C 1 convergence away from the pole and ∂ B1 u 2k dσ = 1 for all k > 0, the multiple of the Green function in the claim is fixed. Thus, the convergence does not depend 123 113 Page 44 of 52 J. M. Gallegos on the sequence rk chosen and is not up to subsequences. For this reason, we assume without loss of generality that rk = 2−k+1 . We will see now that the limit   lim log2 H u (x, 2r ) − log2 H u (x, r ) r ↓0 exists and by Lemma 8.2 this coincides with the order of vanishing at zero. If r ∈ (1/2k , 1/2k−1 ), then    H u k−1 (0, 2k−1 r ) log2 H u (0, 2r ) − log2 H u (0, r ) = log2  . H u k−1 (0, 2k−2 r ) Since r ∈ (1/2k , 1/2k−1 ), this happens in the region where u k converges to ũ in C 1 norm. Thus H u k (0, r ) →k→∞ H ũ (0, r ) for 1/8 < r < 1. By Lemma 8.2 this implies that both u and ũ have the same vanishing order.   The general elliptic case is a direct consequence of the previous lemma following the same proof as in the harmonic case. 9 Example with dimH ({x ∈ 6 | limr↓0 !(B(x, r))r 1−d = 0}) > d − 1 −  The following example by Xavier Tolsa shows that in a Lipschitz domain  (even with small Lipschitz constant) there is no hope for a (non-trivial) Hausdorff dimension bound of the set of points of  where ∂ν u(x) = 0. Remark 9.1 We are interested in the normal derivative, but it may not exist at every point x ∈ . Nonetheless, since we are in a Lipschitz domain  for every point x ∈  we can consider a non-tangential cone Cτ (x) contained in the domain. Thus, we can consider a non-tangential approach to the normal derivative. That is, for x ∈ , we define |u(y) − u(x)| ∇nt u(x) = lim sup . y→x |y − x| y∈Cτ (x) We will show that the set {x ∈  | ∇nt u(x) = 0} can be very large (in terms of Hausdorff dimension). The following lemma shows that non-tangential derivatives for positive harmonic functions are closely related with the density of harmonic measure. Lemma 9.2 Assume u is positive and harmonic in  and vanishes in . Then, for x ∈ , ∇nt u(x) = 0 if and only if limr ↓0 ω(B(x,r r d−1 )) = 0 where ω is the harmonic measure for the Laplacian in . Proof of Remark 9.1 Fix a point p ∈  far from  and consider the Green function g with pole at p. Then, in a neighborhood of , the boundary Harnack inequality (see [11]) implies that the non-tangential derivative of u is 0 if and only if the non-tangential derivative of g is 0 wherever one of the two exists (if one exists and is 0 the other exists too by boundary Harnack inequality). This does not depend on the pole p chosen. Now, the non-tangential derivative of g is 0 if and only if limr ↓0 ω(B(x,r r d−1 )) = 0. This is a consequence of [8, Lemma 1] which shows r d−2 g(x + Cr ed ) ≈ ω p (B(x, r )) for some C depending on the dimension and the Lipschitz constant of the domain.   