Lipschitz continuity of the eigenfunctions on optimal sets for functionals with variable coefficients
aa r X i v : . [ m a t h . A P ] M a r LIPSCHITZ CONTINUITY OF THE EIGENFUNCTIONS ON OPTIMALSETS FOR FUNCTIONALS WITH VARIABLE COEFFICIENTS
BAPTISTE TREY
Abstract.
This paper is dedicated to the spectral optimization problemmin (cid:8) λ (Ω) + · · · + λ k (Ω) + Λ | Ω | : Ω ⊂ D quasi-open (cid:9) where D ⊂ R d is a bounded open set and 0 < λ (Ω) ≤ · · · ≤ λ k (Ω) are the first k eigenvalues onΩ of an operator in divergence form with Dirichlet boundary condition and H¨older continuouscoefficients. We prove that the first k eigenfunctions on an optimal set for this problem arelocally Lipschtiz continuous in D and, as a consequence, that the optimal sets are open sets.We also prove the Lipschitz continuity of vector-valued functions that are almost-minimizers ofa two-phase functional with variable coefficients. Contents
1. Introduction and main results 12. Lipschitz continuity of quasi-minimizers 42.1. Continuity and H¨older continuity 52.2. Bound of the Lipschitz constant in { u > } Introduction and main results
Let D be a bounded open subset of R d and Λ be a positive constant. We consider the spectraloptimization problemmin (cid:8) λ (Ω) + · · · + λ k (Ω) + Λ | Ω | : Ω ⊂ D quasi-open (cid:9) (1.1)where 0 < λ (Ω) ≤ · · · ≤ λ k (Ω) denote the first k eigenvalues, counted with the due multiplicity,of the operator in divergence form − b ( x ) − div ( A x ∇· ). This means that for every λ i (Ω) there isan eigenfunction u i ∈ H (Ω) such that (cid:26) − div( A ∇ u i ) = λ i (Ω) b u i in Ω u i = 0 on ∂ Ω . (1.2)The aim of the present paper is twofold. From one side, we prove a Lipschitz regularity resultfor vector-valued functions which are almost-minimizers of a two-phase functional with variablecoefficients (Theorem 1.2). On the other hand, we show that if Ω ∗ is an optimal set for (1.1),then the vector U = ( u , . . . , u k ) of the first k eigenfunctions on Ω ∗ satisfies the almost-minimalitycondition of Theorem 1.2, and hence that the eigenfunctions u , . . . , u k are Lipschitz continuous.We first state our Lipschitz regularity result for eigenfunctions on optimal sets for (1.1). Date : April 1, 2020.
Theorem 1.1.
Let D ⊂ R d be a bounded open set and let Λ > . Let A : D → Sym + d be a matrixvalued function satisfying (1.5) and (1.6) and let b ∈ L ∞ ( D ) be a function satisfying (1.7) (seebelow). Then the spectral optimization problem (1.1) admits a solution Ω ∗ . Moreover, the first k eigenfunctions on any optimal set Ω ∗ are locally Lipschitz continuous in D . As a consequence,every optimal set for (1.1) is an open set. In [3], Briancon, Hayouni and Pierre proved the Lipschitz continuity of the first eigenfunc-tion on an optimal set which minimizes the first eigenvalue of the Dirichlet Laplacian among allsets of prescribed volume included in a box. Their proof, which is inspired by the pioneeringwork of Alt and Caffarelli in [1] on the regularity for a free boundary problem, relies on thefact that the first eigenfunction is the minimum of a variational problem. For spectral optimiza-tion problems involving higher eigenvalues, the study of the regularity of the optimal sets andthe corresponding eigenfunctions is more involved due to the variational characterization of theeigenvalue λ k through a min-max procedure. In [4] the authors considered the spectral func-tionals F ( λ (Ω) , . . . , λ k (Ω)) which are bi-Lipschitz with respect to each eigenvalue λ i (Ω) of theDirichlet Laplacian, a typical example being the sum of the first k eigenvalues. In particular,they proved the Lipschitz continuity of the eigenfunctions on optimal sets minimizing the sum λ (Ω) + · · · + λ k (Ω) among all shapes Ω ⊂ R d of prescribed measure (see [4, Theorem 6.1]). Thepresent paper extends this result to the case of an operator with variable coefficients, but with acompletely different proof.Concerning spectral optimization problems involving an operator with variable coefficients, aregularity result has been obtained in [15], where the authors consider the problem of minimizingthe first eigenvalue of the operator with drift − ∆ + ∇ Φ · ∇ , Φ ∈ W , ∞ ( D, R d ), under inclusionand volume constraints. We stress out that our result also applies to this operator with drift sinceit corresponds to the special case where A = e − Φ Id and b = e − Φ . We would like also to mention arecent work of Lamboley and Sicbaldi in [11] where they prove an existence and regularity resultfor Faber-Krahn minimizers in a Riemanninan setting.Let us highlight that the Lipschitz regularity of the eigenfunctions in Theorem 1.1 turned outto be a quite difficult question due to both the min-max nature of the eigenvalues and the presenceof the variable coefficients, but it is an important first step for the analysis of the regularity ofthe free boundary of the optimal shapes for (1.1) which we study in [17].As already pointed out, the proof of Theorem 1.1 goes through the study of the Lipschitzregularity of vector-valued almost-minimisers for a two-phase functional with variable coefficients.Our approach is to reduce from the non-constant coefficients case to the constant coefficients-one by a change of variables and is inspired by [16], where the authors prove free boundaryregularity of almost-minimizers of the one-phase and two-phase functionals in dimension 2 usingan epiperimetric inequality. The second contribution which was a strong inspiration for ourwork is of David and Toro in [8]. They in particular prove the Lipschitz regularity of almost-minimizers of the one-phase and the two-phase functionals with constant coefficients (see also [7]for free boundary regularity results).We have the following result for almost-minimizers of the two-phase functional. Theorem 1.2.
Let D ⊂ R d be a bounded open set and let Ω ⊂ D be a quasi-open set. Let A : D → Sym + d be a matrix valued function satisfying (1.5) and (1.6) . Let f = ( f , . . . , f k ) ∈ L ∞ ( D, R k ) . Assume that U = ( u , . . . , u k ) ∈ H (Ω , R k ) is a vector-valued function such that • U is a solution of the equation − div( A ∇ U ) = f in Ω , (1.3) • U satisfies the following quasi-minimality condition: for every C > , there exist con-stants ε ∈ (0 , and C > such that Z D A ∇ U · ∇ U + Λ |{| U | > }| ≤ (cid:0) C k U − ˜ U k L (cid:1) Z D A ∇ ˜ U · ∇ ˜ U + Λ |{| ˜ U | > }| , (1.4) for every ˜ U ∈ H ( D, R k ) such that k U − ˜ U k L ≤ ε and k ˜ U k L ∞ ≤ C . IPSCHITZ CONTINUITY OF THE EIGENFUNCTIONS 3
Then the vector-valued function U is locally Lipschitz continuous in D . Remark 1.3 (On the assumption (1.4) of Theorem 1.2) . The quasi-minimality in Theorem 1.2is not local but naturally arises from the shape optimization problem (1.1) (see Proposition 3.4).We stress out that our conclusion also holds, with exactly the same proof, if the quasi-minimalityproperty (1.4) is replaced by its ”local” version, namely: for every C >
0, there exist constants r ∈ (0 ,
1) and
C > x ∈ D and every r ≤ r such that B r ( x ) ⊂ D we have Z B r ( x ) A ∇ U ·∇ U +Λ |{| U | > }∩ B r ( x ) | ≤ (cid:0) C k U − ˜ U k L (cid:1) Z B r ( x ) A ∇ ˜ U ·∇ ˜ U +Λ |{| ˜ U | > }∩ B r ( x ) | , for every ˜ U ∈ H ( D, R k ) such that U − ˜ U ∈ H ( B r ( x ) , R k ) and k ˜ U k L ∞ ≤ C . Remark 1.4.
We point out that we will only use the assumption (1.3) to prove that U is boundedand to get an almost-monotonicity formula (see Proposition 2.15 and Corollary 2.16).In [8, Theorem 6.1], David and Toro proved an almost-monotonicity formula for quasi-minimizersin the case of the Laplacian. It is natural to expect that the same holds for an operator with vari-able coefficients, but we will not address this question in the present paper since we are mainlyinterested in the Lipschitz continuity of the eigenfunctions on optimal shapes for the problem(1.1) for which the equation (1.3) is already known.However, soon before the present paper was published online, a new preprint of the sameauthors, in collaboration with Engelstein and Smit Vega Garcia (see [6]), appeared on Arxiv.They prove a regularity result for functions satisfying a suitable quasi-minimality condition foroperators with variable coefficients. We stress that the present paper and the work in [6] weredone in a completely independent way. We notice that our main result neither directly impliesnor is directly implied by the main result from [6]. Notations.
Let us start by setting the assumptions on the coefficients of the operator that wewill use throughout this paper. The matrix-valued function A = ( a ij ) ij : D → Sym + d has H¨oldercontinuous coefficients and is uniformly elliptic, where Sym + d denotes the family of all real positivesymmetric d × d matrices. Precisely, there exist positive constants δ A , c A > λ A ≥ | a ij ( x ) − a ij ( y ) | ≤ c A | x − y | δ A , for every i, j and x, y ∈ D ; (1.5)1 λ A | ξ | ≤ ξ · A x ξ = d X i,j =1 a ij ( x ) ξ i ξ j ≤ λ A | ξ | , for every x ∈ D and ξ ∈ R d . (1.6)The function b ∈ L ∞ ( D ) is positive and bounded away from zero: there exists c b > c − b ≤ b ( x ) ≤ c b for almost every x ∈ D. (1.7)We now fix some notations and conventions. For x ∈ R d and r > B r ( x )to denote the ball centred at x of radius r and we simply write B r if x = 0. We denote by | Ω | the Lebesgue measure of a generic set Ω ⊂ R d and by ω d the Lebesgue measure of the unit ball B ⊂ R d . The ( d − H d − . Moreover, we definethe positive and the negative parts of a function u : R → R by u + = max( u,
0) and u − = max( − u, . For a quasi-open set Ω ∈ R d we denote by H (Ω) the Sobolev space defined as the set offunctions u ∈ H ( R d ) which, up to a set of capacity zero, vanishe outside Ω; that is H (Ω) = { u ∈ H ( R d ) : u = 0 quasi-everywhere in R d \ Ω } . (see e.g. [10] for a definition of the capacity). Notice that if Ω is an open set, then H (Ω) isthe usual Sobolev space defined as the closure of the smooth real-valued functions with supportcompact C ∞ c (Ω) with respect to the norm k u k H = k u k L + k∇ u k L . We denote by H (Ω , R k ) BAPTISTE TREY the space of vector-valued functions U = ( u , . . . , u k ) : Ω → R k such that u i ∈ H (Ω) for every i = 1 , . . . , k , and endowed with the norm k U k H (Ω) = k U k L (Ω) + k∇ U k L (Ω) = k X i =1 (cid:0) k u i k L (Ω) + k∇ u i k L (Ω) (cid:1) . We also define the following norms (whenever it makes sense) k U k L (Ω) = k X i =1 k u i k L (Ω) and k U k L ∞ (Ω) = sup ≤ i ≤ k k u i k L ∞ (Ω) . Moreover, for U = ( u , . . . , u k ) : Ω → R k we set | U | = u + · · · + u k , |∇ U | = |∇ u | + · · · + |∇ u k | and A ∇ U · ∇ U = A ∇ u · ∇ u + · · · + A ∇ u k · ∇ u k . For f = ( f , . . . , f k ) ∈ L (Ω , R k ) we say that U = ( u , . . . , u k ) ∈ H (Ω , R k ) is solution to the equation − div( A ∇ U ) = f in Ω , U ∈ H (Ω , R k )if, for every i = 1 , . . . , k , the component u i is solution to the equation − div( A ∇ u i ) = f i in Ω , u i ∈ H (Ω) , where the PDE is intended is the weak sense, that is Z Ω A ∇ u i · ∇ ϕ = Z Ω f i ϕ for every ϕ ∈ H (Ω) . Moreover, we always extend functions of the spaces H (Ω) and H (Ω , R k ) by zero outside Ω sothat we have the inclusions H (Ω) ⊂ H ( R d ) and H (Ω , R k ) ⊂ H ( R d , R k ).2. Lipschitz continuity of quasi-minimizers
This section is dedicated to the proof of Theorem 1.2. Our approach is to locally freeze thecoefficients to reduce to the case where A = Id . More precisely, for every point x ∈ D , an almost-minimizer of the functional with variable coefficients becomes, in a new set of coordinates near x , an almost minimizer for a functional with constant coefficients. We stress out the dealing withthe dependence of this change of variables with respect to the point x is not a trivial task. Wethen adapt the strategy developed by David and Toro in [8] for almost-minimizers of a functionalinvolving the Dirichlet energy.In this section, u will stand for a coordinate function of the vector U from Theorem 1.2. Insubsection 2.1 we explicit the change of variables for which u becomes a quasi-minimizer of theDirichlet energy (in small balls of fixed center). We then prove that u is continuous and we givean estimate of the modulus of continuity from which we deduce that u is locally H¨older continuousin D .Subsection 2.2 is addressed to the Lipschitz continuity of u in some region where the function u has a given sign. We show, using in particular the H¨older continuity of u , that most of theestimates proved in Subsection 2.1 can be improved provided that u keeps the same sign. In thiscase, we prove that u is Lipschitz continuous and we provide a bound on the Lipschitz constantof u . We also show that u is C ,β -regular for some β ∈ (0 , u in a small ball is big enough, then u keeps the same sign in a smaller ball, which in view of thepreceding analysis implies that u is Lipschitz continuous.In subsection 2.3 we complete the proof of the Lipschitz continuity of u . The main missingstep is to deal with the case where the Dirichlet energy is big and the first assumption of (2.45)in Proposition 2.11 fails. Using an almost-monotonicity formula for operators with variablecoefficients proved by Matevosyan and Petrosyan in [13, Theorem III], we show that in this casethe value of the Dirichlet energy has to decrease at some smaller scale. IPSCHITZ CONTINUITY OF THE EIGENFUNCTIONS 5
Throughout this section we fix u := u i , for some i = 1 , . . . , k , a coordinate function of thevector U = ( u , . . . , u k ) from Theorem 1.2. We start by proving that u is a bounded function in D . Lemma 2.1 (Boundedness) . Let Ω ⊂ D be a (non-empty) quasi-open set, f ∈ L p ( D ) for some p ∈ ( d/ , + ∞ ] and let u ∈ H (Ω) be the solution of − div( A ∇ u ) = f in Ω , u ∈ H (Ω) . Then, there is a dimensional constant C d such that k u k L ∞ ≤ λ A C d / d − / p | Ω | / d − / p k f k L p . Proof.
