NONHOMOGENEOUS POLYHARMONIC ELLIPTIC PROBLEMS AT CRITICAL GROWTH WITH SYMMETRIC DATA. Mónica Clapp. Marco Squassina

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1 COMMUNICATIONS ON Website: PURE AND APPLIED ANALYSIS Volume, Number, June 003 pp NONHOMOGENEOUS POLYHARMONIC ELLIPTIC PROBLEMS AT CRITICAL GROWTH WITH SYMMETRIC DATA Mónica Clapp Instituto de Matemáticas Universidad Nacional Autónoma de México Circuito Exterior, Ciudad Universitaria México Marco Squassina Dipartimento di Matematica e Fisica Università Cattolica del Sacro Cuore Via Musei 41, 511, Brescia, Italy Abstract. We show the existence of multiple solutions of a perturbed polyharmonic elliptic problem at critical growth with Dirichlet boundary conditions when the domain and the nonhomogenous term are invariant with respect to some group of symmetries. 1. Introduction and Main Result. Let K 1 and let be a bounded smooth domain in R N with N > K. In this paper we consider the polyharmonic elliptic problem ( K u = u K u + f in, ( j (P, f ν u = 0, j = 0,..., K 1, N N K where f H K ( and = denotes the critical exponent for the Sobolev embedding H0 K ( L (. If f = 0 this problem is invariant under dilations. Lac of compactness in elliptic problems which are invariant under dilations is nown to produce quite interesting phenomena. It often gives rise to solutions of small perturbations of such problems. This behavior has been extensively studied for K = 1 (and 1 =, we refer to [5], [7] and [30] for a detailed discussion. For K > 1 perturbations of problem (P, 0 by adding a subcritical term have been considered by many authors; we refer to the wor of Gazzola [14] and the references therein. For K = perturbations of the domain giving rise to solutions were also recently considered by Gazzola, Grunau and one of the authors [15]. Adding a nonhomogeneous term produces a similar effect. For K = 1 it was shown by Tarantello [9] that, if f 0 and f H 1 is small enough, problem (P, f has at least two nontrivial solutions. This result was extended to the case K = by Deng and Wang [11]. One consequence of the main result in this paper is that this is true for every K 1. We shall show that the following holds Mathematics Subject Classification. 31B30, 58E05. Key words and phrases. Polyharmonic problems, symmetric domains, multiplicity of solutions. The first author was partially supported by CONACyT, Mexico under Research Project E. The second author was partially supported by MURST (40% 1999 and by GNAFA. 171

2 17 MÓNICA CLAPP AND MARCO SQUASSINA Corollary 1.1. There exists a κ > 0 such that, if f 0 and f H K < κ, then problem (P, f has at least solutions. It is well nown that the presence of symmetries gives rise, in many cases, to additional solutions. The impact of symmetries on problem (P, f for K = 1 has been studied recently by Kavian, Ruf and one of the authors [10]. The main goal of this paper is to study the effect of symmetries on the number and on the type of solutions of problem (P, f for arbitrary K 1. We consider domains which are invariant under the action of some closed subgroup G of the group O(N of orthogonal transformations of R N, that is, gx for every g G, x. We assume that f is G invariant, that is, f(gx = f(x for every g G, x, and loo for additional solutions of problem (P, f which are also G invariant. Recall that G is said to act freely on if g 1 x g x for all g 1 g G, x. As a consequence of our main result we obtain the following. Corollary 1.. If G 1 acts freely on then there exists a κ > 0 with the property that, for every f 0 which is G invariant and such that f H K < κ, problem (P, f has at least 3 solutions one of which is G invariant and one of which is not. For example, if is symmetric with respect to the origin (i.e. x if x and 0 then, for every even function f 0 with f H K small enough, problem (P, f has at least 3 solutions, one of which is even and one of which is not. We write (u, v K, = q u q v dx if K = q, (1.1 q u q v dx if K = q + 1 for the usual scalar product in the Sobolev space H0 K (, and denote by S K the best Sobolev constant for the embedding H0 K ( L K (, S K = inf u K, : u H0 K (, u dx = 1. We write G j for the cardinality of G j. Our main result is the following. Theorem 1.3. Let 1 = G 1,..., G m be closed subgroups of O(N acting freely on such that G 1 < < G m and G m 1 G m. Then at least one of the following assertions holds: (a m > 1 and for f = 0 problem (P, 0 has a nontrivial solution u such that u K, ( G m 1(S K N/K ; (b m 1 and there exists a κ > 0 with the property that, if f is G i invariant for each i = 1,..., m, f 0 and f H K < κ, then problem (P, f has at least m + 1 solutions u 0, u 1,..., u m such that u i is G i invariant but not G i+1 invariant for i = 1,..., m 1, and u m is G m invariant. We recall that a wea solution of (P, f belongs to C K,α ( if is of class C K,α and f C 0,α ( [0]. Whether has symmetries or not, Theorem 1.3 (with m = 1 asserts the existence of at least two solutions of problem (P, f for f 0 and f H K small enough. This is Corollary 1.1. Moreover, if and f have appropriate symmetries, Theorem 1.3 provides an additional solution. Indeed, since (P, 0 has no ground

