Normalized solutions for a class of nonlinear Choquard equations

Abstract

We prove the existence of a least energy solution to the problem

$$\begin{aligned} -\Delta u-(I_{\alpha }*F(u))f(u)=\lambda u\ \text { in }\ {\mathbb {R}}^{N},\quad \int _{{\mathbb {R}}^N}u^2(x)dx = a^2, \end{aligned}$$

where $N\ge 1$, $\alpha \in (0,N)$, $F(s):=\int _{0}^{s}f(t)dt$, and $I_{\alpha }:{\mathbb {R}}^{N}\rightarrow {\mathbb {R}}$ is the Riesz potential. If f is odd in u then we prove the existence of infinitely many normalized solutions.

1 Introduction

We consider the equation

$$\begin{aligned} -\Delta u-(I_{\alpha }*F(u))f(u)=\lambda u\quad \text { in }\ {\mathbb {R}}^{N}, \end{aligned}$$

(1.1)

where $N\ge 1$, $\alpha \in (0,N)$, $F(s):=\int _{0}^{s}f(t)dt$, and $I_{\alpha }:{\mathbb {R}}^{N}\rightarrow {\mathbb {R}}$ is the Riesz potential defined by

$$\begin{aligned} I_{\alpha }(x) = \frac{\Gamma \left( \frac{N-\alpha }{2}\right) }{\Gamma \left( \frac{\alpha }{2} \right) \pi ^{N/2}2^{\alpha }\left| x\right| ^{N-\alpha }}\quad \text { for }\ x\in {\mathbb {R}}^{N}\backslash \{0\}. \end{aligned}$$

Equation (1.1) is usually called the nonlinear Choquard equation. It is a semilinear elliptic equation with a nonlocal nonlinearity. For the physical case $N=3$, $\alpha =2$ and $f(s)=s$, (1.1) is the Choquard–Pekar equation which goes back to the 1954’s work by S. Pekar on quantum theory of a polaron at rest [28]. The Choquard equation was also introduced by P. Choquard in 1976 in the modelling of a one-component plasma [20]. Mathematically, in [20,21,22], Lieb and Lions opened the way to an intensive study of (1.1). The existence and qualitative properties of solutions of the Choquard equation (1.1) have been studied for a few decades by variational methods; see [9, 10, 17, 25,26,27] and further references therein, for instance. Almost all papers deal with the case of fixed $\lambda$.

From a physical point of view, it is interesting to find solutions of (1.1) with prescribed $L^{2}$-norms, $\lambda$ appearing as Lagrange multiplier. Solutions of this type are often referred to as normalized solutions. The present paper is devoted to such solutions. We study the existence and multiplicity of solutions to the problem

$$\begin{aligned} \left\{ \begin{array}{ll}-\Delta u-(I_{\alpha }*F(u))f(u)=\lambda u&{}\quad \text {in }\ {\mathbb {R}}^{N},\\ \int _{{\mathbb {R}}^N}u^2dx=a^2,&{}\quad u\in H^1({\mathbb {R}}^N),\ \ \lambda \in {\mathbb {R}}, \end{array}\right. \end{aligned}$$

(1.2)

where a is a positive constant. By a solution we mean a couple $(\lambda ,u)$ which satisfies (1.2) with $\lambda$ being a Lagrangian multiplier.

The existence of normalized solutions $(\lambda ,u)$ to the semilinear elliptic equation

$$\begin{aligned} -\Delta u-g(u)=\lambda u\quad \text {in}\ {\mathbb {R}}^{N},\ \lambda \in {\mathbb {R}}\end{aligned}$$

(1.3)

has achieved considerable attention recently. Mathematically, this is a more challenging issue than the existence of solutions of (1.3) with a prescribed frequency $\lambda$, and is much less understood. Normalized solutions can be obtained as critical points of a functional constrained to an $L^2$ sphere. The difficulty here is that even for subcritical nonlinearities the functional does not satisfy the Palais-Smale condition. In fact, a Palais-Smale sequence ((PS) sequence for short in the following) does not even need to be bounded. In order to overcome the difficulties in dealing with normalized solutions, L. Jeanjean in [12] introduced a stretched functional whose critical points are solutions of (1.3) and whose (PS) sequences carry information of the Pohozaev identity, which can be used to prove boundedness. Then another difficulty appears, namely that a weak limit of a bounded (PS) sequence need not lie on the prescribed $L^2$ sphere. In order to control this it is important to bound the Lagrange multipliers. Using a mountain pass structure for the stretched functional L. Jeanjean in [12] proved the existence of at least one normalized solution of (1.3). The existence of infinitely many normalized solutions of (1.3) was later proved by T. Bartsch and S. de Valeriola in [1] using a new linking geometry for the stretched functional (see also the papers by T. Bartsch and N. Soave [4, 5]). More results on normalized solutions for scalar equations and systems can be found in [2, 3, 6, 7, 13,14,15, 19, 24, 30, 31].

We know only a few papers dealing with the existence of normalized solutions of the Choquard equation. In the case $N\ge 3$, G. Li and H. Ye in [19] obtained a ground state solution $(\lambda _a,u_a)$ of (1.2) under a set of assumptions on f, which when f takes the special form $f(s)=C_1|s|^{r-2}s+C_2|s|^{p-2}s$ requires that $\frac{N+\alpha +2}{N}<r\le p<\frac{N+\alpha }{N-2}$. For a monomial nonlinearity $f(s)=|s|^{p-2}s$, qualitative properties including existence and nonexistence of minimizers of the functional associated to the nonlinear Choquard equation were discussed by H. Ye in [32].

The goal of this paper is to first prove the existence of a least energy solution of (1.2) in all dimensions $N\ge 1$, including $N=1,2$. We believe that our proof is simpler and more transparent than the one from [19]. In addition we prove the existence of infinitely many solutions of (1.2) if f is odd. This will be based on a linking argument which in turn depends on a certain topological intersection property. This has been proved in [1] using a cohomological index theory. Here we present a new and more elementary proof of this property using only the classical Borsuk-Ulam theorem.

We now present our results. Let $2_{\alpha }^{*}:=(N+\alpha )/(N-2)$ if $N\ge 3$, and $2_{\alpha }^{*}:=+\infty$ if $N=1,2$. We assume the following hypotheses on the nonlinearity.

$(f_{1})$:

$f\in C^0({\mathbb {R}})$ and there exist r, $p\in {\mathbb {R}}$ satisfying

$$\begin{aligned} \frac{N+\alpha +2}{N}<r\le p<2_{\alpha }^{*} \end{aligned}$$

such that

$$\begin{aligned} 0<rF(s)\le f(s)s\le pF(s)\quad \text { for }s\ne 0. \end{aligned}$$

$(f_{2})$:

The function $\widetilde{F}(s):=f(s)s-\frac{N+\alpha }{N}F(s)$ satisfies:

$$\begin{aligned} \frac{\widetilde{F}(s)}{|s|^{(N+\alpha +2)/N}}\ \text { is nonincreasing in } (-\infty ,0) \text { and nondecreasing in } (0,+\infty ). \end{aligned}$$

Let

$$\begin{aligned} J(u)=\frac{1}{2}\int _{{\mathbb {R}}^{N}}\left| \nabla u\right| ^{2}dx- \frac{1}{2}\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u))F(u)dx \end{aligned}$$

(1.4)

be the corresponding variational functional of (1.2) defined on the constraint

$$\begin{aligned} S(a)=\{u\in H^{1}({\mathbb {R}}^{N}):\ \left\| u\right\| _{L^{2}({\mathbb {R}}^{N})}=a\}. \end{aligned}$$

Setting

$$\begin{aligned} I(u)=\int _{{\mathbb {R}}^{N}}\left| \nabla u\right| ^{2}dx+\frac{ N+\alpha }{2}\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u))F(u)dx-\frac{N}{2} \int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u))f(u)u\,dx \end{aligned}$$

(1.5)

every solution of (1.2) lies on the Pohozaev manifold

$$\begin{aligned} V(a)=\{u\in S(a):\ I(u)=0\}. \end{aligned}$$

Our first main result states that

$$\begin{aligned} m(a):=\underset{u\in V(a)}{\inf }J(u) \end{aligned}$$

is achieved by a solution.

Theorem 1.1

If $(f_{1})$ and $(f_{2})$ hold then, for any $a>0$, problem (1.2) possesses a solution $(\lambda _a,u_a)\in {\mathbb {R}}\times H^{1}({\mathbb {R}}^{N})$ such that $\lambda _a<0$ and $J(u_a)=m(a)$.

Remark 1.2

• (a) Due to the radial symmetry of (1.2) we may also work on the space $E:=H^1_{rad}({\mathbb {R}}^N)$ of radial functions. A critical point of $J|_{S(a)\cap E}$ is also a critical point of $J|_{S(a)}$ by the principle of symmetic criticality. It is even simpler to obtain a solution $(\lambda _a^{rad},u_a^{rad})\in {\mathbb {R}}\times E$ such that

  $$\begin{aligned} J(u_a^{rad}) = m_{rad}(a) := \inf _{u\in V(a)\cap E} J(u) \end{aligned}$$

  because E embeds compactly into $L^q({\mathbb {R}}^N)$ for $2<q<2^*$. It is an open problem whether $m(a)=m_{rad}(a)$. Observe that V(a) is not invariant under symmetrization.

• (b) Replacing f by $f^+:=\max \{f,0\}$ one obtains by a symmetrization argument a least energy solution $(\widetilde{\lambda }_a,\widetilde{u}_a)$ of

  $$\begin{aligned} \left\{ \begin{array}{ll}-\Delta u-(I_{\alpha }*F^+(u))f^+(u)=\lambda u&{}\quad \text { in }\ {\mathbb {R}}^{N},\\ \int _{{\mathbb {R}}^N}u^2dx=a^2,\ \ u\in H^1({\mathbb {R}}^N),&{}\quad \lambda \in {\mathbb {R}}, \end{array}\right. \end{aligned}$$

  with $\widetilde{\lambda }_a<0$ and $\widetilde{u}_a>0$ being radial, where $F^+(s):=\int _0^sf^+(t)dt$. This is also a solution of (1.2). Similarly (1.2) has a solution $(\widehat{\lambda }_a,\widehat{u}_a)$ such that $\widehat{\lambda }_a<0$ and $\widehat{u}_a<0$ being radial. This suggests of course that there should exist a third solution of mountain pass type on V(a), i.e. of Morse index 2 on S(a). To obtain this one may work in the space of radial functions. If f is odd then $\widetilde{u}_a=-\widehat{u}_a$ are least energy solutions. In general the relation between $\widetilde{u}_a$, $\widehat{u}_a$ and the least energy solution $u_a$ from Theorem 1.1 is not clear but we conjecture that $u_a$ does not change sign.