Up to arguing with the positive and the negative parts of f , we can assume that f is anon-negative function. By the maximum principle (see [9, Theorem 8.1]) we have u ≥ u is a minimum of the following functional J ( ϕ ) := 12 Z Ω A ∇ ϕ · ∇ ϕ − Z Ω f ϕ, ϕ ∈ H (Ω) . We consider, for every 0 < t < k u k L ∞ and ε >
0, the test function u t,ε = u ∧ t +( u − t − ε ) + ∈ H (Ω).Then, by ellipticity of the matrices A x and the inequality J ( u ) ≤ J ( u t,ε ) we get that12 λ A Z { tt } f ≤ ε k f k L p |{ u > t }| p − p , The end of the proof now follows precisely as in [15, Lemma 5.3]. (cid:3)
Continuity and H¨older continuity.
We change the coordinates and reduce to the case A = Id using in particular the H¨older continuity of the coefficients of A , and we then prove that u is locally H¨older continuous in D . Let us first introduce few notations that we will use throughoutthis section. For x ∈ D we define the function F x : R d → R d by F x ( ξ ) := x + A / x [ ξ ] , ξ ∈ R d . Moreover, we set u x = u ◦ F x for every x ∈ D . Remark 2.2.
For M ∈ Sym + d we denote by M / the square root matrix of M . We recall thatif M ∈ Sym + d , then there is an orthogonal matrix P such that P M P t = diag( λ , . . . , λ d ), where P t is the transpose of P and diag( λ , . . . , λ d ) is the diagonal matrix with eigenvalues λ , . . . , λ d .The matrix M / is then defined by M / := P t DP where D = diag( √ λ , . . . , √ λ d ). Remark 2.3 (Notation of the harmonic extension) . One of the main ingredient in the proof ofTheorem 1.2 is based on small variations of the function u x . Precisely, we will often compare u x in some ball B r with the harmonic extension of the trace of u x to ∂B r . This function will often bedenoted by h x,r , or more simply h r if there is no confusion, and is defined by h r = h x,r ∈ H ( B r )and ∆ h r = 0 in B r , u x − h r ∈ H ( B r ) . We notice that h r is a minimizer of the Dirichlet energy in the ball B r , that is Z B r |∇ h r | ≤ Z B r |∇ v | for every v ∈ H ( B r ) such that h r − v ∈ H ( B r ) . We now prove that the function u x is in some sense a quasi-minimizer for the Dirichlet energyin small balls centred at the origin. BAPTISTE TREY
Proposition 2.4.
There exist constants r ∈ (0 , and C > such that, if x ∈ D and r ≤ r satisfy B λ A r ( x ) ⊂ D , then we have Z B r |∇ u x | ≤ (1 + Cr δ A ) Z B r |∇ ˜ u | + Cr d , (2.1) for every ˜ u ∈ H ( R d ) ∩ L ∞ ( R d ) such that u x − ˜ u ∈ H ( B r ) and k ˜ u k L ∞ ≤ k u k L ∞ .Proof. Let v ∈ H ( D ) be such that ˜ u = v ◦ F x and set ˜ U = ( u , . . . , v, . . . , u k ) ∈ H ( D, R k ),where v stands at the i -th position. Set ρ = λ A r and note that F x ( B r ) ⊂ B ρ ( x ) ⊂ D . Then,using ˜ U as a test function and observing that u − v ∈ H ( F x ( B r )), we get Z F x ( B r ) A ∇ u · ∇ u ≤ Z F x ( B r ) A ∇ v · ∇ v + C k u − v k L Z D A ∇ ˜ U · ∇ ˜ U + Λ | B ρ | , where C is the constant from Theorem 1.2. Together with Z D A ∇ ˜ U · ∇ ˜ U ≤ Z D A ∇ U · ∇ U − Z D A ∇ u · ∇ u + Z D A ∇ v · ∇ v ≤ Z D A ∇ U · ∇ U + Z F x ( B r ) A ∇ v · ∇ v this yields Z F x ( B r ) A ∇ u · ∇ u ≤ (1 + ˜ Cr d ) Z F x ( B r ) A ∇ v · ∇ v + ˜ Cr d , (2.2)for some constant ˜ C . On the other hand, using the H¨older continuity and the ellipticity of A weestimate Z B r |∇ u x | = det( A − / x ) Z F x ( B r ) A x ∇ u · ∇ u ≤ det( A − / x )(1 + dc A λ A ρ δ A ) Z F x ( B r ) A ∇ u · ∇ u. (2.3)Similarly, we have the following estimate from below Z B r |∇ ˜ u | ≥ det( A − / x )(1 − dc A λ A ρ δ A ) Z F x ( B r ) A ∇ v · ∇ v. (2.4)Now, combining (2.3), (2.2) and (2.4) we get Z B r |∇ u x | ≤ (1 + dc A λ A ρ δ A ) (cid:20) Cr d − dc A λ A ρ δ A Z B r |∇ ˜ u | + λ d A ˜ Cr d (cid:21) . which gives (2.2). (cid:3) We now prove that the function u is continuous in D . In the sequel we will often use thefollowing notation: for x ∈ D and r > ω ( u, x, r ) = − Z B r ( x ) |∇ u | ! / and ω ( u x , r ) = (cid:18) − Z B r |∇ u x | (cid:19) / . Proposition 2.5.
The function u is continuous in D . Moreover, there exist r > and C > such that, if x ∈ D and r ≤ r satisfy B r ( x ) ⊂ D , then we have | u ( y ) − u ( z ) | ≤ C (cid:16) ω ( u, x, r ) + log r | y − z | (cid:17) | y − z | for every y, z ∈ B r/ ( x ) . (2.5)The next Lemma shows that ω ( u x , r ) cannot grow too fast as r tends to zero and will be usefulthroughout the proof of the Lipschitz continuity of u . IPSCHITZ CONTINUITY OF THE EIGENFUNCTIONS 7
Lemma 2.6.
There exist constants r > and C > such that, if x ∈ D and r ≤ r satisfy B λ A r ( x ) ⊂ D , then we have ω ( u x , s ) ≤ Cω ( u x , r ) + C log (cid:16) rs (cid:17) for every < s ≤ r. (2.6) If, moreover, x is a Lebesgue point for u , then we have (cid:12)(cid:12)(cid:12)(cid:12) u ( x ) − − Z B r u x (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cr (1 + ω ( u x , r )) . (2.7) Proof.