3 CRITICAL POLYHARMONIC PROBLEMS ON SYMMETRIC DOMAINS 173 state solution, Theorem 1.3 includes Corollary 1. as a special case (a detailed argument is given in Section 4 below. Theorem 1.3 asserts the existence of many solutions of problem (P, f provided that it has many symmetries and that the unperturbed problem (P, 0 has no nontrivial solution below a certain energy level. Little is nown about nonexistence of solutions of problem (P, 0. For K = 1 Pohožaev s identity [3] implies nonexistence of solutions in starshaped domains. But for K > 1, even though Oswald did show that there are no positive solutions in domains of this ind [1], as far as we now there is no result excluding sign changing ones apart from the case K = where the existence of radial solutions on a ball has been ruled out [15]. In any case, notice that the condition that G acts freely on is quite strong. It implies that 0 /, which excludes starshaped domains. It also implies that has nontrivial topology. For K = 1 a well nown result of Bahri and Coron [1] asserts the existence of a solution of problem (P, 0 if has nontrivial topology. A similar result for any K 1 was recently obtained by Bartsch, Weth and Willem []. Moreover, even in some contractible domains, solutions of (P, f are nown to exist [], [15]. Quite recently, however, Ben Ayed, El Mehdi and Hammami [3] obtained a nonexistence result for problem (P, 0 on thin annuli. They showed that, for K = 1, problem (P, 0 has no positive solution below a given energy level if the annular domain is thin enough. This fact, together with Theorem 1.3 and some stronger results of this ind, provides multiple solutions of problem (P, f for K = 1 and small f 0 on thin annuli [10]. For f 0 and K = 1 there is an effect of the domain topology [6] together with its symmetries [10] on the number of solutions of (P, f. Also more general group actions are allowed in this case. This is a consequence of the fact that, for K = 1, least energy solutions are positive if f 0 and small enough. For K > 1 this positivity preservation property does not hold in general, due to the lac of maximum principles for ( K [17]. Finally we would lie to mention that for nonhomogeneous polyharmonic problems at (small enough subcritical growth with homogeneous or nonhomogeneous Dirichlet boundary conditions much stronger results hold for arbitrary K 1 [18]. This paper is organized as follows: in Section we describe the variational setting associated to problem (P, f in the presence of symmetries. In Section 3 we give a compactness condition for this problem and obtain a first G invariant solution. In Section 4 a further G invariant solution is provided, and Theorem 1.3 is proved. As in the case K = 1 [9], [10], the proof of Theorem 1.3 for K 1 relies, on one hand, on the nowledge of the first G invariant noncompactness level for the unperturbed problem (P, 0. On the other hand, it requires fine estimates similar to those obtained by Brézis and Nirenberg in [8]. These questions will be handled in Sections 5 and 6 respectively.. The Variational Framewor. Let G be a closed subgroup of O(N and assume that and f are G invariant. Consider the problem ( K u = u K u + f in, ( j ν u = 0, j = 0,..., K 1, (P, G f u(gx = u(x for g G.

