The Brézis-Nirenberg Result for the Fractional Elliptic Problem with Singular Potential

arxiv:1705.08387v1 [math.ap] 23 May 2017 The Brézis-Nirenberg Result for the Fractional Elliptic Problem with Singular Potential Lingyu Jin, Lang Li and Shaomei Fang Department of Mathematics, South China Agricultural University, Guangzhou 510642, P. R. China Abstract In this paper, we are concerned with the following type of fractional problems: ( s u µ u x 2s λu = u 2 s 2 u+f(x,u, in, ( u = 0 in R N \ where s (0,1, 2 s = 2N/(N 2s is the critical Sobolev exponent, f(x,u is a lower order perturbation of critical Sobolev nonlinearity. We obtain the existence of the solution for (* through variational methods. In particular we derive a Brézis- Nirenberg type result when f(x,u = 0. Key words and phrases. Fractional Laplacian, positive solution, critical Sobolev nonlinearity. AMS Classification: 35J10 35J20 35J60 1 Introduction Problems involving critical Sobolev nonlinearity have been an interesting topic for a long time. In the celebrated paper [2], Brézis and Nirenberg investigated the problem { u λu = u 2 2 u, in, (1.1 u = 0 on The research was supported by the Natural Science Foundation of China (11101160,11271141 and the China Scholarship Council (201508440330 1

where R N is a bounded domain with smooth boundary, N 3, 2 = 2N. They proved N 2 the following existence and nonexistence results. Let λ 1 is the first eigenvalue of with homogeneous Dirichlet boundary conditions. Then, A if N 4, problem (1.1 has a positive solution for 0 < λ < λ 1 ; B if N = 3, there exists a constant λ (0,λ 1 such that problem (1.1 has a positive solution for λ (λ,λ 1 ; C If N = 3 and is a ball, then λ = 1 4 λ 1, and problem (1.1 has no solution for λ (0,λ. Also, Jannelli in [11] obtained the Brézis-Nirenberg existence and nonexistence results for a class of problem with the critical Sobolev nonlinerity and the Hardy term as follows u µ u x λu = 2 u in, 2 u 2 (1.2 u = 0 on. For more results on (1.2, please see [4, 5, 6, 7, 11, 12] and the references therein. Recently Servadei and Valdinoci [14] extended the results of [2](part A to the following nonlocal fractional equation { ( s u λu = u 2 s 2 u, in, (1.3 u = 0 in R N \ where s (0,1, ( s is the fractional operator, which may be defined ( s u(x u(y u = 2c N,s P.V. R x y N+2sdy, x RN (1.4 N with c N,s = 2 2s 1 π N 2 Γ(N+2s 2 Γ( s (see [3]. Motivated by the above papers, in this paper we consider the following nonlocal fractional equation with singular potential ( s u µ u x λu = 2 2s u 2 s u+f(x,u, in, (1.5 u = 0 in R N \ where 0 R N is a bounded domain with smooth boundary, N 3, 0 < s < 1, ( s is the fractional operator defined in (1.4, 2 = 2N/(N 2s is the critical Sobolev exponent, and f(x, u is a lower order perturbation of critical Sobolev nonlinearity. Problem (1.5 can be considered as doubly critical due to the critical power u 2 s 2 in the semilinear term 2