Let t ≤ r and use h t as a test function in (2.1), where h t = h x,t denotes the harmonicextension in B t of the trace of u x to ∂B t , to get Z B t |∇ ( u x − h t ) | = Z B t |∇ u x | − Z B t |∇ h t | ≤ Ct δ A Z B t |∇ h t | + Ct d ≤ Ct δ A Z B t |∇ u x | + Ct d , (2.8)where in the last inequality we have used that h t is a minimizer of the Dirichlet energy on B t .Moreover, since |∇ h t | is subharmonic on B s for every s ≤ t , we have − Z B s |∇ h t | ≤ − Z B t |∇ h t | for every s ≤ t. (2.9)Therefore, the triangle inequality, (2.9) and (2.8) give for every s ≤ t ≤ r ω ( u x , s ) ≤ (cid:18) − Z B s |∇ ( u x − h t ) | (cid:19) / + (cid:18) − Z B s |∇ h t | (cid:19) / ≤ (cid:16) ts (cid:17) d/ (cid:18) − Z B t |∇ ( u x − h t ) | (cid:19) / + (cid:18) − Z B t |∇ h t | (cid:19) / ≤ (cid:16) ts (cid:17) d/ C (cid:16) t δ A / ω ( u x , t ) + 1 (cid:17) + ω ( u x , t ) (2.10) ≤ (cid:16) C (cid:16) ts (cid:17) d/ t δ A / (cid:17) ω ( u x , t ) + C (cid:16) ts (cid:17) d/ . We then use the estimate (2.10) with the radii r i = 2 − i r , i ≥
0, and we get ω ( u x , r i ) ≤ (cid:16) Cr δ A / i − (cid:17) ω ( u x , r i − ) + C, i ≥ . This, with an iteration, implies that for every i ≥ ω ( u x , r i ) ≤ ω ( u x , r ) i − Y j =0 (cid:16) Cr δ A / j (cid:17) + C i − X j =1 i − Y l = j (cid:16) Cr δ A / l (cid:17) + C ≤ Cω ( u x , r ) + Ci, (2.11)where we used that the product Q ∞ j =0 (cid:0) Cr δ A / j (cid:1) is bounded by a constant depending on r . Thefirst estimate of the Lemma now follows from (2.11). Indeed, choose i ≥ r i +1 < s ≤ r i and note that we have ω ( u x , s ) ≤ d/ ω ( u x , r i ). If i = 0, this directly implies (2.6); otherwise, i ≥ i ≥ m i = − R B ri u x . By the Poincar´e inequalityand (2.11) we have − Z B ri | u x − m i | ! / ≤ Cr i ω ( u x , r i ) ≤ Cr i ( ω ( u x , r ) + i ) . (2.12) BAPTISTE TREY
Furthermore, 0 is a Lebesgue point for u x since x is a Lebesgue point for u and that for every s ≤ r we have λ − d A − Z B λ − A s ( x ) | u − u ( x ) | ≤ − Z B s | u x − u x (0) | = − Z F x ( B s ) | u − u ( x ) | ≤ λ d A − Z B λ A s ( x ) | u − u ( x ) | . In particular, it follows that m i converges to u x (0) = u ( x ) as i → + ∞ . Therefore, this with theCauchy-Schwarz inequality and (2.12) give | u ( x ) − m i | ≤ + ∞ X j = i | m j +1 − m j | ≤ + ∞ X j = i − Z B rj +1 | u x − m j |≤ d + ∞ X j = i − Z B rj | u x − m j | ≤ d + ∞ X j = i − Z B rj | u x − m j | ! / ≤ C + ∞ X j = i r j ( ω ( u x , r ) + j ) ≤ Cr i ( ω ( u x , r ) + i + 1) , where in the last inequality we used that P + ∞ j = i i − j j ≤ C ( i + 1). Then, observe that (2.7) isprecisely the above inequality with i = 0 to conclude the proof. (cid:3) Proof of Proposition . Let y, z ∈ B r/ ( x ) and notice that it is enough to prove (2.5) when y and z are Lebesgue points for u . Set δ = | y − z | . We first assume that 4 λ A δ ≤ r . Observe thatwe hence have the inclusions F z ( B λ − A δ ) ⊂ F y ( B λ A δ ) ⊂ B r ( x ) ⊂ D . Using a change of variables,the Poincar´e inequality and then the ellipticity of A , we estimate (cid:12)(cid:12)(cid:12)(cid:12) − Z B λ A δ u y − − Z B λ − A δ u z (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) − Z F y ( B λ A δ ) u − − Z F z ( B λ − A δ ) u (cid:12)(cid:12)(cid:12)(cid:12) ≤ − Z F z ( B λ − A δ ) (cid:12)(cid:12)(cid:12) u − − Z F y ( B λ A δ ) u (cid:12)(cid:12)(cid:12) ≤ d λ d A − Z F y ( B λ A δ ) (cid:12)(cid:12)(cid:12) u − − Z F y ( B λ A δ ) u (cid:12)(cid:12)(cid:12) ≤ Cδ − Z F y ( B λ A δ ) |∇ u | ! / (2.13) ≤ Cδλ A − Z F y ( B λ A δ ) A y ∇ u · ∇ u ! / ≤ Cδω ( u y , λ A δ ) . On the other hand, since F z ( B λ − A δ ) ⊂ F y ( B λ A δ ) we have ω ( u z , λ − A δ ) = − Z F z ( B λ − A δ ) A z ∇ u · ∇ u / ≤ λ A − Z F z ( B λ − A δ ) |∇ u | / ≤ d/ λ d +1 A − Z F y ( B λ A δ ) |∇ u | ! / ≤ d/ λ d +2 A − Z F y ( B λ A δ ) A y ∇ u · ∇ u ! / (2.14) ≤ Cω ( u y , λ A δ ) . We now apply (2.7) to get (cid:12)(cid:12)(cid:12)(cid:12) u ( y ) − − Z B λ A δ u y (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cδ ( ω ( u y , λ A δ ) + 1) (2.15)and (cid:12)(cid:12)(cid:12)(cid:12) u ( z ) − − Z B λ − A δ u z (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cδ ( ω ( u z , λ − A δ ) + 1) ≤ Cδ ( ω ( u y , λ A δ ) + 1) , (2.16) IPSCHITZ CONTINUITY OF THE EIGENFUNCTIONS 9 where we used (2.14) in the last inequality. Therefore, combining the triangle inequality, (2.15),(2.13) and (2.16) we get that | u ( y ) − u ( z ) | ≤ Cδ ( ω ( u y , λ A δ ) + 1) . (2.17)Moreover, by (2.6) (recall that we assumed 4 λ A δ ≤ r ) we have ω ( u y , λ A δ ) ≤ Cω ( u y , (2 λ A ) − r ) + C log r λ A δ . (2.18)By the ellipticity of A and since F y ( B (2 λ A ) − r ) ⊂ B r ( x ), we have the following estimate ω ( u y , (2 λ A ) − r ) = − Z F y ( B (2 λ A ) − r ) A y ∇ u · ∇ u ! / ≤ λ A − Z F y ( B (2 λ A ) − r ) |∇ u | ! / ≤ d/ λ d +1 A − Z B r ( x ) |∇ u | ! / ≤ Cω ( u, x, r ) . (2.19)Finally, combine (2.17), (2.18) and (2.19) to get | u ( y ) − u ( z ) | ≤ Cδ (cid:16) ω ( u, x, r ) + log r λ A δ (cid:17) (2.20) ≤ C | y − z | (cid:16) ω ( u, x, r ) + log r | y − z | (cid:17) , which is (2.5).Now, if the assumption 4 λ A | y − z | ≤ r is not satisfied, choose n points y = y, y , . . . , y n = z in B r ( x ) such that 4 λ A η = | y − z | , where we have set η = | y i − y i +1 | , i = 1 , . . . , n . Then we have4 λ A η ≤ r . We notice that we can assume the y i to be Lebesgue points for u . Moreover, observethat we can bound the number of points by n ≤ λ A + 2. Therefore, applying the estimate (2.20)to each pair ( y i , y i +1 ) we have | u ( y ) − u ( z ) | ≤ n − X i =1 | u ( y i ) − u ( y i +1 ) | ≤ C n − X i =1 η (cid:16) ω ( u, x, r ) + log r λ A η (cid:17) ≤ nC | y − z | λ A (cid:16) ω ( u, x, r ) + log r | y − z | (cid:17) , which concludes the proof. (cid:3) We are now in position to prove the H¨older continuity of u . Proposition 2.7.
The function u is locally α -H¨older continuous in D for every α ∈ (0 , , thatis, for every compact set K ⊂ D , there exist r K > and C K > such that for every x ∈ K wehave | u ( y ) − u ( z ) | ≤ C K | y − z | α for every y, z ∈ B r K ( x ) . (2.21) Proof.
Let x ∈ K and set 4 r K = r = min { r , dist( K, D c ) } where r is given by Proposition2.5. Since the function r r − α log( r /r ) is non-decreasing on (0 , c α ) for some constant c α > α and r , it follows from Proposition 2.5 that, if y, z ∈ B r / ( x ) are such that | y − z | ≤ c α , we have | u ( y ) − u ( z ) | ≤ C (cid:16) r − α (1 + ω ( u, x, r )) + c − αα log r c α (cid:17) | y − z | α ≤ C (1 + ω ( u, x, r )) | y − z | α (2.22)If now | y − z | > c α , then choose n points y = y, . . . , y n = z in B r / ( x ) such that | y i − y i +1 | = c α r − | y − z | , with n bounded by some constant depending on α and r . Then apply (2.22) toeach pair ( y i , y i +1 ) to prove that u is α -H¨older continuous in the ball B r / ( x ) with a modulusof continuity depending on ω ( u, x, r ). Now, (2.21) follows by a compactness argument with the constant C K depending on max { ω ( u, x i , r ) , i = . . . N } , where the x i ’s are given by somesubcovering of K ⊂ ∪ Ni =1 B r K ( x i ). (cid:3) Bound of the Lipschitz constant in { u > } . We prove that u is Lipschitz continuousand even C ,β -regular in the regions where u keeps the same sign. We also provide in this casean estimate of the Lipschitz constant of u in terms of ω ( u, x, r ) (see Proposition 2.8. Then, weshow that under suitable conditions, u keeps the same sign and is therefore Lipschitz continuous(see Proposition 2.11). Proposition 2.8.
Let K ⊂ D be a compact set. There exist constants r K > and C K > suchthat, if x ∈ K and r ≤ r K satisfyeither u x > a.e. in B r or u x < a.e. in B r , (2.23) then u is Lipschitz continuous in B r/ ( x ) and we have | u ( y ) − u ( z ) | ≤ C K (1 + ω ( u, x, r )) | y − z | for every y, z ∈ B r/ ( x ) . (2.24) Moreover, u is C ,β in the ball B r/ ( x ) where β = δ A d + δ A +2 and we have |∇ u ( y ) − ∇ u ( z ) | ≤ C K r − δ A d +2 (1 + ω ( u, x, r )) | y − z | β for every y, z ∈ B r/ ( x ) . (2.25)In the next Lemma we compare the Dirichlet energy of u x and of its harmonic extension insmall balls where u x has a given sign. The estimate (2.26) in Lemma 2.9 below is similar to (2.1)but with a smaller error term. Thanks to this improvement, the strategy developed in the proofof Lemma 2.6 will lead to a sharper result than estimate (2.6), namely (2.24). Lemma 2.9.
Let K ⊂ D be a compact set and let α ∈ (0 , . There exist constants r K > and C > such that, if x ∈ K and r ≤ r K are such that (2.23) holds, then the function u x = u ◦ F x satisfies Z B r |∇ u x | ≤ (1 + Cr δ A ) Z B r |∇ h r | + Cr d + α , (2.26) where h r stands for the harmonic extension of the trace of u x to ∂B r .Proof. Set ρ := λ A r for some r > B ρ ( x ) ⊂ D . We define v ∈ H ( D )by h r = v ◦ F x in B r and v = u elsewhere so that we have u − v ∈ H ( F x ( B r )). Set ˜ U =( u , . . . , v, . . . , u k ) ∈ H ( D, R k ) and observe that |{| ˜ U | > }| = |{| U | > }| by (2.23) and because v > F x ( B r ). Then, we use ˜ U as a test function in (1.3) to get Z F x ( B r ) A ∇ u · ∇ u ≤ Z F x ( B r ) A ∇ v · ∇ v + C k u − v k L Z D A ∇ ˜ U · ∇ ˜ U , where C is the constant from Theorem 1.2. Now, since u is locally α -H¨older continuous, we havethe bound k u − v k L ≤ C d C K r d + α , where the constant C K is given by Proposition 2.7. Moreoverwe have the estimate Z D A ∇ ˜ U · ∇ ˜ U ≤ Z D A ∇ U · ∇ U + Z F x ( B r ) A ∇ v · ∇ v. Altogether this gives Z F x ( B r ) A ∇ u · ∇ u ≤ (1 + ˜ Cr d + α ) Z F x ( B r ) A ∇ v · ∇ v + ˜ Cr d + α , for some constant ˜ C which involves R D A ∇ U · ∇ U . Finally, using the H¨older continuity and theellipticity of A as in the proof of Proposition 2.4, we get Z B r |∇ u x | ≤ (1 + dc A λ A ρ δ A ) (cid:20) Cr d + α − dc A λ A ρ δ A Z B r |∇ h r | + λ d A ˜ Cr d + α (cid:21) . which gives (2.26). (cid:3) IPSCHITZ CONTINUITY OF THE EIGENFUNCTIONS 11
Next Lemma is analogue to Lemma 2.6 with a better estimate of the error term. Its proof isquite similar but we nonetheless sketch the argument since there are small differences.
Lemma 2.10.
Let K ⊂ D be a compact set and α ∈ (0 , . There exist constants r K > and C > such that, for every x ∈ K and every r ≤ r K such that (2.23) holds, we have ω ( u x , s ) ≤ Cω ( u x , r ) + Cr α/ for every < s ≤ r. (2.27) If, moreover, x is a Lebesgue point for u , we have (cid:12)(cid:12)(cid:12)(cid:12) u ( x ) − − Z B r u x (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cr ( ω ( u x , r ) + r α/ ) . (2.28) Proof.
