GROUND STATES FOR FRACTIONAL KIRCHHOFF EQUATIONS WITH CRITICAL NONLINEARITY IN LOW DIMENSION

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1 GROUND STATES FOR FRACTIONAL KIRCHHOFF EQUATIONS WITH CRITICAL NONLINEARITY IN LOW DIMENSION ZHISU LIU, MARCO SQUASSINA, AND JIANJUN ZHANG Abstract. We study the existence of ground states to a nonlinear fractional Kirchhoff equation with an external potential V. Under suitable assumptions on V, using the monotonicity trick and the profile decomposition, we prove the existence of ground states. In particular, the nonlinearity does not satisfy the Ambrosetti-Rabinowitz type condition or monotonicity assumptions. Contents. Introduction and results.. Overview.. Main results 3.3. Main difficulties 4. Variational setting 5 3. The perturbed functional 6 4. Upper estimate of c λ and limit problems An energy estimate The limit problem 5. Behaviour of Palais-Smale sequences 5.. Splitting lemmas Profile decomposition 6 6. Proof of the main results Nontrivial critical points of I λ Completion of the proof References 4. Introduction and results.. Overview. In this paper we are concerned with the existence of positive ground state solutions to the following nonlinear fractional Kirchhoff equation ( a + b ( u dx ( u + V (xu = f(u in, (K R N u H (, u > 0 in, where a, b are positive constants, (0, and N >. The operator ( is the fractional Laplacian defined as F ( ξ F (u, where F denotes the Fourier transform on. When a = and b = 0, then (K reduces to the following fractional Schrödinger equation (. ( u + V (xu = f(u in, 000 Mathematics Subject Classification. 35Q55, 35Q5, 53C35. Key words and phrases. Fractional Kirchhoff type problems, ground state solutions, profile decomposition. Z. Liu is supported by the NSFC (667. M. Squassina is member of the Gruppo Nazionale per l Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA of the Istituto Nazionale di Alta Matematica (INdAM. J. J. Zhang was partially supported by the Science Foundation of Chongqing Jiaotong University (5JDKJC-B033.

2 Z. S. LIU, M. SQUASSINA, AND J. J. ZHANG which has been proposed by Laskin [8] in fractional quantum mechanics as a result of extending the Feynman integrals from the Brownian like to the Lévy like quantum mechanical paths. For such a class of fractional and nonlocal problems, Caffarelli and Silvestre [8] expressed ( as a Dirichlet-Neumann map for a certain local elliptic boundary value problem on the half-space. This method is a valid tool to deal with equations involving fractional operators to get regularity and handle variational methods. We refer the readers to [4, 3] and to the references therein. Investigated first in [,] via variational methods, there has been a lot of interest in the study of the existence and multiplicity of solutions for (. when V and f satisfy general conditions. We cite [0, 9, 30] with no attempts to provide a complete list of references. If =, then problem (K formally reduces to the well-known Kirchhoff equation ( (. a + b u dx u + V (xu = f(u in, related to the stationary analogue of the Kirchhoff-Schrödinger type equation u ( t a + b u dx u = f(t, x, u, Ω where Ω is a bounded domain in, u denotes the displacement, f is the external force, b is the initial tension and a is related to the intrinsic properties of the string. Equations of this type were first proposed by Kirchhoff [7] in 883 to describe the transversal oscillations of a stretched string. Besides, we also point out that such nonlocal problems appear in other fields like biological systems, where u describes a process depending on the average of itself. We refer readers to Chipot and Lovat [9], Alves et al. []. However, the solvability of the Kirchhoff type equations has been well studied in a general dimension by various authors only after J.-L. Lions [] introduced an abstract framework to such problems. For more recent results concerning Kirchhoff-type equations we refer e.g. to [4, 5, 4, 6, 35, 37]. In [0], by using a monotonicity trick and a global compactness lemma, Li and Ye proved that for f(u = u p u and p (3, N/(N, problem (. has a positive ground state. Subsequently, Liu and Guo [3] extended the above result to p (, N/(N. Fiscella and Valdinoci, in [3], proposed the following stationary Kirchhoff variational equation with critical growth ( M ( u dx ( u = λf(x, u + u u in Ω, (.3 R N u = 0 in \ Ω, which models nonlocal aspects of the tension arising from measurements of the fractional length of the string. They obtained the existence of non-negative solutions when M and f are continuous functions satisfying suitable assumptions. Autuori, Fiscella and Pucci [3] considered the existence and the asymptotic behavior of non-negative solutions of (.3. Pucci and Saldi [7] established multiplicity of nontrivial solutions. Via a three critical points theorem, Nyamoradi [5] studied the subcritical case of (.3 and obtained three solutions. See also [8, 36] for related results. To the best of our knowledge, there are few papers in the literature on fractional Kirchhoff equations in. Recently, Ambrosio and Isernia [] considered the fractional Kirchhoff problem ( (.4 a + b ( u dx ( u = f(u in, where f is an odd subcritical nonlinearity satisfying the well known Berestycki-Lions assumptions [6]. By minimax arguments, the authors establish a multiplicity result in the radial space H rad (RN

3 GROUND STATES FOR FRACTIONAL CRITICAL KIRCHHOFF EQUATIONS 3 when the parameter b is sufficiently small. As in [0], Teng [33] also searched for ground state solutions for the fractional Schrödinger-Poisson system in R 3 with critical growth { ( u + V (xu + φu = µ u q u + u u in R 3, ( t φ = u in R 3. We point out that, in [0,33] the corresponding limit problems play an important rǒle. In order to get the existence of ground state solutions of the limit problems, the authors used a constrained minimization on a manifold M obtained by combining the Nehari and Pohožaev manifolds... Main results. Motivated by the works above, in this paper we aim to study the existence of positive ground state solutions to the fractional Kirchhoff equation with the Berestycki-Lions type conditions of critical type, firstly introduced in [38].... Assumptions on V. On the external potential we assume the following: (V V C (, R and, setting W (x := max{x V (x, 0}, we assume ( u dx W N < as R L (R, S := inf N N ( u D, ( /, u 0 u := N N ; dx (V there exists V R such that V (x lim V (y = V, for all x ; y (V 3 the operator a( + V (x : H ( H ( satisfies ( a ( u + V (xu dx R inf N > 0. u H ( u dx u 0... Assumptions on f. We assume that f(t = 0 for all t 0 and (f f C (R +, R and lim t = 0; f(t (f lim = ; t t (f 3 there are D > 0 and < q < such that f(t t + Dt q for any t 0. Now we state our first result. t 0 f(t Theorem.. Assume (V -(V 3, (f -(f 3 and N = with (, or N = 3 with ( 3 4,. (i If q (,, there is D > 0 such that, for D D, (K admits a positive ground state solution. (ii If q ( 4 N,, for any D > 0, (K admits a positive ground state solution. We point out that without any symmetry assumption on V, the ground state solution obtained above maybe is not radially symmetric. In the following, we impose a monotonicity assumption of V and show that (K admits a radially symmetric solution. Assume now that V is radially symmetric and increasing, that is (V 4 for all x, y : x y V (x V (y. Theorem.. Under the assumptions of Theorem. and (V 4, (K admits a radially symmetric positive solution at the global (unrestricted to radial paths mountain pass energy level.