4 174 MÓNICA CLAPP AND MARCO SQUASSINA The action of G on induces an orthogonal G action on H K 0 ( given by (gu(x := u(g 1 x, that is, (u, v K, = (gu, gv K, for all g G, u, v H0 K (. We write p for the L p norm. The energy functional E f (u = 1 u K, 1 u K fu dx is G invariant, that is, E f (gu = E f (u for all g G and u H K 0 (. Wea solutions of problem (P G, f are critical points of the restriction of E f to the space of fixed points H K 0 ( G = u H K 0 ( : u(gx = u(x for all g G. They lie on the Nehari set Nf G = u H0 K ( G : DE f (uu = 0 = u H0 K ( G : u K, u From now on we assume that the following condition holds: (H 1 For every v H0 K ( G with v K = 1, fv dx < b N,K v N+K K K,, b N,K = 4K N K fu dx = 0 ( N K N + K N+K 4K. This condition is the same one given by Tarantello [9] for K = 1 and by Deng and Wang [11] for K =. Observe that condition (H 1 holds provided that f H K < b N,K S N/4K K. Let us recall some properties satisfied by the Nehari set. Proposition.1. If condition (H 1 holds then Nf G has the following properties: (a if f 0 then Nf G is a C 1 submanifold of H0 K ( G ; if f = 0 then N0 G \0 is a C 1 submanifold of H0 K ( G ; (b for every 0 u N G f the line Ru is transversal to N G f at u ; (c N G f has two components ( (Nf G + = u Nf G : u N + K K, N K ( (Nf G = u Nf G : u N + K K, N K u K 0, ; u K < 0 (d (N G f is radially diffeomorphic to the unit sphere in H K 0 ( G ; (e E f (u < 0 for u (N G f + with u 0 ; (f E f is bounded below on N G f. The proof is easy and similar to the one for K = 1 [9], [10]. Details are left to the reader. We define c G f,0 := inf u N G f E f = inf u (N G f + E f, c G f,1 := inf u (N G f E f.

5 CRITICAL POLYHARMONIC PROBLEMS ON SYMMETRIC DOMAINS 175 These infima are natural candidates for being critical values. We shall show that they are actually achieved. Notice that Proposition.1 implies that < c G f,0 c G f,1 < + and that cg f,0 < 0 if f 0. If f = 0 we write N G = (N G 0, c G 1 = inf N G E The G invariant Compactness Range. Definition 3.1. A sequence (u H K 0 ( G such that E f (u c, DE f (u H K 0, will be called a G PS sequence for E f at the level c. E f will be said to satisfy the G Palais Smale condition (P S G c at the level c if every such sequence has a convergent subsequence. Let us set ( µ G := min x Gx K N (S K N/K, where Gx denotes the cardinality of the G orbit Gx = gx : g G of x. The following Proposition, which extends a result of P.L. Lions [19] to any K 1, will be proved in Section 5. Proposition 3.. E 0 satisfies (P S G c at every c < µ G. In particular, if every G orbit in is infinite, then E 0 satisfies (P S G c at every c R. Corollary 3.3. E f satisfies (P S G c at every c < c G f,0 + µg. In particular, if every G orbit in is infinite, then E f satisfies (P S G c at every c R. Proof. Let (u be G PS sequence for E f with E f (u c. It is readily seen that (u is bounded. Hence a subsequence converges wealy in H K 0 ( G to a wea solution u of (P G, f. Let v := u u. A standard argument shows that E f (u = E f (u + E 0 (v + o(1 o(1 = DE f (u = DE f (u + DE 0 (v + o(1 = DE 0 (v + o(1 as +. Therefore (v is a G PS sequence for E 0 such that v 0 wealy in H0 K ( G and E 0 (v c E f (u c c G f,0 < µ G. It follows from Proposition 3. that, up to a subsequence, v 0 strongly in H0 K ( G and, hence, u u strongly in H0 K ( G. Easy consequences of Corollary 3.3 are the following. Theorem 3.4. If f satisfies assumption (H 1 then c G f,0 is achieved at a point ug f,0 (Nf G+ which is a critical point of E f on H0 K ( G. Moreover, f H K 0 = u G f,0 K, 0. Proof. If f = 0 tae u G f,0 = 0. For f 0 let (u n be a minimizing sequence for E f on (Nf G+. Eeland s variational principle [30] allows us to assume that (u n is a Palais Smale sequence for E f on Nf G. Since cg f,0 < 0, we may also assume that u n 0. Hence Ru n is transversal to Nf G at u n and therefore (u n is a G PS sequence for E f at the level c G f,0. Corollary 3.3 above asserts that a subsequence of

