Ground state solutions for fractional scalar field equations under a general critical nonlinearity

In this paper we study existence of ground state solution to the following problem $$ (- \Delta)^{\alpha}u = g(u) \ \ \mbox{in} \ \ \mathbb{R}^{N}, \ \ u \in H^{\alpha}(\mathbb R^N) $$ where $(-\Delta)^{\alpha}$ is the fractional Laplacian, $\alpha\in (0,1)$. We treat both cases $N\geq2$ and $N=1$ with $\alpha=1/2$. The function $g$ is a general nonlinearity of Berestycki-Lions type which is allowed to have critical growth: polynomial in case $N\geq2$, exponential if $N=1$.


Introduction
In the present paper, we are interesting in the existence of ground state solution for a class of nonlocal problem of the following type where N ≥ 1, α ∈ (0, 1), (−∆) α denotes the fractional Laplacian operator and g is a C 1 −function verifying some conditions which will be mentioned later on.
The main motivation for this paper comes from the papers Berestycki and Lions [4] and Berestycki, Gallouet and Kavian [5] which have studied the existence of solution for (P) in the local case α = 1, that is, for a class of elliptic equations like where N ≥ 2, ∆ denotes the Laplacian operator and g is a continuous function verifying some conditions. In [4], Berestycki and Lions have assumed N ≥ 3 and the following conditions on g: where G(s) = s 0 g(t) dt. In [5], Berestycki, Gallouet and Kavian have studied the case where N = 2 and the nonlinearity g possesses an exponential growth of the type lim sup s→0 + g(s) e βs 2 = 0, ∀β > 0.
In the two papers above mentioned, the authors have used the variational method to prove the existence of solution for (P). The main idea is to solve the minimization problem min 1 2 R N |∇u| 2 dx : R N G(u) dx = 1 and min 1 2 R N |∇u| 2 dx : for N ≥ 3 and N = 2 respectively. After that, the authors showed that the minimizer functions of the above problem are in fact ground state solutions of (1.1). By a ground state solution, we mean a solution u ∈ H 1 (R N ) which satisfies E(u) ≤ E(v) for all nontrival solution v of (1.1), where E : H 1 (R N ) → R is the energy functional associated to (P) given by After, Jeanjean and Tanaka in [12] showed that the mountain pass level of E is a critical level and it is indeed the lowest critical level.
In the above mentioned papers, the nonlinearity does not have critical growth. Motivated by this fact, Alves, Montenegro and Souto in [1] have studied the existence of ground state solution for (P) by supposing that g(s) = f (s) − s and that f may have critical growth, more precisely, the following condition were considered: there is τ > 0 and q ∈ (2, 2 * ) if N ≥ 3 and q ∈ (2, +∞) if N = 2 such that f (s) ≥ τ s q−1 , ∀s ≥ 0.
By using the variational method, the authors in [1] give a unified approach in order to deal with subcritical and critical case. However, we would like to point out that the Concentration Compactness Principle of Lions [14] was crucial for the case N ≥ 3. For the case N = 2, as in the previous references, a Trudinger-Moser inequality due to Cao [6] was the main tool used. A similar study was made for the critical case and N ≥ 3 in Zhang and Zou [18].
After a review bibliographic, we have observed that there is no a version of the paper [1] for the fractional Laplacian operator. Motivated by this fact, we have decide to study this class of problem. However, we would like point out that some estimates made in [1] are not immediate for fractional Laplacian operator. For example, there is some restriction to use Concentration Compactness Principle of Lions [14] as mentioned in Palatucci and Pisante [16,Theorem 1.5] for the dimension N ≥ 2 and α ∈ (0, 1). To overcome this difficulty, we use a new approach which do not use the Concentration Compactness Principle of [16]. For the dimension N = 1 and α = 1/2, we use a Trudinger-Moser inequality due to Ozawa [17] which also permits to apply variational methods in this case. Here, it is very important to mention that Zhan, do O and Squassina [19,Theorem 4.1] studied the existence of ground state solution for (P) for N ≥ 2, by supposing that g satisfies lim s→+∞ g(s) s 2 * α −1 = b > 0. These condition is not assumed in our paper, and so, our results complete the studied made in that paper.
Before stating our main results, we must fix some notations. We will look for weak solutions of (P) hence the natural setting involves the fractional Sobolev spaces H α (R N ) defined as endowed with scalar product and (squared) norm given by It is well known that H α (R N ) is a Hilbert space with the above scalar product. We are denoting with |u| p = ( R N |u| p dx) 1/p the L p −norm of u, and by (−∆) α the fractional Laplacian, which is the pseudodifferential operator defined via the Fourier transform of the following way It is known that H α (R N ) has continuos embedding into L q (R N ) for suitable q depending on N : we will denote by C q > 0 the embedding constant. It is useful to introduce also the homogeneous fractional Sobolev space It is well known that the following inequality holds for some positive S > 0. For these facts and the relation between the fractional Laplacian and the fractional Sobolev space H α (R N ), we refer the reader to classical books on Sobolev space, and to the monograph [8].