4 4 Z. S. LIU, M. SQUASSINA, AND J. J. ZHANG As a main tool to prove Theorem. we shall give the profile decomposition of the Palais-Smale sequences by which we can derive some compactness and get a positive ground states for (K. The main tool for the proof of Theorem. is a symmetric version of the monotonicity trick [3]. We recall that Zhang and Zou [39] studied the critical case for Berestycki-Lions theorem of the Schrödinger equation u + V (xu = f(u. They obtained positive ground state solutions when V satisfies similar assumptions as (V -(V 3, f satisfies (f -(f 3 and 4 (f 4 f (t C( + t N, for t > 0 and some C > 0. We should mention that in the present paper, (f 4 is removed..3. Main difficulties. We mention the difficulties and the idea in proving Theorem.. Firstly, without the Ambrosetti Rabinowitz condition, it is difficult to get the boundedness of Palais-Smale sequences. In order to overcome this difficulty, inspired by [0], we will use the monotonicity trick developed by Jeanjean [6], introduce a family of functionals I λ and obtain a bounded (PS cλ sequence for I λ for almost all λ in an interval J, where c λ is given in Section 3. Secondly, by the presence of the Kirchhoff term, one obstacle arises in getting the compactness of I λ, even in the subcritical case. Precisely, this does not hold in general: for any ϕ C0 (RN, ( un dx ( un ( ϕdx ( u dx ( u( ϕdx, where {u n } n N is a (PS-sequence of I λ satisfying u n u in H (. Then, even in the subcritical case, it is not clear that weak limits are critical points of I λ. In [], for (.4 the compactness was recovered by restricting I λ to the radial space Hrad (RN, which is compactly embedded L s ( for all s (,. For the related works in the bounded domains, see e.g. [3, 7, 36]. In the present paper, we do not impose any symmetry and just consider (K in H (. So the arguments mentioned above cannot be applied. Inspired by [0], in place of I λ, we consider a family of related functionals J λ, whose corresponding problem is a non Kirchhoff equation. Thirdly, the critical exponent makes the problem rather tough. The (PS-condition does not hold in general and to overcome this difficulty, we show that the mountain pass level c λ is strictly less than some critical level c λ. For u + V (xu = λf(u with critical growth, if S is the best constant of D, ( L (, one can show that [7] c λ = N S N λ N. For ( a + b R 3 u dx u + V (xu = λf(u in R 3 involving critical growth [9, ] c λ = ab 4λ S3 + [b S λaS] 4λ + b3 S 6 4λ. However, for fractional Kirchhoff equations, to give the exact value of c λ is complicated, since one cannot solve precisely a fractional order algebra equation. A careful analysis is needed at this stage. With an estimate of c λ, inspired by [39], we establish a profile decomposition of the Palais-Smale sequence {u n } n N (Lemma 5.4 related to J λ. Thanks to this result, for almost every λ [/, ] we obtain a nontrivial critical point u λ of I λ at the level c λ. Finally, choosing a sequence λ n [/, ] with λ n, thanks to the Pohožaev identity we obtain a bounded (P S c -sequence of the original functional I. Using the idea above again, we obtain a nontrivial solution of problem (K. Throughout this paper, C will denote a generic positive constant. The paper is organized as follows. In Section, the variational setting and some preliminary lemmas are presented.

5 GROUND STATES FOR FRACTIONAL CRITICAL KIRCHHOFF EQUATIONS 5 In Section 3, we consider a perturbation of the original problem (K. Then using the monotonicity trick developed by Jeanjean, we obtain the bounded (PS cλ -sequence {u n } n N for almost all λ. In Section 4, an upper estimate of the mountain pass value is obtained and the limit problem is discussed. In Section 5, we give the profile decomposition of {u n } n N. In Section 6, Theorem. and. are finally proved.. Variational setting In this section we outline the variational framework for (K and recall some preliminary lemmas. For any (0,, the fractional Sobolev space H (R 3 is defined by { } H ( := u L ( u(x u(y : L (. x y N+ It is known that where RN u(x u(y dxdy = C(n, x y N+ ( C(n, = cos ζ ζ N+ ( u dx, dζ. We endow the space H ( with the norm ( / u H ( := u dx + ( u dx. H ( is also the completion of C0 (RN with H ( and it is continuously embedded into L q ( for q [, ]. The homogeneous space D, ( is { } D, ( := u L ( u(x u(y : L (, x y N+ and it is also the completion of C0 (RN with respect to the norm ( / u D, ( := ( u dx. Lemma. (Norm equivalence. Assume (V -(V 3. Then there exist ε 0 > 0 and ω > 0 with ( (a ε ( u + V (xu dx ω u dx, for every u H ( and all ε (0, ε 0. Proof. By contradiction, for ε 0 = ω = /n there exist ε n 0 and {u n } n N H ( with ( (a εn ( un + V (xu n dx u n dx. n Then, up to a standard nomalization, we may assume that u n H ( = and ( (a εn ( un + V (xu n dx n. In view of (V 3, we get u n 0, which implies from (V and the above inequality that {u n } n N goes to zero in D, (. Therefore u n 0 in H (, which contradicts the normalization.