6 176 MÓNICA CLAPP AND MARCO SQUASSINA (u n converges strongly to u G f,0 (N f G+. Hence, E f (u G f,0 = cg f,0. The last assertion follows immediately from the inequality 0 E f (u G f,0 = K N u f,0 K, N + K fu f,0 dx N K u G f,0 N N + K K, N f H u G (N + K K f,0 K, 16NK f H. K For K = 1 this result is due to Tarantello [9] if G = 1 and for arbitrary G it was proved in [10]. A further consequence of Corollary 3.3 is the following. Proposition 3.5. If condition (H 1 holds then c G f,0 < cg f,1. Proof. If c G f,0 = cg f,1 then, arguing as in Theorem 3.4, we obtain a ug f,1 (N G f such that E f (u G f,1 = cg f,1. Let t 0 0 be the largest real number with E f (t 0 u G f,1 (N G f +. Assumption (H 1 implies that This is a contradiction. c G f,0 E f (t 0 u G f,1 < E f (u G f,1 = c G f,1. 4. A Second G invariant Solution. We wish to give conditions for c G f,1 to be achieved by E f on (Nf G. Corollary 3.3 immediately gives the following. Theorem 4.1. Assume that condition (H 1 holds. If every G orbit of is infinite, then c G f,1 is achieved at a point ug f,1 (N f G which is a critical point of E f on H0 K ( G. Next we consider the case when the domain has a finite G orbit. We assume throughout that condition (H 1 holds and also (H G is a finite group acting freely on. For ε > 0 and y R N we consider the ground state solutions ( ε N K K T ε,y (x = c N,K ε + x y, c N,K = (N j j=1 K of the problem ( K u = u u in R N (see [8]. It satisfies T ε,y K, = (S K N/K = T ε,y K. For y we consider the multi bump function w ε,y = g G ϕ gy T ε,gy H K 0 ( G N K 4K where ϕ C (R N is radially symmetric, 0 ϕ 1, ϕ = 1 on B(0, 1 and ϕ = 0 outside B(0,, and ϕ gy (x = ϕ(ρ 1 (x gy with 0 < ρ < min 1 dist(y,, 1 4 gy g y : g, g G, g g. Hence supp(ϕ gy and supp(ϕ gy supp(ϕ g y = if g g. If f 0 then u G f,0 0 and we may assume without loss of generality that ug f,0 > 0 in some set Σ of positive measure. The following lemma, which extends a result of Brézis and Nirenberg [8] to any K 1, will be proved in Section 6.

7 CRITICAL POLYHARMONIC PROBLEMS ON SYMMETRIC DOMAINS 177 Lemma 4.. For each K 1 and t 0 and for a.e. y Σ the following holds u G f,0 + tw ε,y = u G f,0 dx + t w ε,y dx + t K 1 u G f,0 wε,y K 1 dx + t u G f,0 K u G f,0 w ε,y dx + R ε + S ε with R ε = o ( ε (N K/ and S ε = o ( ε (N K/ as ε 0. Proposition 4.3. Assume that f 0 and that conditions (H 1 and (H hold. Then for every t 0 and a.e. y Σ it results for each ε > 0 sufficiently small. E f (u G f,0 + tw ε,y < c G f,0 + µ G Proof. For every t 0 and each y it results E f (u G f,0 + tw ε,y = 1 ug f,0 K, + t(u G f,0, w ε,y K, + t w ε,y K, 1 u G f,0 + tw ε,y K K fu G f,0 dx t fw ε,y dx. In view of Lemma 4., this equality yields E f (u G f,0 + tw ε,y = E f (u G f,0 + tde f (u G f,0w ε,y + t w ε,y K, dx t t K 1 u G f,0 wε,y K 1 dx + o ( ε (n K/ = c G f,0 + t w ε,y K, dx t t 1 w ε,y w ε,y K u G f,0 w 1 ε,y dx + o ( ε (n K/ (4.1 as ε 0 for a.e. y Σ. Arguing as in [14], one finds C > 0 such that ϕ y T ε,y K, S N/K K + Cε N K, ϕ y T ε,y S N/K K Cε N, (4. as ε 0. Since t t K N and since G acts freely on, it follows that E f (u G f,0 + tw ε,y c G f,0 + ( G K N SN/K K t 1 u G f,0 w 1 ε,y dx + o ( ε (N K/ (4.3 as ε 0 for a.e. y Σ. On the other hand, u G f,0 w 1 ε,y dx = ( G u G f,0ϕ 1 y T 1 ε,y dx ε (N+K/ = ( GD N,K u G f,0(x ϕ 1 y (x R (ε N + x y = ( GD N,K ε (N K/ (N+K/ dx u G f,0(x ϕ K 1 y (x 1 ( x y R ε N ψ ε N dx