123 Size of the zero set of solutions of elliptic PDEs near the… Page 45 of 52 113 Now we can start setting up an appropriate domain. Consider, for λ ∈ (0, 1), the λ-Cantor set defined as ∞ % Cλ = E nλ n=1 where E 0λ = [0, 1] and E kλ = 1−λ λ 2 E k−1 ∪ ( 2 + 1+λ 1−λ λ 2 E k−1 ) for k ≥ 1. We will refer to the λ intervals in [0, 1]\E k by gaps. Remark 9.3 The set Cλ has Hausdorff dimension between 0 and 1 depending on λ, but by making λ small enough we can obtain a set with dimension arbitrarily close to 1. From now, on we will denote it by s = s(λ) = dimH Cλ . Remark 9.4 If λ = 1/(2k + 1) for some k ∈ N, then the set Cλ coincides with the set of real numbers in [0, 1] such that its decimal expansion in basis 2k + 1 does not contain the digit k. Instead of choosing λ, we will choose k large enough so that the Hausdorff dimension s of Cλ is as close as we want to 1. Theorem 9.5 Let λ = 1/(2k + 1) for some k ∈ N, and s = s(λ) = dimH (Cλ ). Then, Hs - a.e. x ∈ Cλ satisfies that each possible digit {0, 1, . . . , k − 1, k + 1, . . . , 2k} in its decimal representation in basis 2k + 1 appears with the same asymptotic frequency 1/(2k). The proof of the theorem is analogous to the proof of Borel’s normal number theorem. Also, we will say that the points x ∈ Cλ that satisfy this are Cλ -normal. Fix a ∈ (0, 1) and let  λ := B(0, 2) ∩ {Ca ( p)} ⊂ R2 , p∈Cλ where Ca ( p) are vertical open cones of aperture a with vertex at p. Thus, λ is the union of all the open cones with aperture a centered at a point of the Cantor set Cλ . The domain λ clearly has Lipschitz boundary and its Lipschitz constant can be made arbitrarily small as it depends only on the aperture of the cones a. Remark 9.6 Let the interval (ξ1 , ξ2 ) be a gap in [0, 1]\E λj . Then ∂ (Ca (ξ1 ) ∪ Ca (ξ2 )) ∩ {(x, y) ∈ R2 | x ∈ (ξ1 , ξ2 )} ⊂ ∂λ . In other words, the boundary of ∂λ looks like a cone (pointing upwards) where the Cantor set Cλ has gaps. Note that, if λ = 1/(2k + 1), the gaps correspond to subintervals of [0, 1] where all the numbers have the digit k in the position j of the decimal expansion in base 2k + 1. Now we will prove that every Cλ -normal point (x, 0) ∈ Cλ ⊂ R ∩ λ satisfies that the density of harmonic measure at (x, 0) is 0. Proposition 9.7 Let λ be the Lipschitz domain defined above. Let (x, 0) be a Cλ -normal point in Cλ . Then limr ↓0 ω(B(x,r r d−1 )) = 0 in (x, 0) where ω is the harmonic measure of λ with pole in (x, 1). Proof Theorem 6.2. in [15, page 149] (called dx/θ (x) estimate) tells us ˆ 1 8 ds ω(B((x, 0), r )) ≤ exp −π (9.1) π r s(s) 123 113 Page 46 of 52 J. M. Gallegos for r small enough where s(s) is the length of the connected arc in  ∩ ∂ B((x, 0), s) that separates (x, 0) and the pole (x, 1). We want to prove that ω(B((x, 0), r )) lim sup =0 r →0 r which is a consequence of ˆ 1 1 1 ds lim sup −π − → −∞ r →0 r (s) π s by inequality (9.1). Since the domain is contained in the upper half plane and x ∈ R we have that 0 < (s) ≤ π. This implies that we can change the lim sup by a lim. Also we can rewrite the integral as  n ˆ (2k+1)1− j 1 1 ds −π lim − . n→∞ (2k+1)− j (s) π s j=1 Now we will use that x is Cλ -normal. This implies that, asymptotically, for a fixed proportion of digits in the decimal expansion of x in basis 2k + 1 the digit is k − 1 or k + 1. If the j-th digit in the decimal expansion of x in basis 2k + 1 is either k − 1 or k + 1, then for s ∈ [2.1(2k + 1)− j , (2k + 1)1− j ] we have that (s) < π −  for some  > 0 depending only on the aperture of the cones a. This is because the arc of circumference centered at (x, 0) with radius s meets with a cone Ca ( p) corresponding to a gap in [0, 1]\E λj (see Remark 9.6) where p is the truncation to j digits of x in basis 2k + 1. Thus, (s) < π − . Since x is Cλ -normal, this happens for a positive proportion of digits, the integral tends to −∞, and limr ↓0 ω(B(x,r r d−1 )) = 0.   