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1 SIAM J. CONTROL OPTIM. Vol. 44, No. 5, pp c 2005 Society for Industrial and Applied Mathematics REGULARITY OF THE FREE BOUNDARY IN AN OPTIMIZATION PROBLEM RELATED TO THE BEST SOBOLEV TRACE CONSTANT JULIÁN FERNÁNDEZ BONDER, JULIO D. ROSSI, AND NOEMI WOLANSKI Abstract. In this paper we study the regularity properties of a free boundary problem arising in the optimization of the best Sobolev trace constant in the immersion H 1 () L q ( ) for functions that vanish in a subset of. This problem is also related to a minimization problem for Steklov eigenvalues. Key words. Sobolev trace constant, free boundaries, eigenvalue optimization problems AMS subject classifications. 35J20, 35P30, 49K20 DOI / Introduction. The study of Sobolev inequalities and of optimal constants is a subject of interest in the analysis of PDEs and related topics. It has been widely studied in the past by many authors and is still an area of intensive research. See, for instance, the book [1] and, for recent developments in this field, see the articles [6, 11, 9, 17] and the survey [7] among others. The optimal Sobolev constant and its corresponding extremals (if they exist) are related to eigenvalue problems. In the case of the best Sobolev trace embedding H 1 () L q ( ), where is a bounded smooth domain in R N, the best constant and the extremal (that exists for 1 q<2 =2(N 1)/(N 2) since the immersion is compact) give rise to the following elliptic problem with nonlinear boundary conditions: { Δu + u = 0 in, u ν = λuq 1 on. The constant λ depends on the normalization of the extremal u. For instance, if u is chosen so that u L q ( ) = 1, then λ = S, the best Sobolev trace constant. In the linear case, q = 2, this problem becomes an eigenvalue problem that is known as the Steklov eigenvalue problem [19]. In this paper we are interested in the best Sobolev trace constant among functions that vanish in a subset of. We try to optimize this best constant when varying the subset in the class of measurable sets with prescribed positive measure α. In a previous article [10], we proved that there exists an optimal set. In this paper we focus our attention on regularity properties of these optimal sets. Received by the editors August 18, 2004; accepted for publication (in revised form) February 24, 2005; published electronically November 22, Supported by ANPCyT PICT , , and , CONICET PIP0660/98 and PEI6388/04, UBA X052 and X066, and Fundación Antorchas J. D. Rossi and N. Wolanski are members of CONICET. Departamento de Matemática, FCEyN, UBA (1428) Buenos Aires, Argentina (jfbonder@dm. uba.ar, jfbonder; wolanski@dm.uba.ar, wolanski). Consejo Superior de Investigaciones Científicas (CSIC), Serrano 117, Madrid, Spain. On leave from Departamento de Matemática, FCEyN, UBA (1428) Buenos Aires, Argentina (jrossi@dm. uba.ar, jrossi). 1614

2 REGULARITY IN AN OPTIMIZATION PROBLEM 1615 More precisely, in [10] we studied the following problem. Let J (v) = v 2 + v 2 dx, A α = {v H 1 () / v L q ( ) = 1 and {v >0} = α}. Then the problem is as follows: (P α ) Find φ 0 A α such that S(α) := inf v A α J (v) =J (φ 0 ). In [10] we proved that there exists a solution φ 0 to (P α ), but the approach in [10] does not give any regularity properties of φ 0 or of the hole {φ 0 =0}. In this paper we consider a different approach. Instead of minimizing J (v) over A α we penalize the functional and minimize without the measure restriction. This approach has been used with great success by many authors starting with the work [2] (see also, for instance, [3, 15, 16, 20]). Thus, let (1.1) J ε (v) = v 2 + v 2 dx + F ε ( {v >0} ), where 1 (s α) if s α, F ε (s) = ε ε(s α) if s<α. The penalized problem is to minimize J ε over the class K 1 = {v H 1 () / v L q ( ) =1}. For technical reasons, it is better to minimize in the class K = {v H 1 () / v L q (Γ N ) =1,v= ϕ 0 on Γ D }, where Γ N, Γ D = \ Γ N is the closure of a relatively open set of the boundary and ϕ 0 H 1 (), ϕ 0 c 0 > 0onΓ D. We will only need to assume that Γ D at the end of our arguments. See section 4, Lemma 4.3. So the penalized problem is as follows: (P ε ) Find u ε Ksuch that J ε (u ε ) = inf v K J ε(v). Observe that minimizing J ε over K gives a problem with mixed boundary conditions. We believe that this problem has independent interest. The main idea is to prove that for ε small any minimizer u ε of J ε in K satisfies {u ε > 0} = α; therefore, the penalization term F ε vanishes, and hence we have a minimizer of our original problem. This allows us to avoid the passage to the limit (as ε 0) where uniform bounds are needed. Proving regularity of the minimizers of J ε and their free boundaries, {u ε > 0}, is easier than the original problem, thanks to the results of [4]. The main theorem in this article is the following. Theorem 1.1. For every ε>0 there exists a solution u ε Kto (P ε ). Moreover, any such solution is a locally Lipschitz continuous function and the free boundary