8 178 MÓNICA CLAPP AND MARCO SQUASSINA where D N,K > 0 and ψ(ξ = (1 + ξ (N+K/. Since by [13, Theorem 8.15] u G f,0(x ϕy K 1 (x 1 ( x y R ε N ψ dx u G ε f,0(y ψ(ξ dξ N R N as ε 0 for a.e. y Σ it follows that for some D N,K > 0 E f (u G f,0 + tw ε,y c f,0 + µ G t 1 DN,K ( G u G f,0(y ε (N K/ + o ( ε (N K/. (4.4 as ε 0. Finally, since u G f,0 (y > 0 for a.e. y Σ, the result follows. Notice that if we new that u G f,0 > 0 in all of, a similar argument would yield ( K E f (u G f,0 + tw ε,y < c G f,0 + µ G = c G f,0 + Gx N (S K N/K min x even if the action of G on is not free (see [10, Proposition 18]. But, since all we now is that u G f,0 > 0 in some set Σ of positive measure it might very well be that, if the action is not free, no G orbit Gy with y Σ has minimum cardinality. This is why we consider free actions. Proposition 4.3 and Theorem 4.4 below are still true if instead of (H we assume that has only one G orbit type, that is, all G orbits in are G isomorphic. As a consequence of Proposition 4.3 we obtain the following result. Theorem 4.4. Assume that f 0 and that conditions (H 1 and (H hold. Then c G f,1 is achieved at a point ug f,1 (N G f which is a critical point of E f on H K 0 ( G. Proof. Since the ray u G f,0 + tw ε,y : t 0 crosses (N G f, it follows from Proposition 4.3 above that c G f,1 = inf E f < c G u (Nf G f,0 + µ G. By Corollary 3.3 the value c G f,1 is achieved by E f on (N G f. For K = 1 this result was proved by Tarantello [9] for the trivial group and extended to arbitrary groups in [10]. For the proof of Theorem 1.3 we require the following easy Lemma. Lemma 4.5. For every α > 0, f H K ( N K cg 1 1/ α = c G 1 α c G f,1. Proof. Let ε > 0 and let v (Nf G be such that E f (v < c G f,1 + ε. Let t 0 > 0 be such that u = t 0 v (N0 G. Then E 0 (u = K N u K, cg 1 and, therefore, ( N 1/ c G f,1 + ε > E f (u = E 0 (u fu dx E 0 (u f H K 0(u K E E 0 (u ( c G 1/ 1 α (E0 (u 1/ c G 1 α because the function t t (c G 1 1/ α t 1/ is increasing for t c G 1.