• (c) For $N\ge 3$, a similar result has been proved in [19]. Here we include the dimensions $N=1,2$. The idea of the argument in [19] is as follows. The stretched functional method from [12] was first used to find a (PS) sequence $\{v_n\}$ for $J|_{S(a)}$ at the level m(a) such that $I(v_n)\rightarrow 0$. Then it was proved that there exists $\lambda _a\in {\mathbb {R}}$ such that $\langle J'(v_n),v_n\rangle /\Vert v_n\Vert _{L^2({\mathbb {R}}^N)}^2\rightarrow \lambda _a$. After that, the key step was to prove that $\lambda _a<0$. In order to achieve this, it was proved that $\{v_n\}$ is a (PS) sequence of the functional $\Psi$ defined by

  $$\begin{aligned} \Psi (u)=J(u)-\frac{1}{2}\lambda _a\int _{{\mathbb {R}}^N}u^2dx \end{aligned}$$

  and that m(a) has a strict subadditivity property. This strict subadditivity property together with decomposition properties of (PS) sequences of $\Psi$ make it possible to prove $\lambda _a<0$ and that there exists $u_a\in H^1({\mathbb {R}}^N){\setminus }\{0\}$ such that $(\lambda _a,u_a)$ is a solution of (1.2). We will also use the stretched functional method from [12] to obtain a (PS) sequence $\{v_n\}$ for $J|_{S(a)}$ at the level m(a) with the property $I(v_n)=o(1)$. But unlike [19], we use a concentration compactness argument to find a solution $(\lambda _a,u_a)$ with $\lambda _a<0$ and $u_a\ne 0$ of (1.1) from the (PS) sequence $\{v_n\}$ directly. Then we prove that m(a) is strictly decreasing in a. This property combined with the fact that $I(u_a)=\lim _{n\rightarrow \infty }I(v_n)=0$ makes it possible for us to prove that $(\lambda _a,u_a)$ is indeed a solution of (1.2) and $J(u_a)=m(a)$. The proof presented here is shorter, easier, and more transparent than the one in [19].

Our second main result deals with odd nonlinearities.

Theorem 1.3

If $N\ge 2$, $(f_{1})-(f_{2})$ hold and f is odd, then for any $a>0$, problem (1.2) possesses an unbounded sequence of pairs of radial solutions $(\lambda _k,\pm u_k)$ with $\lambda _k<0$ and $J(u_k) \rightarrow \infty$ as $k\rightarrow \infty$.

When we say that $(\lambda ,u)$ is a radial solution, we mean that $u\in E=H^1_{rad}({\mathbb {R}}^N)$.

Remark 1.4

For the semilinear equation (1.3) infinitely many normalized solutions have been obtained via two different approaches in [1] and in [4]. In [1], the stretched functional method incorporating a new linking structure of the associated functional produces a bounded (PS) sequence and leads to normalized solutions at a sequence of energy levels $c_n\rightarrow \infty$ which is constructed by a minimax procedure. In [4], see also [5] for systems, the authors considered the functional corresponding to J, constrained to the Pohozaev manifold corresponding to V(a). In this way they avoided the introduction of the stretched functional. In the present paper we extend the stretched functional method to deal with (1.2).

The paper is organized as follows. In Sect. 2 we prove Theorem 1.1. We first show that m(a) is the mountain pass level of $J|_{S(a)}$, and we use the stretched functional method to obtain a (PS) sequence $\{v_n\}$ for $J|_{S(a)}$, which satisfies $J(v_n)\rightarrow m(a)$ and $I(v_n)\rightarrow 0$. Then we use a concentration compactness argument, which does not rely on the compactness from radial symmetry [16, 26], to show that, up to translations, $v_n\rightarrow u_a$ weakly in $H^1({\mathbb {R}}^N)$ for some $u_a\ne 0$ and that there exists $\lambda _a<0$ such that $(\lambda _a,u_a)$ is a weak solution of (1.1). Next we show that m(a) is strictly decreasing. This fact is used to show that $\Vert u_a\Vert _{L^2({\mathbb {R}}^N)}=a$ and $J(u_a)=m(a)$. Therefore, $(\lambda _a,u_a)$ is a weak solution of (1.2). We prove Theorem 1.3 in Sect. 3, working in the subspace $H^1_\text {rad}({\mathbb {R}}^N)$ of $H^1({\mathbb {R}}^N)$ consisting of radially symmetric functions. Here we first present a new and more elementary proof of the intersection lemma from [1]. This and a suitable equivariant pseudogradient vector field for the stretched functional will be used to construct an unbounded sequence of minimax values $c_k$ for $J|_{S(a)}$ together with (PS)$_{c_k}$ sequences $\{v_{k,n}\}_{n=1}^\infty$ at each minimax value $c_k$ satisfying $\lim _{n\rightarrow \infty }I(v_{k,n})=0$, which is used to show that $\{v_{k,n}\}_{n=1}^\infty$ is bounded. Then we use the compactness of the imbedding $H^1_\text {rad}({\mathbb {R}}^N)\hookrightarrow L^q(\mathbb R^N)$ for $q\in (2,2^*)$ to show that $\{v_{k,n}\}_{n=1}^\infty$ converges along a subsequence strongly in E to some $u_k\in S(a)$ as $n\rightarrow \infty$.

2 Proof of Theorem 1.1

Recall the definition of J and I in (1.4) and (1.5), respectively. Observe that $(f_{1})$ implies, for $s\in {\mathbb {R}}$,

$$\begin{aligned} t^{p}F(s)\le F(ts)\le t^{r}F(s)\quad \text { if }\ 0\le t\le 1, \end{aligned}$$

(2.1)

and

$$\begin{aligned} t^{r}F(s)\le F(ts)\le t^{p}F(s)\quad \text { if }\ t>1. \end{aligned}$$

(2.2)

These inequalities will be used frequently in what follows.

For each $u\in S(a)$ and $t\in {\mathbb {R}}$, set

$$\begin{aligned} u^{t}(x)=e^{\frac{Nt}{2}}u(e^tx). \end{aligned}$$

It is clear that $u^t\in S(a)$ if $u\in S(a)$ and $t\in {\mathbb {R}}$. The following lemma asserts that on the curve $t\mapsto u^t$ there exists a unique point belonging to V(a), at which $J(u^t)$ achieves its maximum.

Lemma 2.1

• (a) For every $u\in S(a)$, there exists a unique $t(u)\in {\mathbb {R}}$ such that $u^{t(u)}\in V(a)$ and $J(u^{t(u)})={\max _{t\in \mathbb R}}J(u^{t})$. The map

  $$\begin{aligned} S(a) \rightarrow V(a)\times {\mathbb {R}},\quad u \mapsto \left( u^{t(u)},t(u)\right) \end{aligned}$$

  is a homeomorphism with inverse

  $$\begin{aligned} V(a)\times {\mathbb {R}}\rightarrow S(a),\quad (u,t) \mapsto u^{-t}. \end{aligned}$$

• (b) $\displaystyle I(u)= \frac{d}{dt}\bigg |_{t=0} J(u^t)$.

Proof

For $u\in S(a)$ and $t\in {\mathbb {R}}$, we have

$$\begin{aligned} J(u^{t})=\, & {} \frac{1}{2}\int _{{\mathbb {R}}^{N}}\left| \nabla u^{t}\right| ^{2}dx-\frac{1}{2}\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u^{t}))F(u^{t})dx \nonumber \\= \, & {} \frac{e^{2t}}{2}\int _{{\mathbb {R}}^{N}}\left| \nabla u\right| ^{2}dx-\frac{1}{2e^{(N+\alpha )t}}\int _{{\mathbb {R}}^{N}}\left( I_{\alpha }*F\left( e^{\frac{Nt}{2}}u\right) \right) F\left( e^{\frac{Nt}{2}}u\right) dx. \end{aligned}$$

(2.3)

Then

$$\begin{aligned} \frac{d}{dt}J(u^{t})= \, & {} e^{2t}\int _{{\mathbb {R}}^{N}}\left| \nabla u\right| ^{2}dx+\frac{N+\alpha }{2e^{(N+\alpha )t}}\int _{{\mathbb {R}}^{N}}\left( I_{\alpha }*F\left( e^{\frac{Nt}{2}}u\right) \right) F\left( e^{\frac{Nt}{2}}u\right) dx \nonumber \\&-\frac{N}{2e^{(N+\alpha )t}}\int _{{\mathbb {R}}^{N}}\left( I_{\alpha }*F\left( e^{\frac{Nt}{2}}u\right) \right) f\left( e^{\frac{Nt}{2}}u\right) e^{\frac{Nt}{2}}udx \nonumber \\= \, & {} e^{2t}\left( \int _{{\mathbb {R}}^{N}}\left| \nabla u\right| ^{2}dx-\frac{N}{2}\psi (t)\right) , \end{aligned}$$

(2.4)

where

$$\begin{aligned} \psi (t)=\int _{{\mathbb {R}}^{N}}\left( I_{\alpha } *\frac{F(e^{\frac{Nt}{2}}u)}{(e^{\frac{Nt}{2}})^{\frac{N+\alpha +2}{N}}}\right) \frac{\widetilde{F}(e^{\frac{Nt}{2}}u) }{(e^{\frac{Nt}{2}})^{\frac{N+\alpha +2}{N}}}dx. \end{aligned}$$

From (2.4) we obtain the result b).