3 1616 J. FERNÁNDEZ BONDER, J. D. ROSSI, AND N. WOLANSKI {u ε > 0} is locally a C 1,β surface up to a set of zero H N 1 measure. In the case N =2the free boundary is locally a C 1,β surface. Moreover, if Γ D, for ε small we have {u ε > 0} = α. Outline of the paper. In section 2 we begin our analysis of problem (P ε ) for fixed ε. First we prove the existence of a minimizer, local Lipschitz regularity, and nondegeneracy near the free boundary (Theorem 2.1). Then we prove that a minimizer u ε of (P ε ) is a weak solution to the free boundary problem Δu + u =0 in{u >0}, u ν = λ ε on {u >0}, where λ ε is a positive constant (Theorem 2.6). In section 3, again for fixed ε, we analyze the regularity of the free boundary and show that, up to a set of zero H N 1 measure, {u ε > 0} is locally a C 1,β surface and, in the case N = 2, the free boundary has no exceptional points (Theorem 3.1). The proof of this result follows almost exactly the lines in [4], so we note only the significant differences and refer the reader to [4] for further details. In section 4 we analyze the behavior of the solutions to (P ε ) for small ε. We prove that if Γ D, the positivity set of the minimizer u ε has measure α (Theorem 4.1). Finally, in section 5, we go back to our original problem and show, under some mild assumptions on the solutions φ 0 to (P α ), that they are also solutions to (P ε ) for small ε, so they inherit the properties of the solutions to (P ε ) (Theorem 5.1). These extra assumptions are satisfied, for instance, if is a ball (Corollary 5.1). In the general case, without the assumption that Γ D, we prove that the set of α s for which there is a solution to (P α ) with smooth free boundary is dense in (0, ) (Theorem 5.2). Then, we show that the minimizers of (P ε ) converge (up to a subsequence) to a solution to (P α ) (Theorem 5.3). We believe that this last result might be of interest in numerical approximations. 2. The penalized problem. In this section, we consider the penalized problem (P ε ) stated in the introduction and prove the existence of a minimizer and some regularity properties. Theorem 2.1. There exists a solution to the problem (P ε ). Moreover, any such solution u ε has the following properties: (1) u ε is locally Lipschitz continuous in. (2) For every D, there exist constants C, c > 0 such that for every x D {u ε > 0}, c dist(x, {u ε > 0}) u ε (x) C dist(x, {u ε > 0}). (3) For every D, there exists a constant c>0 such that for x {u >0} and B r (x) D, c B r(x) {u ε > 0} B r (x) 1 c. The constants may depend on ε. The proof will be divided into a series of steps for the reader s convenience. Proof of existence. Let (u n ) Kbe a minimizing sequence for J ε. Then J ε (u n ) is bounded and so u n H 1 () C. Therefore there exists a subsequence (that we still

4 REGULARITY IN AN OPTIMIZATION PROBLEM 1617 call u n ) and a function u ε H 1 () such that Thus, Hence u ε Kand u n u ε weakly in H 1 (), u n u ε strongly in L q ( ), u n u ε a.e.. u ε L q (Γ N ) =1, u ε = ϕ 0 on Γ D, {u ε > 0} lim inf {u n > 0}, n u ε H 1 () lim inf u n H n 1 (). and J ε (u ε ) lim inf n J ε(u n ) = inf v K J ε(v); therefore u ε is a minimizer of J ε in K. Remark 1. Any minimizer u ε of J ε satisfies the inequality (2.1) Δu u 0 in. In fact, this can be seen by performing one-side perturbations. Namely, we let v = u ε tϕ with t>0 and ϕ C0 (), ϕ 0, to get u ε ϕ + u ε ϕ 0. In the remainder of the section we will remove the subscript ε from the solution of (P ε ). For the proof of properties (1) (3), we apply the ideas developed in [4]. To this end, we need a series of lemmas. Lemma 2.1. Let u K be a solution to (P ε ). There exists a constant C = C(N,,ε) such that for every ball B r 1 r u C implies u>0 in B r. B r Proof. The idea is similar to that of Lemma 3.2 in [4]. Let v be the solution to { v = u in \ B r, (2.2) Δv = v in B r. Then v K, v>0inb r. We claim that (2.3) In fact, u v 2 H 1 () = u 2 H 1 () v 2 H 1 (). B r v (v u)+v(v u) dx =0

5 1618 J. FERNÁNDEZ BONDER, J. D. ROSSI, AND N. WOLANSKI since v u H0 1 (B r ). This implies (2.4) u v + uv dx = B r v 2 + v 2 dx. B r This equality implies the claim since u = v in \ B r. By (2.1), u v in B r. Now, by (2.3), since u is a minimizer and u = v in \ B r, we have (2.5) (u v) 2 +(u v) 2 dx F ε ( {u >0} )+F ε ( {v >0} ) C ε {u =0} B r. Now, as in [4], the idea is to control {u =0} B r from above by the left-hand side of (2.5). By replacing u(x) byu(x 0 + rx)/r we can assume that B r = B 1 (0). For z 1 2 we consider the change of variables from B 1 into itself such that z becomes the new origin. We call u z (x) =u ( (1 x )z + x ), v z (x) =v ( (1 x )z + x ) and define { r ξ = inf r such that 1 } 8 r 1 and u z(rξ) =0 if this set is nonempty. Observe that this change of variables leaves the boundary fixed. Now, for almost every ξ B 1 we have 1 d v z (r ξ ξ)= dr (u z v z )(rξ) dr ( ) 1/2 1 (2.6) 1 r ξ (u z v z )(rξ) 2 dr. r ξ (2.7) Let us see that r ξ v z (r ξ ξ) C(N,)(1 r ξ ) u. B 1 In fact, v z (r ξ ξ)=v ( (1 r ξ )z + r ξ ξ ), and if (1 r ξ )z + r ξ ξ 3 4, by the Harnack inequality applied to a solution to Δv r 2 v =0inB 1 with r 1, v z (r ξ ξ) C N v(0). Clearly (2.7) follows from (2.8) v(0) α(n) v = α(n) B 1 B 1 u. But (2.8) is a consequence of the mean value property of solutions to the Schrödinger equation Δv r 2 v = 0, namely, v(0) = 1 J(r) v, where J(r) =Γ(N/2)( r 2 )1 N 2 I N 2 (r) and I N B 1(0) is the Bessel function. In particular, J(0)=1.