As a consequence of Proposition 9.7 and the previous claim, we can construct a domain where limr ↓0 ω(B(x,r r d−1 )) = 0 in a set of dimension s and s can be made as close to 1 as desired. This implies that the non-tangential derivative of a harmonic function (the Green function of the domain) can be 0 in a set of Hausdorff dimension arbitrarily close to 1. Moreover, we can obtain a domain where this set has Hausdorff dimension 1 by concatenating domains of the form λ considered before. 10 S6 (u) = S6 (u) in the C 1,Dini case Recall that for a harmonic function u in a Lipschitz domain that vanishes in a relatively open subset of the boundary  ⊂ ∂, we define S (u) = {x ∈  | |∇u(x)| = 0} and  S (u) = {x ∈  | u −1 ({0}) ∩ B(x, r ) ∩   = ∅, ∀r > 0}. Remark 10.1 Note that S (u) is well defined in the C 1,Dini case since u ∈ C 1 ( ∪ ) thanks to the results of [9]. The proof of Proposition 1.9 follows from a local expansion of u as the sum of a homogeneous harmonic polynomial and an error term of higher degree (see [22, Theorem 1.1]). 123 Size of the zero set of solutions of elliptic PDEs near the… Page 47 of 52 113 Proof of Proposition 1.9 By [22, Theorem 1.1], for every x ∈  there exists a positive radius R = R(x) and a positive integer N = N (x) such that u(y) = PN (y − x) + ψ(y − x), in B R (x) ∩  where PN is a non-trivial homogeneous harmonic polynomial of degree N and the error term ψ satisfies lim |ψ(y)||y|−N = 0 and lim |∇ψ(y)||y|−N +1 = 0. y→0 y→0 We will show that N (x) = 1 implies x ∈ / S (u) and x ∈/ S  (u) and that N (x) > 1  implies x ∈ S (u) and x ∈ S (u). Case N = 1: We have that ∇u(x) = ∇ P1 (0) + ∇ψ(0) = ∇ P1 (0)  = 0 since P1 is a non- trivial linear function, thus x ∈/ S (u). Also, we know that u(y) = 0 for y ∈ ∂ ∩ B R (x). Suppose there exist (z n )n ∈  ∩ B R (x) tending to x such that u(z n ) = 0. Then, by Rolle’s theorem, we get a contradiction since the derivative of u in the direction ∇ P1 /|∇ P1 | does not vanish in a neighborhood of x (because lim y→0 |∇ψ(y)| = 0) but u vanishes on (z n )n and on ∂ ∩ B R (x). Thus x ∈ / S (u). Case N > 1: Clearly, ∇ PN (0) = 0 if N > 1 which implies that x ∈ S (u). Assume that x ∈/ S  (u) and, without loss of generality, u is positive near x. Then, by the generalized Hopf principle of [31], we get that ∂ν u(x) < 0 contradicting x ∈ S (u).   Acknowledgements Part of this work was carried out while the author was visiting the Hausdorff Research Institute for Mathematics in Bonn during the research trimester Interactions between Geometric measure theory, Singular integrals, and PDE. The author thanks this institution and its staff for their hospitality. The author is also grateful to Xavier Tolsa for his guidance and advice, to Jaume de Dios for some useful discussions, and to the anonymous referee that helped improve the readability of the paper. Funding Open Access Funding provided by Universitat Autonoma de Barcelona. