Remarks on the Clark theorem Guosheng Jiang1, Kazunaga Tanaka2, Chengxiang Zhang3 7 1 School of Mathematical Sciences, Capital Normal University 1 0 Beijing 100048, China 2 2 Department of Mathematics, School of Science and Engineering n a Waseda University, 3-4-1 Ohkubo, Shinjuku-ku, Tokyo 169-8555, Japan J 3 Chern Institute of Mathematics and LPMC, Nankai University 3 1 Tianjin 300071, China ] P A . Abstract. The Clark theorem is important in critical point theory. For a class of h t a even functionals it ensures the existence of infinitely many negative critical values m converging to 0 and it has important applications to sublinear elliptic problems. [ We study the convergence of the corresponding critical points and we give a 1 v characterization of accumulation points of critical points together with examples, 0 7 in which critical points with negative critical values converges to non-zero critical 5 point. Our results improve the abstract results in Kajikiya [Ka1] and Liu-Wang 3 0 [LW]. . 1 0 7 1 : 1. Introduction and main results v i X The Clark theorem is one of the most important results in critical point theory (Clark r a [Cl], see also Heinz [H]). It was successfully applied to sublinear elliptic problems with odd symmetry and the existence of infinitely many solutions which accumulate to 0 was shown. To state the Clark theorem, we need some terminologies: let (X,k·k ) be a Banach X space and I ∈ C1(X,R). (i) For c ∈ R we say that I(u) satisfies the (PS) condition if any sequence (u )∞ ⊂ X c j j=1 with I(uj) → c, kI′(uj)kX∗ → 0 has a convergent subsequence. (ii) Let E be the family of sets A ⊂ X \ {0} such that A is closed and symmetric with respect to 0. For A ∈ E, the genus γ(A) is introduced by Krasnosel’skii [Kr] (c.f. Coffman [Co], Rabinowitz [R]) as the smallest integer n such that there exists an odd 1 continuous map ζ ∈ C(A,Rn\{0}). When there does not exist such a map, we set γ(A) = ∞. See Rabinowitz [R] for fundamental properties of the genus. Now we give a variant of the Clark theorem due to Heinz [H]. Theorem 1.1 (Heinz [H]). Let (X,k·k ) be a Banach space and suppose that I(u) ∈ X C1(X,R) satisfies the following conditions: (A1) I(0) = 0. I(u) is even in u and bounded from below; (A2) I(u) satisfies (PS) for all c < 0; c (A3) For any k ∈ N, there exists A ∈ E such that γ(A) ≥ k and supI(u) < 0. u∈A Then I(u) has a sequence (c )∞ of critical values of I(u) such that j j=1 c < 0 for all j ∈ N, j c → 0 as j → ∞. j Here c = inf supI(u). (1.1) j A∈E,γ(A)≥ju∈A Remark 1.2. In [H], it was assumed that (A2’) I(u) satisfies (PS) for all c ∈ R. c From its proof, we can easily see that (PS) just for c < 0 is enough for the existence of c critical values. By Theorem 1.1, there exists a sequence (u )∞ of critical points of I(u) such that j j=1 I(u ) = c → −0 as j → ∞. Thus it is natural to ask whether u → 0 holds or not. More j j j generally, the existence of a sequence of non-zero critical points (u )∞ (or critical points j j=1 withnegativecriticalvalues)satisfyingu → 0isofinterest. Thisquestionhasbeenstudied j by Kajikiya [Ka1] and Liu-Wang [LW] together with applications to sublinear elliptic problems. We note that Liu-Wang [LW] also studied periodic solutions of Hamiltonian systems. More precisely, under the assumptions of (A1), (A2’) and (A3), Kajikiya [Ka1] showed either (C1) There exists a sequence (u )∞ such that j j=1 I′(u ) = 0, I(u ) < 0 and u → 0 as j → ∞. j j j 2 or (C2) There exists two sequences (u )∞ and (v )∞ such that j j=1 j j=1 I′(u ) = 0, I(u ) = 0, u 6= 0 and u → 0 as j → ∞ j j j j and I′(v ) = 0, I(v ) < 0 and v converges to a non-zero limit. j j j holds. Liu-Wang [LW] assumed (A1), (A2’) and the following (A3’), which is stronger than (A3), (A3’) For any k ∈ N there exists a k-dimensional subspace Xk of X and ρ > 0 such that k sup{I(u); u ∈ Xk, kuk = ρ } < 0 X k and they showed either (C1) above or (C3) There exists r > 0 such that for any 0 < a < r there exists a critical point u such that kuk = a and I(u) = 0. X In what follows, we denote by K the connected component of K = {u ∈ X; I′(u) = 0 0 0, I(u) = 0} including 0. b Remark 1.3. From their proof of their main result, Liu-Wang [LW] claimed that (C3) can be strengthened as (C3’) There exists r > 0 such that K ∩{u ∈ X; kuk = r} =6 ∅. 0 X b The aim of this paper is to show the following Theorem 1.4 and Theorem 1.6; In Theorem 1.4, we give a new characterization of accumulation points of critical points with negative critical values and unifies the results of Kajikiya and Liu-Wang. On the other hand, in Theorem 1.6 we answer a natural question concerning (C1), which is stated below. We believe that Theorems 1.4 and 1.6 give us a better understanding of the Clark theorem. First we give our Theorem 1.4. 3 Theorem 1.4. Let (X,k·k ) be a Banach space and suppose I ∈ C1(X,R) satisfies (A1), X (A3) and (A2”) I(u) satisfies (PS) for all c ≤ 0. c Then there exists a sequence (u )∞ ⊂ X of critical points of I(u) such that j j=1 I(u ) < 0 for all j ∈ N, (1.2) j I(u ) → 0 as j → ∞, (1.3) j dist(u ,K ) ≡ inf{ku −vk ; v ∈ K } → 0 as j → ∞. j 0 j X 0 (cid:16) (cid:17) b b As an immediate corollary to our Theorem 1.4, we have Corollary 1.5. Under the assumptions of Theorem 1.4, assume that (C1) does not take place. Then K 6= {0}. 0 Since K 6= {0b} implies (C2) and (C3), Corollary 1.5 covers the results of Kajikiya [Ka1] 0 and Liu-Wang [LW]. b Next we study a question concerning (C1). In many applications of the Clark theorem to sublinear elliptic equations, there exist sequences (u )∞ of solutions with (1.2), (1.3) j j=1 and u → 0 as j → ∞. (1.4) j So (C1) may be expected under the assumption of Theorem 1.4 and a natural question is to ask whether (C1) always takes place under the assumption of Theorem 1.4 or not. Our Theorem 1.6 answers this question negatively. Theorem 1.6. Conditions (A1), (A2”), (A3’) do not imply (C1). In particular, under the assumptions of Theorem 1.4, (C1) does not hold in general. Remark 1.7. An example related to our Theorem 1.6 was given in Example 1.3 of [Ka1] (c.f. [Ka2]). It shows that there exists a functional I ∈ C1(X,R) which satisfies (A1), (A2”), (A3) and the following property: There exists an r > 0 independent of j such that 0 I′(u) = 0 and I(u) = c imply kuk ≥ r . j X 0 Here c is given in (1.1) and c satisfies c < 0 and c → 0 as j → ∞. Thus a special j j j j case of (C1) does not hold for I. In Section 3.1 we give another example I ∈ C1(ℓ2,R) 4 for which we give an explicit description of all critical points of I(u) and no critical points with negative critical values do not exist in a neighborhood of 0. Especially (C1) does not hold for our I(u). Our example also shows a typical situation of our Theorem 1.4. Finally we remark that in our Theorem 1.4, (A2”), especially (PS) is important. In 0 fact, we have Theorem 1.8. Under the assumptions of Theorem 1.1, especially without (PS) , the 0 conclusion of Theorem 1.4 does not hold in general. In the following Section 2, we give a proof to our Theorem 1.4. Here estimates of I′(u) play important roles. In Section 3, we give two examples which show Theorems 1.6 and 1.8. 