6 REGULARITY IN AN OPTIMIZATION PROBLEM 1619 See Theorem 9.9 in [18] for this result. Now, if (1 r ξ )z + r ξ ξ 3 4, we prove by a comparison argument that inequality (2.7) also holds. In fact, first observe that we can assume that B1 v = B1 u =1. Then, by (2.8), v C N α in B 3/4. Let w(x) =e λ x 2 e λ. There exists λ = λ(n,α) such that Δw w in B 1 \ B 3/4, w C N α in B 3/4, w =0 in B 1, so that, since Δv v, there holds that v w C(1 x ) inb 1 \ B 3/4. Therefore, ( ) v z (r ξ ξ) C 1 (1 r ξ )z + r ξ ξ u C(1 r ξ ) B 1 B 1 u since z 1 2. Thus (2.7) holds for every r ξ 1 8. By (2.6) and (2.7) we have c ( ) 1/2 1 1 r ξ u (u z v z ) B 2 (rξ) dr. 1 r ξ Hence ( 2 1 c B 2 (1 r ξ ) ds ξ u) (u z v z ) 1 B 2 (rξ) dr ds ξ 1 B 1 r ξ C (u z v z ) 2 dx. B 1 Since (1 r ξ ) ds ξ χ {u=0} dx, B 1 B 1\B 1/4 (z) we have ( 2 c 2 {x B 1 \ B 1/4 (z) /u(x) =0} u) C (u z v z ) B 2 dx 1 B 1 K (u v) 2 dx. B 1 Finally, we integrate over z B 1/2 (0) and use (2.5) to obtain ( 2 (2.9) B 1 {u =0} u) K (u v) B 2 dx 1 B 1 KC ε B 1 {u =0}. Therefore we either have u>0 a.e. in B 1 or else u KC ε. B 1 Hence we deduce that if u KC ε = C(N,,ε), B 1

7 1620 J. FERNÁNDEZ BONDER, J. D. ROSSI, AND N. WOLANSKI then B 1 {u =0} = 0. Thus by (2.5) u = v>0inb 1. Now we can prove the Lipschitz continuity of the minimizer u. Proof of Theorem 2.1(1). The proof follows as in [4, Lemma 3.3]. In fact, let D D and x D. Let r>0bethe largest number such that B r (x) {u> 0} D. As in [4] we prove by using Lemma 2.1 that {u >0} is open and 1 r u C B r(x) with C independent of either u or x. Since u>0inb r (x), it is a solution to Δu = u in B r (x). In fact, let v be the solution to Δv = v in B r (x), v = u on \ B r (x). Then, 0 u v 2 H 1 () = u 2 H 1 () v 2 H 1 () = J ε(u) J ε (v) 0. Thus, u = v in B r (x). Hence, there is a universal constant such that u(x) C {r u L (Br(x)) + 1r } u. B r(x) Now, since u is subharmonic in and D, there holds that u is bounded in D by a constant that depends on the H 1 norm of u in, which is bounded by a constant that depends only on and ε. Therefore, u(x) C with C depending only on N,, ε,d, and D. In order to prove the nondegeneracy of u we need the following lemma (see [4, Lemma 3.4]). Lemma 2.2. Let u Kbe a solution to (P ε ). For 0 <κ<1there exists a constant c = c(κ, N,,ε) such that for every ball B r (x 0 ), 1 r u c implies that u =0 in B κr. B r Proof. As in [4, Lemma 3.4], we consider the function ( ( ) N 2 s s 1) for N 3, N 2 x (2.10) φ N s (x) = s log s for N =2, x s x for N =1. For simplicity let us take ū(x) = 1 r u(x 0 + rx), 1 ( s α ) F ε (s) = ε( r N ε s α ) r N if s> α r N, if s α r N,

8 REGULARITY IN AN OPTIMIZATION PROBLEM 1621 and J ε (w) = w 2 + r 2 w 2 + F ε ( {w >0} ), r where r = 1 r ( x 0). Thus, J ε (u) =r N Jε (ū). Now, let v(x) = γ κ φ N κ ( κ) max( φn κ (x), 0), where, since ū is subharmonic, γ := 1 sup ū C 1 (N,κ) ū = C 1 (N,κ) 1 κ B 1 r u. B κ Hence, v ū on B κ, and therefore if there holds that w = { min(ū, v) in B κ, ū in r \ B κ, B r(x 0) ū 2 + r 2 ū 2 dx + B κ {ū>0} B κ = J ε (ū) ū 2 + r 2 ū 2 dx + B κ {ū>0} F ε ( {ū >0} ) r \B κ J ε (w) ū 2 + r 2 ū 2 dx + B κ {ū>0} F ε ( {ū >0} ) r \B κ = w 2 + r 2 w 2 dx B κ \Bκ ū 2 + r 2 ū 2 dx + B κ {ū>0} B κ \Bκ + F ε ( {w >0} ) F ε ( {ū >0} ) w 2 + r 2 w 2 dx B κ \Bκ ū 2 + r 2 ū 2 dx +(1 ε) B κ {ū>0}, B κ \Bκ since w =0inB κ, w =ū in r \ B κ. We have also used that F ε (A) F ε (B) ε(a B) ifa B and {w >0} {ū>0}. This inclusion follows from the fact that w ū. Thus, ū 2 + r 2 ū 2 dx + ε B κ {ū>0} B κ w 2 + r 2 w 2 dx ū 2 + r 2 ū 2 B κ \Bκ B κ \Bκ = ū (ū v) + 2 ū 2 dx + r 2 (ū (ū v) + ) 2 ū 2 dx B κ \Bκ B κ \Bκ = (ū v) + (ū + v) dx r 2 (ū v) + (ū + v) dx B κ \Bκ B κ \Bκ = (ū v) + ūdx r 2 (ū v) + ūdx B κ \Bκ B κ \Bκ (ū v) + vdx r 2 (ū v) + vdx B κ \Bκ B κ \Bκ