For t ≤ r ≤ r K we have by Lemma 2.9 Z B t |∇ ( u x − h t ) | = Z B t |∇ u x | − Z B t |∇ h t | ≤ Ct δ A Z B t |∇ u x | + Ct d + α , (2.29)since h t is a minimizer of the Dirichlet energy on B t . Now, for s ≤ t ≤ r we use (2.9) and (2.29)to estimate as in (2.10) ω ( u x , s ) ≤ (cid:16) C (cid:16) ts (cid:17) d/ t δ A / (cid:17) ω ( u x , t ) + C (cid:16) ts (cid:17) d/ t α/ , which, applied to s = 2 − i r and t = 2 − ( i − r , gives ω ( u x , r i ) ≤ (cid:16) Cr δ A / i − (cid:17) ω ( u x , r i − ) + Cr α/ i − , i ≥ , where we have set r i = 2 − i r . Iterating the above estimate we get for every i ≥ ω ( u x , r i ) ≤ ω ( u x , r ) i − Y j =0 (cid:16) Cr δ A / j (cid:17) + C i − X j =1 (cid:18) r α/ j − i − Y l = j (cid:16) Cr δ A / l (cid:17)(cid:19) + Cr α/ i − ≤ Cω ( u x , r ) + Cr α/ , since Q ∞ j =0 (cid:0) Cr δ A / j (cid:1) is bounded by a constant depending on r K . This proves (2.27).Finally, (2.28) is proved in the same way than (2.7) but with (2.12) replaced by the estimate − Z B ri | u x − m i | ! / ≤ Cr i ω ( u x , r i ) ≤ Cr i ( ω ( u x , r ) + r α/ ) . (cid:3) Proof of Proposition . Let us first prove (2.24). We follow the proof of Proposition 2.5 and weonly detail the few differences. Let y, z ∈ B r/ ( x ) be Lebesgue points for u and set δ = | y − z | .We first assume that 4 λ A δ ≤ r . By (2.28) we have (cid:12)(cid:12)(cid:12)(cid:12) u ( y ) − − Z B λ A δ u y (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cδ ( ω ( u y , λ A δ ) + δ α/ ) , (2.30)and, using also (2.14), (cid:12)(cid:12)(cid:12)(cid:12) u ( z ) − − Z B λ − A δ u z (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cδ ( ω ( u z , λ − A δ ) + r α/ ) ≤ Cδ ( ω ( u y , λ A δ ) + δ α/ ) . (2.31)Moreover, by (2.27) we have ω ( u y , λ A δ ) ≤ Cω ( u y , (2 λ A ) − r ) + Cr α/ . (2.32)Then, combining (2.30), (2.13), (2.31) and then (2.32) and (2.19) we have | u ( y ) − u ( z ) | ≤ Cδ ( ω ( u y , λ A δ ) + δ α/ ) ≤ Cδ (1 + ω ( u, x, r ) + r α/ ) ≤ C (1 + ω ( u, x, r )) | y − z | . (2.33) Finally, if 4 λ A δ > r , we argue as in the proof of Proposition 2.5 and choose a few number ofpoints which connect y and z to prove (2.33).We now prove the estimate (2.25). Let y ∈ B r/ ( x ) and ¯ r ≤ λ − A r/
4. We set m ( u y , ρ ) = − R B ρ ∇ u y for ρ ≤ ¯ r and m = − R B ¯ r ∇ h y, ¯ r = ∇ h y, ¯ r (0), where h y, ¯ r denotes the harmonic extension of the traceof u y to ∂B ¯ r . Let η ∈ (0 , / − Z B η ¯ r |∇ u y − m ( u y , η ¯ r ) | ≤ − Z B η ¯ r |∇ u y − m | ≤ − Z B η ¯ r |∇ ( u y − h y, ¯ r ) | + 2 − Z B η ¯ r |∇ h y, ¯ r − m | . (2.34)Firstly, by (2.29) we have − Z B η ¯ r |∇ ( u y − h y, ¯ r ) | ≤ C ( η ¯ r ) − d − Z B ¯ r |∇ ( u y − h y, ¯ r ) | ≤ C ( η ¯ r ) − d (cid:16) ¯ r δ A Z B ¯ r |∇ u y | + ¯ r d + α (cid:17) ≤ Cη − d ¯ r δ A ω ( u y , ¯ r ) + Cη − d ¯ r α . (2.35)Moreover, (2.33) says that for almost every z ∈ B r/ ( y ) ⊂ B r/ ( x ) we have |∇ u ( z ) | ≤ C (1 + ω ( u, x, r )), which implies that ω ( u y , ¯ r ) = − Z B ¯ r |∇ u y | ≤ λ d +1) A − Z B r/ ( y ) |∇ u | ≤ C (1 + ω ( u, x, r )) . (2.36)On the other hand, by estimates on harmonic functions (see [9, Theorem 3.9]), the Cauchy-Schwarz inequality and (2.36) we have for every ξ ∈ B η ¯ r |∇ h y, ¯ r ( ξ ) − m | = |∇ h y, ¯ r ( ξ ) − ∇ h y, ¯ r (0) | ≤ η ¯ r sup B η ¯ r |∇ h y, ¯ r | ≤ Cη sup B η ¯ r |∇ h y, ¯ r |≤ Cη (cid:18) − Z B ¯ r |∇ h y, ¯ r | (cid:19) ≤ Cη (cid:18) − Z B ¯ r |∇ h y, ¯ r | (cid:19) / ≤ Cη (cid:18) − Z B ¯ r |∇ u y | (cid:19) / (2.37) ≤ Cη ω ( u y , ¯ r ) ≤ Cη (1 + ω ( u, x, r )) , where ∇ h y, ¯ r stands for the Hessian matrix of h y, ¯ r . Therefore, combining (2.34), (2.35), (2.36)and (2.37) we get − Z B η ¯ r |∇ u y − m ( u y , η ¯ r ) | ≤ Cη − d ¯ r δ A (1 + ω ( u, x, r )) + Cη − d ¯ r α + Cη (1 + ω ( u, x, r )) ≤ C (1 + ω ( u, x, r )) h η − d ¯ r δ A + η − d ¯ r α + η i . (2.38)We set α = δ A (recall that α ∈ (0 ,
1) was arbitrary). Moreover, we set β = δ A d + δ A +2 and η = ¯ r δ A d +2 so that we have η − d ¯ r δ A = η = ( η ¯ r ) β . Notice also that η ¯ r = ¯ r ε , where ε = δ A d +2 . Therefore,(2.38) implies that for every y ∈ B r/ ( x ) and every ρ ≤ (cid:0) r λ A (cid:1) ε we have − Z B ρ |∇ u y − m ( u y , ρ ) | ≤ C (1 + ω ( u, x, r )) ρ β . (2.39)Now, let y, z ∈ B r/ ( x ) be Lebesgue points for u and set δ = | x − y | . We first assume that2 λ A δ ≤ (cid:0) r λ A (cid:1) ε . Setting δ i = 2 − i δ , i ≥
0, we have using (2.39) |∇ u y (0) − m ( u y , δ ) | ≤ + ∞ X i =0 | m ( u y , δ i +1 ) − m ( u y , δ i ) | ≤ + ∞ X i =0 − Z B δi +1 |∇ u y − m ( u y , δ i ) |≤ d + ∞ X i =0 − Z B δi |∇ u y − m ( u y , δ i ) | ≤ d + ∞ X i =0 (cid:18) − Z B δi |∇ u y − m ( u y , δ i ) | (cid:19) / ≤ C (1 + ω ( u, x, r )) δ β . (2.40)Similarly, we have |∇ u z (0) − m ( u z , λ A δ ) | ≤ C (1 + ω ( u, x, r )) δ β . (2.41) IPSCHITZ CONTINUITY OF THE EIGENFUNCTIONS 13
Moreover, using that F − z ◦ F y ( B δ ) ⊂ B λ A δ we have | m ( u y , δ ) − m ( u z , λ A δ ) | ≤ − Z B δ |∇ u y − m ( u z , λ A δ ) |≤ − Z F − z ◦ F y ( B δ ) | A / y A − / z ∇ u z − m ( u z , λ A δ ) | (2.42) ≤ (2 λ A ) d − Z B λ A δ | A / y A − / z ∇ u z − m ( u z , λ A δ ) | . Notice that the matrices A / have H¨older continuous coefficients with exponent δ A / | A / y A − / z − Id | ≤ λ A | A − / y − A − / z | ≤ Cδ δ A / ≤ Cδ β (because β ≤ δ A / − Z B λ A δ | A / y A − / z ∇ u z − ∇ u z | ≤ Cδ β − Z B λ A δ |∇ u z | ≤ Cδ β ω ( u z , λ A δ ) ≤ Cδ β ( ω ( u x , λ − A r ) + r δ A ) ≤ C (1 + ω ( u, x, r )) δ β . (2.43)Furthermore, the triangle inequality in (2.42) together with (2.43), (2.39) and Cauchy-Schwarz’sinequality give | m ( u y , δ ) − m ( u z , λ A δ ) | ≤ C (1 + ω ( u, x, r )) δ β . (2.44)Now, (2.40), (2.44) and (2.41) infer |∇ u y (0) − ∇ u z (0) | ≤ |∇ u y (0) − m ( u y , δ ) | + | m ( u y , δ ) − m ( u z , λ A δ ) | + |∇ u z (0) − m ( u z , λ A δ ) |≤ C (1 + ω ( u, x, r )) δ β . Since |∇ u y (0) | ≤ λ A |∇ u ( y ) | ≤ C (1 + ω ( u, x, r )) for almost every y ∈ B r/ ( x ) by (2.33), we get |∇ u ( y ) − ∇ u ( z ) | = | A − / y ∇ u y (0) − A − / z ∇ u z (0) |≤ | A − / y ∇ u y (0) − A − / z ∇ u y (0) | + | A − / z ∇ u y (0) − A − / z ∇ u z (0) |≤ | A − / y − A − / z ||∇ u y (0) | + | A − / z ||∇ u y (0) − ∇ u z (0) |≤ C (1 + ω ( u, x, r )) δ β . If | y − z | ≥ (cid:0) r λ A (cid:1) ε , then we can connect y and z through less than λ A (cid:0) λ A r (cid:1) ε + 2 points. Thisshows (2.25) and concludes the proof. (cid:3) The strategy to prove Theorem 1.2 is to show that ω ( u x , r ) cannot become too big as r getssmall. In the next Proposition we prove, under some condition (see the first inequality in (2.45)below), that if ω ( u x , r ) is big enough then u keeps the same sign near the point x and is henceLipschitz continuous by Proposition 2.8. The case where ω ( u x , r ) is big and this condition failsis treated in the next subsection. We set for x ∈ D and r > b ( u x , r ) = − Z ∂B r u x d H d − and b + ( u x , r ) = − Z ∂B r | u x | d H d − . Proposition 2.11.
Let K ⊂ D be a compact set and let γ > . There exists constants r K , C K > and κ > such that, if x ∈ K and r ≤ r K satisfy γr (1 + ω ( u x , r )) ≤ | b ( u x , r ) | and κ ≤ ω ( u x , r ) , (2.45) then there exists a constant c > (independent from x and r ) such that u is Lipschitz continuousin B cr/ ( x ) and we have | u ( y ) − u ( z ) | ≤ C K (1 + ω ( u, x, r )) | y − z | for every y, z ∈ B cr/ ( x ) . (2.46) Moreover, u is C ,β in B cr/ ( x ) where β = δ A d + δ A +2 and we have |∇ u ( y ) − ∇ u ( z ) | ≤ C K r − δ A d +2 (1 + ω ( u, x, r )) | y − z | β for every y, z ∈ B cr/ ( x ) . (2.47) Roughly speaking, the condition (2.45) says that the absolute value of the trace of u x to ∂B r is big. This will in fact ensure that u x has, in some smaller ball, the same sign than (the averageof) u x on ∂B r . Lemma 2.12.