2. Proof of Theorem 1.4 In what follows, we use the following notation for δ > 0 B (u) = {x ∈ X; kx−uk < δ} for u ∈ X, δ X N (D) = {x ∈ X; dist(x,D) < δ} for D ⊂ X, δ where dist(x,D) = inf kx−yk . X y∈D We note that N (D) = B (y). δ y∈D δ 2.1. A fundamental fSact from topology To show our Theorem 1.4, we need the following characterization of connected components of compact sets. Lemma 2.1. Let D ⊂ X be a compact set such that 0 ∈ D. For δ > 0, let O be the δ connected component of N (D) including 0. Then we have δ O = D, δ δ>0 \ b where D is the connected component of D including 0. Proof. By the definition of O and D, it is clear that D ⊂ O for all δ > 0. Thus b δ δ b b D ⊂ O ⊂ O . δ δ δ>0 δ>0 \ \ b 5 By the compactness of D, we also have D = N (D). δ>0 δ We set T A = O ⊂ D. δ δ>0 \ It suffices to show that A is connected. For δ > 0 we also set D = O ∩D. Then we have δ δ A = D , (2.1) δ δ>0 \ O = B (u), (2.2) δ δ u[∈Dδ δ < δ implies D ⊂ D . (2.3) 1 2 δ1 δ2 Arguing indirectly, we suppose that A is not connected. Then there exist two compact sets A , A ⊂ X such that A ∩A = ∅, A ∪A = A. We set 1 2 1 2 1 2 1 β = dist(A ,A ) > 0. 1 2 2 For each δ > 0, since O is a connected set including A ∪A , we have δ 1 2 O ∩{x ∈ X; dist(x,A ) = β} =6 ∅. δ 1 By (2.2), we can see that for any x ∈ O there exists u ∈ D such that x ∈ B (u). Thus δ δ δ D ∩{x ∈ X; dist(x,A ) ∈ [β −δ,β +δ]} =6 ∅. δ 1 Since D ∩{x ∈ X; dist(x,A ) ∈ [β −δ,β +δ]} has the finite intersection property δ 1 δ>0 by (2.3(cid:16)), we have (cid:17) A∩{x ∈ X; dist(x,A ) = β} 1 = D ∩{x ∈ X; dist(x,A ) ∈ [β −δ,β +δ]} 6= ∅, δ 1 δ\>0(cid:16) (cid:17) which contradicts with the choice of β > 0. Thus A is a connected set. 2.2. A gradient estimate Suppose that I(u) ∈ C1(X,R) satisfies the assumptions of Theorem 1.4. We use the following notation: K = {u ∈ X; I′(u) = 0}, K = {u ∈ K; I(u) = 0}, 0 K = {u ∈ X; there exists (v )∞ ⊂ K such that − j j=1 I(v ) < 0 for all j and I(v ) → 0, v → u as j → ∞}. j j j 6 By (PS) , we have K ⊂ K . We also use notation for a < b 0 − 0 [I ≤ a] = {u ∈ X; I(u) ≤ a}, [a ≤ I ≤ b] = {u ∈ X; a ≤ I(u) ≤ b}. It is clear that 0 ∈ K . We denote by K the connected component of K including 0. To 0 0 0 show our Theorem 1.4 it suffices to prove b K ∩K 6= ∅. (2.4) − 0 For δ > 0, let O be the connected componenbt of N (K ) including 0. By Lemma 2.2, we δ δ 0 have K = O . 0 δ δ>0 \ b Thus to prove (2.4) it suffices to show O ∩K 6= ∅ for all δ > 0. δ − We argue indirectly and suppose for some δ > 0 0 O ∩K = ∅. (2.5) δ0 − Under the assumption (2.5), we set K = O ∩K , K = K \O . 0,i δ0 0 0,e 0 δ0 Then K and K are disjoint compact sets such that K = K ∪K and 0,i 0,e 0 0,i 0,e dist(K ,K ) ≥ 2δ , (2.6) 0,i 0,e 0 K ⊂ K . (2.7) − 0,e We note that (2.7) follows from (2.5). First we have Lemma 2.2. Assume (2.5). Then for any r > 0 there exist ρ > 0 and ν > 0 such that [−ρ ≤ I ≤ 0]∩K ⊂ N (K ), (2.8) r 0 [−ρ ≤ I < 0]∩K ⊂ N (K ) ⊂ N (K ), (2.9) r − r 0,e kI′(u)kX∗ ≥ ν for all u ∈ [−ρ ≤ I ≤ 0]\Nr(K0). (2.10) 7 Moreover for any ε ∈ (0,ρ) there exists ν ∈ (0,ν] such that ε kI′(u)kX∗ ≥ νε for all u ∈ [−ρ ≤ I ≤ −ε]\Nr(K0,e). (2.11) Proof. Using (PS) and the definition of K , we can check (2.8)–(2.10) easily for small 0 − ρ and ν > 0. We show (2.11). Suppose that for r, ρ, ν > 0, (2.8)–(2.10) hold. If (2.11) does not hold, we can find ε ∈ (0,ρ) and a sequence (u )∞ such that j j=1 I(u ) ∈ [−ρ,−ε], kI′(u )k → 0, u 6∈ N (K ), j j j r 0,e By (PS), we can extract a subsequence (u ) such that u → u for some u ∈ [−ρ ≤ I ≤ jk jk 0 0 −ε]∩K. By (2.9), we have u 6∈ N (K ) for large k, which is a contradiction. Thus we jk r 0,e have (2.11). 