9 1622 J. FERNÁNDEZ BONDER, J. D. ROSSI, AND N. WOLANSKI 2 (ū v) + vdx 2r 2 B κ \Bκ (ū v) + vdx B κ \Bκ 2 ū vηds C 2 (N,κ) γ B κ ū. B κ Therefore, (2.11) ū 2 + r 2 ū 2 dx + ε B κ {ū>0} C 2 (N,κ) γ B κ B κ ū. Here we have used that min(ū, v) =ū (ū v) +,Δv =0inB κ \ B κ, v =0on B κ, and (ū v) + =0on B κ. Recall that γ is controlled by 1 r u;thusγ will be small if 1 Br(x 0) r u is Br(x 0) small. On the other hand, by standard estimates, ū C 3 (N,κ) ū +ūdx B κ B κ { 1 C 3 (N,κ) ū 2 dx + 1 } 2 B κ 2 B κ {ū>0} + γ B κ {ū>0} { } C 3 (N,κ) ū 2 + r 2 ū 2 dx + B κ {ū>0} B κ if γ 1/2. Thus, by (2.11), if γ is small enough (γ 1/2 and C 2 (N,κ)C 3 (N,κ)γ < 1), we deduce that B κ {ū>0} = 0. That is, u =0inB rκ (x 0 ) and the lemma is proved. We can now prove the nondegeneracy of u. Proof of Theorem 2.1(2). Let x {u>0} and r = dist (x, {u =0}). As we proved in (2.8), since Δu = u in B r (x), there holds that u(x) α(n) u. B r(x) Since u(x) > 0, 1 r B r(x) u c, where c is the constant in Lemma 2.2 for κ =1/2. Thus, u(x) cα r. The upper bound clearly follows from the Lipschitz continuity of u. Hence (2) is proved. Proof of Theorem 2.1(3). In order to prove the uniform positive density of {u >0} and {u =0} at every free boundary point we proceed as in [4, Lemma 3.7]. The only difference is that the function v that we have to take is the one in (2.2). This ends the proof of Theorem 2.1.

10 REGULARITY IN AN OPTIMIZATION PROBLEM 1623 Corollary 2.1. Let u Kbe a solution to (P ε ). Let D. There exist constants c, C > 0 depending only on N,,D, and ε such that for B r (x) D and x {u >0}, (2.12) c 1 r u C. B r(x) Proof. The proof follows easily from Lemmas 2.1 and 2.2. Lemma 2.3. Let u Kbe a solution to (P ε ). Then u satisfies, for every ϕ C0 () such that supp ϕ {u>0}, (2.13) u ϕ + uϕ dx =0. Moreover, the application λ(ϕ) := u ϕ + uϕ dx from C0 () into R defines a nonnegative Radon measure with support on {u > 0}. Proof. The proof follows exactly as in [4, Lemma 4.2]. Theorem 2.2. Let u Kbe a solution to (P ε ). Let D. Then there exist constants C, c > 0 such that for B r (x) D and x {u >0}, cr N 1 dλ Cr N 1. B r(x) Proof. Forn large enough, let u n = u ρ n, where ρ n are the standard mollifiers. Then λ ρ n dx = Δu n u n dx = u n νds u n B r(x) B r(x) B r(x) B r(x) ω N 1 sup B r(x) u n r N 1 Cr N 1 since u n u C for a certain constant C depending on D. By taking the limit for n we get dλ Cr N 1. B r(x) The other inequality follows as in the proof of Theorem 4.3 in [4] by taking as G y (z) the (positive) Green function of Δ+Id with homogeneous Dirichlet boundary conditions in the ball B r (x). Then for 0 <κ<1/2 and y B κr (x) one uses the inequality v(y) Cv(x) Cα u B r(x) for v the solution to Δv v =0inB r (x), v = u on B r (x), which follows from the Harnack inequality and (2.8). Theorem 2.3 (representation theorem). Let u Kbe a solution to (P ε ). Then

11 1624 J. FERNÁNDEZ BONDER, J. D. ROSSI, AND N. WOLANSKI (1) H N 1 (D {u >0}) < for every D. (2) There exists a Borel function q u such that Δu u = q u H N 1 {u >0}. (3) For D there are constants 0 <c C< depending on N,,D, and the constants in (2.12) such that for B r (x) D and x {u >0}, c q u (x) C, c r N 1 H N 1 (B r (x) {u >0}) Cr N 1. Proof. The proof follows exactly as in Theorem 4.5 in [4]. Remark 2. Let u Kbe a solution to (P ε ) and let D. Then D {u >0} has finite perimeter. Thus, the reduced boundary red {u>0} is defined as well as the measure theoretic normal ν(x) for x red {u>0}. See [8]. If the free boundary {u >0} is a regular surface, then q u = ν u. In Theorem 2.4 it is shown that this is true for almost all points in the reduced boundary. Proposition 2.1. Let u Kbe a solution to (P ε ) and let B ρk (x k ) be a sequence of balls with ρ k 0, x k x 0, and u(x k )=0.Let u k (x) := 1 ρ k u(x k + ρ k x). We call u k a blow-up sequence with respect to B ρk (x k ). Since u is locally Lipschitz continuous, there exists a blow-up limit u 0 : R N R satisfying (2.12) with the same constants, when x k {u >0} and such that for a subsequence, u k u 0 in Cloc(R α N ) for every 0 <α<1, u k u 0 weakly in L loc(r N ), {u k > 0} {u 0 > 0} locally in Hausdorff distance, χ {uk >0} χ {u0>0} in L 1 loc(r N ), Δu 0 =0 in {u 0 > 0}. Moreover, if x k {u >0}, then 0 {u 0 > 0}. Proof. The proof follows as in [4, section 4.7], observing that Δu k ρ 2 k u k =0in {u k > 0}. Theorem 2.4 (identification of q u ). Let u Kbe a solution to (P ε ). Then, for almost every x 0 red {u>0}, u(x 0 + x) =q u (x 0 ) x, ν(x 0 ) + o( x ) for x 0 with ν(x 0 ) the outward unit normal de {u >0} in the measure theoretic sense. Proof. The proof follows exactly as in Theorem 4.8 and Remark 4.9 in [4]. Remark 3. Observe that by Theorem 2.1(3) H N 1 ( {u >0}\ red {u>0}) =0. See [8]. Now we get a more precise identification of q u. Theorem 2.5. Let u Kbe a solution to (P ε ) and let q u be the function in Theorem 2.4. Then there exists a constant λ u such that (2.14) (2.15) lim sup u(x) = λ u for every x 0 {u >0}, x x 0 u(x)>0 q u (x 0 )=λ u, H N 1 a.e. x 0 {u >0}.