We will study (P) by variational methods: its solutions will be found as critical points of a C 1 functional I : H α (R N ) → R. Actually our results concern the existence of ground state solutions, that is a solution u ∈ H α (R N ) such that I(u) ≤ I(v) for every nontrivial solution v ∈ H α (R N ) of (P). In view of this, we make the following assumptions on the nonlinearity f . More precisely we assume that f : R → R is a C 1 -function satisfying • If N = 1, we assume q > 2 and (f5) there exist ω ∈ (0, π) and β 0 ∈ (0, ω], such that lim s→+∞ f (s) e βs 2 = 0, ∀β > β 0 , and lim s→+∞ f (s) e βs 2 = +∞, ∀β < β 0 . As we can see, a critical growth for the function f is allowed. Note also that a weaker condition than the usual Ambrosetti-Rabinowitz condition is imposed on f , see condition (f3).
Our main results are the following one. Theorem 1.1. Suppose that N ≥ 2 and f satisfies (f1)-(f4). Then problem (P) admits a ground state solution which is non-negative, radially symmetric and decreasing. Theorem 1.2. Suppose that N = 1 and f satisfies (f1), (f4) and (f5). Then problem (P) admits a ground state solution a ground state solution which is non-negative, radially symmetric and decreasing.
The plan of the paper is the following: In Section 2 we study the case N ≥ 2. We first introduce the variational framework, then give some preliminaries results and Lemmas which will be useful to prove Theorem 1.1. In Section 3 we consider the case N = 1 and α = 1/2, where again, after some preliminaries, the proof of Theorem 1.2 is given.
Before concluding this introduction, we would like to cite some papers involving the fractional Laplacian operator where the problem is related to the problem (P) in some sense, see for example, Ambrosio [2], Barrios, Colorado, de Pablo and Sánchez [3], Frank and Lenzmann [9], Felmer, Quass and Tan [10], Iannizzotto and Squassina [11], Zhang, doÓ and Squassina [19] and their references.
Notations As a matter of notations, we will use in all the paper the letter C, C, C ′ , . . . to denote suitable positive constants whose exact value is insignificant for our purpose.
2. The case N ≥ 2 2.1. The variational framework. The energy functional I : H α (R N ) → R associated to equation (P) is defined as follows Under assumptions (f1) and (f2), I ∈ C 1 (H α (R N ), R) with Frechét derivative given by Hence the critical points are easily seen to be weak solutions to (P).
We remark two inequalities which will be frequently used in the sequel. From (f1) and (f2), for any ε > 0 there exists C ε > 0 such that and, then by integration, Once we intend to find nonnegative solution, we will assume that f (s) = 0 for every s ≤ 0. Let us consider the set of non-zero critical points of I, that is non trivial solution of (P), the so called ground state level.
In particular It is worth to point out that if we define the C 1 functional it holds from (f3): The last information will be used later on. In addition, we define the min-max level associated to the functional I which is not empty since I has a Mountain Pass Geometry. Let us define also the set, usually called Pohozaev manifold, which, according to [

2.2.
Some preliminary stuff. At this point we establish some preliminary results which will be useful in order to prove Theorem 1.1.
where b is the min-max level of I defined in (2.6).
Proof. Indeed, from [13, Lemma 2.3], for each γ ∈ Γ with from where it follows that p ≤ b and the result follows from (2.7).
The next result is standard. We recall the proof for the reader's convenience.
is not empty and a C 1 manifold.
The next steps consists in proving the boundedness of the minimizing sequences in H α (R N ) for the problem Proof. Let {u n } ⊂ M be a minimizing sequence for T , then Then (2.9) 1 2 R N |(−∆) α/2 u n | 2 dx ≤ C for all n ∈ N and for some constant C > 0 By using (2.1) with ε = 1/4, we get Then, for every n ∈ N, by using (2.9), it follows Consequently {u n } is bounded also in L 2 (R N ) and this ensures its boundedness in H α (R N ).
By the Ekeland Variational Principle we can assume that the minimizing sequence {u n } is also a Palais-Smale sequence, that is, there exists a sequence of Lagrange multipliers {λ n } ⊂ R such that In the remaining part of this section, {λ n } will be the associated sequence of Lagrange multipliers. At this point it is useful to establish some properties of the levels D and b. Proof. Clearly by definition D ≥ 0. Suppose, by contradiction, that D = 0. If {u n } is a minimizing sequence for D = 0, then Then, for any ε > 0, see (2.2), By choosing ε = 1/2, we obtain This contradiction concludes the proof.