9 CRITICAL POLYHARMONIC PROBLEMS ON SYMMETRIC DOMAINS 179 Using the results above and arguing as in the case K = 1 [10] one can now prove Theorem 1.3. We give the details for the reader s convenience. Proof of Theorem 1.3. Assume that problem (P, 0 has no solution u N such that E 0 (u = K N u K, ( G m 1 K N (S K N/K. Then, by Theorem 4.1 and Proposition 3., c Gi 1 = µ G i = ( G i K N (S K N/K < + for i = 1,..., m 1. If c Gm 1 is not achieved by E 0 on N G m then c G m 1 = µ G m too. On the other hand if it is achieved then necessarily c G m 1 > c G m 1 1, because c G m 1 N Gm 1 1 is not achieved by E 0 on N Gm. So, since G i 1 < G i, we obtain in any case that c G i 1 > c G i 1 1 for all i =,..., m. Let α := min c G i 1 c G i 1 1 : i =,..., m > 0. By Lemma 4.5 above there is a κ > 0 independent of f, such that, if f H 1 κ, then f satisfies assumption (H 1 and c Gi 1 α c Gi f,1 for all i = 1,..., m. This, together with Proposition 4.3, implies that c G 1 f,0 < cg 1 f,1 < < cg i 1 1 c G i f,1 < cg i 1 < c G m 1 1 c G m f,1. Theorem 3.4 and Theorem 4.4 provide m + 1 different solutions, u G1 G1 f,0 (Nf +, u G i f,1 (N G i f, with E f (u G 1 f,0 = cg 1 f,0 and E f (u G i f,1 = cg i f,1 for i = 1,..., m. Since c G i+1 f,1 is the least possible energy of a G i+1 invariant solution on N f, ug i f,1 is not G i+1 invariant for i = 1,..., m 1. Proof of Corollary 1.. Since problem (P,0 is invariant under dilations, the best Sobolev constant S K is independent of the domain. It follows that the infimum c 1 1 = K N (S K N/K is not achieved by E 0 on N 1 H K 0 (. Indeed, if it were achieved at some point u 1 N 1, then extending u 1 by 0 outside of would give a minimum of E 0 on the Nehari manifold in H K (R N, that is, a solution of the problem ( K u = u K u in R N which vanishes outside of, contradicting the unique continuation property [4]. As a consequence, assertion (b of Theorem 1.3 must hold. Observe that Theorem 1.3 is still true if we assume that every G m orbit of is infinite instead of asing that G m acts freely on. The proof is similar except that, in this case, both c Gm 1 and c Gm f,1 are always achieved (cf. Theorem 4.1. In particular, the following holds. Corollary 4.6. If every G orbit of is infinite, then there exists a κ > 0 with the property that, for every f 0 which is G invariant and such that f H K < κ, problem (P, f has at least three solutions one of which is G invariant and one of which is not. Remar 4.7. These results can be extended to eigenvalue problems ( K u = λu + u K u + f in, ( j u ν = 0, j = 0,..., K 1, (P,λ,f provided that 0 λ < λ 1, where λ 1 is the first eigenvalue of ( K with Dirichlet boundary conditions (see [11]. For f = 0 and = B R (0 this problem has been

10 180 MÓNICA CLAPP AND MARCO SQUASSINA studied by many authors (see e.g. [4, 5], mainly in dealing with the so called Pucci Serrin conjecture which says that the dimensions for which there is 0 < Λ < λ 1 such that (P,λ,0 admits no solution for λ < Λ are precisely K + 1,..., K + j,..., 4K 1. Even though a full proof of the Pucci Serrin conjecture seems out of reach, a wea version (for positive solutions has been given by Grunau in [16]. 5. Proof of Proposition 3.. In this section we prove Proposition 3.. Let us first prove the following result. Lemma 5.1. Assume that (v H K (R N is such that v 0. Then as +. h C c (B : hv K, = (v, h v K, + o(1 Proof. We consider the case K = q. The case K = q + 1 can be treated in a similar fashion. Setting for each 1 q q α v A = c α h j,α x α 1 j 1 x α q x α 1 j q j 1 x α, q j q it results Moreover, setting B = j 1,..., jq =1 α 0,1, q, α 0 q (hv = h q v + ha q v + A. (5.1 q j 1,..., jq =1 α 0,1, q, α 0 c j,α x α 1 j 1 q α v x αq j q x α 1 j 1 α h x αq j q, it follows that q v q (h v = h q v + B q v. (5. By combining (5.1 and (5. yields q (hv = q v q (h v + A ( B ha q v. (5.3 Since D j v 0 in L loc (RN for j = 0,..., q 1, being α 0, it results A 0, B ha 0 in L (supt h as +. In particular equation (5.3 yields the assertion. Proof of Theorem 3.. Let (u H0 K ( G be a G PS sequence for E 0 such that E 0 (u c < µ G. We wish to show that a subsequence of (u converges strongly in H0 K (. Since PS sequences for E 0 are bounded in H0 K (, u K, = N K E 0(u N K K DE 0(u u N K c as +. Thus c 0. We may assume that u u wealy in H0 K ( and that u u a.e. in. It is easy to see that DE 0 (u = 0 and that v := u u is a PS sequence for E 0 such that v 0 wealy in H0 K ( and E 0 (v = E 0 (u E 0 (u + o(1 = c E 0 (u + o(1. Let d := c E 0 (u. Thus 0 d c. If d = 0 then u u strongly in H K 0 ( and we are done.