u and the spectral anomaly of the Hardy potential x 2s. The fractional framework introduces nontrivial difficulties which have interest in themselves. In this paper we investigate the Brézis-Nirenberg type result for (1.5. Firstly we prove the existence of solutions for (1.5 through variational methods under some natural assumptions on f(x, u. Secondly we derive a Brézis-Nirenberg type result for the case f(x,u = 0. For other existence results involving (1.5, you can refer to [8, 9, 16]. is where Notice that equation (1.5 has a variational structure. The variational functional of (1.5 J λ (u = 1 (c R2N u(x u(y 2 N,s dxdy λ 2 x y N+2s 1 u(x 2 2 s dx F(x,udx, s F(x,t = t 0 u(x 2 u(x 2 dx µ x dx 2s (1.6 f(x,τdτ. (1.7 We say u(x is a weak solution of (1.5 if u(x satisfy that (u(x u(y(φ(x φ(y u(xφ(x c N,s dxdy µ dx λ u(xφ(xdx R x y 2N N+2s x 2s = u(x 2 s 2 u(xφ(xdx+ f(x,uφdx, φ H s (R N with φ = 0, a.e. in R N \, u H s (R N with u = 0 a.e. in R N \. As a result of Hardy inequality [8], it is easy to see that (1.8 ( u(x 2 RN Γ N,s R x dx u(x u(y 2 c N 2s N,s dy, u C x y N+2s 0 (RN (1.9 where Γ N,s = 2 2sΓ2 ( N+2s 4 Γ 2 ( N 2s,c N,s = 2 2s 1 π N Γ( N+2s 2 2 Γ( s, (1.10 4 for 0 < µ < Γ N,s, we can define the first eigenvalue λ 1,µ of the following problem ( s u µ u = λu in, x 2s (1.11 u = 0 in R N \ as in (2.7. In this paper, we consider problem (1.5 in the case 0 < λ < λ 1,µ. As in the classical case of Laplacian, the main difficulty for nonlocal elliptic problems with critical nonlinearity is the lack of compactness. This would cause the functional J λ not to satisfy the Palais-Smale condition. Nevertheless, we are able to use the Mountain-Pass Theorem 3

without the Palais-Smale condition (as given in [2], Theorem 2.2 to overcome the lack compactness of Sobolev embedding. Before introducing our main results, we give some notations and assumptions. Notations and assumptions: Denote c and C as arbitrary constants. Let B r (x denote a ball centered at x with radius r, B r denote a ball centered at 0 with radius r and B C r = RN \B r. Let o n (1 be an infinitely small quantity such that o n (1 0 as n. Let f(x,u in (1.5 be a Carathéodary function f : R R satisfying the following conditions: f(x,u is continuous with respect to u; (1.12 f(x,u lim = 0 uniformly in x ; u 0 u (1.13 f(x,u lim = 0 uniformly in x. u u 2 s 1 (1.14 (1.12-(1.14 ensure that f(x, u is a lower order perturbation of the critical nonlinearity. From Lemma 5 and Lemma 6 in [14], for any ε > 0 there exist constants M(ε > 0,δ(ε > 0 such that for any u R,x, f(x,u ε u 2 1 +M(ε; f(x,u δ(ε u 2 1 +ε u. (1.15 We denote by H s (R N the usual fractional Sobolev space endowed with the norm u H s (R N = u L 2 (R N +(c N,s R 2N u(x u(y 2 x y N+2s dxdy 1/2. (1.16 Let X 0 be the functional space defined as with the norm u Xµ = X 0 = {u H s (R N, u = 0 a.e. in R N \} (1.17 (c R2N u(x u(y 2 u(x 2 1/2 N,s dxdy µ x y N+2s x dx, 2s which is equivalent to the norm H s (R N for 0 < µ < Γ N,s (see in Lemma 2.1. Define u(x u(y c 2 N,s dxdy µ u(x 2 dx R S = inf 2N x y N+2s R N x 2s u H s (R N \{0} (, (1.18 u(x R N 2 s dx 2/2 s the Euler equation associated to (1.18 is as follows ( s u µ u x 2s = u 2 s 2 u in R N. (1.19 4