2.3. Deformation argument The aim of this section is the following Proposition 2.3. Assume (2.5). Then for any r ∈ (0,δ /3] there exists d > 0 with the 0 following property: for any ε ∈ (0,d/2] there exists an odd continuous map η : [I < 0] → ε [I < 0] such that η ([I ≤ −ε]) ⊂ [I ≤ −d]∪N (K ). ε 3r 0,e Proof. First we define an ODE in X to define η . ε For a given r ∈ (0,δ /3], let ρ, ν > 0 be constants given in Lemma 2.2. We set 0 1 d = min{ρ,νr} > 0. (2.12) 3 Then again by Lemma 2.2, for any given ε ∈ (0,d] there exists ν > 0 with the property ε (2.11). By (2.9), we have I′(u) 6= 0 for all u ∈ [−3d ≤ I < 0]\N (K ). Thus there exists a r 0,e locally Lipschitz odd vector field V(u) : [−3d ≤ I < 0]\N (K ) → X such that r 0,e kV(x)k ≤ 1 for all u ∈ [−3d ≤ I < 0]\N (K ), X r 0,e I′(u)V(u) > 0 for all u ∈ [−3d ≤ I < 0]\N (K ), (2.13) r 0,e ν I′(u)V(u) ≥ for all u ∈ [−3d ≤ I < 0]\N (K ), (2.14) r 0 2 ν I′(u)V(u) ≥ ε for all u ∈ [−3d ≤ I ≤ −ε]\N (K ). (2.15) r 0,e 2 Let φ (u), φ (u) : X → [0,1] be even Lipschitz continuous functions such that 1 2 1 for u ∈ [−d ≤ I], 1 for u ∈ X \N (K ), φ (u) = φ (u) = 2r 0,e 1 0 for u ∈ [I ≤ −2d], 2 0 for u ∈ N (K ). r 0,e (cid:26) (cid:26) 8 We set V(u) = φ (u)φ (u)V(u) 1 2 and we note that V(u) is well-defined on [I < 0]. For u ∈ [I < 0] we consider e dη e = −V(η), dt  η(0,u) = u.  e We have for all (t,u)  η(t,−u) = −η(t,u), (2.16) dη k k ≤ 1. (2.17) X dt Since d dη I(η(t,u)) = I′(η(t,u)) = −I′(η)V(η), dt dt it follows from (2.13)–(2.15) that e d I(η(t,u)) ≤ 0 if η(t,u) ∈ [I < 0], (2.18) dt d ν I(η(t,u)) ≤ − if η(t,u) ∈ [−d ≤ I < 0]\N (K ), (2.19) 2r 0 dt 2 d ν ε I(η(t,u)) ≤ − if η(t,u) ∈ [−d ≤ I ≤ −ε]\N (K ). (2.20) 2r 0,e dt 2 By (2.17) and (2.18), we note that for any u ∈ [I < 0], η(t,u) exists globally, that is, η(t,u) : [0,∞)×[I < 0] → [I < 0] is well-defined. For a latter use, we note that N (K )\N (K ) ⊂ X \N (K ). 3r 0,e 2r 0,e 2r 0 Thus by (2.19) d ν I(η(t,u)) ≤ − if η(t,u) ∈ [−d ≤ I < 0]∩(N (K )\N (K )). (2.21) 3r 0,e 2r 0,e dt 2 Next we claim that Claim. Let T = 2d. Then ε νε η(T ,u) ∈ [I ≤ −d]∪N (K ) for any u ∈ [I ≤ −ε]. (2.22) ε 3r 0,e To prove (2.22), it suffices to show that if u ∈ [I ≤ −ε] satisfies η(T ,u) 6∈ [I ≤ −d], (2.23) ε 9 then η(T ,u) ∈ N (K ). (2.24) ε 3r 0,e We note that under the condition (2.23) η(t,u) ∈ [−d ≤ I ≤ −ε] for all t ∈ [0,T ]. (2.25) ε Step 1: Assume (2.23), i.e., (2.25). Then η([0,T ],u)∩N (K ) 6= ∅. (2.26) ε 2r 0,e In fact, if (2.26) does not hold, it follows from (2.20) that d ν ε I(η(s,u)) ≤ − for all s ∈ [0,T ]. ε ds 2 Thus, by the definition of T , ε Tε d I(η(T ,u)) ≤ I(u)+ I(η(s,u))ds ε ds Z0 ν ε ≤ −ε− T ε 2 < −d, which is in contradiction with (2.23). Step 2: Assume (2.23), i.e., (2.25). Then (2.24) holds. Assume (2.24) does not hold. Then η(T ,u) 6∈ N (K ) and by (2.26) the orbit η(t,u) ε 3r 0,e enters in N (K ) for some t ∈ [0,T ]. Thus there exists an interval [t ,t ] ⊂ [0,T ] such 2r 0,e ε 0 1 ε that η(t ,u) ∈ ∂N (K ), 0 2r 0,e η(t ,u) ∈ ∂N (K ), 1 3r 0,e η(t,u) ∈ N (K )\N (K ) for all t ∈ [t ,t ]. 3r 0,e 2r 0,e 0 1 By (2.17), we have t1 d r ≤ kη(t ,u)−η(t ,u)k ≤ k η(s,u)k ds 1 0 X X ds Zt0 ≤ t −t . 1 0 10

Remarks on the Clark theorem

Similar Remarks on the Clark theorem