11 CRITICAL POLYHARMONIC PROBLEMS ON SYMMETRIC DOMAINS 181 So let us assume that d > 0. Since (v is a PS sequence for E 0, it is bounded in H K 0 (. Hence v = N K E 0(v N K DE 0(v v N K d > 0. Let δ = min Nd K, ( S K N K and denote by B(x, ϱ the closed ball in R N with center x and radius ϱ. The Levy concentration function Φ (ϱ = sup v K dξ x R N B(x,ϱ satisfies Φ (0 = 0 and Φ (+ > δ for large enough. Hence we may choose y and ε > 0 such that v dξ = v dξ = δ > 0. (5.4 sup x R N B(x,ε B(y,ε Observe that, being bounded, the sequence (ε is bounded. We define v (z = ε N K v (ε z + y H K (R N. It results v K, = v K, and v K = v K. In particular, (v is a bounded sequence in H K (R N and, up to a subsequence, v v wealy in H K (R N, D j v D j v a.e. in R N, j = 0,..., K 1, D j v D j v in L loc(r N, j = 0,..., K 1, as +. We wish to show that v 0. Assume by contradiction that v = 0. It follows from Lemma 5.1 and equality (5.4 above, and from Sobolev and Hölder inequalities that, as +, S K hv hv K, = ( v, h v K, + o(1 ( = h v K dx + DE 0 (v h ( y v + o(1 R ε N ( K/N ( /K v dx hv dx + o(1 B(z,1 R N δ K/N hv + o(1 S K hv + o(1 for each z R N and h Cc (B(z, 1. Hence v 0 in L K loc (RN. This is a contradiction to (5.4. Therefore, v 0. Since is bounded and v 0 in H0 K (, up to a subsequence, y y and ε 0. If (ε 1 dist(y, is bounded, we may assume that lim + ε 1 dist(y, = b. It is then easy to verify that, up to a rotation of R N, the sets = z R N : ε z + y

12 18 MÓNICA CLAPP AND MARCO SQUASSINA satisfy + n=1 ( + = H N := =n Hence, v is a nontrivial solution of the equation On the other hand, if (z 1,..., z N R N : z N b. ( K u = u u in H N. ε 1 dist(y, + then v is a nontrivial solution of the equation ( K u = u u in R N. In both cases we obtain that E 0 (v = K N v K, K N (S K N/K. Let Γ = g G : gy = y be the isotropy subgroup of y. Thus, the G orbit Gy of y is G homeomorphic to the homogeneous space of right cosets G/Γ [1]. Let S = g 1,..., g m be a finite subset of G whose elements represent pairwise distinct cosets [g 1 ],..., [g m ] in G/Γ. Since y y and ε 0, it follows that ε 1 g iy g j y + for each i j. Hence, since v is G invariant, it results m ( v ε K N vg 1 gi y i i=1 ε m = ε N K v (ε + g 1 y i=1 vg 1 i ( K, + g 1y g i y = v g1 1 vg1 1 ( vg 1 i + g 1y g i y ε i 1 K, = v g1 1 ( vg 1 i + g 1y g i y vg 1 1 ε K, + o(1 i 1 K, = v ( ε K N vg 1 gi y i v ε K, + o(1 i 1 K, and, inductively, v K, = v m i=1 ε K N m(s K N/K + o(1 vg 1 i ( gi y ε K, ε K, + m v K, + o(1 as + for all m G/Γ. Since v K, is bounded it follows that Gy = G/Γ is finite. If c < µ G, then v K, N K d < N K µg. It follows that ( Gy < Gx. min x This is a contradiction. Hence E 0 satisfies (P S G c for each c < µ G. For a complete description of all G PS sequences for K = 1 we refer to [9].