For any $s\in {\mathbb {R}}$, $s\ne 0$, from $(f_1)$ and $(f_2)$ we see that both the functions $\frac{F(ts)}{t^r}$ and $\frac{\widetilde{F}(ts)}{t^{(N+\alpha +2)/N}}$ are nondecreasing in $t\in (0,\infty )$. Moreover, since $r>\frac{N+\alpha +2}{N}$ and

$$\begin{aligned} \frac{F(ts)}{t^{(N+\alpha +2)/N}}= \frac{F(ts)}{t^r}t^{r-\frac{N+\alpha +2}{N}}, \end{aligned}$$

we deduce that $\frac{F(ts)}{t^{(N+\alpha +2)/N}}$ is strictly increasing in $t\in (0,\infty )$. This implies $\psi (t)$ is strictly increasing in $t\in {\mathbb {R}}$ and there is at most one $t\in {\mathbb {R}}$ such that $\frac{d}{dt}J(u^t)=0$. By (2.1) and (2.2), we have

$$\begin{aligned} \lim _{t\rightarrow 0}\frac{F(ts)}{t^{(N+\alpha +2)/N}}=0\quad \text { and }\quad \lim _{t\rightarrow +\infty }\frac{F(ts)}{ t^{(N+\alpha +2)/N}}=+\infty . \end{aligned}$$

(2.5)

Since

$$\begin{aligned} \left( r-\frac{N+\alpha }{N}\right) F(s)\le {\widetilde{F}}(s)\le \left( p-\frac{N+\alpha }{N}\right) F(s) \end{aligned}$$

and since $(N+\alpha +2)/N<r\le p$, we also have

$$\begin{aligned} \lim _{t\rightarrow 0}\frac{{\widetilde{F}}(ts)}{t^{(N+\alpha +2)/N}}=0\quad \text { and }\quad \lim _{t\rightarrow +\infty }\frac{\widetilde{F}(ts)}{ t^{(N+\alpha +2)/N}}=+\infty . \end{aligned}$$

(2.6)

From (2.5) and (2.6), we deduce $\psi (t)\rightarrow 0$ as $t\rightarrow -\infty$ using the Lebesgue dominated convergence theorem, and we have $\psi (t)\rightarrow +\infty$ as $t\rightarrow \infty$ by the Fatou Lemma. As a consequence of (2.4), there exists exactly one $t=t(u)\in {\mathbb {R}}$ such that $\frac{d}{dt}J(u^{t})|_{t=t(u)}=0$, and moreover, $\frac{d}{dt}J(u^{t})>0$ for $t<t(u)$ and $\frac{d}{dt}J(u^{t})<0$ for $t>t(u)$. Therefore, there exists a unique $t(u)\in {\mathbb {R}}$ such that $u^{t(u)}\in V(a)$ and $J(u^{t(u)})={\max }_{t\in {\mathbb {R}}}J(u^{t})$. Now assume that $\{u_n\}\subset S(a)$ and $u_n\rightarrow u$ in $H^1({\mathbb {R}}^N)$, and let $t_n=t(u_n)$. Then

$$\begin{aligned} \frac{2}{N}\int _{{\mathbb {R}}^{N}}\left| \nabla u_n\right| ^{2}dx=\int _{{\mathbb {R}}^{N}}\left( I_{\alpha } *\frac{F\left( e^{\frac{Nt_n}{2}}u_n\right) }{\left( e^{\frac{Nt_n}{2}}\right) ^{\frac{N+\alpha +2}{N}}}\right) \frac{\widetilde{F}\left( e^{\frac{Nt_n}{2}}u_n\right) }{\left( e^{\frac{Nt_n}{2}}\right) ^{\frac{N+\alpha +2}{N}}}dx. \end{aligned}$$

Using the Lebesgue dominated convergence theorem and the Fatou Lemma again, we see that the sequence $\{t_n\}$ is bounded. Assume $t_n\rightarrow t^*\in {\mathbb {R}}$, passing to a subsequence if necessary. Then passing to the limit in the last equation yields

$$\begin{aligned} \frac{2}{N}\int _{{\mathbb {R}}^{N}}\left| \nabla u\right| ^{2}dx=\int _{{\mathbb {R}}^{N}}\left( I_{\alpha } *\frac{F\left( e^{\frac{Nt^*}{2}}u\right) }{\left( e^{\frac{Nt^*}{2}}\right) ^{\frac{N+\alpha +2}{N}}}\right) \frac{\widetilde{F}\left( e^{\frac{Nt^*}{2}}u\right) }{\left( e^{\frac{Nt^*}{2}}\right) ^{\frac{N+\alpha +2}{N}}}dx, \end{aligned}$$

which together with the uniqueness of t(u) implies that $t(u)=t^*=\lim _{n\rightarrow \infty }t(u_n)$. Hence t(u) is continuous in u, hence the map

$$\begin{aligned} S(a) \rightarrow V(a)\times {\mathbb {R}},\quad u \mapsto \left( u^{t(u)},t(u)\right) \end{aligned}$$

is a homeomorphism because $u^t$ is continuous in (u, t). This proves a). $\square$

In the next lemma, we show that J has a mountain pass structure on S(a) as in [12, 19], and we use the stretched functional method as in [12] to find a (PS) sequence $\{v_n\}$ of $J|_{S(a)}$ at the mountain pass level such that $I(v_n)\rightarrow 0$ as $n\rightarrow \infty$.

Lemma 2.2

• (a) Let $D_{k}:=\{u\in S(a):\ \int _{{\mathbb {R}}^{N}}\left| \nabla u\right| ^{2}dx\le k\}$. There exist $0<k_{1}<k_{2}$ such that

  $$\begin{aligned} 0<\underset{u\in D_{k_{1}}}{\sup }J(u)<\underset{u\in \partial D_{k_{2}}}{ \inf }J(u)\quad \text { and }\quad J(u)>0\quad \text { for }\quad u\in D_{k_{2}}. \end{aligned}$$

• (b) Setting

  $$\begin{aligned} \Gamma (a):=\{\gamma \in C([0,1],\ S(a)):\ \gamma (0)\in D_{k_{1}},J(\gamma (1))<0\}, \end{aligned}$$

  there holds

  $$\begin{aligned} m(a)=\underset{\gamma \in \Gamma (a)}{\inf }\ \underset{t \in [0,1]}{\sup }J(\gamma (t)) > 0. \end{aligned}$$

  Moreover, there exists a sequence $\{v_{n}\}\subset S(a)$ such that, as $n\rightarrow +\infty$,

  $$\begin{aligned} J(v_{n}) \rightarrow m(a), \quad \big (J|_{S(a)}\big )^{\prime }(v_{n}) \rightarrow 0, \quad I(v_{n}) \rightarrow 0. \end{aligned}$$

Proof

• (a) We first claim that J has a mountain pass geometry on S(a). Let $u\in S(a)$. From (2.1) and (2.2), we have, for $s\in {\mathbb {R}}$,

  $$\begin{aligned} F(s)\le (F(-1)+F(1))(\left| s\right| ^{r}+\left| s\right| ^{p}), \end{aligned}$$

  and then

  $$\begin{aligned} \int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u))F(u)dx \le (F(-1)+F(1))^2\int _{{\mathbb {R}}^{N}}(I_{\alpha }*(\left| s\right| ^{r}+\left| s\right| ^{p}))(\left| s\right| ^{r}+\left| s\right| ^{p})dx. \end{aligned}$$

  Let $C>0$ be a constant depending only on $N,\alpha ,a,r$, and p, which may change from line to line. Using the Hardy-Littlewood-Sobolev inequality, the Gagliardo-Nirenberg inequality and the Sobolev embedding inequality, we obtain for $u\in S(a)$

  $$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u))F(u)dx \\&\quad \le C\left[ \int _{{\mathbb {R}}^{N}}(I_{\alpha }*\left| u\right| ^{r})\left| u\right| ^{r}dx+2\int _{{\mathbb {R}}^{N}}(I_{\alpha }*\left| u\right| ^{r})\left| u\right| ^{p}dx+\int _{{\mathbb {R}}^{N}}(I_{\alpha }*\left| u\right| ^{p})\left| u\right| ^{p}dx\right] \\&\quad \le C\left( \left\| u\right\| _{L^{2Nr/(N+\alpha )}({\mathbb {R}}^{N})}^{2r}+\left\| u\right\| _{L^{2Np/(N+\alpha )}({\mathbb {R}}^{N})}^{2p}\right) \\&\quad \le C\left( \left\| \nabla u\right\| _{L^{2}({\mathbb {R}}^{N})}^{rN-N-\alpha }+\left\| \nabla u\right\| _{L^{2}({\mathbb {R}}^{N})}^{pN-N-\alpha }\right) . \end{aligned} \end{aligned}$$

  (2.7)

  This implies that

  $$\begin{aligned} J(u)\ge \frac{1}{2}\int _{{\mathbb {R}}^{N}}\left| \nabla u\right| ^{2}dx-\frac{C}{2}\left\| \nabla u\right\| _{L^{2}({\mathbb {R}}^{N})}^{rN-N-\alpha }-\frac{C}{2}\left\| \nabla u\right\| _{L^{2}({\mathbb {R}}^{N})}^{pN-N-\alpha }. \end{aligned}$$

  On the other hand, we have

  $$\begin{aligned} J(u)\le \frac{1}{2}\int _{{\mathbb {R}}^{N}}\left| \nabla u\right| ^{2}dx. \end{aligned}$$

  Since $pN-N-\alpha \ge rN-N-\alpha >2$, there exist $0<k_{1}<k_{2}$ small enough such that

  $$\begin{aligned} 0<\underset{u\in D_{k_{1}}}{\sup }J(u)<\underset{u\in \partial D_{k_{2}}}{ \inf }J(u)\quad \text { and }\quad J(u)>0\quad \text { for }\quad u\in D_{k_{2}}. \end{aligned}$$

• (b) Clearly,

  $$\begin{aligned} \lim _{t\rightarrow -\infty }\int _{{\mathbb {R}}^{N}}\left| \nabla u^{t}\right| ^{2}dx=0\quad \text { and }\quad \lim _{t\rightarrow +\infty }\int _{{\mathbb {R}}^{N}}\left| \nabla u^{t}\right| ^{2}dx=+\infty . \end{aligned}$$

  Using (2.3), we rewrite $J(u^t)$ as

  $$\begin{aligned} J(u^t)=\frac{e^{2t}}{2}\left( \int _{{\mathbb {R}}^{N}}\left| \nabla u\right| ^{2}dx-\varphi (t)\right) , \end{aligned}$$

  where

  $$\begin{aligned} \varphi (t)=\int _{{\mathbb {R}}^{N}}\left( I_{\alpha }*\frac{F\left( e^{\frac{Nt}{2}}u\right) }{\left( e^{\frac{Nt}{2}}\right) ^{\frac{N+\alpha +2}{N}}}\right) \frac{F\left( e^{\frac{Nt}{2}}u\right) }{\left( e^{\frac{Nt}{2}}\right) ^{\frac{N+\alpha +2}{N}}}dx. \end{aligned}$$