In particular it has been showed in Theorem 1.2 of [8] that for any solution u(x H s (R N of (1.19, there exist two positive constants c,c such that c ( N 2s x 1 ηµ (1+ x 2ηµ 2 u(x C (, in R N \{0} (1.20 N 2s x 1 ηµ (1+ x 2ηµ 2 where η µ = 1 2α µ N 2s, (1.21 and α µ (0, N 2s 2 is a suitable parameter whose explicit value will be determined as the unique solution to the following equation ϕ s,n (α µ = 2 2sΓ(αµ+2s Γ( N αµ 2 2 Γ( N αµ 2s Γ( αµ = µ (1.22 2 2 and ϕ is strictly increasing. Obviously any achieve function U(x of S also satisfies (1.20. Define the constant u(x u(y c 2 N,s dxdy µ u(x 2 dx λ R S λ = inf 2N x y N+2s x 2s u(x 2 dx u X 0 \{0} ( sdx 2 u(x 2 2 s (1.23 Obviously, S λ S because λ > 0. Just as in [2] and [11], we will actually focus on the case when the strict inequality occurs. The first results of the present paper is the following one: Theorem 1.1. Let s (0,1,N > 2s, be an open bounded set of R N, λ 1,µ is the first eigenvalue of (1.11, 0 < µ < Γ N,s (Γ N,s is defined in (1.10 and 0 < λ < λ 1,µ. Let f be a Carathéodory function verifying (1.12-(1.14, assume that there exists u 0 H s (R N \{0} with u 0 0 a.e. in such that supj λ (tu 0 < s t 0 N S N 2s, (1.24 problem (1.5 admits a solution u H s (R N, which is not identically zero, such that u = 0 a.e. in R N \. If assumption (1.24 is satisfied, then it follows Theorem 1.2. Theorem 1.2. Let s (0,1,N > 2s, be an open bounded set of R N, λ 1,µ is the first eigenvalue of (1.11, and µ ϕ 1 ( N 4s (ϕ is defined in (1.22. For any 0 < λ < λ 2 1,µ and f(x,u 0, problem (1.5 admits a solution u H s (R N, which is not identically zero, such that u = 0 a.e. in R N \. 5

Recall the results in [11] as the follows. Let λ 1 denote the first eigenvalue of the operator 1 x 2 in with Dirichlet boundary conditions and µ = ( N 2 2 2. a If µ µ 1, problem (1.2 has a positive solution for 0 < λ < λ 1 ; b if µ 1 < µ < µ, there exists a constant λ (0, λ 1 such that problem (1.2 has a positive solution for λ (λ, λ 1 ; c If µ 1 < µ < µ and is a ball, then problem (1.2 has no solutions for λ λ. As for the results A, B, C for problem (1.3, the space dimension N plays a fundamental role when one seeks solutions of (1.3. In particular in [11] Jannelli point out the N = 3 is a critical dimension for (1.3, while for (1.2 it is only a matter of how µ close to µ. Theorem 1.2 in our paper is an extension of the results of [11] (part a. Here ϕ 1 ( N 4s = µ 1 when 2 s = 0. This paper is organized as follows. In Section 2, we give some preliminary lemmas. In Section 3 we prove the main results Theorem 1.1 and Theorem 1.2 by variational methods. In fact, we first prove Theorem 1.1 by Mountain-Pass Theorem. Then we reduce Theorem 1.2 to Theorem 1.1 by choosing a test function to verify the assumption of Theorem 1.1. 2 Preliminary lemmas In this section we give some preliminary lemmas. Lemma 2.1. Assume 0 < µ < Γ N,s. Then there exist two positive constants C and c such that for all u X 0 R2N C u 2 H s (R N c u(x u(y 2 u(x 2 N,s dxdy µ x y N+2s x dx 2s c u 2 H s (R N. (2.1 that is u Xµ = (c R2N u(x u(y 2 u(x 2 1/2 N,s dxdy µ x y N+2s x dx 2s is a norm on X 0 equivalent to the usual one defined in (1.16. (2.2 Proof. On one hand, from (1.9, for all 0 < µ < Γ N,s, u 2 H s (R N c N,s R2N u(x u(y 2 dxdy µ x y N+2s u(x 2 dx. (2.3 x 2s 6