Lemma 2.5. The sequence of Lagrange multipliers {λ n } associated to the minimizing sequence {u n } is bounded. More precisely, we have that Hence, for some subsequence, still denoted by {λ n }, we can assume that λ n → λ * , for some λ * ∈ (0, D]. Proof. By (2.11), Since {u n } is a bounded minimizing sequence, it is easy to see that |J ′ (u n )[u n ]| = | R N g(u n )u n | ≤ C, and then by (2.12) and the fact that 2T (u n ) → 2D > 0, we infer that lim inf n→+∞ λ n > 0.
The proof is thereby completed.
In the sequel, we will show that a minimizing sequence for (2.8) can be choose nonnegative and radially symmetric around the origin. Note that for our proof we do not need to consider the "odd extension" of the nonlinearity, as it is usually done in the literature to show that the minimizing sequence can be replaced by the sequence of the absolute values. In fact we will prove that the minimizing sequence can be replaced, roughly speaking, with the sequence of the positive parts.
Lemma 2.6. Any minimizing sequence {u n } for (2.8) can be assumed radially symmetric around the origin and nonnegative.
Proof. To begin with, we recall that F (s) = 0 for all s ≤ 0. Thus, F (u n ) = F (u + n ) for all n ∈ N with u + n = max{0, u n }. From this, the equality Defining the function h n : the conditions on f yield that h is continuous with h n (1) ≥ 1. Once u + n = 0 for all n ∈ N, the condition (f1) ensures that h n (t) < 0 for t close to 0. Thus there is t n ∈ (0, 1] such that h n (t n ) = 1, that is, implying that t n u + n ∈ M. On the other hand, we also know that Once t n ∈ (0, 1], the last inequality gives t n u + n ∈ M and T (t n u + n ) → D, showing that {t n u + n } is a minimizing sequence for T . Thereby, without lost of generality, we can assume that {u n } is a nonnegative sequence.
Moreover, by noticing that where u * n is the Schwartz symmetrization of u n , any minimizing sequence can be assumed radially symmetric, non-negative and decreasing in r = |x|.
In what follow, we will use that the embedding (2.13) H α rad (R N ) ֒→ L p (R N ) is compact for all p ∈ (2, 2 * α ), see Lions [15] for more details. Due to the boundedness in H α (R N ) of the (non-negative and radial symmetric) minimizing sequence {u n } (see Lemma 2.3) we can assume that {u n } has a weak limit in H α (R N ) denoted hereafter with u. Observe also that, by the boundedness in L 2 (R N ) we have the uniform decay |u n (x)| ≤ C|x| −N/2 , see [2,Lemma 1]. Therefore, passing to a subsequence, if necessary, we deduce that the weak limit u is non-negative, radially symmetric and decreasing.
It turns out that the weak limit u is a solution of the minimizing problem (2.4) we were looking for. Before to see this some preliminary lemmas are in order to recover some compactness. Then Proof. First of all, we recall the limit T ′ (u n ) − λ n J ′ (u n ) → 0 as n → +∞ gives Using standard arguments, it is possible to prove that and Using the growth conditions on f , fixed q ∈ (2, 2 * α ) and given ε > 0, there exists C = C(ε, q) > 0 such that f (t)t ≤ εt 2 + C|t| q + (1 + ε)|t| 2 * α , ∀t ∈ R. From this, Now, using the definition of S, see (1.2), we get (2.14) Passing to the limit in (2.14), recalling that {v n } is bounded, and that u n → 0 in L q (R N ) (see (2.13)), we find By the arbitrariety of ε, we derive L ≤ D (L/S) 2 * α /2 , or equivalently, On the other hand (2.15) and the proof is finished.
In the next result the condition τ > τ * given in (f4) plays a crucial role.
This gives (by the definition of τ * ) exactly the conclusion.
Lemma 2.9. If u n ⇀ u in H α (R), then u n → u in D α,2 (R N ). In particular, u n → u in Proof. Of course v n = u n − u ⇀ 0 in H α (R). Suppose by contradiction that u n → u in D α,2 (R N ). Thereby, On the other hand, from Lemma 2.1 from which, using (2.17), it follows that This contradicts Lemma 2.8 and finishes the proof.

2.3.