  From (2.5), a similar discussion as in the proof of Lemma 2.1 shows that $\varphi (t)\rightarrow 0$ as $t\rightarrow -\infty$, and $\varphi (t)\rightarrow \infty$ as $t\rightarrow \infty$, hence

  $$\begin{aligned} \lim _{t\rightarrow -\infty }J(u^{t})=0\quad \text { and }\quad \lim _{t\rightarrow +\infty }J(u^{t})=-\infty . \end{aligned}$$

  (2.8)

  Then for $u\in V(a)$, there exist $t_{1}<0$ and $t_{2}>0$ such that $u^{t_{1}}\in D_{k_{1}}$ and $J(u^{t_{2}})<0$. So, if we set $g(t)=u^{(1-t)t_{1}+tt_{2}}$ for $t\in [0,1]$, then $g\in \Gamma (a)$ and $\underset{t\in [0,1]}{\sup }J(g(t))=J(u)$. This implies

  $$\begin{aligned} m(a)\ge \underset{\gamma \in \Gamma (a)}{\inf }\ \underset{t \in [0,1]}{\sup }J(\gamma (t))\ge \underset{u\in \partial D_{k_{2}}}{\inf }J(u)>0. \end{aligned}$$

  (2.9)

  On the other hand, from Lemma 2.1 we see that $S(a){\setminus } V(a)$ has precisely two components given by $S^\pm (a):=\{u\in S(a):\ \pm I(u)>0\}$. The result a) of this lemma implies that $D_{k_1}\subset S^+(a)$. By $(f_1)$, if $J(u)<0$ then $I(u)<0$. This shows $J^0:=\{u\in S(a):\ J(u)<0\}\subset S^-(a)$. Hence any path in $\Gamma (a)$ must intersect V(a). This property together with (2.9) implies

  $$\begin{aligned} m(a)=\underset{\gamma \in \Gamma (a)}{\inf }\ \underset{t \in [0,1]}{\sup }J(\gamma (t))>0. \end{aligned}$$

  Now we recall the stretched functional introduced first in [12]:

  $$\begin{aligned} \widetilde{J}:H^{1}({\mathbb {R}}^{N})\times {\mathbb {R}}\rightarrow {\mathbb {R}}, \, \quad (u,t)\mapsto J(u^{t}) \end{aligned}$$

  and define

  $$\begin{aligned} {\widetilde{\Gamma }}(a):=\{g\in C([0,1], S(a)\times {\mathbb {R}}):\ g(0)\in D_{k_{1}}\times \{0\},\ g(1)\in J^{0}\times \{0\}\}. \end{aligned}$$

  Since for $\gamma \in \Gamma (a)$,

  $$\begin{aligned} g(\cdot ):=(\gamma (\cdot ),0)\in {\widetilde{\Gamma }}(a)\quad \text { and }\quad \widetilde{J}(g(t))=J(\gamma (t))\ \text { for }\ t\in [0,1] \end{aligned}$$

  and for $g=(g_1,g_2) \in \widetilde{\Gamma }(a)$,

  $$\begin{aligned} \gamma (\cdot ):=g_{1}(\cdot )^{g_{2}(\cdot )}\in \Gamma (a)\quad \text { and }\quad J(\gamma (t))=\widetilde{J}(g(t))\ \text { for }\ t\in [0,1], \end{aligned}$$

  we have

  $$\begin{aligned} \underset{g\in \widetilde{\Gamma }(a)}{\inf }\ \underset{t\in [0,1]}{\sup }\widetilde{J}(g(t))=m(a). \end{aligned}$$

  Then, using the Ekeland variational principle as in [12, Lemma 2.3], it follows that there exists a sequence $\{(v_n,t_n)\}\subset S(a)\times {\mathbb {R}}$ such that, as $n\rightarrow +\infty$,

  $$\begin{aligned} t_n\rightarrow 0,\quad \widetilde{J}(v_n,t_n) \rightarrow m(a), \quad \left( \widetilde{J}|_{S(a)\times {\mathbb {R}}}\right) '(v_n,t_n)\rightarrow 0. \end{aligned}$$

  Without loss of generality, we may assume that $t_{n}=0$ because $\widetilde{J} (v_{n},t_n)=\widetilde{J}(v_{n}^{t_{n}},0)$ and

  $$\begin{aligned} \left( \widetilde{J}|_{S(a)\times {\mathbb {R}}}\right) '(v_{n},t_n)[\varphi ,s] = \left( \widetilde{J}|_{S(a)\times {\mathbb {R}}}\right) '(v_{n}^{t_n},0)[\varphi ^{t_n},s] \end{aligned}$$

  for all $s\in {\mathbb {R}}$ and $\varphi \in H^1({\mathbb {R}}^N)$ with $\int _{{\mathbb {R}}^N}v_n\varphi dx=0$. This implies that (see [12] for details), as $n\rightarrow +\infty$,

  $$\begin{aligned} J(v_{n})\rightarrow m(a),\quad \left( J|_{S(a)}\right) '(v_{n})\rightarrow 0\quad \text { and }\quad \partial _{t}\widetilde{J}(v_{n},0)\rightarrow 0. \end{aligned}$$

  Since $\partial _{t}\widetilde{J}(v_{n},0)=I(v_{n})$ the proof is complete.

$\square$

Now we study the (PS) sequence $\{v_n\}$ of $J|_{S(a)}$ obtained in Lemma 2.2. With the help of the additional information $I(v_n)\rightarrow 0$, we will see that $\{v_n\}$ is bounded. We will use the concentration compactness principle to prove that, up to a translation and a subsequence, $v_n\rightarrow u_a$ weakly in $H^1(\mathbb R^N)$ for some $u_a\ne 0$. The fact that $I(v_n)\rightarrow 0$ will also be used to show that $\lim _{n\rightarrow \infty }\langle J'(v_n),v_n\rangle /a^2=\lambda _a$ for some $\lambda _a<0$ and that $(\lambda _a,u_a)$ is a weak solution of (1.1).

Lemma 2.3

If $\{v_n\}$ is the sequence obtained in Lemma 2.2, then there exists $u_a\in H^{1}({\mathbb {R}}^{N})\backslash \{0\}$ such that, up to a subsequence and a translation, $v_{n}\rightarrow u_a$ weakly in $H^{1}({\mathbb {R}}^{N})$. Moreover, there exists $\lambda _a<0$ such that $(\lambda _a,u_a)$ is a weak solution of (1.1) and $I(u_a)=0$.

Proof

For $n\in {\mathbb {N}}$,

$$\begin{aligned} 2J(v_{n})-I(v_{n})=\frac{N}{2}\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(v_{n}))\left[ f(v_{n})v_{n}-\frac{N+\alpha +2}{N}F(v_{n})\right] dx. \end{aligned}$$

(2.10)

Then by $(f_{1})$ there exist $C_{1}>0$ and $C_{2}>0$ such that

$$\begin{aligned} C_{1}\le \int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(v_{n}))F(v_{n})dx\le C_{2}. \end{aligned}$$

(2.11)

Then we see that $\{v_{n}\}$ is bounded in $H^{1}({\mathbb {R}}^{N})$ from

$$\begin{aligned} \int _{{\mathbb {R}}^{N}}\left| \nabla v_{n}\right| ^{2}dx=2J(v_{n})+\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(v_{n}))F(v_{n})dx. \end{aligned}$$

Next we claim that

$$\begin{aligned} \underset{n\rightarrow +\infty }{\underline{\lim }}\ \underset{y\in {\mathbb {R}}^{N}}{\sup }\int _{B_{1}(y)}\left| v_{n}\right| ^{2}>0. \end{aligned}$$

(2.12)

If this is false, we obtain $v_{n}\rightarrow 0$ in $L^{q}({\mathbb {R}}^{N})$ for $q\in (2,2^{*})$ by Lions’ vanishing lemma [23], and then, by the second inequality in (2.7), $\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(v_{n}))F(v_{n})dx\rightarrow 0$ as $n\rightarrow +\infty$, contrary to (2.11). Hence (2.12) is true. Consequently, there is a sequence $\{y_n\}\subset {\mathbb {R}}^N$ and a $u_a\in H^{1}( {\mathbb {R}}^{N})\backslash \{0\}$ such that, up to a subsequence, $v_{n}(\cdot -y_n)\rightarrow u_a$ weakly in $H^{1}({\mathbb {R}}^{N})$ and a.e. in ${\mathbb {R}}^N$. Replacing $v_{n}(\cdot -y_n)$ by $v_n$, we may assume $y_n=0$, and $v_{n}\rightarrow u_a$ weakly in $H^{1}({\mathbb {R}}^{N})$ and a.e. in ${\mathbb {R}}^N$. By $(f_1)$, $\{F(v_n)\}$ is bounded in $L^{2N/(N+\alpha )}({\mathbb {R}}^{N})$ and $F(v_{n})\rightarrow F(u_a)$ a.e. in ${\mathbb {R}}^{N}$. This in particular implies $F(v_{n})\rightarrow F(u_a)$ weakly in $L^{2N/(N+\alpha )}({\mathbb {R}}^{N})$, and therefore $I_{\alpha }*F(v_{n})\rightarrow I_{\alpha }*F(u_a)$ weakly in $L^{2N/(N-\alpha )}({\mathbb {R}}^{N})$ since $I_{\alpha }*:\ L^{2N/(N+\alpha )}({\mathbb {R}}^{N})\rightarrow L^{2N/(N-\alpha )}({\mathbb {R}}^{N})$ is a bounded linear operator.