On the other hand, c N,s u(x u(y 2 u(x 2 dxdy µ R x y 2N N+2s x R2N u(x u(y 2 (1 µ Γ N,s c N,s 2s dx x y N+2s dxdy Cc N,s R2N u(x u(y 2 x y N+2s dxdy (2.4 and ( u(x 2 dx (2 s 2/2 s u 2 s dx 2 u(x u(y 2 2 s c dxdy (2.5 R x y 2N N+2s then from (2.4 and (2.5 u H s (R N c ( c N,s u(x u(y 2 dxdy µ R x y 2N N+2s Collecting with (2.3 and (2.6, the proof is complete. u(x 2 x 2s dx 1/2. (2.6 Lemma 2.2. Let s (0,1,N > 2s, be an open bounded set of R N and 0 < µ < Γ N,s. Then problem (1.11 admits an eigenvalue λ 1,µ which is positive and that can be characterized as c N,s λ 1,µ = min u X 0 \{0} R 2N u(x u(y 2 dxdy µ x y N+2s u(x 2 dx u(x 2 x 2s dx Proof. It is remain to show that λ 1,µ can be attained and λ 1,µ > 0. Denote I(u = 1 2 Obviously (2.7 is equivalent to M = {u X 0, u L 2 ( = 1}, ( R2N u(x u(y 2 u(x 2 cn,s dxdy µ x y N+2s x dx = 1 2s 2 u 2 X µ. λ 1,µ = min u M (c N,s R2N u(x u(y 2 dxdy µ x y N+2s Let us take a minizing sequence u n M such that > 0. (2.7 u(x 2 dx > 0. (2.8 x 2s I(u n inf I(u 0 as n. (2.9 u M From Lemma 2.1, u n H s (R N is bounded. Then there exists u 0 such that (up to a subsequence, still denoted by u n u n u 0 in H s (R N,u n u 0 in L 2 loc(r N, and u n u 0 a.e. in R N as n. (2.10 7

So u 0 L 2 ( = 1 and u 0 = 0 a.e. in R N \, that is u 0 M. Denote v n = u n u 0, we have v n 0 in H s (R N,v n 0 in L 2 loc(r N, and v n 0 a.e. in R N as n, (2.11 then by Brézis-Lieb Lemma in [1], it follows R2N u n (x u n (y 2 R2N v n (x v n (y 2 R2N u 0 (x u 0 (y 2 dxdy = dxdy+ dxdy+o x y N+2s x y N+2s x y N+2s n (1, (2.12 u n (x 2 v n (x 2 u 0 (x 2 dx = dx+ dx+o x 2s x 2s x 2s n (1, (2.13 which implies that I(u n = 1 2 ( R2N u n (x u n (y 2 u n (x 2 cn,s dxdy µ dx x y N+2s x 2s = 1 2 v n 2 X µ + 1 2 u 0 2 X µ +o n (1 inf u M I(u(1+ v n 2 L 2 ( where the last inequality follows from (2.7 and (2.8. Then (2.14 1 inf I(u = lim u M n 2 v n Xµ + 1 2 u 0 2 X µ inf I(u. (2.15 u M Thus inf I(u = I(u 0, (2.16 u M that is, u 0 is a minimizer of I(u. Since u 0 L 2 ( = 1, it is easy to obtain that λ 1,µ = I(u 0 > 0. The proof is complete. 3 The proof of main results In this section, we study the critical points of J λ which are solutions of problem (1.5. In order to find these critical points, we will apply a variant of Mountain-Pass Theorem without the Palais-Smale condition (refer [2]. Due to the lack of compactness in the embedding X 0 L 2 s (, the functional Jλ does not verify the Palais-Smale condition globally, but only in the energy range determined by the best fractional critical Sobolev constant S given in (1.18. First of all, we prove the functional J λ has the Mountain-Pass geometric structure. Proposition 3.1. Let λ (0,λ 1,µ and f be a Carathédory function satisfying conditions (1.12-(1.14. Then there exists ρ > 0, β > 0, e X 0 such that 1 for any u X 0 with u Xµ = ρ, J λ (u β; 2 e 0 a.e. in R N, e Xµ > ρ and J λ (e < β. 8