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. Appendix A. Existence of non-tangential limits for ∇u We will closely follow the ideas of Appendix A of [33] in order to prove the L 2 convergence of the non-tangential limits of ∇u in . Let  ⊂ Rd be a Lipschitz domain, B a ball centered on ∂,  = B ∩ ∂, and σ denote the surface measure on . Without loss of generality, we assume  is locally above  in the direction of ed = (0, . . . , 0, 1). For σ -a.e. x ∈ , the outer unit normal vector ν(x) is well defined. For a parameter a ∈ (0, 1) and x ∈ , we consider the inner cone and outer cone X a+ (x) = {y ∈ Rd |(x − y, ν(x)) > a|y − x|}, X a− (x) = {y ∈ Rd | − (x − y, ν(x)) > a|y − x|}, respectively. For a function f defined on Rd \, we define the non-tangential limits f +,a (x) = lim f (y), f −,a (x) = lim f (y), X a+ (x)$y→x X a− (x)$y→x 123 113 Page 48 of 52 J. M. Gallegos when they exist. We prove the following theorem about the convergence of the non-tangential limits of the gradient of the solution of an elliptic PDE (of the type of Sect. 2.1). Theorem A.1 Let  ⊂ Rd be a Lipschitz domain, B be an open ball centered in ∂, and  = B ∩ ∂ be a Lipschitz graph such that  ∩ B is above . Let u be a solution of div(A∇u) = 0 in  with A as in Sect. 2.1. Assume that u is continuous up to the boundary and that it vanishes continuously on . Then, for all a ∈ (0, 1) large enough, (∇u)+,a exists σ -a.e. and belongs to L 2loc (σ ). Further, (∇u)+,a has vanishing tangential component. That is, (∇u)+,a = (∇u, ν)ν. Further, lim ∇u (· + εen ) → (∇u, ν) ν in L 2loc (σ ). ε→0 Proof We extend u by 0 out of  and denote u + = max(u, 0), u − = max(−u, 0). Both u + and u − are continuous and subsolutions (that is − div(A∇u ± ) ≤ 0 in B). Claim In the sense of distributions, div(A∇u) = div(A∇u + ) − div(A∇u − ) restricted to B is a signed Radon measure supported on . Proof of the claim To see this, take arbitrary φ ∈ Cc∞ (B). For 0 <  % r (B), denote  =  + ed and  =  + ed , where ed = (0, . . . , 0, 1). Let ψδ be a bump function in an δ-neighborhood of  such that ∇ψδ ∞  δ −1 and ∇ 2 ψδ ∞  δ −2 . Then, writing φ = φψδ + φ(1 − ψδ ) for φ ∈ Cc∞ (B), we get ˆ  ˆ  ˆ          A∇u∇φdx  ≤     A∇u∇(φψδ )dx  +  A∇u∇(φ(1 − ψδ ))dx  .  B∩  B∩   B∩  A B We directly obtain B = 0 since u is a weak solution of the PDE in . We have, because u ∈ W 1,2 (B) and the divergence theorem, that ˆ ˆ A∇u∇(φψδ )dx = lim A∇u∇(φψδ )dx B∩ →0 B∩ ˆ ˆ = lim u(A∇(φψδ ), ν)dσ − u div(A∇(φψδ ))dx →0 ∂ ∩B  ∩B     C D The term C converges uniformly to 0 as  → 0 as u is 0 in . On the other hand, | div(A∇(φψδ ))| ≤| div(A∇ψδ )φ| + | div(A∇φ)ψδ | + |2(A∇φ, ∇ψδ )| 1 1  2 φ L ∞ (supp ψδ ) + ∇ 2 φ∞ + ∇φ∞ δ δ with constants depending on the Lipschitz and ellipticity constants of A. We can bound D as ˆ     u div(A∇(φψ ))dx   δ  ∩B ˆ 1 1  φ L ∞ (supp ψδ ) + ∇ 2 φ∞ + ∇φ∞ |u|dx. (A.1) δ 2 δ supp ψδ ∩ 