6 6 Z. S. LIU, M. SQUASSINA, AND J. J. ZHANG Let { } H := u H ( : V (xu dx < be the Hilbert space equipped with the inner product u, v H := a ( u( v dx + V (xuv dx, R N and the corresponding induced norm ( / u := a ( u dx + V (xu dx. From Lemma., it easily follows that the above norm is equivalent to H. A function u H is a (weak solution to problem (K if, for every ϕ H, we have ( a + b ( u dx ( / u( / ϕdx + V (xuϕdx = f(uϕdx. We stress that, under assumptions (V -(V 3 and (f -(f 3, if u is a weak solution to the above problem, then u is globally bounded and Hölder regular allowing the pointwise reppresentation of ( u by the results of []. In particular u > 0 a.e. wherever u 0. Lemma. (Lions lemma, see [9]. Assume that {u n } n N is bounded in H and lim sup u n dx = 0, n y B r(y for some r > 0. Then u n 0 in L s ( for all s (,. The energy functional associated with (K, I : H R, is defined as I(u = u + b ( ( u dx F (udx, u H, 4 with F (u = u 0 f(tdt. Obviously I C (H and its critical points are weak solutions to (K. 3. The perturbed functional Since we do not impose the well-known Ambrosetti-Rabinowitz condition, the boundedness of the Palais-Smale sequence becomes complicated. To overcome this difficulty, we adopt a monotonicity trick due to Jeanjean [6]. Theorem 3. (Monotonicity trick [6]. Let (E, be a real Banach space with its dual space E and J R + an interval. Consider the family of C functionals on E I λ = A(u λb(u, λ J, with B nonnegative and either A(u + or B(u + as u, satisfying I λ (0 = 0. We set Γ λ := {γ C([0, ], E γ(0 = 0, I λ (γ( < 0}, for all λ J. If for every λ J, Γ λ is nonempty and c λ = inf γ Γ λ max I λ(γ(s > 0, s [0,] then for almost any λ J, I λ admits a bounded Palais-Smale sequence {u n } n N E, namely sup n N u n <, I λ (u n c λ and I λ (u n 0 in E. Moreover λ c λ is left continuous.

7 GROUND STATES FOR FRACTIONAL CRITICAL KIRCHHOFF EQUATIONS 7 Set J := [, ], E := H and A(u := u + b ( ( u dx, B(u := F (udx. 4 We consider the family of functionals I λ : H R defined by I λ (u = A(u λb(u, that is I λ (u = u + b ( ( u dx λ F (udx. 4 It is easy to see that B(u 0 for all u H and A(u + as u. In the following H denotes a closed half-space of containing the origin, 0 H. We denote by H the set of closed half-spaces of containing the origin. We shall equip H with a topology ensuring that H n H as n if there is a sequence of isometries i n : such that H n = i n (H and i n converges to the identity. Given x, the reflected point σ H (x will also be denoted by x H. The polarization of a nonnegative function u : R + with respect to H is defined as { max{u(x, u H u(σh (x}, for x H, (x := min{u(x, u(σ H (x}, for x \ H. Given u, the Schwarz symmetrization u of u is the unique function such that u and u are equimeasurable and u (x = h( x, where h : (0, R + is a continuous and decreasing function. We set H + := {u H : u 0}. Now we state a symmetric version of Theorem 3.. Lemma 3. (Symmetric monotonicity trick [3]. Under the assumptions of Theorem 3. for E = H, assume that I λ ( u I λ (u for any λ J and u H and I λ (u H I λ (u, for any λ J, u H + and H H. Then, for any p [, ], I λ has a bounded Palais-Smale sequence {u n } n N H with u n u n p 0. Lemma 3.3 (Uniform Mountain-Pass geometry. Assume that (f -(f 3 and (V -(V 3 hold. Furthermore let N = with (, or N = 3 with ( 3 4,. Then we have: ( Γ λ, for every λ J; ( there exist r, η > 0 independent of λ, such that u = r implies I λ (u η. In particular c λ η. Proof. ( For every ϕ H + \ {0}, taking into account of (f 3, we have I λ (ϕ ϕ + b ( ( ϕ dx D ϕ q dx 4 q ϕ dx. Under the assumptions on N and, it follows that > 4. Then there exists t 0 > 0 sufficiently large, independent of λ J, such that I λ (t 0 ϕ < 0. Setting w := t 0 ϕ H, we have I λ (w < 0 and we can define the corresponding Γ λ. Then, setting γ(t := tw, we have γ Γ λ for every λ J. (ii (f -(f imply that, for any ε > 0, there exists C ε > 0 such that Then there exist σ, σ > 0 such that F (s ε s + C ε s, for all s R. I λ (ϕ σ ϕ σ ϕ, for every ϕ H. Hence there exist r, η > 0, independent of λ, such that for u = r, I λ (u η (and I λ (ϕ > 0 as soon as ϕ r with ϕ 0. Now fix λ J and γ Γ λ. Since γ(0 = 0 and I λ (γ( < 0, certainly γ( > r. By continuity, we conclude that there exists t γ (0, such that γ(t γ = r. Therefore, for every λ J, we conclude c λ inf γ Γλ I λ (γ(t γ η > 0.

8 8 Z. S. LIU, M. SQUASSINA, AND J. J. ZHANG Lemma 3.4 (I λ decreases upon polarization. Assume (V 4 holds. Then for any λ J, for all u H + and H H there holds I λ (u H I λ (u. Proof. It is known (see [5, Theorem ] that RN u H (x u H (y RN u(x u(y x y N+ dxdy x y N+ dxdy, for all u H+. Furthermore, we have (see [34] F (u H dx = F (udx, for all u H +, and, by the monotonicity assumptions on V, V (x(u H dx V (xu dx, for all u H +, which concludes the proof by the definition of I λ. Assume (V -(V 3 and (f -(f 3. As a consequence we now get the following result. Corollary 3.5 (Bounded Palais-Smale with sign. For almost every λ J, there is a bounded sequence {u n } n N H + such that I λ (u n c λ, I λ (u n 0. Furthermore, u n u n 0 if (V 4 is assumed. Proof. For a.a. λ J, a bounded (PS-sequence {u n } n N H for I λ is provided by combining Theorem 3. with Lemmas 3.3 and 3.4. Furthermore, if (V 4 holds, using Lemma 3. in place of Theorem 3., we also get u n u n 0. Next we show that we can assume that u n is nonnegative. Indeed, we know that I λ (u n, u n = µ n, u n with µ n 0 in H as n, with u n = min{u n, 0}, namely (f(s = 0 for s 0 ( a + b ( un dx ( u n u n dx + V (x u n dx = µ n, u n. As it is readily checked, for all x, y, we have (u n (x u n (y(u n (x u n (y (u n (x u n (y, which yields that C(n, R ( u n u (u n (x u n (y(u n (x u n (y 3 n dx = R N x y N+ dxdy RN (u n (x u n (y x y N+ dxdy = C(n, u n D,. Thus u n = o n (, which also yields that {u + n } n N is bounded. We can now prove that I λ (u + n c λ and I λ (u+ n 0. Of course u n = u + n + o n ( and ( ( ( un dx = ( u + n dx + o n (. Notice that from (f -(f, we get F (u n dx F (u + n dx R C ( u n + u n u N n C u n + C u n = o n (.