13 CRITICAL POLYHARMONIC PROBLEMS ON SYMMETRIC DOMAINS Proof of Lemma 4.. By [8, Lemma 4] there exists C N,K > 0 such that for all α, β R α + β K α K β K αβ ( α K + β K α β 1 C N,K α K 1 β for α β, for α β if N 6K, and for all α, β R α + β α β αβ ( α + β C N,K ( α β + α β provided that N K + 1,..., 6K 1. (6.1 In the following, it is understood that u G f,0 = 0 outside. Case N 6K ; by (6.1, to prove the assertion on R ε it suffices to estimate the right hand side of the inequality R ε C N,K u G f,0 (tw ε,y K 1 dx. u G f,0 tw ε,y Splitting the integration into x y < ϱ/ and ϱ/ x y, one gets RN R ε C N,K,t ( Gε γ u G (N K/ f,0 (x 1+γ 1 x y dx γ (N K +C N,K,t ( G ε (N+K/ where γ 1, γ > 0 satisfy γ 1 + γ = 1 and γ < N/(N K. Note that u G f,0 1+γ1 L 1 (R N, x γ(n K L 1 (R N since 1 + γ 1 < and γ (N K < N. In particular, by [13, p.3], u G 1 f,0(x 1+γ1 dx < + R x y N γ (N K for a.e. y Σ (as convolution of two L 1 functions. It follows that R ε = O ( ε Nϑ/ for each ϑ < 1. In a similar fashion, estimating the right hand side of S ε C N,K u G f,0 K 1 tw ε,y dx u G f,0 <twε,y yields S ε = O ( ε Nϑ/ for each ϑ < 1 as ε 0. Case N K + 1,..., 6K 1 ; it results N = K + j, We distinguish three cases: j = 1,..., K 1, j = K, j = K + 1,..., 4K 1. = 4K + j, j = 1,..., 4K 1. j

14 184 MÓNICA CLAPP AND MARCO SQUASSINA Case j = 1,..., K 1 ; by (6. for each j it results R ε C N,K,t ε j ( G u G f,0(x 4K/j 1 dx, R x y j N RN S ε C N,K,t ε j ( G u Gf,0(x ε K j (ε + x y dx. K The first estimate gives R ε = O ( ε j since the integral, again by [13, p.3], is finite for a.e. y Σ. The second estimate gives R ε = O ( ε j as well, since the ernels ε K j x (ε + x y K = 1 1 ε N ( 1 + x y K correspond to that of a mollifier (up to a constant and thus RN u Gf,0(x ε K j (ε + x y K dx cug f,0(y as ε 0 for some c > 0 and a.e. y Σ (see [13, Theorem 8.15]. Case j = K ; by (6. it results RN R ε, S ε C N,K,t ( G u Gf,0(x ε K (ε + x y dx. K If γ 1, γ > 0 satisfy γ 1 + γ = K, then (ε + x y K ε γ1 x y, γ which implies R ε, S ε C N,K,t ε K γ 1 ( G u G f,0(x 1 dx. R x y N γ This yields R ε = S ε = O(ε Kϑ for all ϑ < 1. Case j = K + 1,..., 4K 1 ; by (6., one has RN R ε C N,K,t ε K ( G u Gf,0(x 4K/j ε j K (ε + x y j dx, S ε C N,K,t ε K ( G u G f,0(x 1 dx. R x y 4K N Taing into account that u G f,0(x 1 dx < + R x y 4K N a.e. y Σ and that the ernels ε j K x (ε + x y j = 1 1 ε N ( 1 + x y j correspond to that of a mollifier (up to a constant, arguing as before one gets R ε = S ε = O ( ε K. Therefore, putting the previous conclusions together, we have ε ε

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