Proof. For all u X 0, by (2.7 and (1.15 we get for ε > 0 J λ 1 (c R2N u(x u(y 2 N,s dxdy λ u(x 2 dx µ 2 x y N+2s 1 u(x 2 2 s dx ε u 2 dx δ(ε u(x 2 s dx s 1 2 (1 λ (c R2N u(x u(y 2 N,s dxdy µ λ 1,µ x y N+2s ε 2 s 2 2 s ( u 2 s dx 2 s /2 (δ(ε+ 1 u(x 2s 2 s dx ( 1 2 (1 λ λ 1,µ ε 2 u(x 2 x 2s dx s 2 2 s S u 2 X µ (δ(ε+ 1 2 ss 2 s /2 u 2 s /2 Choose ε > 0 small enough, there exist α > 0 and ν > 0 such that u(x 2 x 2s dx X µ. J λ (u α u 2 X µ ν u 2 s X µ. (3.1 Now let u X 0, u Xµ = ρ > 0, since 2 s > 2, choose ρ small enough, then there exists β > 0 such that inf J λ β > 0. (3.2 u X 0, u Xµ =ρ Fix any u 0 X 0 with u 0 L 2 s( > 0. Without loss of generality, we can assume u 0 0 a.e. in R N (otherwise, we can replace any u 0 X 0 with its positive part, which belong to X 0 too, thanks to Lemma 5.2 in [13]. Since for t > 0 large enough it follows that Take J λ (u 0 ct 2 u 0 2 X µ t2 s F(x,tu 0 t2 s u 0 2 s, (3.3 22 s 22 s u 0 2 s L 2 s as t. (3.4 e = t 0 u 0 (3.5 and t 0 sufficiently large, we have J(t 0 u 0 < 0 < β. The proof is complete. As a consequence of Proposition 3.1, J λ (u satisfies the geometry structure of Mountain- Pass Theorem. Set c =: inf sup J λ (γ(t, (3.6 γ Γ t [0,1] where Γ = {γ C([0,1],X 0 : γ(0 = 0,γ(1 = e X 0 }. Proof of Theorem 1.1 for all γ Γ, the function t γ(t Xµ is continuous, then there exists t (0,t such that γ( t Xµ = ρ. It follows that sup J λ (γ(t J λ (γ( t inf J λ (u β, (3.7 t [0,1] u X 0, u Xµ =ρ 9

then c β. Since te Γ, we have that 0 < β c = inf sup γ Γ t [0,1] sup J λ (te t [0,1] J λ (γ(t supj λ (ζu 0 < s ζ 0 N S N 2s. (3.8 By Theorem 2.2 in [2] and Proposition 3.1, there exists a Palais-Smale sequence {u n } X 0 a.e., J λ (u n c, J λ (u n 0. (3.9 From (1.15 and the definition of J λ, it follows which implies u n 2 s L 2 s( c +1 J λ (u n 1 2 < J λ(u n,u n > = ( 1 2 1 2s u n 2 s + (1 L 2 s( 2 f(x,u nu n F(x,u n dx ( 1 2 1 2s u n 2 s ε u L 2 s( n 2 s dx c u n dx c u n 2 s L 2 s( 2ε u n 2 s L 2 s( C is bounded for ε > 0 small enough. So c +1 J λ (u n 1 2 (1 λ u n 2 X λ µ 1 u 1,µ 2 n 2 s F(x,u L 2 s( n dx s 1 2 (1 λ u n 2 X λ µ 1 u 1,µ 2 n 2 s ε u L 2 s( n 2 L 2 ( C u n 2 s L 2 s( s 1 λ+2ε (1 u n 2 X 2 λ µ c. 1,µ (3.10 (3.11 Thus u n Xµ is bounded for ε > 0 small enough. Next we claim that there exists a solution u 0 X 0 of (1.5. Since u n is bounded in X 0, by Lemma 2.1, up to a subsequence, still denoted by u n, as n u n u 0 in H s (R N, (3.12 u n u 0 in L p loc (, for all p [1,2 s, (3.13 u n u 0 a.e. in R N. (3.14 Thus u 0 = 0 in R N \, u 0 X 0 and for all φ C0 (, as n (u n (x u n (y(φ(x φ(y (u 0 (x u 0 (y(φ(x φ(y c N,s dxdy c R x y 2N N+2s N,s dxdy, R x y 2N N+2s 10 (3.15