Let γ and τ be two positive constants. There exist r , η ∈ (0 , and κ > suchthat, if x ∈ D and r ≤ r satisfy B λ A r ( x ) ⊂ D , γ (1 + ω ( u x , r )) ≤ r | b ( u x , r ) | and κ ≤ ω ( u x , r ) , (2.48) then there exist ρ ∈ ( ηr , ηr ) such that τ (1 + ρ δ A / ω ( u x , ρ )) ≤ ρ | b ( u x , ρ ) | and b + ( u x , ρ ) ≤ | b ( u x , ρ ) | . (2.49) Moreover, b ( u x , r ) and b ( u x , ρ ) have the same sign.Proof. We first prove the second inequality in (2.49). Let us recall that h r = h x,r denotesthe harmonic extension of the trace of u x to ∂B r . We want to estimate both − R ∂B ρ | h r | and − R ∂B ρ | u x − h r | in terms of | b ( u x , r ) | for some ρ ∈ ( ηr , ηr ) defined soon (by (2.52)). If η ≤ / |∇ h r | in B r we have that for every ξ ∈ B ηr |∇ h r ( ξ ) | ≤ − Z B r/ ( ξ ) |∇ h r | ≤ d − Z B r |∇ h r | ≤ d ω ( u x , r ) . Moreover, b ( u x , r ) = b ( h r , r ) = h r (0) since h r is harmonic and hence, choosing η such that η d/ ≤ γ/
4, we get | b ( u x , r ) − h r ( ξ ) | = | h r (0) − h r ( ξ ) | ≤ ηr k∇ h r k L ∞ ( B ηr ) ≤ ηr d/ ω ( u x , r ) ≤ γr ω ( u x , r ) ≤ | b ( u x , r ) | , (2.50)where in the last inequality we used the first estimate of (2.48). This gives (because ρ < ηr )34 | b ( u x , r ) | ≤ − Z ∂B ρ | h r | ≤ | b ( u x , r ) | . (2.51)On the other hand, we now fix some ρ = ρ x ∈ ( ηr , ηr ) such that Z ∂B ρ | u x − h r | ≤ ηr Z ηrηr/ ds Z ∂B s | u x − h r | . (2.52)By Cauchy-Schwarz’s inequality, Poincar´e’s inequality and (2.1) applied to the test function h r ,it follows that we have Z ∂B ρ | u x − h r | ≤ ηr Z B ηr | u x − h r | ≤ C ( ηr ) d − Z B ηr | u x − h r | ! / ≤ C ( ηr ) d − (cid:18)Z B r | u x − h r | (cid:19) / ≤ C ( ηr ) d − r (cid:18)Z B r |∇ ( u x − h r ) | (cid:19) / ≤ C ( ηr ) d − r (cid:18) r δ A Z B r |∇ h r | + r d (cid:19) / ≤ Cη d − r d (cid:0) r δ A ω ( u x , r ) + 1 (cid:1) / ≤ Cη d − r d ( r δ A / ω ( u x , r ) + 1)In view of the two hypothesis in (2.48) we then get − Z ∂B ρ | u x − h r | ≤ Cη − d r ( r δ A / ω ( u x , r ) + 1) ≤ Cη − d r ω ( u x , r ) (cid:16) r δ A / + 1 κ (cid:17) ≤ Cη − d γ − | b ( u x , r ) | (cid:16) r δ A / + 1 κ (cid:17) ≤ | b ( u x , r ) | , (2.53) IPSCHITZ CONTINUITY OF THE EIGENFUNCTIONS 15 where the last inequality holds if we choose r small enough and κ > η ) such that Cγ − (cid:16) r δ A / + 1 κ (cid:17) ≤ η d/ . (2.54)Now, using (2.51) and (2.53) we have b + ( u x , ρ ) = − Z ∂B ρ | u x | ≤ − Z ∂B ρ | h r | + − Z ∂B ρ | u x − h r | ≤ | b ( u x , r ) | , and, using also that h r keeps the same sign on ∂B ρ by (2.50), we have | b ( u x , ρ ) | ≥ (cid:12)(cid:12)(cid:12)(cid:12) − Z ∂B ρ h r (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) − Z ∂B ρ ( u x − h r ) (cid:12)(cid:12)(cid:12)(cid:12) ≥ − Z ∂B ρ | h r | − − Z ∂B ρ | u x − h r | ≥ | b ( u x , r ) | . (2.55)This proves the second inequality in (2.48). Moreover, (2.53) and (2.50) imply that | b ( u x , ρ ) − b ( u x , r ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12) b ( u x , ρ ) − − Z ∂B ρ h r (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) − Z ∂B ρ h r − b ( u x , r ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ − Z ∂B ρ | u x − h r | + − Z ∂B ρ | h r − b ( u x , r ) | ≤ | b ( u x , r ) | , which shows that b ( u x , r ) and b ( u x , ρ ) have the same sign.For the first estimate in (2.49), by (2.55) and the first hypothesis in (2.48), we have1 ρ | b ( u x , ρ ) | ≥ ρ | b ( u x , r ) | ≥ ηr | b ( u x , r ) | ≥ γ η (1 + ω ( u x , r )) , which using (2.6) gives (notice that we assumed that B λ A r ( x ) ⊂ D )1 + ρ δ A / ω ( u x , ρ ) ≤ ω ( u x , ρ ) ≤ C (cid:18) ω ( u x , r ) + log rρ (cid:19) ≤ C (cid:0) ω ( u x , r ) + | log η | (cid:1) ≤ C (cid:0) | log η | (cid:1) (1 + ω ( u x , r )) ≤ C (cid:0) | log η | (cid:1) ηγρ | b ( u x , ρ ) | . Finally, observe that with η small enough (and also r small enough and κ large enough so that(2.54) still holds) we have τ C (cid:0) | log η | (cid:1) ηγ ≤ . This completes the proof. (cid:3)
We continue with a self-improvement lemma whose strategy is similar to the one followed inthe previous lemma, the main difference being that we now consider u x with different points x . Lemma 2.13.
There exist constants r ∈ (0 , and τ ≥ with the following property: if x ∈ D , τ ≥ τ and ρ ≤ r satisfy B λ A ρ ( x ) ⊂ D , τ (1 + ρ δ A / ω ( u x , ρ )) ≤ ρ | b ( u x , ρ ) | and b + ( u x , ρ ) ≤ | b ( u x , ρ ) | , (2.56) then for every y ∈ B ερ ( x ) , where ε = τ − /d , there exists ρ ∈ ( ερ , ερ ) such that τ (1 + ρ δ A / ω ( u y , ρ )) ≤ ρ | b ( u y , ρ ) | and b + ( u y , ρ ) ≤ | b ( u y , ρ ) | . (2.57) Moreover, b ( u x , ρ ) and b ( u y , ρ ) have the same sign.Proof. Firstly, if ε is small enough so that ¯ ε := 2 λ A ε ≤ /
4, then by standard estimates onharmonic functions (see [9, Theorem 3.9]) h ρ = h x,ρ satisfies k∇ h ρ k L ∞ ( B ¯ ερ ) ≤ Cρ k h ρ k L ∞ ( B ρ/ ) ≤ Cρ − Z ∂B ρ | h ρ | = Cρ b + ( u x , ρ ) . Using that b ( u x , ρ ) = b ( h ρ , ρ ) = h ρ (0) by harmonicity and the second hypothesis in (2.56), itfollows that for every ξ ∈ B ¯ ερ we have | b ( u x , ρ ) − h ρ ( ξ ) | ≤ ¯ ερ k∇ h ρ k L ∞ ( B ¯ ερ ) ≤ εCb + ( u x , ρ ) ≤ τ − d C | b ( u x , ρ ) | ≤ | b ( u x , ρ ) | , (2.58)where the last inequality holds if τ is big enough. This implies that34 | b ( u x , ρ ) | ≤ | h ρ ( ξ ) | ≤ | b ( u x , ρ ) | for every ξ ∈ B ¯ ερ . (2.59)Moreover, by (2.1) applied to h ρ (and since ¯ ερ ≤ ρ for τ large enough) we have Z B ¯ ερ | u x − h ρ | ≤ C (¯ ερ ) d Z B ¯ ερ | u x − h ρ | ! / ≤ C ( ερ ) d Z B ρ | u x − h ρ | ! / ≤ C ( ερ ) d ρ Z B ρ |∇ ( u x − h ρ ) | ! / ≤ C ( ερ ) d ρ ρ δ A Z B ρ |∇ h ρ | + ρ d ! / (2.60) ≤ Cε d ρ d +1 (cid:0) ρ δ A ω ( u x , ρ ) + 1 (cid:1) / ≤ Cε d ρ d +1 ( ρ δ A / ω ( u x , ρ ) + 1) . We now fix some y ∈ B ερ ( x ). Let F : B ερ ⊂ R d → B ¯ ερ ⊂ R d be the function defined by F ( z ) = F − x ◦ F y ( z ). Then the coarea formula gives (and because ∂F ( B s ) = ∂ {| F − | > s } ) λ − A Z ερερ/ ds Z ∂F ( B s ) | u x − h ρ | d H d − ≤ Z ερερ/ ds Z ∂F ( B s ) | u x − h ρ ||∇| F − || d H d − = Z { ερ/ ≤| F − |≤ ερ } | u x − h ρ | (2.61) ≤ Z F ( B ερ ) | u x − h ρ | ≤ Z B ¯ ερ | u x − h ρ | . We now choose ρ ∈ ( ερ , ερ ) such that Z ∂F ( B ρ ) | u x − h ρ | d H d − ≤ ερ Z ερ/ ερ ds Z ∂F ( B ρ ) | u x − h ρ | d H d − , so that (2.60), (2.61) and the first hypothesis in (2.56) imply − Z ∂F ( B ρ ) | u x − h ρ | d H d − ≤ Cε − d ρ ( ρ δ A / ω ( u x , ρ ) + 1) ≤ Cε − d τ − | b ( u x , ρ ) |≤ Cτ − / | b ( u x , ρ ) | ≤ λ − d − A | b ( u x , ρ ) | , (2.62)where the last inequality holds for τ is large enough. Moreover, because the functions F and F − are Lipschitz continuous with Lipschitz constants bounded by λ A , we have for every set E ⊂ R d (see [12, Proposition 3.5]) λ − d − A H d − ( E ) ≤ F H d − ( E ) ≤ λ d − A H d − ( E ) , (2.63) IPSCHITZ CONTINUITY OF THE EIGENFUNCTIONS 17 where F H d − stands for the pushforward measure of H d − along F . Therefore, by (2.59) (andsince ∂F ( B ρ ) ⊂ B ¯ ερ ), (2.63) and (2.62) we have b + ( u y , ρ ) = − Z ∂B ρ | u y | d H d − = 1 H d − ( ∂B ρ ) Z ∂F ( B ρ ) | u x | dF H d − ≤ H d − ( ∂B ρ ) Z ∂F ( B ρ ) | h ρ | dF H d − + Z ∂F ( B ρ ) | u x − h ρ | dF H d − ! ≤ | b ( u x , ρ ) | + λ d − A − Z ∂F ( B ρ ) | u x − h ρ | d H d − (2.64) ≤ | b ( u x , ρ ) | . On the other hand, we have by (2.62) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) b ( u y , ρ ) − H d − ( ∂B ρ ) Z ∂F ( B ρ ) h ρ dF H d − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ H d − ( ∂B ρ ) Z ∂F ( B ρ ) | u x − h ρ | dF H d − ≤ λ d − A − Z ∂F ( B ρ ) | u x − h ρ | d H d − (2.65) ≤ | b ( u x , ρ ) | . Moreover, by (2.58) and since ∂F ( B ρ ) ⊂ B ¯ ερ we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) H d − ( ∂B ρ ) Z ∂F ( B ρ ) h ρ dF H d − − b ( u x , ρ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:26) max ξ ∈ ∂F ( B ρ ) h ρ ( ξ ) − b ( u x , ρ ) , b ( u x , ρ ) − min ξ ∈ ∂F ( B ρ ) h ρ ( ξ ) (cid:27) ≤ | b ( u x , ρ ) | . (2.66)Therefore, using the triangle inequality, (2.65) and (2.66) we get | b ( u y , ρ ) − b ( u x , ρ ) | ≤ | b ( u x , ρ ) | . This proves that b ( u x , ρ ) and b ( u y , ρ ) have the same sign and also implies that | b ( u y , ρ ) | ≥ | b ( u x , ρ ) | − | b ( u y , ρ ) − b ( u x , ρ ) | ≥ | b ( u x , ρ ) | (2.67)Finally, (2.64) and (2.67) gives b + ( u y , ρ ) ≤ | b ( u x , ρ ) | ≤ | b ( u y , ρ ) | , which is the second inequality in (2.57).We now prove the first inequality in (2.57). By (2.67) and the first hypothesis in (2.56) wehave | b ( u y , ρ ) | ≥ | b ( u x , ρ ) | ≥ τ ρ ρ δ A / ω ( u x , ρ )) , (2.68)We then apply Lemma 2.6 (notice that we have B λ A ρ ( x ) ⊂ D and 2 λ A ρ ≤ ρ ) and eventuallychoose τ bigger (depending only on d and δ A ) to get1 + ρ δ A / ω ( u y , ρ ) ≤ ρ δ A / λ A ω ( u x , λ A ρ ) ≤ ρ δ A / C ( ω ( u x , ρ ) + log(2¯ ε − )) ≤ C (1 + ρ δ A / ω ( u x , ρ ) + ρ δ A / log(2¯ ε − )) ≤ C (1 + ρ δ A / ω ( u x , ρ ) + τ − δ A / d log( τ )) ≤ C (1 + ρ δ A / ω ( u x , ρ )) . Therefore, with (2.68) this gives1 ρ | b ( u y , ρ ) | ≥ ερ τ ρ ρ δ A / ω ( u x , ρ )) ≥ τ /d C τ (1 + ρ δ A / ω ( u y , ρ )) , which, choosing τ big enough so that τ /d ≥ C , completes the proof. (cid:3) We are now in position to prove Proposition 2.11 using the results from Lemmas 2.12, 2.13and Proposition 2.8.