9 GROUND STATES FOR FRACTIONAL CRITICAL KIRCHHOFF EQUATIONS 9 This shows that I λ (u + n c λ. We claim that I λ (u+ n 0. Setting w n := I λ (u n I λ (u+ n, it is enough to prove that w n 0 in H. For any ϕ H with ϕ H, we have ( w n, ϕ = a + b ( un dx ( / u n ( / ϕdx R ( N a + b ( u + n dx ( / u + n ( / ϕdx + V (xu n ϕdx λ (f(u n f(u + n ϕdx ( = a + b ( u + n dx ( / u n ( / ϕdx + V (xu n ϕdx λ f(u n ϕdx + ξ n, ϕ, for some ξ n 0 in H. Then, by (f -(f, w n, ϕ C u n H + ξ n H, proving the claim. Observe now that by the triangular inequality and the contractility property of the Schwarz symmetrization in L p -spaces (i.e. w z p w z p for all w, z L p ( with w, z 0, we get u + n (u + n u n u n u n + ((u + n u n u n + (u + n u n u n + u + n u n = u n C u n H = o n (. Since u n u n 0, we have u + n (u + n 0 as n. This ends the proof. 4. Upper estimate of c λ and limit problems In this section, we give an upper estimate of the mountain pass value c λ. corresponding limit problem is discussed. Moreover, the 4.. An energy estimate. Next we provide a crucial energy estimate for c λ. Lemma 4. (Energy estimate. Suppose that (f -(f 3 and (V -(V 3 hold. For any λ [, ], assume that 4 q ( N, 4 or q (, ] with D large enough. N Then we have c λ < c λ, c λ := as T N + bs 4 T N 4 λ T N, where T = T (λ > 0 continuously depends on λ. Proof. Let η C0 (R3 be a cut-off function with support in B (0 such that η on B (0 and η [0, ] on B (0. It is well known that S is achieved by T (x := κ (µ + x x 0 N for arbitrary κ R, µ > 0 and x 0. Then, taking x 0 = 0, we can define v ε (x := η(xu ε (x, u ε (x = ε N u (x/ε, u (x := T ( x/s /(. T

10 0 Z. S. LIU, M. SQUASSINA, AND J. J. ZHANG Then ( u ε = u ε u ε and ( u ε = u ε = S N. As in [30], we have (4. A ε := On the other hand, for any q [,, we obtain v ε q dx ε N RN ( vε (x dx = S N + O(ε N. (N q κ q T q S N S N/( εs /( 0 r N (µ + r (N q where S N is the unit sphere in. Observe that, as ε 0, εs /( r N c (0,, if q > N N, dr = O(log ( N 0 (µ + r (N q ε, if q = N, = O(ε (N q N, if q < N Then (4. Since < N N, Similar as above, So that we have (4.3 N. (N q N O(ε, if q > N C ε := v ε q N, dx O(log ( (N q ε εn, if q = N N,, if q < N O(ε (N q v ε dx ε κ T S N S N/( As can be seen in [30], it holds v ε dx O(ε N. εs /( 0 B ε := v ε dx = O(ε N. D ε := RN v ε dx = S N + O(ε N. N. dr, r N (µ + r N dr O(εN. Step. For any ε > 0 small there exists t 0 > 0 such that I λ (γ ε (t 0 < 0, where γ ε (t := v ε ( /t. Indeed, by (V and (f 3, for any t > 0, I λ (γ ε (t a ( γε (t dx + b ( ( γε (t dx 4 + V [ γ ε (t γε (t dx λ + D γ ε(t q ] dx q = aa ( ε tn + ba ε V B ε 4 tn 4 + λd ε λdc ε (4.4 t N. q Noting that < N < 4, we have 0 < N 4 < N. Then by (4.3, V B ε λd ε λs N, as ε 0.

11 GROUND STATES FOR FRACTIONAL CRITICAL KIRCHHOFF EQUATIONS So it follows from (4. that for any ε > 0 small enough, I λ (γ ε (t as t +. Then there exists t 0 > 0 such that I λ (γ ε (t 0 < 0. Step. Notice that, as t 0 +, we have [ ( γε (t + γ ε (t ] dx = t N A ε + t N B ε 0 uniformly for ε > 0 small. We set γ ε (0 = 0. Then γ ε (t 0 Γ λ, where Γ λ is as in Theorem 3. and c λ sup I λ (γ ε (t. t 0 Recalling that c λ > 0, by (4.4, there exists t ε > 0 such that sup I λ (γ ε (t = I λ (γ ε (t ε. t 0 By (4., (4. and (4.4, we get I λ (γ ε (t 0 + as t 0 + and I λ (γ ε (t as t + uniformly for ε > 0 small. Then there exist t, t > 0 (independent of ε > 0 such that t t ε t. Let then J ε (t := aa ε tn + ba ε 4 tn 4 λd ε t N, ( V B ε c λ sup J ε (t + t 0 By formula (4., for any q (,, we have Then by (4.3, we conclude that (N q N C ε O(ε. λdc ε t N ε q c λ sup J ε (t + O(ε N (N q N O(Dε. t 0 Noting that N > 0 and N (N q/ > 0, we have sup t 0 J ε (t c λ / uniformly for ε > 0 small. As above, there are t 3, t 4 > 0 (independent of ε > 0 such that sup t 0 J ε (t = sup t [t3,t 4 ] J ε (t. By (4., (4.5 c λ sup t 0 where Observe that for t > 0, K (t = K(S t + O(ε N (N q N O(Dε, K(t = as tn + bs 4 tn 4 λ t N. (N tn K(t, where K(t := as + bs t N λt, and K (t = t N (bs (N λt 4 N. Since 4 > N, there is a unique T > 0 such that K(t > 0 if t (0, T and K(t < 0 if t > T. Hence, T is the unique maximum point of K. Then by (4.5, (4.6 c λ K(T + O(ε N (N q N O(Dε.

