Ground state for Choquard equation with doubly critical growth nonlinearity

In this paper we consider nonlinear Choquard equation −∆u + V(x)u = (Iα ∗ F(u)) f (u) in RN , where V ∈ C(RN), Iα denotes the Riesz potential, f (t) = |t|p−2t + |t|q−2t for all t ∈ R, N > 5 and α ∈ (0, N − 4). Under suitable conditions on V, we obtain that the Choquard equation with doubly critical growth nonlinearity, i.e., p = (N + α)/N, q = (N + α)/(N − 2), has a nonnegative ground state solution by variational methods.


Introduction and main results
In this paper we consider nonlinear Choquard equation where N 5, α ∈ (0, N − 4), I α is the Riesz potential given by Γ denotes the Gamma function, F(t) = |t| p /p + |t| q /q, f (t) = |t| p−2 t + |t| q−2 t for all t ∈ R, and the potential function V ∈ C(R N ) and satisfies (V) there exist V 0 , V ∞ > 0 such that V 0 V(x) V ∞ for all x ∈ R N , and lim |x|→∞ V(x) = V ∞ .
In view of the Hardy-Littlewood-Sobolev inequality, see Lemma 2.1 below, it can be shown that the energy functional corresponding to (1.2), for every α ∈ (0, N), is well defined on H 1 (R N ) and belongs to where (N + α)/N is called the lower critical exponent and (N + α)/(N − 2) is called the upper critical exponent.V. Moroz For the more content of the equation (1.2), we refer the interested reader to the guide [14].
When V is a positive constant, F ∈ C 1 and satisfies (F 1 ) there exists a positive constant C such that Moroz and Van Schaftingen [13] proved the existence of ground state to the equation (1.1).J. Seok [17] acts against the subcriticality condition (F 2 ), and consider that F is doubly critical, i.e., For the related critical problems involving only a single critical exponent, we refer to [1,12,16].However, few work concerns the case that F is doubly critical.J. Seok cleverly estimated the energy, overcome the lack of compactness, and proved that the equation (1.1) admits a nontrivial solution under appropriate assumptions on α and N in radial space H 1 r (R N ).Two natural questions arise.Does the solution has the least energy among nontrivial solutions of equation (1.1) in H 1 (R N )?Furthermore, does the equation (1.1) has ground state solution in H 1 (R N ) if V is not a constant?To the best of our knowledge, there are no results on these questions.The present paper is devoted to these aspects and answers these questions.Our main result is as follows.
We say that a function u ∈ H 1 (R N ) is a solution to (1.1) if J (u) = 0, for the definition of J, see (2.2) below.The solution u obtained in Theorem 1.1 is a ground state solution in the sense that it minimizes the corresponding energy functional J among all nontrivial solutions.
Since the appearance of the potential V breaks down the invariance under translations in R N , we cannot use the translation invariant argument directly.To overcome this challenge, we need use the comparison arguments between the minimax level of the energy functional corresponding to (1.1) and that to the limit equation Thus, we first need to study the existence of ground state solution to the equation (1.3).The result is stated as follows.
The proof of Theorem 1.2 relies on two ingredients: the nontrivial nature of solution to the equation (1.3) up to translation under the strict inequality obtained by a concentration-compactness argument (Lemma 3.2) and the proof of the latter strict inequality (Lemma 3.1).The rest of this paper is organized as follows.We give some preliminaries in Section 2. Theorems 1.2 and 1.1 are proved in Sections 3 and 4, respectively.
Throughout this paper we always use the following notations.The letters C i , i = 1, 2, . . .and C are positive constants which may change from line to line.