u n (xφ(x dx x 2s u n (xφ(xdx u 0 (xφ(x x 2s dx, (3.16 u 0 (xφ(xdx. (3.17 Denote g(u = u 2 s 2 u. From (3.12-(3.14, (1.15 and f(x,u,g(u is continuous in u, applying Lemma A.2 in [15], f(x,u n f(x,u 0 L 1 ( c u n u L q ( 0, g(u n g(u 0 L 1 ( c u n u L q ( 0, (3.18 where q = 2 s 1. Then it implies for all φ C 0 (, as n ( u n 2 s 2 u n u 0 2 s 2 u 0 φdx c g(u n g(u 0 L 1 ( 0, (3.19 (f(x,u n f(x,u 0 φdx c f(x,u n f(x,u 0 L 1 ( 0. (3.20 Thus u 0 X 0 is a solution of (1.5. To complete the proof of Theorem 1.1, it suffices to show that u 0 0. Suppose, in the contrary, u 0 0. From (3.12-(3.14 and (1.15 lim sup n f(x,u n u n dx εlimsup n u n 2 s dx+climsup n u n dx cε (3.21 where ε > 0 is an arbitrary constant. Letting ε 0, it follows that lim sup n f(x,u n u n dx 0. (3.22 Similar as (3.22, it follows Since < J λ (u u n (x u n (y 2 n,u n > = c N,s dxdy µ R x y 2N N+2s u n (x 2 s dx f(x,u n u n dx, thanks to (3.22 and (3.23, we have u n (x u n (y 2 c N,s dxdy µ R x y 2N N+2s lim sup F(x,u n dx 0. (3.23 n 11 u n (x 2 dx λ u x 2s n (x 2 dx (3.24 u n (x 2 dx u x 2s n (x 2 s dx 0. (3.25

Denote which implies and c N,s u n (x u n (y 2 dxdy µ R x y 2N N+2s u n (x 2 s dx L, It is easy to see that L > 0. From (1.18 we have u n (x 2 x 2s dx L 0 < β c = lim J λ (u n = ( 1 n 2 1 L. (3.26 2 s L SL 2 2 s (3.27 then c s N S N 2s (3.28 which contradicts (3.8. Hence u 0 0 in. The proof of Theorem 1.1 is complete. For the proof of Theorem 1.2, the key step is to check the condition (1.24. Firstly we choose the test function v ε (x of (1.24 as the following. Let u(x = U(x U(x L 2 s(r N,u ε (x = ε N 2s 2 u( x. (3.29 ε Thus u ε L 2 s(r N = 1, and u ε(x is the achieve function of S (defined in (1.18. Let us fix δ > 0 such that B 4δ, and let η C 0 (B 2δ be such that 0 η 1 in R N, η 1 in B δ. For every ε > 0, define v ε (x = u ε (xη(x,x R N. (3.30 Obviously v ε X 0. Secondly we make some estimates for v ε. The main strategy of the estimations for v ε (x is similar to [14]. Proposition 3.2. 1 For all x B c r, it follows where r > 0 and c r is a positive constant depending on r. 2 For all y B C δ,x RN with x y < δ/2, it follows for all x,y Bδ C, it follows v ε u ε c r ε (N 2sηµ 2, (3.31 v ε c r ε (N 2sηµ 2 (3.32 v ε (x v ε (y c x y ε (N 2sηµ 2 ; (3.33 v ε (x v ε (y cmin{ x y,1}ε (N 2sηµ 2. (3.34 12

Proof. 1From the definition of v ε (x and u ε (x, for x B c r, we have v ε (x u ε (x cε (N 2s/2 ( N 2s x/ε 1 ηµ (1+ x/ε 2ηµ 2 = cε ηµ(n 2s/2 ( N 2s x 1 ηµ (ε 2ηµ + x 2ηµ 2 cε ηµ(n 2s/2 r (1+ηµ(N 2s/2 c r ε ηµ(n 2s/2 (3.35 where c r is a positive constant depending on r. Since u ε cε (N 2s/2( x ε 1 ηµ (1+ x ε 2ηµ N 2s 2 1 ( x ε ηµ + x ε ηµ 1 ε = cε (N 2s/2( x (1+ x N 2s ε 1 ηµ ε 2ηµ 2 ( x + x ε ηµ ε ηµ 1 x (1+ x ε 1 ηµ ε 2ηµ ε ( x + x ε c u ε (x ηµ ε ηµ 1 x (1+ x ε 1 ηµ ε 2ηµ ε c u ε(x ( 1 x + ε2ηµ x 2ηµ+1 c r u ε (x (3.36 and v ε (x = η(xu ε (x+ u ε (xη(x c( u ε (x + u ε (x, (3.37 collecting (3.35-(3.37, we obtain (3.31 and (3.32. 2 Similar to Claim 10 in [14], it follows (3.33-(3.34. In fact, for all x R N,y B C δ, x y < δ 2, v ε (x v ε (y = v ε (ξ x y (3.38 where ξ R N with ξ = tx+(1 ty for some t (0,1. Since taking r = δ, from (3.32 it follows (3.33. 2 For all x B C δ,y BC δ, if x y δ 2, ξ = y +t(x y y t x y > δ 2, (3.39 v ε (x v ε (y v ε (x + v ε (y cmin{1, x y }ε ηµ(n 2s/2 ; if x y δ, (3.34 follows from (3.33. 2 Proposition 3.3. Let s (0,1, and N > 2s, then the following estimates holds v ε (x v ε (y 2 v ε (x 2 1 c N,s dxdy µ dx S +O(ε (N 2sηµ ; (3.40 R x y 2N N+2s x 2s 13