12 Z. S. LIU, M. SQUASSINA, AND J. J. ZHANG If q > 4/(N, then 0 < N (N q/ < N, which implies by (4.6 that for any fixed D > 0, c λ < K(T for ε > 0 small. If < q 4/(N, for ε > 0 small and D ε (N q/, then also in this case c λ < K(T, which completes the proof. 4.. The limit problem. Note that V (x V as x. For any λ [/, ], we consider the problem ( a + b ( u dx ( u + V u = λf(u in, R N u H (, u > 0 in, whose energy functional is defined by Iλ (u := (a ( u + V u dx + b ( ( u dx λ F (udx. 4 We will use of the following Pohožaev type identity, whose proof is similar as in [0]. Lemma 4. (Pohožaev identity. Let u be a critical point of Iλ in H for λ [, ]. Then P λ (u = 0, P λ (u := N a ( u dx + N ( b ( u dx R (4.7 N + N V u dx Nλ F (udx. Notice that P λ (u = d dt I λ (u( /t t=. Lemma 4.3. For λ [, ], if w λ H \ {0} solves P λ (w λ = 0, then there exists γ λ C([0, ], H such that γ λ (0 = 0, I λ (γ λ( < 0, w λ γ λ ([0, ], 0 γ λ ((0, ] and max t [0,] I λ (γ λ (t = Iλ (w λ. Proof. Note that Iλ (w λ ( /t = tn RN ( a ( wλ dx + btn 4 ( wλ dx 4 R N + tn V w λdx t N λ F (w λ dx = 0, which, by (4.7, yields lim t I λ (w λ ( /t < 0. Then there is t 0 > 0 such that Iλ (w λ( /t 0 < 0. Let γ λ (t = w λ ( /tt 0 for 0 < t and γ λ (0 = 0. Then γ λ C([0, ], H, w λ γ λ ([0, ] and max t [0,] Iλ (γ λ(t = Iλ (w λ as t = t 0 is the unique maximum point of t Iλ (γ λ(t by Lemma Behaviour of Palais-Smale sequences By Corollary 3.5, for almost every λ [/, ], there exists a bounded Palais-Smale sequence {u n } n N H for I λ at the level c λ. Then there exists a subsequence of {u n } n N, still denoted by {u n } n N, such that u n u 0 in H and u n u 0 a.e. in as n. Let v n := u n u 0. Then v n 0 in H and v n 0 a.e. Notice that, since u n 0, the dominated convergence theorem implies that (v n 0 in L q ( for any q.

13 GROUND STATES FOR FRACTIONAL CRITICAL KIRCHHOFF EQUATIONS Splitting lemmas. Let us set g(t := f(t (t +, G(t := t 0 g(sds. In order to get the profile decomposition of {u n } n N, we state the following splitting lemmas. Lemma 5. (Splitting lemma I. We have (5. (g(u n g(u 0 g(v nϕdx o n( ϕ, where o n ( 0 as n, uniformly for any ϕ C 0 (RN. Proof. For each n, there exists θ n (0, such that (5. g(u n g(v n g (v n + θ n u 0 u 0. In view of (f -(f 3, for any ε > 0, there exists D > 0 such that (5.3 g(t ε t, for t D/. Let Ω n ( D := {x : u n (x D} and for r > 0, B r := {x : x < r}, Br c := \ B r (0. Since u 0 H, we have BR c { u 0(x D/} 0 as R. Then for ε given as above, there exist R > 0 and Ω R with Ω R Λ ε such that u 0 (x < D/ for x BR c \ Ω R, where Λ ε > 0 will be chosen later small enough. Then, by Hölder s inequality, (5. and (5.3, we have g(u n g(vn ϕ dx (5.4 BR c \Ω R (B c R \Ω R Ω n( D εc( u n g(u n g(vn ϕ dx + g(u n g(vn ϕ dx (BR c \Ω R Ω c n( D + vn ( / ϕ. ϕ + max g (t u 0(xdx t D It follows from (f and (f that, for ε > 0 given, there exists C ε = C ε (f > 0 such that g(u n g(vn ϕ dx Ω R (5.5 ε ( u n + vn ϕ dx + C ε ( u n + vn ϕ dx Ω R Ω R εc( u n + vn ϕ + C ε Ω R N ( un + vn ϕ. By (5.4 and (5.5, by choosing Λ ε such that C ε Λ /N ε ε, there exists C > 0 with (5.6 g(u n g(vn ϕ dx Cε ϕ. Moreover, (5.7 B c R B c R g(u 0 ϕ dx C B c R ( C B c R B c R u 0 ϕ dx + u 0 ϕ dx BR c / ϕ ( ( u 0 dx + C u 0 / dx ϕ. B c R

14 4 Z. S. LIU, M. SQUASSINA, AND J. J. ZHANG It follows from (5.6 and (5.7 that, for ε > 0 above, we choose R > 0 above large enough such that (5.8 (g(u n g(u 0 g(vnϕdx Cε ϕ, B c R where C is independent of n, ε and ϕ C0 (RN. On the other hand, ( ( g(u n g(u 0 ϕ dx g(u n g(u 0 /( / ( / dx ϕ. B R B R B R Observe that g /( (t /( (t lim = lim = 0. t + t g t 0 + t /( Then g(u n g(u 0 /( 0 in L (B R. Hence, we deduce (5.9 g(u n g(u 0 ϕ dx o n ( ϕ. B R Similarly, we also obtain that (5.0 g(vn ϕdx o n ( ϕ, B R for any ϕ C 0 (RN. It follows from (5.8, (5.9 and (5.0 that (5. holds. Lemma 5. (Splitting lemma II. We have ( u n u n u 0 u 0 vn vn ϕdx o n( ϕ, where o n ( 0 as n, uniformly for any ϕ C 0 (RN. Proof. For any ε > 0, there exists R = R(ε > 0 such that ( u n u n u 0 u 0 vn v n ϕdx \B R (0 ( (5. u n u n vn v n ϕdx \B R (0 + ( C u n + vn u 0 ϕ dx + \B R (0 \B R (0 \B R (0 u 0 u 0 ϕdx u 0 ϕ dx Cε ϕ. On the other hand, for every r > 0, we have ( u n u n u 0 u vn vn ϕdx B R (0 u n u n u 0 u 0 vn v B R (0 { vn r} n ϕdx + u n u n u 0 u 0 vn vn ϕdx =: I + I. B R (0 { v n r} Now, there exists r = r(r such that r B R (0 / ε. Therefore, we have ( (5. I C u n + u 0 + vn vnϕ dx B R (0 { v n r} Cr B R (0 / ϕ Cε ϕ.