12 REGULARITY IN AN OPTIMIZATION PROBLEM 1625 Moreover, if B is a ball contained in {u =0} touching the boundary {u >0} at x 0, then (2.16) lim sup x x 0 u(x)>0 u(x) dist(x, B) = λ u. Proof. We follow the ideas of [2, Theorem 3] and [16, Theorem 5.1 and Lemma 5.2]. Let x 0,x 1 {u >0} and ρ k 0 +. For i =0, 1, let x i,k x i with u(x i,k )=0 such that B ρk (x i,k ) and such that the blow-up sequence u i,k (x) = 1 ρ k u(x i,k + ρ k x) has a limit u i (x) =λ i x, ν i, with 0 <λ i < and ν i a unit vector. We will prove that λ 0 = λ 1. From this, the theorem will follow as in [16]. Assume that λ 1 <λ 0. Then we will perturb the minimizer u near x 0 and x 1 and get an admissible function with less energy, which is a contradiction. We perform a perturbation that increases the measure of the positivity set in a neighborhood of x 0,k and decreases its measure in a neighborhood of x 1,k. We perform this perturbation in such a way that we change the measure of the positivity set in an amount of essentially order o(ρ N k ). To this end, we take a nonnegative C0 symmetric function Φ supported in the unit interval and for t>0 small, we define ( x x0,k x + tρ k Φ ρ k )ν 0 for x B ρk (x 0,k ), ( τ k (x) = x x1,k x tρ k Φ ρ k )ν 1 for x B ρk (x 1,k ), x which is a diffeomorphism if t is small enough. Now, let v k (x) =u(τ 1 k (x)), elsewhere, which are admissible functions. Moreover, since Dτ 1 k C independent of k for t small enough, there holds that independent of k. Also, we have (2.17) v k L C F ε ( {v k > 0} ) F ε ( {u >0} ) =o(t) ρ N k + o(ρ N k ). In fact, v k = u in \ (B ρk (x 0,k ) B ρk (x 1,k )) and {v k > 0} B ρk (x i,k ) {u>0} B ρk (x i,k ) = ( =( 1) i ρ N k t Φ( y ) dhy N 1 B 1 {y 1=0} ) + o i (t) + o(ρ N k ),

13 1626 J. FERNÁNDEZ BONDER, J. D. ROSSI, AND N. WOLANSKI since Φ( y ) is radially symmetric and χ {ui,k >0} χ { x,νi <0} in L 1 loc (RN ). Similar computations that also involve the development of v k in terms of u and Dτ k give (2.18) v k 2 dx = ρ N k ( (λ 2 1 λ 2 0) t u 2 dx B 1(0) {y 1=0} Φ( y ) dh N 1 y ) + o(t) + o(ρ N k ). See [2] or [16] for detailed computations. It remains to estimate the difference of the L 2 norms. Since u(x i,k ) = 0 there holds that On the other hand, Thus, u(x) Cρ N k in B ρk (x i,k ). 0=u(x i,k )=v k (τ k (x i,k )) = v k (x i,k +( 1) i tρ k Φ(0)ν i ). v k (z) C z xi,k ( 1) i tρ k Φ(0)ν i Kρk if z B ρk (x i,k ). Therefore, (2.19) vk 2 dx u 2 dx = o(ρ N k ). Thus, we get from (2.17), (2.18), and (2.19), for t small enough and k large enough, that J ε (v k ) < J ε (u), a contradiction. Summing up, we have the following theorem, Theorem 2.6. Let u Kbe a solution to (P ε ). Then u is a weak solution to the free boundary problem Δu + u =0 in {u >0}, u ν = λ u on {u >0}, where λ u is the constant in Theorem 2.5. More precisely, H N 1 a.e. point x 0 {u > 0} belongs to red {u>0} and u(x 0 + x) =λ u x, ν(x 0 ) + o( x ) for x 0. Finally, we get an estimate of the gradient of u that will be needed in order to get the regularity of the free boundary. Theorem 2.7. Let u Kbe a solution to (P ε ). Given D, there exist constants C = C(N, ε, D), r 0 = r 0 (N,D) > 0, and γ = γ(n, ε, D) > 0 such that if x 0 D {u >0} and r<r 0, then sup u λ u (1 + Cr γ ). B r(x 0)

14 REGULARITY IN AN OPTIMIZATION PROBLEM 1627 Proof. The proof follows the lines of the proof of Theorem 4.1 in [5]. Let U k = ( u λ u 1 k ) + and U0 = ( u λ u ) +. By (2.14) we know that Uk vanishes in a neighborhood of the free boundary. Also, the support of U k is contained in {u >0}. Therefore U k satisfies ΔU k U k in {u>0} and vanishes in a neighborhood of the free boundary. We extend U k by zero into {u =0} and set h k (r) = sup U k, h 0 (r) = sup U 0 B r(x 0) B r(x 0) for any r<r 0 = dist (D, ) and x 0 D {u >0}. Then, h k (r) U k is a supersolution of Δv = v in the ball B r (x 0 ) and h k (r) U k 0 in B r (x 0 ) = h k (r) in B r (x 0 ) {u =0}. Applying the weak Harnack inequality (see [12, p. 246]) with 1 p<n/(n 2), we get ( ) inf hk (r) U k cr N/p h k (r) U k L p (B r(x 0)) ch k (r), B r/2 (x 0) since, by Theorem 2.1(3), B r (x 0 ) {u =0} cr N. Taking now k we obtain ( ) inf h0 (r) U 0 ch0 (r) B r/2 (x 0) for some 0 <c<1, which is the same as sup U 0 (1 c)h 0 (r). B r/2 (x 0) Therefore ( r ) h 0 (1 c)h 0 (r), 2 from which it follows that h 0 (r) Cr γ for some C>0, 0 <γ<1, and now the conclusion of the theorem follows. 3. Regularity of the free boundary. At this point we have that our minimizer u ε meets the conditions of the regularity theory developed in [4], the only difference being the equation satisfied by u ε in {u ε > 0}. We recall some definitions and point out the only significant difference with [4]. The rest of the proof of the regularity then follows as sections 7 and 8 of [4] with only minor modifications. Throughout this section we remove the subscript ε. Definition 3.1 (flat free boundary points). Let 0 <σ +,σ 1 and τ>0. We say that u is of class F (σ +,σ ; τ) in B ρ = B ρ (0) if