2 v ε (x 2 s = L 2 s( 1+O(εηµN ; (3.41 cε ηµ(n 2s cε 2s, if η µ < 2s, N 2s 3 v ε (x 2 dx c logε ε 2s, if η µ = 2s N 2s, (3.42 cε ηµ(n 2s +cε 2s, if η µ > 2s Proof. 1 Define N 2s. D = {(x,y R N R N x B δ,y B C δ, x y δ 2 } (3.43 E = {(x,y R N R N x B δ,y Bδ C, x y < δ }. (3.44 2 Then v ε (x v ε (y 2 c N,s dxdy R x y 2N N+2s u ε (x u ε (y 2 = c N,s dxdy +c B δ B δ x y N+2s N,s u ε (x v ε (y 2 +c N,s dxdy +c x y N+2s N,s E D B C δ BC δ u ε (x v ε (y 2 dxdy x y N+2s v ε (x v ε (y 2 x y N+2s dxdy. (3.45 From (3.33 and E B δ B 2δ, we have c N,s E u ε (x v ε (y 2 dxdy c x y N+2s N,s B δ B 2δ ε (N 2sηµ x y N+2s 2dxdy c δε (N 2sηµ, (3.46 where c δ > 0 is a constant depending on δ. Similarly, from (3.34 c N,s B C δ BC δ u ε (x v ε (y 2 dxdy c x y N+2s N,s Bδ C BC δ c δ ε (N 2sηµ. ε (N 2sηµ min{1, x y 2 } dxdy x y N+2s (3.47 where c δ > 0 is a constant depending on δ. Now, it remains to estimate the integral on D of (3.45, u ε (x v ε (y 2 c N,s dxdy D x y N+2s u ε (x u ε (y 2 c N,s dxdy +c x y N+2s N,s D +c N,s D 2 v ε (y u ε (y u ε (x u ε (y dxdy x y N+2s D u ε (y v ε (y 2 dxdy x y N+2s (3.48 14

Now we estimate the last term in the right hand side of (3.48. Once more by (3.31, u ε (y v ε (y 2 c N,s dxdy D x y N+2s u ε (y 2 + v ε (y 2 c N,s dxdy x y N+2s (3.49 D c x y δ/2 where c δ > 0 is a constant depending on δ. And for (x,y D, ε (N 2sηµ x y N+2sdxdy c δε (N 2sηµ. u ε (xv ε (y u ε (xu ε (y cε (N 2sηµ/2 ε (N 2s/2 ( N 2s x/ε 1 ηµ (1+ x/ε 2ηµ 2 c (, N 2s x 1 ηµ (1+ x/ε 2ηµ 2 (3.50 let x = x,z = y x, it follows ε u ε (xu ε (y c N,s dxdy D x y N+2s 1 c ( dxdy N 2s D x 1 ηµ (1+ x/ε 2ηµ 2 x y N+2s cε N 1 ( d xdz x δ/ε, z > δ N 2s 2 ε x 1 ηµ (1+ x 2ηµ 2 z N+2s 1 cε N (N 2s(1 ηµ 2 d x ( cε N (N 2s(1 ηµ 2 ( cε N (N 2s(1 ηµ 2 ( cε N (N 2s(1 ηµ 2 = O(ε (N 2sηµ. x δ/ε x 1 r 1 r 1 ( N 2s x 1 ηµ (1+ x 2ηµ 2 1 ( d x+ N 2s x 1 ηµ (1+ x 2ηµ 2 r (N 1 (1 ηµn 2s 2 dr + N 2s (1+r 2ηµ 2 1 r δ/ε 1 x δ/ε 1 ( d x N 2s x 1 ηµ (1+ x 2ηµ 2 r (N 1 (1+ηµ(N 2s 2 dr r (N 1 (1 ηµn 2s 2 dr +ε N+(1+ηµ(N 2s 2 +c (3.51 15