15 GROUND STATES FOR FRACTIONAL CRITICAL KIRCHHOFF EQUATIONS 5 For such r, R fixed above, u n converges to u in measure in B R (0, i.e. B R (0 { v n r} 0 for n. Therefore, for n large, ( (5.3 I C u n + vn u 0 ϕ dx + B R (0 { vn r} B R (0 { v n r} Then (5., (5. and (5.3 yield the assertion. u 0 ϕ dx Cε ϕ. Lemma 5.3 (Splitting lemma III. We have f(u n u n dx = f(vnv ndx + f(u 0 u 0 dx + o n (, where o n ( 0 as n. Furthermore F (u n dx = F (vndx + F (u 0 dx + o n (. Proof. Since f(t = g(t + t for t 0, by the standard Brezis Lieb lemma, it suffices to prove g(u n u n dx = g(vnv ndx + g(u 0 u 0 dx + o n (, where o n ( 0 as n. Fixed ε > 0, there exists C ε > 0 such that (5.4 g(t εt + C ε t, t 0. Then there exists R = R(ε > 0 large enough such that g(u 0 v ndx g(u 0 vn dx R B + g(u 0 vn dx (5.5 N R BR c (ε u 0 + C ε u 0 vn dx + ε( vn + vn Cε + C ε o n (. B R and g(v nu 0 dx g(vnu R B 0 dx + g(vnu 0 dx N R BR c (5.6 (ε vn + C ε vn u 0 dx + (ε vn + C ε vn u 0 dx B R Cε + C ε o n (. It follows from (5.5, (5.6 and Lemma 5. that (g(u n u n g(u 0 u 0 g(vnv ndx R N (g(u n g(u 0 g(vnu n dx + g(vnu 0 dx + g(u 0 vn dx o n ( u n + Cε + C ε o n (. Letting n and ε 0 + completes the proof of the first assertion. The second assertion follows from the standard Brezis Lieb lemma and G(u n dx = G(vndx + G(u 0 dx + o n (, whose proof is left to the reader. B c R

16 6 Z. S. LIU, M. SQUASSINA, AND J. J. ZHANG 5.. Profile decomposition. In the following, we give the profile decomposition of {u n } n N, which plays a crucial role in getting the compactness. Since c λ > 0, for some B > 0 we have ( un dx B, as n. Now, for any u H, let J λ (u := a + b B ( u dx + V (x u dx λ F (udx and Jλ (u := a + b B ( u dx + V u dx λ F (udx, which are respectively the corresponding functional of the following problems (a + b B ( u + V (xu = f(u, (a + b B ( u + V u = f(u, u H. Here we point out that in contrast with the original problem (K, the problems above are both non Kirchhoff. Now we take advantage of this to get the profile decomposition of {u n } n N. Lemma 5.4 (Profile decomposition. Let {u n } n N H be the sequence mentioned above and assume that conditions (V -(V 3, (f -(f 3 hold and N < 4. Then J λ (u 0 = 0 with u 0 0, and there exist a number k N {0}, nontrivial positive critical points w,..., w k H ( of J λ which decay polynomially at infinity as w j (x x N+ = O(, such that (i yn j +, yn j yn i + if i j, i, j k, n +, (ii c λ + b B 4 4 = J λ (u 0 + k Jλ (wj, j= (iii u n u 0 k w j ( yn j 0, j= (iv B = ( u 0 + k ( w j. j= Moreover, we agree that in the case k = 0 the above holds without w j. In addition, if (V 4 holds, then k = 0 and u 0 H rad (RN. Proof. Observe that, from I λ (u n = c λ + o n ( and I λ (u n 0 in H, we obtain J λ (u n = c λ + b B o n(, J λ(u n 0 in H. Then, it is standard to get J λ (u 0ϕ = 0 for all ϕ H. From Lemma 5.3, we get F (vndx = F (u n dx F (u 0 dx + o n (, R N f(vnv ndx = f(u n u n dx f(u 0 u 0 dx + o n (. It follows that (5.7 (5.8 J λ (u n = J λ (v n + J λ (u 0 + o n (, J λ(v nv n = J λ(u n u n J λ(u 0 u 0 + o n ( = o n (.

17 GROUND STATES FOR FRACTIONAL CRITICAL KIRCHHOFF EQUATIONS 7 On the other hand, by a slight variant of [0, Proposition 4.], u 0 satisfies the Pohǒzaev identity N (a + b B ( u0 dx + V (x x u 0dx + N V (xu 0dx Nλ F (u 0 dx = 0. Then by (V and N < 4, we have NJ λ (u 0 = (a + b B ( u0 dx V (x x u 0dx R N (a + b B ( u0 dx W N u 0 (a + b B ( u0 dx a ( u0 dx = b B ( u0 dx > 0, which implies that (5.9 J λ (u 0 b B 4 ( u0 dx. We claim that one of the following conclusions holds for vn: (v vn 0 in H, or (v there exist r > 0, σ > 0 and a sequence {yn} n N such that (5.0 lim inf v n n dx σ > 0. B r (yn Indeed, suppose that (v does not occur. Then for any r > 0, we have lim sup v n n dx = 0. y B r(y Therefore, it follows from Lemma. that vn 0 in L s ( for s (,. It follows from (5.4 that for any ε > 0, there exists C ε > 0 such that ( g(vnv n dx ε vn + vn dx + C ε vn q dx. So from vn 0 in L q ( and the arbitrariness of ε, we can easily obtain that f(vnv ndx = ((vn + dx + o n (. Furthermore, from J λ (v nvn = o n ( in (5.8, we have (5. vn + b B ( v n dx = λ (vn + + o n(. In view of conditions (V -(V 3, we can check that V > 0. And so we can also get V (x vn dx = V + (x vn dx + o n (, which, together with the definition of S and (5., implies that ( (5. as vn (R dx 4 + bs N vn dx λ vn dx + o n (.