Proof of Proposition . Set ε = τ − /d and ¯ r = εη r where η and τ are the constants given byLemmas 2.12 and 2.13. Note that in view of the first hypothesis in (2.45) we have b ( u x , r ) = 0.We will prove that if b ( u x , r ) > b ( u x , r ) < u > u < B ¯ r ( x ) .Let y ∈ B ¯ r ( x ) be fixed. We first apply Lemma 2.12. Now, we apply once Lemma 2.13 at x (notice that we have y ∈ B ερ ( x )) and then iteratively at the point y . It follows that there existsa sequence of raddi ρ i > i τ (1 + ρ δ A / i ω ( u y , ρ i )) ≤ ρ i b ( u y , ρ i ) , i ≥ , (2.69)and that b ( u y , ρ i ) has the same sign than b ( u x , r ) for every i ≥
0. Assume that b ( u x , r ) >
0, theproof in the case b ( u x , r ) < h i = h y,ρ i the harmonic extension ofthe trace of the function u y to ∂B ρ i . With the same argument as in (2.58) we get | b ( u y , ρ i ) − h i ( ξ ) | ≤ | b ( u y , ρ i ) | for every ξ ∈ B ερ i . Since b ( u y , ρ i ) >
0, this implies that for every ξ ∈ B ερ i ∩ { u y ≤ } we have | u y ( ξ ) − h i ( ξ ) | ≥ | u y ( ξ ) − b ( u y , ρ i ) | − | b ( u y , ρ i ) − h i ( ξ ) |≥ | b ( u y , ρ i ) | − | b ( u y , ρ i ) | = 34 | b ( u y , ρ i ) | . By the Chebyshev inequality, the Lebesgue measure of B ερ i ∩ { u y ≤ } is estimate as | B ερ i ∩ { u y ≤ }| ≤ | b ( u y , ρ i ) | Z B ερi | u y − h i | . (2.70)On the other hand, by (2.60) in this context we have Z B ερi | u y − h i | ≤ Cε d/ ρ d +1 i (1 + ρ δ A / i ω ( u y , ρ i )) . (2.71)Now, combining (2.70), (2.71) and (2.69) we get | B ερ i ∩ { u y ≤ }|| B ερ i | ≤ ( ερ i ) − d Cε d/ ρ d +1 i (2 i τ ρ i ) − ≤ ε d/ C − i , which implies that | F y ( B ερ i ) ∩ { u ≤ }|| F y ( B ερ i ) | = | B ερ i ∩ { u y ≤ }|| B ερ i | −−−−→ i → + ∞ , where F y ( B ερ i ) = y + ρ i A / y ( B ε ). This shows that the density of the set { u ≤ } at every point y ∈ B ¯ r ( x ) is 0 (see [12, exercise 5.19]), and hence that u > B ¯ r ( x ). Now,we set c = λ − A τ − /d η/
2, where η and τ are the constants given by Lemma 2.12 and 2.13. Then(2.46) and (2.47) follow from Proposition 2.8 and the fact that ω ( u, x, cr ) ≤ c − d/ ω ( u, x, r ). Thisconcludes the proof. (cid:3) IPSCHITZ CONTINUITY OF THE EIGENFUNCTIONS 19
Lipschitz continuity.
In this subsection we prove Theorem 1.2. At this stage, the mainwork is to deal with the case where ω ( u x , r ) is big and the first condition in (2.45) fails. In thiscase, we show in Proposition 2.14 below that the value of ω ( u x , r ) decreases at some smaller scale.Notice that the extra hypothesis (2.72) is almost irrelevant in view of Proposition 2.8. Proposition 2.14.
Let K ⊂ D be a compact set. There exist positive constants r K , γ ∈ (0 , and κ > with the following property: if x ∈ K and r ≤ r K satisfy u x ( ξ ) = 0 for some ξ ∈ B (2 λ A ) − r , (2.72) | b ( u x , r ) | ≤ γr (1 + ω ( u x , r )) and κ ≤ ω ( u x , r ) , (2.73) then we have ω ( u x , r/ ≤ ω ( u x , r ) . (2.74)We will need the following almost-monotonicity formula for operators in divergence form. Werefer to [13, Theorem III] for a proof (see also [2] and [5] for the case of the Laplacian). Let usset for u + , u − ∈ H ( B ) and r ∈ (0 , u + , u − , r ) = (cid:18) r Z B r |∇ u + ( ξ ) | | ξ | d − dξ (cid:19) (cid:18) r Z B r |∇ u − ( ξ ) | | ξ | d − dξ (cid:19) . Proposition 2.15.
Let B = ( b ij ) ij : B → Sym + d be a uniformly elliptic matrix-valued functionwith H¨older continuous coefficients, that is, for every x, y ∈ B and ξ ∈ R d | b ij ( x ) − b ij ( y ) | ≤ c B | x − y | δ B and λ B | ξ | ≤ ξ · B x ξ ≤ λ B | ξ | . Let u + , u − be two non-negative and continuous functions in the unit ball B such that div( B ∇ u ± ) ≥ − in B and u + u − = 0 in B . Then there exist r > and C > , depending only on d, c B , δ B and λ B , such that for every r ≤ r we have Φ( u + , u − , r ) ≤ C (cid:0) k u + + u − k L ( B ) (cid:1) . We now state this almost-monotonicity formula for the functions u ± x . Corollary 2.16.
Let Ω ⊂ D be a quasi-open set and K ⊂ D be a compact set. Let A be amatrix-valued function satisfying (1.5) and (1.6) . Let f ∈ L ∞ ( D ) . Assume that u ∈ H (Ω) is acontinuous function solution of the equation − div( A ∇ u ) = f in Ω . (2.75) Then there exists r K > and C m > , depending only on d, c A , δ A , λ A , k f k L ∞ , | D | and dist ( K, D c ) ,such that for every x ∈ K and every r ≤ r K the function u x satisfies Φ( u + x , u − x , r ) ≤ C m . Proof.
We first prove that we have, in the sense of distributions,div( A ∇ u + ) ≥ f { u> } in D and div( A ∇ u − ) ≥ f { u< } in D. (2.76)Let us define p n : R → R + for n ∈ N by p n ( s ) = 0 , for s ≤ p n ( s ) = ns, for s ∈ [0 , /n ]; p n ( s ) = 1 , for s ≥ /n, and set q n ( s ) = R s p n ( t ) dt . Since p n is Lipschitz continuous, we have p n ( u ) ∈ H (Ω) and ∇ p n ( u ) = p ′ n ( u ) ∇ u . Let ϕ ∈ C ∞ ( D ) be such that ϕ ≥ D . Multiplying the equation (2.75)with ϕp n ( u ) ∈ H (Ω) we get Z D A ∇ q n ( u ) · ∇ ϕ = Z D p n ( u ) A ∇ u · ∇ ϕ ≤ Z D (cid:0) p n ( u ) A ∇ u · ∇ ϕ + ϕp ′ n ( u ) A ∇ u · ∇ u )= Z D A ∇ u · ∇ ( ϕp n ( u )) = Z D f ϕp n ( u ) . Now, the inequality for u + in (2.76) follows by letting n tend + ∞ , because p n ( u ) convergesalmost-everywhere to { u> } and ∇ q n ( u ) converges in L to ∇ u + . The same proof holds for u − .Now, set ρ = λ − A dist( K, D c ) and define for every ξ ∈ B u ± ( ξ ) = ρ − k f k − L ∞ u ± x ( ρξ ) , ˜ f ( ξ ) = f ◦ F x ( ρξ ) , B ξ = A − / x A F x ( ρξ ) A − / x . Then the functions u ± satisfydiv( B ∇ u ± ) ≥ k f k − L ∞ ˜ f { u ± > } ≥ − B . Therefore, by Proposition 2.15 we have for every r ≤ r K := r ρ Φ( u + x , u − x , r ) = ρ k f k L ∞ Φ( u + , u − , r/ρ ) ≤ ρ k f k L ∞ C (cid:0) k u + + u − k L ( B ) (cid:1) ≤ ρ k f k L ∞ C (cid:16) λ d A ρ − d − k f k − L ∞ k u k L ( D ) (cid:17) ≤ ρ k f k L ∞ C (cid:16) λ d A ρ − d − C d | D | /d (cid:17) =: C m . (cid:3) Proof of Proposition . Let us denote as before h r = h x,r the harmonic extension of the traceof u x to ∂B r . Then we have ω ( u x , r/ = − Z B r/ |∇ u x | ≤ − Z B r/ |∇ h r | + 2 − Z B r/ |∇ ( u x − h r ) | . (2.77)By the quasi-minimality property of u x we can estimate the second term in the right hand sideof (2.77) as we did in (2.8), this gives − Z B r/ |∇ ( u x − h r ) | ≤ d − Z B r |∇ ( u x − h r ) | ≤ Cr δ A ω ( u x , r ) + C ≤ C ( r δ A + κ − ) ω ( u x , r ) , (2.78)where in the last inequality we have used the second hypothesis in (2.73). On the other hand,estimates for harmonic functions give k∇ h r k L ∞ ( B r/ ) ≤ Cr k h r k L ∞ ( B r/ ) ≤ Cr − Z ∂B r | h r | = Cr b + ( u x , r ) . (2.79)We now want to estimate b + ( u x , r ) in terms of r ω ( u x , r ). Let us assume that ω ( u + x , r ) ≤ ω ( u − x , r ),the same proof holds if the opposite inequality is satisfied. We first prove that for ξ ∈ B r/ and η < / − Z ∂B r u + x − − Z ∂B ηr ( ξ ) u + x ≤ C d η − d r − Z B r |∇ u + x | . (2.80)Notice that up to considering the function ξ u + x ( rξ ) we can assume that r = 1. Let us definea one to one function F : B \ B η → B \ B η ( ξ ) by F ( ξ ) = ξ + 1 − | ξ | − η ξ . We set v = u + x ◦ F . For every ξ ∈ ∂B we have by the fundamental theorem of the calculus v ( ξ ) − v ( ηξ ) = Z η ddt v ( tξ ) dt ≤ Z η |∇ v ( tξ ) | dt. IPSCHITZ CONTINUITY OF THE EIGENFUNCTIONS 21
Note that F is the identity on ∂B and is simply a translation on ∂B η . Therefore, averaging on ξ ∈ ∂B (and since |∇ v | ≤ C d |∇ u + x ◦ F | ) we get − Z ∂B u + x − − Z ∂B η ( ξ ) u + x = − Z ∂B v − − Z ∂B η ( ξ ) v ≤ dω d Z η dt Z ∂B |∇ v ( tξ ) | d H d − ( ξ ) ≤ η − d dω d Z B \ B η |∇ v | ≤ C d η − d Z B \ B η |∇ u + x ◦ F |≤ C d η − d Z B \ B η ( ξ ) |∇ u + x || det ∇ F − | ≤ C d η − d − Z B |∇ u + x | , which proves (2.80). Now, let ξ ∈ B (2 λ A ) − r be such that u x ( ξ ) = 0 as in (2.72). By Proposition2.5 we have for every ξ ∈ B ηr ( ξ ) (and because F x ( ξ ) , F x ( ξ ) ∈ B (2 λ A ) − r ( x ) if η is small enough) u + x ( ξ ) ≤ | u x ( ξ ) | = | u x ( ξ ) − u x ( ξ ) | = | u ( F x ( ξ )) − u x ( F x ( ξ )) |≤ C | F x ( ξ ) − F x ( ξ ) | (cid:18) ω ( u, x, λ − A r ) + log λ − A r | F x ( ξ ) − F x ( ξ ) | (cid:19) ≤ C | ξ − ξ | (cid:18) ω ( u, x, λ − A r ) + log r | ξ − ξ | (cid:19) (2.81) ≤ Cr (1 + η ω ( u x , r )) , where the last inequality holds for η small enough and since we have ω ( u, x, λ − A r ) = − Z B λ − A r ( x ) |∇ u | ≤ λ A − Z B λ − A r ( x ) A x ∇ u · ∇ u = λ A − Z F − x ( B λ − A r ( x )) |∇ u x | ≤ λ d +1) A ω ( u x , r ) . Moreover, recall that we assumed that ω ( u + x , r ) ≤ ω ( u − x , r ). Using the monotonicity formula inCorollary 2.16 we get ω ( u + x , r ) ≤ ω ( u + x , r ) ω ( u − x , r ) ≤ C d Φ( u + x , u − x , r ) ≤ C d C m , which implies by Cauchy-Schwarz’s inequality − Z B r |∇ u + x | ≤ (cid:18) − Z B r |∇ u + x | (cid:19) / = ω ( u + x , r ) ≤ ( C d C m ) / . (2.82)Therefore, combining (2.81), (2.80), (2.82) and using the first hypothesis in (2.73) we have (andalso since u − x = u + x − u x ) b + ( u x , r ) = − Z ∂B r | u x | = 2 − Z ∂B r u + x − − Z ∂B r u x ≤ k u + x k L ∞ ( ∂B ηr ( ξ )) + 2 (cid:18) − Z ∂B r u + x − − Z ∂B ηr ( ξ ) u + x (cid:19) + | b ( u x , r ) |≤ C (cid:16) ( η + γ ) ω ( u x , r ) + 1 + η − d C / m (cid:17) r (2.83) ≤ C (cid:16) ( η + γ ) + (cid:0) η − d C / m (cid:1) κ − (cid:17) rω ( u x , r ) , where in the last inequality we used the second hypothesis in (2.73). We now return to (2.77).With (2.78), (2.79) and (2.83) we get ω ( u x , r/ ≤ Cr b + ( u x , r ) + C ( r δ A + κ − ) ω ( u x , r ) ≤ C (cid:16) ( η + γ ) + r δ A + (cid:0) η − d C / m (cid:1) κ − (cid:17) ω ( u x , r ) . Therefore, choosing first η, γ and r K small enough and then κ big enough (depending on η ) weobtain (2.74), which concludes the proof. (cid:3) We are now in position to prove Theorem 1.2 using an iterative argument and Propositions2.8, 2.11, 2.14.