15 1628 J. FERNÁNDEZ BONDER, J. D. ROSSI, AND N. WOLANSKI (1) 0 {u >0} and u =0 for x N σ + ρ, u(x) λ(x N + σ ρ) for x N σ ρ; (2) u λ(1 + τ) in B ρ. If the origin is replaced by x 0 and the direction e N by the unit vector ν, we say that u is of class F (σ +,σ ; τ) in B ρ (x 0 ) in direction ν. Observe that the results in section 2 imply that the minimizer u of J ε is in the class F (σ, 1; σ) inb ρ (x 0 ) in direction ν u (x 0 ) for every x 0 red {u>0} with σ = σ(ρ) 0asρ 0. The following lemma (Lemma 7.2 in [4]) is the only one that requires a nonobvious modification. Lemma 3.1. There is a constant C = C(N) such that u F (σ, 1; σ) in B ρ (x 0 ) in direction ν implies u F (2σ, Cσ; σ) in B ρ/2 (x 0 ) in direction ν. Proof. Clearly, by a change of variables, we may assume that x 0 = 0 and ν = e N. Let ū(x) =u(ρx)/λρ; then ū 1+σ and ū F (σ, 1; σ) inb 1. That is, ū =0if x N >σ. Define η(x exp ( 9 x 2 ) )= 1 9 x 2 for x < 1 3, 0 otherwise and choose s 0 maximal with the property that ū =0inx N >σ sη(x ). Now, the proof follows as in Lemma 7.2 of [4] with the only difference being that the comparison function v must be the solution to Δv = ρ 2 ū in D = B 1 {x N < σ sη(x )} instead of a harmonic function. The estimate follows from in D B 1/2 if ρ 2 Cσ, since ν v 1+Cσ [ (v + x N ) C sup(v + x N )+ρ 2] Cσ D Δ(v + x N )=ρ 2 ū in D, v + x N Cσ in D, and ū(x) 2. Once this lemma is established the following regularity result follows. Theorem 3.1. Let u Kbe a solution to (P ε ). Then red {u>0} is a C 1,β surface locally in, and the remainder of the free boundary has zero H N 1 measure. Moreover, if N =2, then the whole free boundary is a C 1,β surface. 4. Behavior of the minimizer for small ε. To complete the analysis of the problem, we now show that if ε is small enough, then {u ε > 0} = α. To this end, we need to prove that the constant λ ε := λ uε is bounded from above and below by positive constants independent of ε. We perform this task in a series of lemmas.

16 REGULARITY IN AN OPTIMIZATION PROBLEM 1629 Lemma 4.1. Let u ε Kbe a solution to (P ε ). Then there exist constants C, c > 0 independent of ε such that (4.1) Hence c {u ε > 0} α + Cε. Proof. AsJ ε (u ε ) is bounded from above uniformly in ε we obtain F ε ( {u ε > 0} ) C. {u ε > 0} α + Cε. For the lower bound, we proceed as follows: by the Sobolev trace embedding, for some 1 <p<2 such that p(n 1)/(N p) >q, 1 u ε Lq ( ) C u ε W 1,p () C u ε H1 () {u ε > 0} θ for some exponent θ that depends only on p. Since u ε H 1 () is uniformly bounded, the lower bound follows. Lemma 4.2. Let u ε Kbe a solution to (P ε ). Then there exists a constant C>0 independent of ε such that λ ε := λ uε C. Proof. Let D smooth, such that ω = D >αand \ D <c, where c is the constant in Lemma 4.1. Then, for ε small enough. On the other hand, D {u ε > 0} α + Cε<ω D {u ε > 0} {u ε > 0} \ D c \ D > 0. Therefore by the relative isoperimetric inequality, we have H N 1 (D {u ε > 0}) c 0 min { D {u ε > 0}, D {u ε =0} } N 1 N c 1 > 0. Now take ϕ C0 () as a test function in Lemma 2.3 such that 0 ϕ 1, ϕ 1inD, and ϕ C = C(dist(D, )) to get, since u ε H 1 () is bounded independently of ε, C u ε ϕdx+ u ε ϕdx= λ ε (ϕ) λ ε H N 1 (D red {u ε > 0}). This completes the proof of the lemma. The proof of the uniform lower bound follows similarly to Lemma 6 in [2]. We only make a sketch of the proof for the reader s convenience. It is at this point where we need the hypothesis that Γ D. Lemma 4.3. Let Γ D be the closure of a relatively open subset of. Let ϕ 0 H 1 () with ϕ 0 c 0 > 0 in Γ D.Letu ε Kbe a solution to (P ε ). Then (1) u ε is positive in a neighborhood of Γ D (depending on ε);

17 1630 J. FERNÁNDEZ BONDER, J. D. ROSSI, AND N. WOLANSKI (2) there exists a constant c>0 independent of ε such that c<λ ε := λ uε. Proof. Let us first prove (1). In fact, arguing as in (2.9), given y 0 Γ D there exists a constant K>0independent of ε such that ( 2 1 r {u =0} u) r K (u v) 2 dx, r r where r = B r (y 0 ) and v is the solution of { Δv = v in r, v = u on r. Therefore, ( c0 ) 2 r {u =0} K( u 2 H r 1 (r) v 2 H 1 (r) ) C ε r {u =0}. Thus, u>0in r for small r depending on ε. In order to see (2) we proceed as in [2, Lemma 6]. Let y 0 Γ D and let D t with 0 t 1 be a family of open sets with smooth boundary and uniformly (in ε and t) bounded curvatures such that D 0 is an exterior tangent ball at y 0, D 1 contains a free boundary point, D t Γ D, and D 0 D t for t>0. Let t (0, 1) be the first time that D t touches the free boundary and let x 0 D t {u ε > 0}. Now, take w as the solution to Δw = w in D t \ D 0 with w = c 0 on D 0 and w =0on D t. Thus w u ε in D t and ν w(x 0 ) cc 0 with c independent of ε; therefore, for r small enough, 1 r u ε 1 B r(x 0) r w cc 0 B r(x 0) with c independent of ε. If v 0 is the solution to then, by (2.9), we have { Δv = v in B r (x 0 ), v = u on B r (x 0 ), ( ) 2 1 c B r (x 0 ) {u ε =0} B r (x 0 ) {u ε =0} r u ε B r(x 0) K (u ε v 0 ) 2 dx B r(x 0) K( u ε 2 H 1 (B v r(x 0)) 0 2 H 1 (B ). r(x 0)) Now let δ r = B r (x 0 ) {u ε =0} and let x 1 {u ε > 0} be such that the free boundary is smooth in a neighborhood of x 1. We perturb {u ε > 0} in a neighborhood of x 1 so that the measure of the perturbed set is increased by an amount δ r (cf. Theorem 2.5).