Now assume $N\ge 3$. By $(f_1)$ again, $\{f(v_n)\}$ is bounded in $L^{2N/(\alpha +2)}({\mathbb {R}}^{N})$ and $f(v_{n})\rightarrow f(u_a)$ strongly in $L^{2N/(\alpha +2)}_\text {loc}({\mathbb {R}}^N)$. Then, for any $\varphi \in C_0^\infty ({\mathbb {R}}^N)$,

$$\begin{aligned} \int _{{\mathbb {R}}^N}(I_\alpha *F(v_n))f(v_n)\varphi \rightarrow \int _{{\mathbb {R}}^N}(I_\alpha *F(u_a))f(u_a)\varphi . \end{aligned}$$

(2.13)

Since $\{(I_\alpha *F(v_n))f(v_n)\}$ is bounded in $L^{ \frac{2N}{N+2}}({\mathbb {R}}^{N})$, which is seen by the Hölder inequality

$$ \int_{{\mathbb{R}^{N} }} | (I_{\alpha } *F(v_{n} ))f(v_{n} )|^{{\frac{{2N}}{{N + 2}}}} \le \left( {\int_{{\mathbb{R}^{N} }} {\left| {I_{\alpha } *F(v_{n} )} \right|} ^{{\frac{{2N}}{{N - \alpha }}}} } \right)^{{\frac{{N - \alpha }}{{N + 2}}}} \left( {\int_{{\mathbb{R}^{N} }} {\left| {f(v_{n} )} \right|} ^{{\frac{{2N}}{{\alpha + 2}}}} } \right)^{{\frac{{\alpha + 2}}{{N + 2}}}} , $$

and since $C_0^\infty ({\mathbb {R}}^N)$ is dense in $L^\frac{2N}{N-2}({\mathbb {R}}^{N})$, we deduce that

$$\begin{aligned} (I_{\alpha }*F(v_{n}))f(v_{n})\rightarrow (I_{\alpha }*F(u_a))f(u_a)\qquad \text { weakly in }L^\frac{2N}{N+2}({\mathbb {R}}^{N}). \end{aligned}$$

(2.14)

Since $\big (J|_{S(a)}\big )^{\prime }(v_{n})\rightarrow 0$, we have (see [8, Lemma 3]), for $v\in H^1({\mathbb {R}}^N)$,

$$\begin{aligned} \int _{{\mathbb {R}}^{N}}\nabla v_n\cdot \nabla vdx-\mu _n\int _{{\mathbb {R}}^{N}}v_nvdx-\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(v_n))f(v_n)vdx=o(1)\Vert v\Vert , \end{aligned}$$

(2.15)

where

$$\begin{aligned} \mu _{n}:=\frac{1}{a^2}\langle J^{\prime }(v_{n}),v_{n}\rangle =\frac{1}{a^2}\left( \int _{{\mathbb {R}}^{N}}\left| \nabla v_{n}\right| ^{2}dx-\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(v_{n}))f(v_{n})v_{n}dx\right) . \end{aligned}$$

Using the definition of I in (1.5) and the fact that $\lim _{n\rightarrow \infty }I(v_n)=0$, $\mu _n$ can be expressed as

$$\begin{aligned} \mu _{n}&=\frac{1}{a^2}\left[ I(v_{n})-\frac{N+\alpha }{2} \int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(v_{n}))F(v_{n})dx+\frac{N-2}{2} \int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(v_{n}))f(v_{n})v_{n}dx\right] \\&=o(1)+\frac{1}{2a^2}\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(v_{n}))\left[ (N-2)f(v_{n})v_{n}-(N+\alpha )F(v_{n})\right] dx. \end{aligned}$$

By $(f_{1})$ and (2.11), for n large, $\mu _n$ is negative, bounded below and bounded away from 0. Therefore, there exists $\lambda _a<0$ such that, up to a subsequence, $\mu _{n}\rightarrow \lambda _{a}$. Moreover, it follows from (2.14) and (2.15) that, for any $v\in H^1({\mathbb {R}}^N)$,

$$\begin{aligned} \int _{{\mathbb {R}}^{N}}\nabla u_a\cdot \nabla vdx-\lambda _{a}\int _{{\mathbb {R}}^{N}}u_avdx-\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u_a))f(u_a)vdx=0. \end{aligned}$$

(2.16)

Then $(\lambda _a,u_a)$ is a weak solution of (1.1). By [26, Theorem 3],

$$\begin{aligned} (N-2)\int _{{\mathbb {R}}^{N}}\left| \nabla u_a\right| ^{2}dx-N\lambda _{a}\int _{{\mathbb {R}}^{N}}\left| u_a\right| ^{2}dx-(N+\alpha )\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u_a))F(u_a)dx=0. \end{aligned}$$

(2.17)

Letting $v=u_a$ in (2.16) and using (2.17) yields $I(u_a)=0$.

If $N=1, 2$, then by $(f_1)$, $\{|f(v_n)|^s\}$ is bounded in $L^1({\mathbb {R}}^{N})$ for any $s\ge \frac{2N}{\alpha +2}$ and $f(v_{n})\rightarrow f(u_a)$ strongly in $L^{s}_\text {loc}({\mathbb {R}}^N)$ for any $s\ge 1$. This in particular implies that $\{f(v_n)\}$ is bounded in $L^{2N/(N+\alpha )}({\mathbb {R}}^{N})$ and $f(v_{n})\rightarrow f(u_a)$ strongly in $L^{2N/(N+\alpha )}_\text {loc}({\mathbb {R}}^N)$. Then, for any $\varphi \in C_0^\infty ({\mathbb {R}}^N)$, (2.13) is still valid. Since $\{(I_\alpha *F(v_n))f(v_n)\}$ is bounded in $L^2({\mathbb {R}}^{N})$ as seen from the inequality

$$ \int_{{\mathbb{R}^{N} }} {\left| {(I_{\alpha } *F(v_{n} ))f(v_{n} )} \right|} ^{2} \le \left( {\int_{{\mathbb{R}^{N} }} {\left| {I_{\alpha } *F(v_{n} )} \right|} ^{{\frac{{2N}}{{N - \alpha }}}} } \right)^{{\frac{{N - \alpha }}{N}}} \left( {\int_{{\mathbb{R}^{N} }} {\left| {f(v_{n} )} \right|} ^{{\frac{{2N}}{\alpha }}} } \right)^{{\frac{\alpha }{N}}} , $$

and since $C_0^\infty ({\mathbb {R}}^N)$ is dense in $L^2({\mathbb {R}}^{N})$, we deduce that

$$\begin{aligned} (I_{\alpha }*F(v_{n}))f(v_{n})\rightarrow (I_{\alpha }*F(u_a))f(u_a)\qquad \text { weakly in }L^2({\mathbb {R}}^{N}). \end{aligned}$$

Then we use the same argument as above to find $\lambda _a<0$ and see that $(u_a,\lambda _a)$ satisfies (2.16). Since $N=1,2$, we have $u_a\in L^{s}({\mathbb {R}}^{N})$ for $s\ge 2$. Then $F(u_a)\in L^s({\mathbb {R}}^{N})$ for any $s\ge 1$. By the Hardy–Littlewood–Sobolev inequality, we deduce that

$$\begin{aligned} \int _{{\mathbb {R}}^N}|(I_\alpha *F(u_a))g|\le C\Vert F(u_a)\Vert _{L^s({\mathbb {R}}^N)}\Vert g\Vert _{L^t({\mathbb {R}}^N)} \end{aligned}$$

if $s>1$, $t>1$, $\frac{1}{s}+\frac{1}{t}=1+\frac{\alpha }{N}$, and $g\in L^t({\mathbb {R}}^N)$. The range of $t>1$ for which there exists $s>1$ such that $\frac{1}{s}+\frac{1}{t}=1+\frac{\alpha }{N}$ is $1<t<\frac{N}{\alpha }$. Then the range of numbers conjugate to such t is $(\frac{N}{N-\alpha },+\infty )$. Therefore, $I_{\alpha }*F(u_a)\in L^{t}({\mathbb {R}}^{N})$ if $t>\frac{N}{N-\alpha }$. Then for any $\mu >1$,

$$\begin{aligned} \int _{{\mathbb {R}}^N}|(I_{\alpha }*F(u_a))f(u_a)|^\mu \le \Vert I_{\alpha }*F(u_a)\Vert _{L^{\mu N/(N-\alpha )}({\mathbb {R}}^N)}^\mu \Vert f(u_a)\Vert _{L^{\mu N/\alpha }(\mathbb R^N)}^\mu . \end{aligned}$$

Thus, by the Caldéron-Zygmund inequality [11, Theorem 9.9], $u_a\in W^{2,\mu }_\text {loc}({\mathbb {R}}^{N})$. Once we have this regularity of $u_a$, we can obtain (2.17) again in the spirit of the proof of [26, Theorem 3]. This combined with (2.16) yields $I(u_a)=0$. $\square$

Even though we have proved that $(\lambda _a,u_a)$ is a solution of (1.1), we do not know whether $\Vert u_a\Vert _{L^2({\mathbb {R}}^N)}=a$ and $J(u_a)=m(a)$ at this stage. This is not easy to see because the embedding $H^1({\mathbb {R}}^N)\rightarrow L^q({\mathbb {R}}^N)$ is not compact for $q\in (2,2^*)$. Here and in what follows, $2^*=2N/(N-2)$ if $N\ge 3$ and $2^*=+\infty$ if $N=1,2$. In the proof of Theorem 1.1, we need the following lemma, in which the estimate is motivated by the paper [31].

Lemma 2.4

For $a>0$, the map $a\mapsto m(a)$ is strictly decreasing.