Proof of Theorem . Recall that we denote by u any coordinate function of the vector U andthat we have to prove that u is locally Lipschitz continuous in D . Let K ⊂ D be a compact setand let x ∈ K . Let r ≤ r K , where r K is smaller than the constants given by Propositions 2.8,2.11 and 2.14. Set κ = max { κ , κ } where κ , κ are the constants given by Propositions 2.11and 2.14. We consider the following four cases: Case 1: either u x > B (2 λ A ) − r or u x < B (2 λ A ) − r (2.84) Case 2: γr (1 + ω ( u x , r )) ≤ | b ( u x , r ) | and κ ≤ ω ( u x , r ) , (2.85) Case 3: | b ( u x , r ) | ≤ γr (1 + ω ( u x , r )) and κ ≤ ω ( u x , r ) , (2.86) Case 4: ω ( u x , r ) ≤ κ. (2.87)For k ≥ r k = 3 − k r . We denote by k , if it exists, the smallest integer k ≥ x, r k ) satisfies either (2.84) or (2.85), and we set k = + ∞ otherwise. If k >
0, thenfor every k < k we have that: if ( x, r k ) satisfies (2.86) then by Proposition 2.14 we have (noticethat (2.72) holds since u is continuous and that (2.84) is not satisfied) ω ( u x , r k +1 ) ≤ ω ( u x , r k ) , while if ( x, r k ) satisfies (2.87), then we have ω ( u x , r k +1 ) ≤ d/ ω ( u x , r k ) ≤ d/ κ. Therefore, with an induction we get that for every 0 ≤ k ≤ k ω ( u x , r k ) ≤ max { − k ω ( u x , r ) , d/ κ } . (2.88)Assume that k = + ∞ . If x is a Lebesgue point for ∇ u , then 0 is a Lebesgue point of u x and itfollows from (2.88) that |∇ u ( x ) | ≤ λ A |∇ u x (0) | = λ A lim k → + ∞ ω ( u x , r k ) ≤ λ A d/ κ. Assume now that k < + ∞ . Then, by definition of k , the pair ( x, r k ) satisfies either (2.84) or(2.85). If (2.84) holds, then Proposition 2.8 infers that u is C ,β near x and that we have (usingalso (2.88)) |∇ u ( x ) | ≤ C K (1 + ω ( u, x, (2 λ A ) − r k )) ≤ C K (1 + ω ( u, x, λ − A r k )) ≤ C K (1 + ω ( u x , r k )) ≤ C K (1 + max { − k ω ( u x , r ) , d/ κ } ) ≤ C K (1 + ω ( u x , r )) . Moreover, by Proposition 2.11 the same estimate holds if the pair ( x, r k ) satisfies (2.85). There-fore, in all cases it follows that for almost every point x ∈ K and every r ≤ r K we have |∇ u ( x ) | ≤ C K (1 + ω ( u x , r )) . (2.89)Let now x ∈ K . Then, for almost every x ∈ B r K / ( x ), it follows by (2.89) that |∇ u ( x ) | ≤ C K (1 + ω ( u x , r K / ≤ C K (1 + ω ( u x , r K )) . With a compactness argument this proves that u is locally Lipschitz continuous in D and com-pletes the proof. (cid:3) IPSCHITZ CONTINUITY OF THE EIGENFUNCTIONS 23 Lipschitz continuity of the eigenfunctions
This section is dedicated to the proof of Theorem 1.1. Precisely, we prove that the vector U = ( u , . . . , u k ) of the first k eigenfunctions on an optimal set for (1.1) is locally Lipschitzcontinuous in D . Using an idea taken from [14], we show that U is a quasi-minimizer in the senseof (1.4), and we then apply Theorem 1.2 to get the Lipschitz continuity of U .3.1. Preliminaries and existence of an optimal set.
We start with some properties aboutthe spectrum of the operator in divergence form defined in 1.2, and we then prove that theproblem (1.1) admits a solution among the class of quasi-open sets.Let us define the weighted Lebesgue measure m = b dx , where dx stands for the Lebesguemeasure in R d . For a quasi-open set Ω ⊂ R d , we define the spaces L (Ω; m ) = L (Ω) and H (Ω; m ) = H (Ω) endowed respectively with the following norms k u k L (Ω; m ) = (cid:18)Z Ω u dm (cid:19) / and k u k H (Ω; m ) = k u k L (Ω; m ) + k∇ u k L (Ω) . Moreover, if Ω = R d we will simply write k u k L ( m ) = k u k L ( R d ; m ) . We notice that, by thehypothesis (1.7) on the function b , the norms k · k L (Ω; m ) and k · k L (Ω) are equivalent. We stressout that the choice of these norms is natural in view of (1.2) and is motivated by the variationalformulation of the sum of the first k eigenfunctions (see (3.1) below). Now, the Lax-Milgramtheorem and the Poincar´e inequality imply that for every f ∈ L (Ω , m ) there exists a uniquesolution u ∈ H (Ω , m ) to the problem − div( A ∇ u ) = f b in Ω , u ∈ H (Ω , m ) . The resolvent operator R Ω : f ∈ L (Ω; m ) → H (Ω; m ) ⊂ L (Ω; m ) defined as R Ω ( f ) = u is acontinuous, self-adjoint and positive operator. Since the Sobolev space H (Ω; m ) is compactlyembedded into L (Ω; m ) (because we have assumed that b ≥ c b >
0, see (1.7)), the resolvent R Ω isin addition a compact operator. We say that a complex number λ is an eigenvalue of the operator(1.2) in Ω if there exists a non-trivial eigenfunction u ∈ H (Ω; m ) solution of the equation − div( A ∇ u ) = λu b in Ω , u ∈ H (Ω; m ) . The above properties of the resolvent ensure that the spectrum of the operator (1.2) in Ω is givenby an increasing sequence of eigenvalues which are strictly positive real numbers, non-necessarilydistinct, and which we denote by0 < λ (Ω) ≤ λ (Ω) ≤ · · · ≤ λ k (Ω) ≤ · · · The eigenvalues λ k (Ω) are variationnaly characterized by the following min-max formula λ k (Ω) = min V subspace ofdimension k of H (Ω ,m ) max v ∈ V \{ } R Ω A ∇ v · ∇ v dx R Ω v dm . Moreover, we denote by u k the normalized (with respect to the norm k · k L (Ω; m ) ) eigenfunctionscorresponding to the eigenvalues λ k (Ω) and note that the family ( u k ) k form an orthonormalsystem in L (Ω; m ), that is Z Ω u i u j dm = δ ij := (cid:26) i = j, i = j. As a consequence, we have the following variational formulation for the sum of the first k eigen-values on a quasi-open set Ω k X i =1 λ i (Ω) = min n Z Ω A ∇ V · ∇ V dx : V = ( v , . . . , v k ) ∈ H (Ω , R k ) , Z Ω v i v j dm = δ ij o , (3.1)for which the minimum is attained for the vector U = ( u , . . . , u k ). We now deduce from thischaracterization that the minimum in (1.1) is reached. Proposition 3.1 (Existence) . The shape optimization problem (1.1) has a solution.Proof.
Let (Ω n ) n ∈ N be a minimizing sequence of quasi-open sets to the problem (1.1) and denoteby U n = ( u n , . . . , u nk ) the first k eigenfunctions on Ω n . Since the matrices A x , x ∈ D , are uniformlyelliptic, we have the following inequality λ − A Z D |∇ U n | dx ≤ Z D A ∇ U n · ∇ U n dx = k X i =1 λ i (Ω n )which infers that the norm k U n k H is uniformly bounded. Therefore, up to a subsequence, U n converges weakly in H ( D, R k ) and strongly in L ( D, R k ) to some V ∈ H ( D, R k ). Notice that V is an orthonormal vector. Set Ω ∗ := {| V | > } . Then using (3.1), the weak convergence in H of U n to V and the semi-continuity of the Lebesgue measure we have k X i =1 λ i (Ω ∗ ) + Λ | Ω ∗ | ≤ Z D A ∇ V · ∇ V dx + Λ |{| V | > }|≤ lim inf n (cid:16) Z D A ∇ U n · ∇ U n dx + Λ |{| U n | > }| (cid:17) ≤ lim inf n (cid:16) k X i =1 λ i (Ω n ) + Λ | Ω n | (cid:17) which concludes the proof. (cid:3) In the next Lemma we prove that the eigenfunctions are bounded. This result is a consequenceof Lemma 2.1 and we refer to [15, Lemma 5.4] for a proof which is based on an interpolationargument.
Lemma 3.2 (Boundedness of the eigenfunctions) . Let Ω ⊂ R d be a bounded quasi-open set.There exist a dimensional constant n ∈ N and a constant C > depending only on d, λ A , c b and | Ω | , such that the resolvent operator R Ω : L (Ω; m ) → L (Ω; m ) satisfies R n ( L (Ω; m )) ⊂ L ∞ (Ω) and k R n k L ( L (Ω; m ); L ∞ (Ω)) ≤ C. In particular, if u is an eigenfunction on Ω normalized by k u k L ( m ) = 1 , then u ∈ L ∞ (Ω) and k u k L ∞ ≤ Cλ (Ω) n , where λ (Ω) denotes the eigenvalue corresponding to u . To conclude this subsection, we show that the first eigenfunction on an optimal set Ω ∗ keepsthe same sign on every connected component of Ω ∗ . Notice that Ω ∗ may not be connected andhas at most k connected components. Lemma 3.3 (Sign of the principal eigenfunction) . Let Ω ⊂ D be an open and connected set andlet u ∈ H (Ω) be the normalized first eigenvalue on Ω , that is − div( A ∇ u ) = λ (Ω) b u in Ω and Z Ω u dm = 1 . Then u is non-negative in Ω (up to a change of sign).Proof. We assume that u + = 0 (if not, take − u instead of u ) and we set u + = u + / k u + k L ( m ) and u − = u − / k u − k L ( m ) . Since u is variationally characterized by λ (Ω) = Z Ω A ∇ u · ∇ u dx = min n Z Ω A ∇ ˜ u · ∇ ˜ u dx : ˜ u ∈ H (Ω) , Z Ω ˜ u dm = 1 o , (3.2)we have Z Ω A ∇ u · ∇ u dx ≤ Z Ω A ∇ u + · ∇ u + dx and Z Ω A ∇ u · ∇ u dx ≤ Z Ω A ∇ u − · ∇ u − dx. IPSCHITZ CONTINUITY OF THE EIGENFUNCTIONS 25
Then, it follows that the two above inequalities are in fact equalities since otherwise we have Z Ω A ∇ u · ∇ u dx = Z Ω A ∇ u + · ∇ u + dx + Z Ω A ∇ u − · ∇ u − dx> (cid:16) Z Ω ( u + ) dm + Z Ω ( u − ) dm (cid:17) Z Ω A ∇ u · ∇ u dx = Z Ω A ∇ u · ∇ u dx, which is absurd. In view of the minimization characterization (3.2), this ensures that u + issolution of the equation − div( A ∇ u + ) = λ (Ω) u + b in Ω . Then, the strong maximum principle (see [9, Theorem 8.19]) and the connectedness of Ω implythat u + is strictly positive in Ω, which completes the proof. (cid:3) Quasi-minimality and Lipschitz continuity of the eigenfunctions.