Preliminaries
In this section, we give some preliminaries.When V satisfies the condition (V), the following lemmas are all set up.
Let H 1 (R N ) be the usual Sobolev space.According to the conditions of the function V, we can define an equivalent norm on there exists a positive constant C s such that The energy functional J associated to the equation (1.1) is defined by By the Hardy-Littlewood-Sobolev inequality, see Lemma 2.1 below, we know that J is well defined on H 1 (R N ) and belongs to C 1 , and its derivative is given by Therefore, a weak solution of the equation (1.1) corresponds to a critical point of the energy functional J.
We consider the following constraint minimization problem where N denotes the Nehari manifold and N is C 1 .
To study the constraint minimization problem related with (1.1), we need to recall the following well-known Hardy-Littlewood-Sobolev inequality, see [7].
Proof.Assume that {v n } is an arbitrary subsequence of {u n }.Since v n → u in L s loc (R N ) for s ∈ [1, 2 * ), there exists a subsequence {w n } of {v n } such that w n → u a.e. on R N .By Fatou's lemma, we have Φ(u) lim inf n→∞ Φ(w n ).Thus, (2.6) holds.
Proof.For each u ∈ H 1 (R N ) \ {0}, the function g(t) := J(tu) takes the form for all t ∈ R + .By Remark 2.4 below, we see that g has a unique positive critical point t u corresponding to its maximum, i.e., g (t u ) = 0 and g(t u ) = max R + g.Hence, I(t u u) = t u g (t u ) = 0 and J(t u u) = max t∈R + J(tu).
Remark 2.4.Let a, b, c be positive constants.By elementary calculation one obtains that the function has a unique positive critical point t 0 with g (t) > 0 for all t ∈ (0, t 0 ), and g (t) < 0 for all t ∈ (t 0 , ∞).Thus, g takes the maximum at t = t 0 .

Lemma 2.5.
There exist positive constants δ and ρ such that u δ and I (u), u −ρ for all u ∈ N .
Proof.Because of the definition of N , by (2.4), the Hardy-Littlewood-Sobolev inequality and (2.1), we can derive that Since p, q > 1, there exists a positive constant δ such that u δ for all u ∈ N .Furthermore, by (2.4) and (2.10), we have that Set ρ = 2(p − 1)δ 2 .The proof is completed.
To obtain a (PS) c sequence of the energy functional J, we show that the functional has the mountain pass geometry.
Lemma 2.6.The functional J satisfies the mountain pass geometry, that is, (i) there exist r, η > 0 such that J(u) η for all u ∈ ∂B r = {u ∈ H 1 (R N ) : u = r}, and J(u) > 0 for all u with 0 < u r; (ii) there exists u 0 ∈ H 1 (R N ) such that u 0 > r and J(u 0 ) < 0.
Proof.(i) By the Hardy-Littlewood-Sobolev inequality and (2.1), we derive that Then (i) follows if r > 0 is small enough.

We define
where Then it follows from Lemma 2.6 (i) that c 1 > 0. Furthermore, we can show that the minimax value c 1 also can be characterized by c = c 1 = c 2 , where c is defined in (2.3), and According to Lemma 2.6 and [19, Theorem 2.8, p.41], there is a (PS) Because of p > 1, the above inequality induce that {u n } is bounded.
Lemma 2.8.Let {u n } ⊂ H 1 (R N ) be a (PS) c sequence of J and u n u in H 1 (R N ).If u = 0, then the equation (1.1) has a nonnegative ground state solution of the equation (1.1).

Proof. Since u n
u in H 1 (R N ), it follows from (2.7) that J (u) = 0.Because u = 0, we know that u ∈ N is a nonzero critical point of J. Using (2.6), we obtain which implied that J(u) = c.Consider w = |u|.An easy computation shows that w ∈ N and J(w) = J(u) = c.It follows from the Lagrange multiplier theorem that J (w) = λI (w) for some λ ∈ R. Hence, λ I (w), w = J (w), w = 0.By Lemma 2.5, we know that I (w), w −ρ.Thus, λ = 0, which implies that w is a nonnegative solution of the equation (1.1).Since J(w) = c, it is a nonnegative ground state solution to the equation (1.1).

Proof of Theorem 1.2
Before giving the proof of Theorem 1.1, we need give the proof of Theorem 1.2.In this section, V = V ∞ .
It follows from 4 + α < N that q < 2 and Since u n δ for sufficiently large n, we can assume lim n→∞ t n = 1.Thus, we have that which contradict to Lemma 4.1.Thus, u = 0.The proof is completed.