Proof

We fix $a_{2}>a_{1}>0$. By the definition of m(a), there exists $\{u_{n}\}\subset V(a_{1})$ such that

$$\begin{aligned} m(a_1)\le J(u_{n})\le m(a_{1})+\frac{1}{n}. \end{aligned}$$

Since $I(u_{n})=0$, we have

$$\begin{aligned} 2J(u_{n})= & {} \int _{{\mathbb {R}}^{N}}\left| \nabla u _{n}\right| ^{2}dx-\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u_{n}))F(u_{n})dx\nonumber \\= & {} \frac{N}{2}\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u_{n}))\left[ f(u_{n})u_{n}-\frac{N+\alpha +2}{N}F(u_{n})\right] dx. \end{aligned}$$

(2.18)

It follows from (2.18) and $(f_1)$ that

$$\begin{aligned} 0<\underset{n\rightarrow +\infty }{\underline{\lim }}\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u_{n}))F(u_{n})dx\le \underset{n\rightarrow +\infty }{\overline{\lim }}\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u_{n}))F(u_{n})dx<+\infty \end{aligned}$$

(2.19)

and

$$\begin{aligned} 0<\underset{n\rightarrow +\infty }{\underline{\lim }}\int _{{\mathbb {R}}^{N}}\left| \nabla u_{n}\right| ^{2}dx\le \underset{ n\rightarrow +\infty }{\overline{\lim }}\int _{{\mathbb {R}}^{N}}\left| \nabla u_{n}\right| ^{2}dx<+\infty . \end{aligned}$$

(2.20)

Setting $v_{n}(x)=\beta ^{\frac{N-2}{2}}u_{n}(\beta x)$ with $\beta :=a_{1}/a_{2}<1$, we obtain $v_{n}\in S(a_{2})$. From Lemma 2.1 we deduce that there exists $t_n:=t(v_{n})\in {\mathbb {R}}$ such that $v_{n}^{t_n}\in V(a_{2})$. Then we have

$$\begin{aligned} m(a_{2})\le & {} J\left( v_{n}^{t_n}\right) \nonumber \\= \, & {} \frac{1}{2}\int _{{\mathbb {R}}^{N}}\left| \nabla v_{n}^{t_n} \right| ^{2}dx-\frac{1}{2}\int _{{\mathbb {R}}^{N}}\left( I_{\alpha }*F\left( v_{n}^{t_n}\right) \right) F\left( v_{n}^{t_n}\right) dx \nonumber \\= \, & {} \frac{1}{2}\int _{{\mathbb {R}}^{N}}\left| \nabla u_{n}^{t_n} \right| ^{2}dx-\frac{1}{2(\beta e^{t_n})^{N+\alpha }}\int _{{\mathbb {R}}^{N}}\left( I_{\alpha }*F\left( \beta ^{\frac{N-2}{2}}e^{\frac{Nt_n}{2}}u_{n}\right) \right) F\left( \beta ^{\frac{N-2}{2}}e^{\frac{Nt_n}{2}}u_{n}\right) dx \nonumber \\= \, & {} J(u_{n}^{t_n})+\frac{1}{2e^{(N+\alpha )t_n }}\int _{{\mathbb {R}}^{N}}\left( I_{\alpha }*F\left( e^{\frac{Nt_n}{2}}u_{n}\right) \right) F\left( e^{\frac{Nt_n}{2}} u_{n}\right) dx \nonumber \\&-\frac{1}{2(\beta e^{t_n})^{N+\alpha }}\int _{{\mathbb {R}}^{N}}\left( I_{\alpha }*F\left( \beta ^{\frac{N-2}{2}}e^{\frac{Nt_n}{2}}u_{n}\right) \right) F\left( \beta ^{\frac{N-2}{2}}e^{\frac{Nt_n}{2}}u_{n}\right) dx \nonumber \\\le & {} J(u_{n})+\frac{1}{2e^{(N+\alpha )t_n }}\int _{ {\mathbb {R}}^{N}}\Phi \left( e^{\frac{Nt_n}{2}}u_{n}\right) dx \nonumber \\\le & {} m(a_{1})+\frac{1}{n}+\frac{1}{2e^{(N+\alpha )t_n }}\int _{ {\mathbb {R}}^{N}}\Phi \left( e^{\frac{Nt_n}{2}}u_{n}\right) dx, \end{aligned}$$

(2.21)

where

$$\begin{aligned} \Phi (u)=(I_{\alpha }*F(u))F(u)-\frac{1}{\beta ^{N+\alpha }}\left( I_{\alpha }*F\left( \beta ^{\frac{N-2}{2}}u\right) \right) F\left( \beta ^{\frac{N-2}{2}}u\right) . \end{aligned}$$

We first consider the case $N\ge 3$. Since $\beta ^{\frac{N-2}{2}}\le 1$, we have $F(\beta ^{\frac{N-2}{2}}s)\ge \beta ^{\frac{p(N-2)}{2}}F(s)$. This implies

$$\begin{aligned} \Phi (u)\le (1-\beta ^{(N-2)p-(N+\alpha )})(I_{\alpha }*F(u))F(u)\le 0, \end{aligned}$$

(2.22)

because

$$\begin{aligned} \beta ^{(N-2)p-(N+\alpha )}=\left( \frac{a_{2}}{a_{1}}\right) ^{N+\alpha -p(N-2)}>1. \end{aligned}$$

Then we have

$$\begin{aligned} m(a_{2})\le J(v_{n}^{t_n})\le m(a_{1})+\frac{1}{n} \end{aligned}$$

(2.23)

which together with the fact that $v_n^{t_n}\in V(a_2)$, by estimates similar to (2.20), yields

$$\begin{aligned} 0<\underset{n\rightarrow +\infty }{\underline{\lim }}\int _{{\mathbb {R}}^{N}}\left| \nabla v_{n}^{t_n}\right| ^{2}dx\le \underset{n\rightarrow +\infty }{\overline{\lim }}\int _{{\mathbb {R}}^{N}}\left| \nabla v_{n}^{t_n}\right| ^{2}dx<+\infty . \end{aligned}$$

(2.24)

Observing that

$$\begin{aligned} \int _{{\mathbb {R}}^N}|\nabla v_n^{t_n}|^2dx=e^{2t_n} \int_{{\mathbb R^N}}{|\nabla v_n|^2dx=e^{2t_n}\int _{{\mathbb {R}}^N}|\nabla u_n|^2dx}, \end{aligned}$$

combining (2.20) and (2.24) implies the existence of $T>0$ such that

$$\begin{aligned} -T<t_n<T. \end{aligned}$$

(2.25)

Using (2.22) and (2.25) we can estimate the last term in (2.21):

$$\begin{aligned} \frac{1}{2e^{(N+\alpha )t_n }}\int _{{\mathbb {R}}^{N}}\Phi \left( e^{\frac{Nt_n}{2}}u_{n}\right) dx&\le -\frac{\beta ^{(N-2)p-(N+\alpha )}-1}{2e^{(N+\alpha )T}}\int _{{\mathbb {R}}^{N}}\left( I_{\alpha }*F\left( e^{\frac{Nt_n}{2}}u_{n}\right) \right) F(e^{\frac{Nt_n}{2}} u_{n})dx\nonumber \\&\le -\frac{\beta ^{(N-2)p-(N+\alpha )}-1}{2e^{(N+Np+\alpha )T}}\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u_{n}))F(u_{n})dx, \end{aligned}$$

(2.26)

where the second inequality follows from

$$\begin{aligned} F\left( e^{\frac{Nt_n}{2}}u_{n}\right) \ge F\left( e^{-\frac{NT}{2}}u_{n}\right) \ge e^{-\frac{NTp}{2}}F(u_n). \end{aligned}$$

In view of (2.19) and (2.21) we arrive at

$$\begin{aligned} m(a_2)\le m(a_1)+\frac{1}{n}-\delta \end{aligned}$$

(2.27)

for n large and for some $\delta >0$ independent of n. Then $m(a_{2})<m(a_{1})$ follows by letting $n\rightarrow +\infty$.

In the case $N=1,2$, we have $\beta ^{\frac{N-2}{2}}\ge 1$ and $F(\beta ^{\frac{N-2}{2}}s)\ge \beta ^{\frac{r(N-2)}{2}}F(s)$, and we still have the estimates from (2.22) to (2.26) with p replaced by r. Therefore (2.27) is valid and the result follows. $\square$

We are now ready to prove Theorem 1.1, using the previous lemmas. Again, the fact that $\lim _{n\rightarrow \infty }I(v_n)=0=I(u_a)$ plays an important role in the proof.

Proof of Theorem 1.1

Let $\{v_{n}\}$ be the sequence obtained in Lemma 2.2. By Lemma 2.3, there exist $\lambda _a<0$ and $u_a\in H^{1}({\mathbb {R}}^{N})\backslash \{0\}$ such that $(\lambda _a,u_a)$ is a weak solution of (1.1), $I(u_a)=0$ and, up to a subsequence, $v_{n}\rightarrow u_a$ weakly in $H^{1}({\mathbb {R}}^{N})$ and a.e. in ${\mathbb {R}}^N$. Clearly we have

$$\begin{aligned} \int _{{\mathbb {R}}^{N}}\left| v_{n}\right| ^{2}dx=\int _{{\mathbb {R}}^{N}}\left| u_{a}\right| ^{2}dx+\int _{{\mathbb {R}}^{N}}\left| v_{n}-u_{a}\right| ^{2}dx+o(1). \end{aligned}$$

Let $a_1=\Vert u_a\Vert _{L^2({\mathbb {R}}^N)}$ and $a_{2,n}=\Vert v_{n}-u_{a}\Vert _{L^2({\mathbb {R}}^N)}$. Then

$$\begin{aligned} a_{1}>0\quad \text { and }\quad a^2=a_{1}^2+a_{2,n}^2+o(1). \end{aligned}$$

Fatou’s lemma implies that $I_\alpha *F(u_a)\le \liminf _{n\rightarrow \infty }I_\alpha *F(v_n)$. Since $I(u_a)=0$, by $(f_{1})$ and Lemma 2.2 and using Fatou’s lemma again, we have

$$\begin{aligned} J(u_a)= & {} J(u_a)-\frac{1}{2}I(u_a) \nonumber \\= & {} \frac{N}{4}\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u_a))\left[ f(u_a)u_a- \frac{N+\alpha +2}{N}F(u_a)\right] dx \nonumber \\\le & {} \frac{N}{4}\underset{n\rightarrow +\infty }{\underline{\lim }}\int _{ {\mathbb {R}}^{N}}(I_{\alpha }*F(v_{n}))\left[ f(v_{n})v_{n}-\frac{N+\alpha +2}{N} F(v_{n})\right] dx \nonumber \\= & {} \underset{n\rightarrow +\infty }{\underline{\lim }}(J(v_{n})-\frac{1}{2} I(v_{n}))= m(a). \end{aligned}$$

(2.28)

On the other hand, it follows from Lemma 2.4 that

$$\begin{aligned} J(u_a)\ge m(a_{1})\ge m(a). \end{aligned}$$

(2.29)

(2.28) together with (2.29) implies

$$\begin{aligned} J(u_a)= m(a_{1})= m(a). \end{aligned}$$

Now by Lemma 2.4 again, we see that $\Vert u_a\Vert _{L^2({\mathbb {R}}^N)}=a_1=a$. This completes the proof. $\square$

The sequence $\{v_n\}$ converges strongly to $u_a$ in $H^1(\mathbb R^N)$. Indeed, the above proof shows that

$$\begin{aligned} J(u_a)=\underset{n\rightarrow +\infty }{\lim } J(v_{n})=m(a), \end{aligned}$$