We prove thatthe vector U = ( u , . . . , u k ) of normalized eigenfunctions on an optimal set Ω ∗ for the problem(1.1) is a local quasi-minimizer of the vector-valued functional H ( D, R k ) ∋ ˜ U Z D A ∇ ˜ U · ∇ ˜ U dx + Λ |{| ˜ U | > }| in the sense of the Proposition below. The Lipschitz continuity of the eigenfunctions is then aconsequence of Theorem 1.2. We notice that, in view of the variational formulation (3.1), thevector U is solution to the following problemmin (cid:26) Z D A ∇ V ·∇ V dx +Λ |{| V | > }| : V = ( v , . . . , v k ) ∈ H ( D, R k ) , Z D v i v j dm = δ ij (cid:27) . (3.3) Proposition 3.4 (Quasi-minimality of U ) . Let Ω ∗ ⊂ D be an optimal set for the problem (1.1) .Then the vector of orthonormalized eigenfunctions U = ( u , . . . , u k ) ∈ H (Ω ∗ , R k ) satisfies thefollowing quasi-minimality condition: for every C > there exist constants ε ∈ (0 , and C > ,depending only on d, k, C , k U k L ∞ and | D | , such that Z D A ∇ U · ∇ U dx + Λ |{| U | > }| ≤ (cid:0) C k U − ˜ U k L (cid:1) Z D A ∇ ˜ U · ∇ ˜ U dx + Λ |{| ˜ U | > }| , (3.4) for every ˜ U ∈ H ( D, R k ) such that k U − ˜ U k L ≤ ε and k ˜ U k L ∞ ≤ C . The next Lemma, in which we get rid of the orthogonality constraint in (3.3), is similar toLemma 2.5 in [14] with only slight modifications, but we decided to recall the whole proof for asake of completeness.
Lemma 3.5.
Let Ω ⊂ D be a quasi-open set and let U = ( u , . . . , u k ) be the vector of normalizedeigenvalues on Ω . Let δ > . Then there exist ε k ∈ (0 , and C k > , depending only on d, k, δ and | Ω | , such that for every ˜ U = (˜ u , . . . , ˜ u k ) ∈ H ( D, R k ) satisfying ε k := k X i =1 Z D | ˜ u i − u i | dm ≤ ε k and sup i =1 ,...,k n k u i k L ∞ + k ˜ u i k L ∞ o ≤ δ the following estimate holds Z D A ∇ V · ∇ V dx ≤ (1 + C k ε k ) Z D A ∇ ˜ U · ∇ ˜ U dx, (3.5) where V = ( v , . . . , v k ) ∈ H ( D, R k ) is the vector obtained by orthonormalizing ˜ U with the Gram-Schmidt procedure: v i = w i / k w i k L ( m ) where w i = ( ˜ u if i = 1˜ u i − P i − j =1 (cid:0)R D ˜ u i v j dm (cid:1) v j if i = 2 , . . . , k. Proof.
We first prove an estimate of k u k − w k k L ( m ) in terms of ε k . Precisely, we prove byinduction on k that there exist constants ε k ∈ (0 ,
1) and C k > ε k ≤ ε k k u k − w k k L ( m ) ≤ C k ε k , k X i =1 k u i − v i k L ( m ) ≤ C k ε k , max i =1 ,...,k k v i k L ∞ ≤ C k . (3.6)For k = 1 the first estimate obviously holds with C ≥
1. Moreover we have k ˜ u − v k L ( m ) = |k ˜ u k L ( m ) − |k ˜ u k L ( m ) k ˜ u k L ( m ) ≤ |k ˜ u k L ( m ) − |k ˜ u k L ( m ) k ˜ u k L ( m ) = |k u + (˜ u − u ) k L ( m ) − |k u + (˜ u − u ) k L ( m ) k u + (˜ u − u ) k L ( m ) (3.7) ≤ R u | ˜ u − u | dm + k ˜ u − u k L ( m ) − R u | ˜ u − u | dm (cid:16) k u k L ( m ) + k ˜ u − u k L ( m ) (cid:17) ≤ (2 k u k L ∞ + k ˜ u − u k L ∞ ) k ˜ u − u k L ( m ) − k u k L ∞ k ˜ u − u k L ( m ) (cid:16) k u k L ( m ) + k ˜ u − u k L ( m ) (cid:17) ≤ δε − δε (cid:16) | Ω | / + ε (cid:17) ≤ δ | Ω | / ε , where the last inequality holds if ε ≤ min { (4 δ ) − , | Ω | / } . This gives the following L -estimate k u − v k L ( m ) ≤ k u − ˜ u k L ( m ) + k ˜ u − v k L ( m ) ≤ (1 + 12 δ | Ω | / ) ε . Finally, we estimate the infinity norm k v k L ∞ = k ˜ u k L ∞ k ˜ u k L ( m ) = k ˜ u k L ∞ k u + (˜ u − u ) k L ( m ) ≤ k ˜ u k L ∞ (1 − R u | ˜ u − u | dm ) / ≤ k ˜ u k L ∞ − R u | ˜ u − u | dm ≤ δ − δε ≤ δ, which proves the claim for k = 1. Suppose now that the claim holds for 1 , . . . , k −
1. We firstestimate k u k − w k k L ( m ) . Since the functions u i form an orthogonal system of L (Ω , m ) and bythe induction’s hypothesis we have (and also because ε k − ≤ ε k ) k − X i =1 (cid:12)(cid:12)(cid:12)(cid:12) Z D ˜ u k v i dm (cid:12)(cid:12)(cid:12)(cid:12) ≤ k − X i =1 Z D (cid:16) | ˜ u k − u k | u i + | v i − u i | u k + | v i − u i || ˜ u k − u k | (cid:17) dm ≤ (( k − δ + δC k − + ( k − C k − + δ )) ε k =: ˜ C k ε k . (3.8)Therefore, with the triangle inequality we obtain k u k − w k k L ( m ) ≤ k u k − ˜ u k k L ( m ) + k − X i =1 (cid:12)(cid:12)(cid:12)(cid:12) Z D ˜ u k v i dm (cid:12)(cid:12)(cid:12)(cid:12) ( k u i k L ( m ) + k v i − u i k L ( m ) ) ≤ (1 + ˜ C k ( | Ω | / + ε k − )) ε k ≤ (1 + 2 ˜ C k | Ω | / ) ε k . (3.9)We now prove the second estimate in (3.6). Using once again (3.8), we have the following estimateof the L ∞ -norm of w k k w k k L ∞ ≤ k ˜ u k k L ∞ + k X i =1 (cid:12)(cid:12)(cid:12)(cid:12) Z D ˜ u k v i dm (cid:12)(cid:12)(cid:12)(cid:12) k v i k L ∞ ≤ δ + C k − ˜ C k . (3.10) IPSCHITZ CONTINUITY OF THE EIGENFUNCTIONS 27
Moreover, with the same procedure as in (3.7), it follows from (3.9) and (3.10) that k ˜ u k − v k k L ( m ) ≤ (3 k u k k L ∞ + k w k k L ∞ ) k w k − u k k L ( m ) − k u k k L ∞ k w k − u k k L ( m ) (cid:16) k u k k L ( m ) + k w k − u k k L ( m ) (cid:17) ≤ (4 δ + C k − ˜ C k )(1 + 2 ˜ C k | Ω | / )1 − δ (1 + 2 ˜ C k | Ω | / ) ε k ( | Ω | / + (1 + 2 ˜ C k | Ω | / )) ε k . Now, choose ε k ≤ [4 δ (1 + 2 ˜ C k | Ω | / )] − so that with the triangle inequality we have k u k − v k k L ( m ) ≤ k u k − ˜ u k k L ( m ) + k ˜ u k − v k k L ( m ) ≤ h δ + C k − ˜ C k )(1 + 2 ˜ C k | Ω | / )( | Ω | / + (1 + 2 ˜ C k | Ω | / ) i ε k . We then use the inductive hypothesis to get the desired L -estimate. It remains only to estimate k v k k L ∞ . Firstly, notice that we have | k w k k L ( m ) − | ≤ | k w k k L ( m ) − | = | k u k + ( w k − u k ) k L ( m ) − |≤ (cid:12)(cid:12)(cid:12)(cid:12) Z D u k ( u k − w k ) dm + Z D ( u k − w k ) dm (cid:12)(cid:12)(cid:12)(cid:12) ≤ (3 k u k k L ∞ + k w k k L ∞ ) k u k − w k k L ( m ) ≤ (4 δ + C k − ˜ C k )(1 + 2 ˜ C k | Ω | / ) ε k . Thus, with the extra assumption ε k ≤ [(4 δ + C k − ˜ C k )(1 + 2 ˜ C k | Ω | / )] − , it follows that 1 / ≤k w k k L ( m ) ≤ /
2. With (3.10) this gives the following L ∞ -norm of v k k v k k L ∞ = k w k k L ∞ k w k k L ( m ) ≤ δ + C k − ˜ C k )and concludes the proof of the claim.We are now in position to prove the Lemma by induction. For k = 1, we ask that ε ≤ (4 δ ) − ,so that we have Z D A ∇ v · ∇ v dx ≤ k ˜ u k − L ( m ) Z D A ∇ ˜ u · ∇ ˜ u dx ≤ (1 − k u k L ∞ k ˜ u − u k L ( m ) ) − Z D A ∇ ˜ u · ∇ ˜ u dx ≤ (1 + 4 δε ) Z D A ∇ ˜ u · ∇ ˜ u dx ≤ (1 + 12 δε ) Z D A ∇ ˜ u · ∇ ˜ u dx. Suppose now that the Lemma holds for 1 , . . . , k −
1. Thanks to the first estimate in (3.6) of thepreceding claim we have k w k k − L ( m ) ≤ (1 − k u k k L ∞ k u k − w k k L ( m ) ) − ≤ (1 + 4 δC k ε k ) ≤ δC k ε k , where the last inequality holds if ε k ≤ (4 δC k ) − . On the other hand, for every i = 1 , . . . , k − Z D A ∇ v i · ∇ v i dx ≤ k − X j =1 Z D A ∇ v j · ∇ v j dx ≤ (1 + C k − ε k − ) k − X j =1 Z D A ∇ ˜ u j · ∇ ˜ u j dx. Therefore, using the estimate (3.8) we get (cid:18)Z D A ∇ w k · ∇ w k dx (cid:19) / ≤ (cid:18)Z D A ∇ ˜ u k · ∇ ˜ u k dx (cid:19) / + k − X i =1 (cid:12)(cid:12)(cid:12)(cid:12) Z D ˜ u k v i dm (cid:12)(cid:12)(cid:12)(cid:12) (cid:18)Z D A ∇ v i · ∇ v i dx (cid:19) / ≤ (cid:18)Z D A ∇ ˜ u k · ∇ ˜ u k dx (cid:19) / + ˜ C k ε k (1 + C k − ) / k − X j =1 Z D A ∇ ˜ u j · ∇ ˜ u j dx / . We then ask that ε k ≤ (2 ˜ C k ) − (1 + C k − ) − / so that we get Z D A ∇ v k · ∇ v k dx = k w k k − L ( m ) Z D A ∇ w k · ∇ w k dx ≤ (1 + 12 δC k ε k ) (1 + ˜ C k (1 + C k − ) / ε k ) Z D A ∇ ˜ u k · ∇ ˜ u k dx + k − X j =1 Z D A ∇ ˜ u j · ∇ ˜ u j dx . This, using once again the inductive hypothesis, proves (3.5) and concludes the proof. (cid:3)
Proof of Proposition . Let ˜ U be a vector satisfying the hypothesis of Proposition 3.4 and let V ∈ H ( D, R k ) be the vector given by Lemma 3.5 and obtained by orthonormalizing ˜ U . Using V as a test function in (3.3) and then Lemma 3.5, we have Z D A ∇ U · ∇ U dx + Λ |{| U | > }| ≤ Z D A ∇ V · ∇ V dx + Λ |{| V | > }|≤ (1 + C k k U − ˜ U k L ) Z D A ∇ ˜ U · ∇ ˜ U dx + Λ |{| ˜ U | > }| , where we have used in the last inequality that {| V | > } ⊂ {| ˜ U | > } (which holds by constructionof V ). (cid:3) We now conclude the proof of Theorem 1.1.
Proof of Theorem . The proof follows from Proposition 3.4 and Theorem 1.2 (see also Lemma3.2). (cid:3)
Acknowledgments.
This work was partially supported by the French Agence Nationale de laRecherche (ANR) with the projects GeoSpec (LabEx PERSYVAL-Lab, ANR-11-LABX-0025-01)and the project SHAPO (ANR-18-CE40-0013).
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