18 8 Z. S. LIU, M. SQUASSINA, AND J. J. ZHANG Let l 0 be such that v n dx l N. If l > 0, then it follows from (5. that K (N l (l = (as l N + bs l N 4 λl N 0, where K has been defined in Lemma 4.. This also implies that l T (T is the unique maximum point of K. On the other hand, by (5.7 and (5.9, we have c λ + b B 4 ( a + b B ( v n + V (x v n λ ((vn + dx + J λ (u 0 + o n ( 4 R = N ( (a + b B 4 ( v n + V (x v n λ ((vn + dx + b B o n(, which, together with (5. and the definition of S, implies that ( c λ ( a ( v n dx + 4 ( b ( v n dx + o n ( ( ( ( as vn dx + 4 ( 4 bs vn dx + o n (. Thus, combining vn dx l N and l T, K (T = 0, we have ( c λ ( as l N + 4 bsl N 4 ( as T N + ( 4 bs T N 4 = as T N + 4 bs T N 4 λ T N = c λ, contradicting c λ < c λ. Hence, l = 0. It follows from (5. that v n 0, that is, u n u 0 in H. Then Lemma 5.4 hold with k = 0 if (v does not occur. In particular, if we assume (V 4 holds, then by Corollary 3.5, u n u n 0. Obviously, { u n } n N H rad (RN is bounded and u n u n q 0 for q (,. Since { u n } n N has a strongly convergent subsequence in L q ( for q (,, without loss of generality, we assume that u n u 0 in L q ( for q (, and u 0 = u 0. As a consequence, (v does not hold and as above, u n u 0 in H. In the following, otherwise, suppose that (v holds, that is (5.0 holds. Consider v n( + y n. The boundedness of {v n} n N and (5.0 imply that v n( + y n w 0 in H. Thus, it follows from v n 0 in H that {y n} n N is unbounded and, up to a subsequence, y n +. Let us prove that (J λ (w = 0. It suffices to show that (J λ (v n( + y nϕ 0 for any ϕ C 0 (RN. Combining Lemma 5. and Lemma 5., we obtain J λ(u n ϕ J λ(u 0 ϕ J λ(v nϕ o n ( ϕ, ϕ C 0 (, which implies that J λ (v nϕ o n ( ϕ, for all ϕ C0 (RN, as n. Notice that J λ(v nϕ( yn C(n, = (a + b B (v n(x vn(y(ϕ(x yn ϕ(y yn R N x y N+ dxdy + V (xvn(xϕ(x yndx λ g(vn(xϕ(x yndx R N λ ((vn(x + ϕ(x yndx = o n ( ϕ( yn = o n ( ϕ.

19 GROUND STATES FOR FRACTIONAL CRITICAL KIRCHHOFF EQUATIONS 9 Thus, as n, it follows that C(n, (a + b B (v n(x + yn v n (y + yn(ϕ(x ϕ(y R N x y N+ dxdy+ (5.3 V (x + ynv n(x + ynϕ(xdx λ g(vn(x + ynϕ(xdx R N ((vn(x + yn + ϕ(xdx = o n ( ϕ. Since yn and ϕ C0 (RN, we obtain (5.4 (V (x + yn V vn(x + ynϕ(xdx 0. Thus, combining (5.3 and (5.4, we have for any ϕ C0 (RN, (Jλ (vn( + ynϕ C(n, = (a + b B (v n(x + yn v n (y + yn(ϕ(x ϕ(y R N x y N+ dxdy + V vn(x + ynϕ(xdx λ g(vn(x + ynϕ(xdx R N λ ((vn(x + yn + ϕ(xdx = o n (. Then, (J λ (w = 0, w > 0 and w (x x N+ = O( as x. Finally, let us set (5.5 v n(x = v n(x w (x y n, then vn 0 in H. Since V (x V as x and vn 0 strongly in L loc (RN, we have (V (x V (vn dx = o n (. It follows that (5.6 V (x vn dx = V (x vn dx + V (x + yn w (x dx R N V (x + ynv n(x + ynw (xdx R N = V u n dx V u 0 dx V w dx + o n ( R N = V (x u n dx V (x u 0 dx V w dx + o n (, and (it is easy to see that (vn = o n ( also ( vn (5.7 = ( u n ( u 0 ( w + o n(, (vn + = u n u 0 w + o n (, (5.8 G(vndx = G(u n dx G(u 0 dx G(w dx. It is readily checked that we also have (5.9 g(vnϕdx = g(u n ϕdx g(u 0 ϕdx g(w ( ynϕdx + o n ( ϕ,

20 0 Z. S. LIU, M. SQUASSINA, AND J. J. ZHANG for any ϕ C 0 (RN. Combining (5.6, (5.7, (5.8 and (5.9, we deduce that ( J λ (v n = J λ (u n J λ (u 0 J λ (w + o n (, ( J λ(v nϕ = J λ(u n ϕ J λ(u 0 ϕ (J λ (w ( y nϕ + o n ( ϕ = o n ( ϕ, (3 J λ (v n = J λ (v n J λ (w + o n ( for any ϕ C 0 (RN. Thus {v n} n N is a Palais-Smale sequence and we get J λ (v n = c λ + b B 4 4 J λ(u 0 J λ (w + o n ( < c λ + b B 4 4. Remark that one of (v and (v holds for vn. If vn 0 in H, then Lemma 5.4 holds with k =. Otherwise, {vn} is non-vanishing, that is, (v holds for vn. Similarly, we repeat the arguments. By iterating this procedure we obtain sequences of points {yn} j such that yn j +, yn j yn i + if i j as n + and vn j = vn j w j (x yn j (like (5.5 with j such that vn j 0 in H, (Jλ (w j = 0. Using the properties of the weak convergence, we have (5.30 k k (a u n u 0 w j ( yn j = u n u 0 w j ( yn j + o n (, j= j= k (b J λ (u n = J λ (u 0 + Jλ (w j + Jλ (vn k+ + o n (. j= Note that there is ρ > 0 such that w ρ for every nontrivial critical point w of Jλ and {u n} n N is bounded in H. By (5.30(a, the iteration stops at some k. That is, vn k+ 0 in H. We stress that the polynomial decay of the limiting profiles w j can be justified as in [, Theorem 3.4 and Theorem.5]. The proof is now complete. 6. Proof of the main results In order to obtain the existence of ground state solutions of problem (K, our strategy is that we firstly obtain the existence nontrivial solutions of the perturbed problem, then as λ goes to, we get a nontrivial solution of the original problem. Finally, thanks to the profile decomposition of the (PS-sequence, we obtain the existence of ground state solutions of problem (K. 6.. Nontrivial critical points of I λ. Lemma 6.. Assume that (V -(V 3 and (f -(f 3 hold. For almost every λ [/, ], there exists u λ H\{0} such that I λ (u λ = c λ and I λ (u λ = 0. In addition, if (V 4 holds, then u λ H rad (RN. Proof. For almost all λ [/, ], there is a bounded sequence {u n } n N H such that I λ (u n c λ, I λ (u n 0. From Lemma 5.4, up to a subsequence, there exist u 0 H and B > 0 such that u n u 0 in H, ( un dx B, as n and J λ (u 0 = 0. Furthermore, there exist k N {0}, nontrivial critical points w,..., w k of Jλ and k sequences of points {yn} j, j k, such that k (6. u n u 0 w j ( yn j 0, c λ + b B 4 k 4 = J λ(u 0 + Jλ (w j j= j=