123 Size of the zero set of solutions of elliptic PDEs near the… Page 49 of 52 113 Using the boundary Harnack inequality (see [11], for example), we can bound ˆ ˆ |u|dx  g dx  δ 2 supp ψδ ∩ supp ψδ ∩ where g is the Green function with fixed pole far from B. To prove the second inequality, we cover supp ψδ by a finite family of cubes Q i with side length (Q) ≈ δ. We can do this with  d−1 approximately r (B) δ cubes. By standard estimates for elliptic measure, we have ω(4Q) g(x)  ≈ ω(4Q)δ 2−d for all x ∈ Q ∩ . (Q)n−2 where ω is the elliptic measure associated to div(A∇·) for  with respect to a fixed pole p ∈ \B. Finally, we have ˆ ˆ  ˆ g dx ≤ g dx  ω(4Q)δ 2−d dx ≈ δ 2 ω() supp ψδ ∩ Q Q∩ Q Q where we have used the doubling properties of elliptic measure. Summing up, we take the limit as δ → 0 in Eq. (A.1) to obtain ((div(A∇u)), φ) ≤ Cφ L ∞ () where we have used that the supp ψδ →  as δ → 0, Thus div(A∇u)| B is a Radon measure supported on .   Moreover, this Radon measure is absolutely continuous with respect to elliptic measure. Claim In the sense of distributions, we have div(A∇u)| B = ρω| , ∞ () and ω is the elliptic measure associated to div(A∇·) for  with respect where ρ ∈ L loc to a fixed pole p ∈ \B. We can assume that B is small enough so that \2B  = ∅ and p ∈ \2B. Proof of the claim To prove the claim, let B  be an open ball concentric with B such that B  ⊂ B. We will show that that there exists some constant C depending on B  and p such that for any compact set K ⊂  ∩ B  , it holds (div(A∇u), χ K ) ≤ Cω(K ). (A.2) By duality, thisimplies the claim.  Given  ∈ 0, 21 dist(K , Rd \B  ) , let {Q i }i∈I be a lattice of cubes covering Rd such that each Q i has diameter /2. Let {φi }i∈I be a partition of unity of Rn , so that each φi is supported in 2Q i and satisfies ∇ j φi ∞  (Q i )− j , for j = 0, 1, 2. Then, we have   (div(A∇u), χ K ) = (div(A∇u), φi ) − (div(A∇u), φi − χ K ), i∈I  i∈I  where I  is the collection of indices i ∈ I that satisfy 2Q i ∩ K  = ∅. By the regularity properties of Radon measures, we obtain    |(div(A∇u), φi − χ K )| ≤ | div(A∇u)|, χU (K )\K → 0 as  → 0, i∈I  123 113 Page 50 of 52 J. M. Gallegos where U (K ) is the -neighborhood of K . For the other term, we have    1 ˆ |(div(A∇u), φi )| ≤ |(u, div(A∇φi ))|  |u|dm    (Q i )2 2Q i i∈I i∈I i∈I using that A is elliptic and Lipschitz. Since |u| is a continuous subsolution in B that vanishes in B\, by the boundary Harnack inequality, we have again |u(x)|  g(x) for all x ∈ B  ∩, where g is the Green function of  for div(A∇·) with pole in \B. The constant C depends on u, p, A (the ellipticity constant of A), and B  , but not on K . Thus, proceeding as in the proof of the previous claim, we obtain   1 ˆ  |(div(A∇u), φi )|  g(x)dx ≈ ω(4Q i ) = ω(U4(K ) ).   (Q i )2 2Q i  i∈I i∈I i∈I Letting  → 0, we have ω(U4 (K )) → ω(K ) from which Eq. (A.2) follows.   