18 REGULARITY IN AN OPTIMIZATION PROBLEM 1631 Let Φ be a smooth nonnegative function supported in B κ (x 1 ) with κ>0 small. For x B κ (x 1 ) we write x = σ + sν(σ) with σ {u ε > 0} and s R, where ν(σ) is the outer unit normal to the free boundary at σ. We define the change of variables y = x Φ(σ)τν(σ) with τ>0small and the deformed set D δr such that D δr B κ (x 1 )={y/x {u ε > 0} B κ (x 1 )}. Observe that if r is small we can perform this perturbation in such a way that it decreases the measure of {u ε > 0} by exactly δ r. Also, observe that δ r 0asr 0. Now let v r be the solution of Δv = v in D δr, (4.2) v =0 on D δr B κ (x 1 ), v = u ε on B κ (x 1 ) D r. Then v r verifies On the other hand, Thus v r ν = λ ε + o(δ r ). u ε = λ ε δ r + o ε (δ r ) on {v r > 0} B κ (x 1 ). B κ(x 1) v r 2 + vr 2 dx u ε 2 +(u ε ) 2 dx B κ(x 1) = (u ε v r ) 2 +(u ε v r ) 2 dx B κ(x 1) v r = {v r>0} B κ(x 1) ν u ε ds = λ 2 εδ r + o ε (δ r ). Now we extend v r by zero to B κ (x 1 ) \D δr and define v r in B κ (x 1 ), w r = v 0 in B r (x 0 ), u elsewhere. Then {w r > 0} = {u ε > 0} and w r = u ε on ; thus 0 J ε (w r ) J ε (u ε )= w r 2 + wr 2 dx u ε 2 +(u ε ) 2 dx = v v0 2 dx u ε 2 +(u ε ) 2 dx B r(x 0) B κ(x 0) + v r 2 + vr 2 dx u ε 2 +(u ε ) 2 dx B κ(x 1) cδ r + λ 2 εδ r + o ε (δ r ) B κ(x 1) for every r>0 small. Therefore, λ 2 ε c/2.

19 1632 J. FERNÁNDEZ BONDER, J. D. ROSSI, AND N. WOLANSKI Now we are in a position to prove the main result of this section, namely, that for ε small the measure of the positivity set is exactly α. Theorem 4.1. Let Γ D be the closure of a relatively open subset of. Let ϕ 0 H 1 () with ϕ 0 c 0 > 0 in Γ D. Let u ε Kbe a solution to (P ε ). Then, for ε small (4.3) {u ε > 0} = α. Proof. Arguing by contradiction, assume first that {u ε > 0} >α. Let x 1 {u ε > 0} be a regular point. We will proceed as in the proof of the previous lemma. Given δ > 0, we perturb the domain {u ε > 0} in a neighborhood of x 1, B κ (x 1 ), decreasing its measure by δ. We choose δ small so that the measure of the perturbed set is still larger than α. Then we let v be the solution to (4.2) extended by zero to the rest of B κ (x 1 ) and equal to u in the rest of. We have 0 J ε (v) J ε (u ε )= v 2 + v 2 u ε 2 +(u ε ) 2 + F ε ( {v >0} ) F ε ( {u ε > 0} ) λ 2 εδ + o ε (δ) 1 ( ε δ C 2 1 ) δ + o ε (δ) < 0 ε if ε<ε 0, and then δ<δ 0 (ε), a contradiction. Now assume that {u ε > 0} <α. We proceed as in the previous case but this time we perturb in a neighborhood of x 1 the set {u ε > 0}, increasing the measure by δ. Then we construct the function v as before, and if δ is small enough, {v >0} <α. Then 0 J ε (v) J ε (u ε )= v 2 + v 2 u ε 2 +(u ε ) 2 + F ε ( {v >0} ) F ε ( {u ε > 0} ) λ 2 εδ + o ε (δ)+εδ ( c 2 + ε)δ + o ε (δ) < 0 if ε<ε 1, and then δ<δ 0 (ε). Again, a contradiction that ends the proof. As a consequence of the previous theorem, we get the following corollary. Corollary 4.1. Let Γ D be the closure of a relatively open subset of. Let ϕ 0 H 1 () with ϕ 0 c 0 > 0 in Γ D. Then there exists a minimizer u of J (v) in the set K α = {v H 1 () / v Lq (Γ N ) =1,v= ϕ 0 on Γ D, {v >0} = α}. This minimizer can be chosen in such a way that it is locally Lipschitz continuous in and the free boundary {u >0} is locally a C 1,β surface up to a set of zero H N 1 measure. In the case N =2the free boundary is locally a C 1,β surface. Proof. From our previous results we have (4.3) for every ε small enough. Therefore we can take u = u ε and the desired regularity of u and its free boundary follows from the results of sections 2 and Main results. In this section we go back to our original minimization problem related to the best Sobolev trace constant. Here we prove that any extremal is a locally Lipschitz continuous function and the boundary of the hole {u >0} is locally C 1,β up to a set of zero H N 1 measure.