21 GROUND STATES FOR FRACTIONAL CRITICAL KIRCHHOFF EQUATIONS and k (6. B = ( u0 + ( w j. j= Now we claim that if u 0 0, then by N < 4, (6.3 J λ (u 0 > b B Indeed, since J λ (u 0 = 0, similar as in [0], we get 4 ( u0 dx. P λ (u 0 := N (a + b B ( u0 dx + N V (xu 0dx + ( V (x, xu 0dx Nλ F (u 0 dx = 0. By hypothesis (V we have J λ (u 0 = N (a + b B ( u0 dx V (x x u N 0dx > N b B ( u0 dx, which implies that (6.3 holds. For each nontrivial critical point w j, (j =,..., k of J λ, N (a + b B ( w j dx + N Then it follows from (6. that V w j dx Nλ F (w j dx = P λ (w j = 0. ( a(n ( w j b(n dx + ( w j dx + N V w j dx Nλ F (w j dx 0. Then there exists t j (0, ] such that (6.4 at N j (N ( w j dx + bt N 4 ( j (N ( w j dx + NtN j V w j dx Nt N j λ F (w j dx = 0. That is, w j ( /t j satisfies the identity P λ (u = 0 and it follows from Lemma 4.3 that there exists γ λ C([0, ], H such that γ λ (0 = 0, I λ (γ λ( < 0, w j γ λ ([0, ] and I λ (w j ( /t j = max t [0,] I λ (γ λ (t. By hypothesis (V, we have max t [0,] I λ (γ λ(t max t [0,] I λ (γ λ (t, which, by the definition of c λ, implies that I λ (wj ( t j c λ. In particular, if V (x V, then (6.5 I λ (w j ( /t j > c λ.

22 Z. S. LIU, M. SQUASSINA, AND J. J. ZHANG So by (6.4 we have Jλ (w j = Jλ (w j ( N P λ (w j = (a + b B ( w j dx R ( N a ( w j ( x dx t j ( ( ( b ( w j ( x dx + b B ( w j dx t j 4 = Iλ ( w j ( t j N P λ(w j ( + b B ( w j dx t j 4 = Iλ ( w j ( b B + ( w j dx t j 4 and then we conclude that Jλ (w j c λ + b B 4 ( w j dx, where the inequality is strict if V (x V. Then by formulas (6.-(6.3, (6.7 c λ + b B 4 4 = J λ(u 0 + k j= J λ (w j kc λ + b B 4 4, with strict inequality if V (x V or u 0 0. If k = 0, we are done. If condition (V 4 holds, then k = 0 and u 0 is radial. Then it follows that I λ (u 0 = J λ (u 0 b B 4 /4 = c λ and I λ (u 0 = J λ (u 0 = 0. If k = and V (x V or u 0 0, then (6.7 yields a contradiction If k = and V (x V and u 0 = 0, then B = ( w and it follows from (6. that c λ = J λ (w b B 4 as desired. Hence, in any case, the assertion follows. 4 = J λ (w b 4 ( w 4 = I λ (w, I λ(w = 0, 6.. Completion of the proof. Choosing a sequence {λ n } n N [, ] satisfying λ n, we find a sequence of nontrivial critical points {u λn } n N (still denoted by {u n } n N of I λn and I λn (u n = c λn. In particular, if (V 4 holds, then {u n } n N Hrad (RN. Now we show that {u n } is bounded in H. Remark that u n satisfies the Pohožaev identity as follows N a ( un dx + N ( b ( un dx + N V (xu ndx + V (x x u ndx Nλ F (u n dx = 0. It follows that NI λn (u n = a ( un dx + ( N ( b ( un dx V (x x u 4 ndx.

23 GROUND STATES FOR FRACTIONAL CRITICAL KIRCHHOFF EQUATIONS 3 Since c λ is continuous on λ, I λ n (u n = c λn + o n ( < c λ n. It follows from (V that there is a positive number κ (0, a such that W N κs. Hence, ( a κ ( un dx NI λn (u n, which implies that a ( u n dx is bounded from above. By (V 3, (f -(f and I λ n (u n u n = 0, there is ν > 0 such that for any ε > 0, there exists C ε > 0 with ν u ndx a ( un dx + V (xu ndx ε u ndx + C ε u n dx, which yields that {u n } n N is bounded in L (. Then {u n } n N is bounded in H. By Theorem 3., ( lim I(u n = lim I λn (u n + (λ n F (u n dx = lim c λ n n n n = c and for any ϕ C0 (RN, ( lim n I (u n ϕ = lim I n λ n (u n ϕ + (λ n f(u n ϕdx = 0. That is, {u n } n N is a bounded Palais-Smale sequence for I at level c. Then by Lemma 6., there is a nontrivial critical point u 0 H (radial, if (V 4 holds for I and I(u 0 = c. Set ν = inf{i(u : u H \ {0}, I (u = 0}. Of course 0 < ν I(u 0 = c <. By the definition of ν, there is {u n } n N H with I(u n ν and I (u n = 0. We deduce that {u n } n N is bounded in H. Up to a sequence, for some B > 0, ( un dx B. Let us set J(u := J (u and J (u := J (u, for any u H. From Lemma 5.4 there exists u 0 H such that u n u 0 in H and J (u 0 = 0. Furthermore, there exist k N {0}, nontrivial critical points w,..., w k of J and k sequences of points {yn} j n N, j k, such that (6.8 u n u 0 k w j ( yn j 0, ν + b B 4 k 4 = J(u 0 + J (w j j= j= and k B = ( u0 + ( w j. j= If k = 0, we are done. If k, assume by contradiction that u 0 0. Then, as in Lemma 6., (6.9 J(u 0 > b B ( u0 dx, 4 for each j there is t j (0, ] such that I (w j ( /t j c, which is strict if V (x V, and J (w j c + b B ( w j dx, 4 where the inequality is strict if V (x V. Then by formulas (6.8-(6.9 and ν c, we get c + b B 4 4 ν + b B 4 4 = J(u 0 + k j= J (w j > kc + b B 4 4,

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