It follows from (3.48-(3.51 that u ε (x v ε (y 2 c N,s dxdy c x y N+2s N,s D D u ε (x u ε (y 2 x y N+2s dxdy +O(ε (N 2sηµ. (3.52 Thus from (3.46, (3.47 and (3.52, v ε (x v ε (y 2 R2N u ε (x u ε (y 2 c N,s dxdy c R x y 2N N+2s N,s dxdy +O(ε (N 2sηµ. (3.53 x y N+2s Since RN (1 η(x 2 u ε (x 2 µ dx c x 2s = c and µ RN (1 η(x 2 u ε (x 2 x 2s dx c it gives = c c c x δ x δ x 2δ x 2δ x 2δ/ε r 2δ/ε u ε (x 2 x 2s dx ε (N 2s ( N 2s x dx x/ε 1 ηµ (1+ x/ε 2ηµ 2s 1 ( N 2s x dx x 1 ηµ (1+ x 2ηµ 2s r N 1 2s (1+ηµ(N 2s dr = O(ε (N 2sηµ, u ε (x 2 x 2s dx ε (N 2s (3.54 ( N 2s x dx = O(ε x/ε 1 ηµ (1+ x/ε (N 2sηµ, 2ηµ 2s RN v ε (x 2 RN u ε (x 2 RN (1 η(x 2 u ε (x 2 µ dx = µ dx µ dx x 2s x 2s x 2s RN u ε (x 2 = µ dx+o(ε (N 2sηµ. x 2s (3.55 (3.56 From (3.53 and (3.56, we have v ε (x v ε (y 2 RN v ε (x 2 c N,s dxdy µ dx S +O(ε (N 2sηµ. (3.57 R x y 2N N+2s x 2s 2 Similar to (3.56, we have R N v ε (x 2 s dx = R N u ε (x 2 s dx+o(ε Nηµ (3.58 16

3Finally, we want to prove (3.42 v ε (x 2 dx u ε (x 2 dx c x δ x δ = ε 2s c ε (N 2s ( N 2s dx x/ε 1 ηµ (1+ x/ε 2ηµ x δ/ε ( 1 N 2s dx x 1 ηµ (1+ x 2ηµ δ/ε cε 2s r N 1 ( N 2s dr 1 r 1 ηµ (1+r 2ηµ cε ηµ(n 2s cε 2s, if η µ < 2s, N 2s = c logε ε 2s, if η µ = 2s N 2s, cε ηµ(n 2s +cε 2s, if η µ > 2s N 2s. (3.59 Proof of Theorem 1.2 From Theorem 1.1, we only need to verify (1.24. Denote t ε be the attaining point of max J λ(tv ε. Similar as the proof of Lemma 8.1 in [10], let t ε be the t>0 attaining point of max J λ(tv ε, we claim t ε is uniformly bounded for ε > 0 small. In fact, t>0 we consider the function g(t = J λ (tv ε = t2 2 ( v ε(x 2 X µ λ v ε 2 dx t2 s v 2 ε 2 s dx s t2 2 (1 λ (3.60 v ε (x 2 X λ µ t2 s v ε 2 s dx. 1,µ Since lim g(t = and g(t > 0 when t closed to 0, so that maxg(t is attained for t + t>0 t ε > 0. Then g (t ε = t ε ( v ε (x 2 X µ λ From (3.61 and Lemma 3.3, for ε sufficiently small, we have S 2 (1 λ t 2 s 2 ε λ µ which implies t ε is bounded for ε > 0 small enough. 2 s v ε 2 dx t 2 s 1 ε v ε 2 s dx = 0. (3.61 = v ε(x 2 X µ λ v ε 2 dx v ε 2 sdx From the definition of η µ, (1.21 and (1.22, we have < 2S. (3.62 µ ϕ 1 ( N 4s η µ 2s 4 N 2s. (3.63 17

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