20 REGULARITY IN AN OPTIMIZATION PROBLEM 1633 We begin with the following theorem. Theorem 5.1. Let φ 0 be a minimizer for (P α ). Assume that there exists a positive constant c such that φ 0 >cin a ball B 0 (resp., on B 0, where B 0 is a ball centered at ). Then φ 0 is a minimizer of J ε in K 2 = {v H 1 () / v L q ( ) =1,v= φ 0 in B 0 } (resp., φ 0 /k is a minimizer of J ε in K = {v H 1 () / v Lq (Γ N ) =1,v= φ 0 /k on Γ D } with Γ D = B 0, Γ N = \ Γ D ). Here, B 0 is a ball compactly contained in B 0 and k = φ 0 L q (Γ N ). In particular, φ 0 is locally Lipschitz continuous in and the free boundary {φ 0 > 0} is locally a C 1,β surface up to a set of zero H N 1 measure. In the case N =2 the free boundary is locally a C 1,β surface. Proof. We will make the proof for the first case; the second one follows in the same way. Let ε be small enough so that any minimizer u ε of J ε in K 2 verifies that {u ε > 0} = α. Then it follows that φ 0 is one such minimizer and thus the conclusions of the theorem follow. In fact, as φ 0 minimizes (P α )wehave (5.1) J ε (φ 0 )= φ φ 0 2 dx v 2 + v 2 dx for any v H 1 () such that v L q ( ) = 1 and {v >0} = α. In particular (5.1) holds for v = u ε.thus J ε (φ 0 ) J ε (u ε ) = inf J ε (v). v K 2 This ends the proof. In particular, by the symmetry results for minimizers of (P α ) in balls of [10] we have the following corollary. Corollary 5.1. Let =B(x 0,r) be a ball and let φ 0 be a minimizer of (P α ). Then φ 0 is locally Lipschitz continuous in B(x 0,r) and the free boundary {φ 0 > 0} B(x 0,r) is locally a C 1,β surface up to a set of zero H N 1 measure. In the case N =2the free boundary is locally a C 1,β surface. Proof. In [10] it was proved that any minimizer φ 0 of (P α ) in the case that is a ball B r (x 0 ) satisfies that, for any c 0 > 0, {φ 0 c 0 } B r (x 0 ) is a spherical cap. Since φ 0 L q ( ) = 1, there exists c 0 > 0 such that {φ 0 c 0 } B r (x 0 ). Hence the conditions of Theorem 5.1 are satisfied. In the general case, for the problem (P α ) we can prove that the set of α s for which there exist minimizers with smooth free boundary is dense in (0, ). More precisely, we have the following theorem. Theorem 5.2. For any 0 <α< there exists α ε α as ε 0 such that there exists a solution φ ε of (P αε ) which is locally Lipschitz continuous in and has locally a C 1,β free boundary up to a set of zero H N 1 measure. In the case N =2the free boundary is locally a C 1,β surface. Proof. Let u ε be a minimizer of J ε. We already know that α ε := {u ε > 0} α + Cε (see (4.1)). Let us see that α ε α as ε 0. If not, there exists a sequence ε j 0 such that α εj = {u εj > 0} θ<α. Let φ 0 be a minimizer of (P α ). By the strict monotonicity of S(α) (see [10, Remark 2.2]) we have J (φ 0 )=S(α) <S(θ) J(u εj )=J εj (u εj ) F εj (α εj ) J εj (φ 0 ) F εj (α εj )=J(φ 0 ) F εj (α εj ) J(φ 0 )+Cε j,

21 1634 J. FERNÁNDEZ BONDER, J. D. ROSSI, AND N. WOLANSKI a contradiction. Now, taking φ ε = u ε we see that φ ε is a minimizer of (P αε ). In fact, let v be an admissible function for (P αε ); then and therefore J (v)+f ε (α ε )=J ε (v) J ε (φ ε )=J (φ ε )+F ε (α ε ) J (v) J(φ ε ). The theorem is proved. Finally, we have the following result. Theorem 5.3. Let u ε be a minimizer of J ε in K 1. Then there exists φ 0 H 1 (), a solution to (P α ) such that, up to a subsequence, u ε φ 0 in H 1 (). Proof. In the proof of Theorem 5.2 we showed that {u ε > 0} α as ε 0. It is easy to see that J ε (u ε ) is bounded uniformly in ε and so u ε is uniformly bounded in H 1 (). Therefore, passing to a subsequence if necessary, there exists u 0 H 1 () such that Thus, u ε u 0 weakly in H 1 (), u ε u 0 strongly in L q ( ), u ε u 0 a.e.. u 0 L q ( ) =1, {u 0 > 0} α = lim {u ε > 0}, and ε 0 u 0 H1 () lim inf u ε H ε 0 1 (). Let us call φ 0 = u 0 and let us see that φ 0 is a solution to (P α ). v H 1 () be such that {v >0} = α and v L q ( ) = 1. Then J (v) =J ε (v) J ε (u ε ). Now, since lim inf ε 0 F ε ( {u ε > 0} ) 0, there holds that In fact, let (5.2) J (v) lim inf J ε(u ε ) lim inf J (u ε) J(φ 0 ). ε 0 ε 0 It remains to see that {φ 0 > 0} = α. Assume not; then α 1 := {φ 0 > 0} <α. Thus, by the strict monotonicity of S( ), there holds that S(α) <S(α 1 ), but S(α) = inf J (v) J(φ 0) S(α 1 ), v a contradiction. Now taking v = φ 0 in (5.2), J (φ 0 ) lim inf J (u ε) lim inf J ε(u ε ) J(φ 0 ). ε 0 ε 0 Hence, φ 0 H 1 () = lim inf ε 0 u ε H 1 () and thus, by taking a further subsequence if necessary, the convergence is actually strong. Remark 4. We believe that, as in the previous cases, the minimizers u ε of J ε in K 1 will already be solutions to (P α ) for ε small. Nevertheless, despite the fact that the result of Theorem 5.3 does not give regularity of the minimizer φ 0, we believe that it could be of interest in numerical approximations of the solution to (P α ).

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