(2.30)

and

$$\begin{aligned}&\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u_a))\left[ f(u_a)u_a-\frac{ N+\alpha +2}{N}F(u_a)\right] dx\nonumber \\&\quad =\underset{n\rightarrow +\infty }{\lim }\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(v_{n}))\left[ f(v_{n})v_{n}-\frac{N+\alpha +2}{N}F(v_{n})\right] dx. \end{aligned}$$

(2.31)

From (2.31) and the decomposition

$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u)) \left[ f(u)u-\frac{N+\alpha +2}{N}F(u)\right] dx \\&\quad =\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u))[f(u)u-rF(u)]dx + \left( r-\frac{N+\alpha +2}{N}\right) \int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u))F(u)dx, \end{aligned} \end{aligned}$$

using Fatou’s lemma we see that

$$\begin{aligned} \underset{n\rightarrow +\infty }{\lim }\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(v_{n}))F(v_{n})dx=\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u_a))F(u_a)dx. \end{aligned}$$

Then it follows from (2.30) that

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{{\mathbb {R}}^{N}}\left| \nabla v_{n}\right| ^{2}dx = \int _{{\mathbb {R}}^{N}}\left| \nabla u_a\right| ^{2}dx, \end{aligned}$$

which together with the fact ${\Vert v_n\Vert_{L^2(\mathbb R^N)}}={\Vert u_a\Vert _{L^2({\mathbb {R}}^N)}}$ implies $v_n\rightarrow u_a$ strongly in $H^1({\mathbb {R}}^N)$.

3 Proof of Theorem 1.3

In this section, we assume that $N\ge 2$, f is odd and $(f_{1})-(f_{2})$ hold. We adapt arguments from [1, 12] to the problem in question. Let $E:=H_\text {rad}^{1}({\mathbb {R}}^{N})$ be the subspace of $H^1({\mathbb {R}}^N)$ consisting of radially symmetric functions. Let $\Vert \cdot \Vert$ be the usual norm of $H^1({\mathbb {R}}^N)$. We fix a strictly increasing sequence of finite-dimensional linear subspaces $V_k\subset E$ such that $\bigcup _kV_k$ is dense in E. Let $V_k^\perp$ be the orthogonal complement of $V_k$ in E.

Lemma 3.1

For any $k\in {\mathbb {N}}$, there exists $\rho _k>0$ such that $b_k:=\underset{u\in B_k}{\inf }J(u)\rightarrow +\infty$ as $k\rightarrow +\infty$ where

$$\begin{aligned} B_k=\{u\in V_{k-1}^{\perp }\cap S(a):\ \left\| \nabla u\right\| _{L^{2}({\mathbb {R}}^{N})}=\rho _k\}. \end{aligned}$$

Proof

For $q\in (2,2^{*})$, let

$$\begin{aligned} \mu _k(q):=\underset{u\in V_{k-1}^{\perp }}{\inf }\frac{\Vert u\Vert ^2}{ \Vert u\Vert _{L^q({\mathbb {R}}^N)}^2}. \end{aligned}$$

(3.1)

Then $\mu _k(q)\rightarrow +\infty$ as $k\rightarrow \infty$ (see [1, Lemma 2.1]). We see from $(f_1)$ that $2<2Nr/(N+\alpha )\le 2Np/(N+\alpha )<2^*$ and, consequently,

$$\begin{aligned} \nu _k:=\min \left\{ \mu _k\left( \frac{2Nr}{N+\alpha }\right) ,\ \ \mu _k\left( \frac{2Np}{N+\alpha }\right) \right\} \rightarrow \infty \quad \text { as } k\rightarrow \infty . \end{aligned}$$

By (2.7) and (3.1), for $u\in V_{k-1}^{\perp }\cap S(a)$ with k sufficiently large,

$$\begin{aligned} \int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u))F(u)dx \le C\left( \left\| u\right\| _{L^{2Nr/(N+\alpha )}({\mathbb {R}}^{N})}^{2r}+\left\| u\right\| _{L^{2Np/(N+\alpha )}({\mathbb {R}}^{N})}^{2p}\right) \le \frac{C}{\nu _k^{r}}(\Vert u\Vert ^{2r}+\Vert u\Vert ^{2p}). \end{aligned}$$

Since $\Vert u\Vert ^2\ge a^2$, we have $\Vert u\Vert ^{2r}\le a^{2r-2p}\Vert u\Vert ^{2p}$, and it follows that

$$ \int_{{\mathbb{R}^{N} }} {(I_{\alpha } *F(u))F(u)} dx \le \frac{C}{{\nu _{k}^{r} }}\left\| u \right\|^{{2p}} \le \frac{C}{{\nu _{k}^{r} }}\left( {\left( {\int_{{\mathbb{R}^{N} }} {\left| {\nabla u} \right|} ^{2} dx} \right)^{p} + 1} \right), $$

where $C=C(f, N, \alpha , a)>0$ is a constant depending only on f, N, $\alpha$ and a. Now we have

$$\begin{aligned} J(u)= & {} \frac{1}{2}\int _{{\mathbb {R}}^{N}}\left| \nabla u\right| ^{2}dx-\frac{1}{2}\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u))F(u)dx\\\ge & {} \frac{1}{2}\int _{{\mathbb {R}}^{N}}\left| \nabla u\right| ^{2}dx-\frac{C}{2\nu _k^{r}}\left( \left( \int _{{\mathbb {R}}^N}|\nabla u|^2dx\right) ^p+1\right) . \end{aligned}$$

Let

$$\begin{aligned} \rho _k=\left( \frac{\nu _k^r}{2C}\right) ^\frac{1}{2p-2}. \end{aligned}$$

Then, for $u\in B_k$,

$$\begin{aligned} J(u)\ge \frac{1}{2}\rho _k^2-\frac{C\rho _k^{2p}}{2\nu _k^{r}}-\frac{C}{2\nu _k^{r}}=\frac{1}{4}\rho _k^2-\frac{C}{2\nu _k^{r}}. \end{aligned}$$

Since $\nu _k\rightarrow +\infty$ as $k\rightarrow \infty$, $\rho _k\rightarrow +\infty$ as $k\rightarrow \infty$ and the result follows. $\square$

According to Lemma 3.1, there exists $k_0\in {\mathbb {N}}$ such that $b_k>1$ for $k\ge k_0$. For $k\ge k_0$, (2.8) and the compactness of $V_k\cap S(a)$ imply that there exists $t_k>0$ large enough so that

$$\begin{aligned} \Vert \nabla u^{-t_k}\Vert _{L^2({\mathbb {R}}^N)}<\rho _k<\Vert \nabla u^{t_k}\Vert _{L^2({\mathbb {R}}^N)}\quad \text { and }\quad \max _{u\in V_k\cap S(a)} \{J(u^{-t_k}),J(u^{t_k})\}<1. \end{aligned}$$

(3.2)

Now we define

$$\begin{aligned} \Gamma _k=\{\gamma :[0,1]\times (S(a)\cap V_k)\rightarrow S(a):\,&\gamma \text { is continuous, odd in }u, \\&\gamma (0,u)=u^{-t_k},\text { and }\gamma (1,u)=u^{t_k}\}. \end{aligned}$$

A key role in the argument in [1] is played by the intersection property:

$$\begin{aligned} \gamma ([0,1]\times (S(a)\cap V_k))\cap B_k\ne \emptyset \quad \text { for every }\gamma \in \Gamma _k. \end{aligned}$$

In [1, Lemma 2.3] this property was proved using the cohomological index theory for spaces with an action of the group $G=\{-1,1\}$. Here we provide a new and more elementary proof of this property that does not require the cohomological index. We first show the following lemma using the Borsuk-Ulam theorem.

Lemma 3.2

Let $L_1\subsetneq L$ be finite dimensional normed vector spaces. Let $a>0$, $S=\{u\in L:\ \Vert u\Vert =a\}$, $\alpha \in {\mathbb {R}}$, and $H=(H_1,H_2):[0,1]\times S\rightarrow {\mathbb {R}}\times L_1$ be a continuous map such that $H_1(t,u)$ is even in u, $H_2(t,u)$ is odd in u, and

$$\begin{aligned} H_1(0,u)<\alpha <H_1(1,u)\quad \text {for}\ u\in S. \end{aligned}$$

(3.3)

Then there exists $(t,u)\in [0,1]\times S$ such that $H(t,u)=(\alpha ,0)$.

Proof

Assume, by contradiction, that the conclusion is false. Let

$$\begin{aligned} K_1=\{(t,u)\in [0,1]\times S:\ H_1(t,u)\le \alpha ,\ H_2(t,u)=0\}\cup (\{0\}\times S) \end{aligned}$$

and

$$\begin{aligned} K_2=\{(t,u)\in [0,1]\times S:\ H_1(t,u)\ge \alpha ,\ H_2(t,u)=0\}\cup (\{1\}\times S). \end{aligned}$$

Then $K_1$ and $K_2$ are closed subsets of $[0,1]\times S$, and hence compact. According to (3.3) and the assumption of contradiction, $K_1$ and $K_2$ are disjoint, and therefore $\delta :=\frac{1}{4}\text {dist}(K_1,K_2)>0$. Set

$$\begin{aligned} N_\delta (K_i)=\{(t,u)\in [0,1]\times S:\ \text {dist}((t,u),\ K_i)<\delta \},\quad i=1,2. \end{aligned}$$

Since $H_1(t,u)$ is even in u and $H_2(t,u)$ is odd in u, $K_i$ and thus $N_\delta (K_i)$ are symmetric sets with respect to u. Moreover, $N_\delta (K_1)\cap N_\delta (K_2)=\emptyset$ and

$$\begin{aligned} H_2(t,u)\ne 0\quad \text {if}\ (t,u)\in \partial ^{[0,1]\times S}N_\delta (K_i), \end{aligned}$$

(3.4)

where $\partial ^{[0,1]\times S}N_\delta (K_i)$ is the boundary of $N_\delta (K_i)$ in $[0,1]\times S$. Denote $\Omega =\{u\in L:\ \Vert u\Vert <a\}$ and let $M=(M_1,M_2):L\rightarrow (\{0\}\times \Omega )\cup ([0,1)\times S)$ be the homeomorphism induced by the stereographic projection with north pole $(1,0)\in {\mathbb {R}}\times L$. To be more precise,

$$\begin{aligned} M(u)=\left\{ \begin{array}{ll} (0,u)&{}\quad \text {if}\ u\in \Omega ,\\ (1-a\Vert u\Vert ^{-1},\ a\Vert u\Vert ^{-1}u)&{}\quad \text {if}\ u\in L\backslash \Omega . \end{array}\right. \end{aligned}$$

Denote $\Omega _1=M^{-1}((\{0\}\times \Omega )\cup N_\delta (K_1))$. Then $\Omega _1$ is an open bounded symmetric neighborhood of 0 in L and $M(\partial \Omega _1)=\partial ^{[0,1]\times S}N_\delta (K_1)$. Now we define $A:\partial \Omega _1\rightarrow L_1$ as

$$\begin{aligned} A(u)=H_2\circ M(u)=H_2(1-a\Vert u\Vert ^{-1},\ a\Vert u\Vert ^{-1}u)\quad \text {for}\ u\in \partial \Omega _1. \end{aligned}$$

Then A is odd and continuous and, by (3.4), $A(u)\ne 0$ for all $u\in \partial \Omega _1$. But this is in contradiction with the Borsuk–Ulam theorem. $\square$

The following is [1, Lemma 2.3], for which we provide a new proof based on Lemma 3.2.