By the solvability of the Dirichlet problem with L 2 data in Lipschitz domains for diver- gence form elliptic equations with Hölder coeficients (see Remark 1.4 in [20]), we have that ω is a B2 weight with respect to the surface measure σ . In particular, the density function ddωσ 2 (σ ). Therefore, in the sense of distributions, belongs to L loc div(A∇u)| B = hσ, for some h ∈ L loc 2 (σ ). Next, we will show that (∇u)+,a exists σ -a.e. and moreover (∇u)+,a = (∇u, ν)ν ∈ 2 (σ ). Consider an arbitrary open ball B̃ centered in  such that 4 B̃ ⊂ B. Let φ be a C ∞ L loc function which equals 1 on 2 B̃ is supported on 3 B̃, and let v = φu. Observe that ˆ v(x) = E (x, y) div(A∇v)(y) ˆ ˆ = E (x, y)φ(y) div(A∇u)(y) + E (x, y)u(y) div(A∇φ)(y) ˆ + 2 E (x, y)(A∇u(y), ∇φ(y)), (A.3) where E is the fundamental solution of div(A∇·) (we consider a Lipschitz, elliptic extension of A defined in Rd ). Note that ∇u ∈ L 2loc (B), by Caccioppoli’s inequality. For a finite Borel measure η, let T η be the gradient of the single layer potential of η (in the case A ≡ I it coincides with the (d − 1)-dimensional Riesz transform of η). That is, ˆ T η(x) = ∇1 E (x, y)dη(y), in the sense of truncations. From identity (A.3), we obtain for x ∈ / ,    ∇v(x) = cd T (φhσ | )(x) + T (u div(A∇φ)m)(x) + 2T ((A∇u, ∇φ)m (x)  Lebesgue measure in R ). Observe that T (u div(A∇φ)m)(x) and (where m is the d T ((A∇u, ∇φ)m (x) are continuous functions in B̃. On the other hand, the non-tangential limit T (φhσ | )±,a (x) exists for σ -a.e. x ∈ ∂, by the jump formulas for the gradients of single layer potentials transforms (see [20, Theorem 4.4] or [32, Theorem 1] for the constant coefficients case in rectifiable sets, for example). From the fact that ∇v = ∇u in B̃, it follows that (∇u)±,a (x) exists for σ -a.e. x ∈  ∩ B̃. By the L 2 (σ ) boundedness of T on Lipschitz graphs (see [20, Theorem 3.1]), we deduce that (∇u)±,a ∈ L 2 (σ |∩ B̃ ). 123 Size of the zero set of solutions of elliptic PDEs near the… Page 51 of 52 113 Since u ≡ 0 in c , we have (∇u)−,a = 0 in  ∩ B̃. As the tangential component of T (φhσ | )(x) is continuous across ∂ for σ -a.e. x ∈ ∂, again by the jump formulas for T ( [20, Theorem 4.4]), we deduce that the tangential component of (∇u)+,a coincides with the tangential component of (∇u)−,a σ -a.e. in  ∩ B̃, and thus (∇u)+,a = (∇u, ν)ν in  ∩ B̃. We will now prove that (div(A∇u))| B = −(A∇u, ν)σ | in the sense of distributions. Let ψ be a function in Cc∞ ( B̃) . Again, for 0 <  % r ( B̃), consider  =  + ed and  =  + ed , where ed = (0, . . . , 0, 1). Then, we have ˆ ˆ &div(A∇u), ψ' = u div(A∇ψ)dm = lim u div(A∇ψ)dm B̃∩ →0 B̃∩ ˆ ˆ = lim u(A∇ψ, ν)dσ − lim ψ(A∇u, ν)dσ →0 B̃∩∂ →0 B̃∩∂ ˆ =0− ψ(A∇u, ν)dσ. (A.4)  The last identity follows from the uniform convergence of u to 0 as  →  and that ∇u(· + en ) converges to (∇u)+,a in L 2 (σ |∩ B̃ ) (this is proven by arguments analogous to the ones above for the σ -a.e. existence of the limit (∇u)+,a (x) in ). From (A.4), we deduce that div(A∇u) = −(A∇u, ν)σ in B̃, and thus also in B.   References 1. Ancona, A.: On positive harmonic functions in cones and cylinders. Rev. Mat. Iberoam. 28, 201–230 (2010) 2. Adolfsson, V., Escauriaza, L.: C 1,α domains and unique continuation at the boundary. Commun. Pure Appl. Math. 50(10), 935–969 (1997) 3. 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