Proof of Theorem 1.1. At this point we wish to show that D is attained by u, where u is the weak limit of {u n }. First of all, we know that so we just need to prove that u ∈ M. By [2, Lemma 1], there is R > 0 such that B R being the ball of radius R centered in 0. Since and u n → u in L 2 * α (B R ), the above information together with the Fatous' Lemma gives Suppose by contradiction that The growth conditions on f ensure that h(t) < 0 for t close to 0 and h(1) = R N G(u) dx > 1. Then, by the continuity of h, there exists t 0 ∈ (0, 1) such that h(t 0 ) = 1. Then, Consequently, by (2.18) which is absurd. Thus R N G(u) dx = 1, i.e. u ∈ M. The fact that the solution u of the minimizing problem gives rise to a ground state solution, follows by standard arguments; indeed, since u is a solution of the minimizing problem (2.4), i.e. D = T (u) = inf w∈M T (w), then there exists an associated Lagrange multiplier λ such that, in a weak sense, Now by testing the previous equation on the same minimizer u, we deduce that so that it has to be, by Lemma 3.5, T (u) ≥ λ > 0. Setting u σ (x) := u(σx) for σ > 0, we easily see that (−∆) α u σ = λσ 2α g(u σ ).
Choosing σ = λ 1/2α we obtain a solution of (P). Arguing as in [4,Theorem 3], u σ is a ground state solution. Denoting with G(u) = F (u) − u 2 2 , the primitive of g(u) = f (u) − u, we introduce the set it is, again as before, We point out here that, since we will deal with minimizing sequences {u n } for the minimization problem (3.2), as in the previous Section we suppose that u n is non-negative and radially symmetric. Moreover, we again define the min-max level associated to the functional I 3.2. Some preliminary stuff. Let us start with the following important result due to T. Ozawa [17] Theorem 3.1. There exists 0 < ω ≤ π such that, for all r ∈ (0, ω), there exists H r > 0 satisfying for all u ∈ H 1/2 (R) with |(−∆) 1/4 u| 2 2 ≤ 1. At this point we establish some preliminary results which will be useful in order to prove Theorem 1.2. Proof. Consider w ∈ C ∞ 0 (R) with w(x) > 0 and define a function From (f1) for t > 0 small we have For ε < 1, we get h(t) < 0 for t > 0 small. Now using (f4) we obtain Then, h(t) > 0 for t > 0 large and h ′ (t) > 0 for t > 0 large. Then there is a t > 0 such that Now we prove that M is a manifold. Indeed, if w ∈ M, then w = 0. Then, from (f1) and the fact that lim |x|→∞ w(x) = 0, there exists x 0 ∈ R such that g(w(x 0 )) < 0. Thereby, by continuity, there is an open interval B δ (x 0 ) such that g(w(x)) < 0, ∀x ∈ B δ (x 0 ).
As a consequence we can always find a φ ∈ C ∞ 0 ( Then, Proof. Without loss of generality, we can assume that there is v ∈ H 1/2 (R), radial, such that Now, choose ε > 0 small enough such that m = ρ (1−ǫ) 2 ∈ (0, 1) and set t = ω Now, setting P (s) = F (s) and Q(s) = e ts 2 − 1, from (f1), (f4) and the last inequality, we get Consequently the hypotheses of the Compactness Lemma of Strauss [4, Theorem A.I] are fulfilled. Hence P (v n ) converges to P (v) in L 1 (R), and then concluding the proof.
The relation between the ground state level and the minimax level defined in (3.3) is given in the following Proof. Arguing as in Lemma 3.2, given v ∈ H 1/2 (R) with v + = max{v, 0} = 0, there is t 0 > 0 such that t 0 v + ∈ M. Then, On the other hand, since f (s) = 0 for s ≤ 0, if v ∈ H 1/2 (R), v = 0 with v + = 0, then max t≥0 I(tv) = ∞. Hence in any case D ≤ b. and we can assume that there exists v ∈ H 1/2 (R), radial, such that v n ⇀ v in H 1/2 (R). From Lemma 3.3 we get Note that R G(v n ) dx = 0 implies R F (v n ) dx = 1 2 and R F (v) dx = 1 2 . Then v = 0. But implies v = 0 which is an absurd.
Proof. It is sufficient to repeat the same argument of the Lemma 2.8, recalling that now by (f4) it is τ * = q − 2 q (q−2)/2 C q/2 q concluding the proof.
3.3. Proof of Theorem 1.2. At this point we will show that D is attained by u, where u is the weak limit of {u n }. Indeed, since u n ⇀ u in H 1/2 (R) we have As in the previous case N ≥ 2, we just need to prove that u ∈ M, i.e. R N G(u) dx = 0. We again argue by contradiction by supposing that R G(u) dx > 0.
As for the case N ≥ 2, one show that the minimizer u of (3.2) gives rise to a ground state solution of (P).