Lemma 3.3

$\gamma ([0,1]\times (S(a)\cap V_k))\cap B_k\ne \emptyset$ for every $\gamma \in \Gamma _k$.

Proof

Set $L=V_k$ and $L_1=V_{k-1}$ in which we use the $L^2({\mathbb {R}}^N)$ norm. Choose $S=S(a)\cap V_k$ and $\alpha =\rho _k$. Let $P_{k-1}:E\rightarrow V_{n-1}$ be the orthogonal projection and define

$$\begin{aligned} h_k:S(a)\rightarrow {\mathbb {R}}\times V_{k-1},\quad u\mapsto (\Vert \nabla u\Vert _{L^2({\mathbb {R}}^N)},\ P_{k-1}u) \end{aligned}$$

and

$$\begin{aligned} H=(H_1,\ H_2)=h_k\circ \gamma . \end{aligned}$$

Then L, $L_1$, S, $\alpha$, and H satisfy all the conditions of Lemma 3.2. By Lemma 3.2, there exists $(t,u)\in [0,1]\times (S(a)\cap V_k)$ such that $H(t,u)=(\alpha ,0)$. That is, $\gamma (t,u)\in B_k$. $\square$

By Lemmas 3.1 and 3.3, we have

$$\begin{aligned} c_k :=\underset{\gamma \in \Gamma _k}{\inf }\ \ \underset{t\in [0,1],\ u\in S(a)\cap V_k}{\max }J(\gamma (t,u)) \ge b_k\rightarrow +\infty . \end{aligned}$$

We will show that $c_k$, $k\ge k_0$, is a critical value of $J|_{S(a)}$. For this we make use of the stretched functional

$$\begin{aligned} {\widetilde{J}}: \mathbf{E}:=E\times {\mathbb {R}}\rightarrow {\mathbb {R}},\quad {\widetilde{J}}(u,s)=J(u^s) \end{aligned}$$

from the proof of Lemma 2.2, now constrained to the space E of radial functions. On $\mathbf{E}$ we consider the involution

$$\begin{aligned} \tau :\mathbf{E}\rightarrow \mathbf{E},\quad \tau (u,s):=(-u,s), \end{aligned}$$

which preserves the natural inner product on $\mathbf{E}$. Clearly ${\widetilde{J}}$ is invariant under $\tau$ because F is even in u. This implies that $\nabla {\widetilde{J}}:\mathbf{E}\rightarrow \mathbf{E}$ is $\tau$-equivariant, i.e. $\nabla {\widetilde{J}}\circ \tau = \tau \circ \nabla {\widetilde{J}}$.

Now we define

$$\begin{aligned} \widetilde{c}_k :=\underset{g\in \widetilde{\Gamma }_k}{\inf }\ \ \underset{t\in [0,1],\ u\in S(a)\cap V_k}{\max }{\widetilde{J}}(g(t,u)) \end{aligned}$$

where

$$\begin{aligned} \widetilde{\Gamma }_k:=\{g:[0,1]\times (S(a)\cap V_k)\rightarrow S(a)\times {\mathbb {R}}:\,&g\text { is continuous and equivariant},\\&g(0,u)=(u^{-t_k},0),\ \text {and}\ g(1,u)=(u^{t_k},0)\}. \end{aligned}$$

Here a map $g:[0,1]\times (S(a)\cap V_k)\rightarrow S(a)\times {\mathbb {R}}$ is said to be equivariant if $g(t,-u)=\tau g(t,u)$.

Lemma 3.4

$\widetilde{c}_k=c_k$.

Proof

Observe that

$$\begin{aligned} \gamma \in \Gamma _k\ \ \Longrightarrow \ \ g:=(\gamma ,0)\in \widetilde{\Gamma }_k,\ \ {\widetilde{J}}(g(t,u))=J(\gamma (t,u)), \end{aligned}$$

and

$$\begin{aligned} g\in \widetilde{\Gamma }_k\ \ \Longrightarrow \ \ \gamma :=g_1^{g_2}\in \Gamma _k,\ \ J(\gamma (t,u))={\widetilde{J}}(g(t,u)). \end{aligned}$$

Then the result follows. $\square$

Recall that we fixed $k\ge k_0$ so that $b_k>1$. Then Lemmas 3.3 and 3.4 yield

$$\begin{aligned} {\widetilde{c}}_k=c_k\ge b_k>1. \end{aligned}$$

To show that $c_k$ is a critical value of $J|_{S(a)}$, we first prove the following result.

Lemma 3.5

Let $0<\varepsilon <c_k-1$ and $g\in \widetilde{\Gamma }_k$ be such that

$$\begin{aligned} \underset{t\in [0,1],\ u\in S(a)\cap V_k}{\max }\widetilde{J}(g(t,u))\le c_k+\varepsilon . \end{aligned}$$

Then there exists $(v,s)\in S(a)\times {\mathbb {R}}$ such that:

• (i) ${\widetilde{J}}(v,s)\in [c_k-\varepsilon , c_k+\varepsilon ]$,

• (ii) $\min _{t\in [0,1],\ u\in S(a)\cap V_k} \Vert (v,s)-g(t,u)\Vert _{\mathbf {E}}\le \sqrt{\varepsilon }$,

• (iii) ${\Vert {\big ({\widetilde{J}}|_{S(a)\times {\mathbb {R}}}\big)'}(v,s)\Vert} \le {2\sqrt{\varepsilon }}$.

Proof

The proof is an equivariant version of the one of [12, Lemma 2.3]. On $\widetilde{\Gamma }_k$, we define the metric

$$\begin{aligned} d(g,h)=\max _{t\in [0,1],\ u\in S(a)\cap V_k}\Vert g(t,u)-h(t,u)\Vert _\mathbf{{E}} \end{aligned}$$

and consider the continuous function

$$\begin{aligned} \Phi :\widetilde{\Gamma }_k\rightarrow {\mathbb {R}},\quad \Phi (g)=\underset{t\in [0,1],\ u\in S(a)\cap V_k}{\max }\widetilde{J}(g(t,u)). \end{aligned}$$

By Ekeland’s variational principle there exists $h\in \widetilde{\Gamma }_k$ such that

• (i) $\Phi (h)\le \Phi (g)$,

• (ii) $d(h,g)\le \sqrt{\varepsilon }$,

• (iii) $\Phi (h_1)>\Phi (h)-\sqrt{\varepsilon }d(h_1,h) \text { for }h_1\in \widetilde{\Gamma }_k\text { with }h_1\ne h$.

Assume, by contradiction, that the result is not true. For $(u,s)\in S(a)\times {\mathbb {R}}$ the tangent space is denoted by

$$\begin{aligned} T_{(u,s)} := T_{(u,s)}(S(a)\times {\mathbb {R}}) = T_uS(a)\times {\mathbb {R}}= \left\{ (z_1,z_2)\in \mathbf{E}:\ \langle u,z_1\rangle _{L^2}=0\right\} . \end{aligned}$$

For $(t,u)\in [0,1]\times (S(a)\cap V_k)$ with ${\widetilde{J}}(h(t,u))\ge c_k-\varepsilon$, there exists $z(t,u)\in T_{h(t,u)}$ with $\Vert z(t,u)\Vert _{\mathbf{E}}=1$ and such that

$$\begin{aligned} \langle \nabla {\widetilde{J}}(h(t,u)),z(t,u)\rangle _{\mathbf{E}}<-2\sqrt{\varepsilon }. \end{aligned}$$

(3.5)

Denote

$$\begin{aligned} S=\{(t,u)\in [0,1]\times (S(a)\cap V_k):\ \ {\widetilde{J}}(h(t,u))\ge c_k-\varepsilon \}. \end{aligned}$$

Then, using (3.5), we can construct a continuous vector field $U:S\rightarrow \mathbf{E}$ such that

• (i) $\Vert U(t,u\Vert _\mathbf{E}=1$,

• (ii) $U(t,u)\in T_{h(t,u)}$,

• (iii) $\langle \nabla {\widetilde{J}}(h(t,u)),U(t,u)\rangle _{\mathbf{E}} < -2\sqrt{\varepsilon }$.

Thus U is a normalized pseudo-gradient vector field for ${\widetilde{J}}$ along h. The vector field

$$\begin{aligned} V(t,u):=\frac{1}{2}\big (U(t,u)+\tau \circ U(t,-u)\big ) \in T_{h(t,u)} \end{aligned}$$

is equivariant, i.e. $V(t,-u)=\tau V(t,u)$. Moreover (iii) with V instead of U holds because $\nabla {\widetilde{J}}\circ h$ is equivariant. It also follows that $V(t,u)\ne 0$, hence we may pass to the normalized vector field

$$\begin{aligned} W:S\rightarrow \mathbf{E},\quad W(t,u)=\frac{1}{\Vert V(t,u)\Vert _\mathbf{E}} V(t,u), \end{aligned}$$

which is continuous and equivariant. Clearly (i)-(iii) from above hold with U replaced by W. Let $\Psi :[0,1]\times (S(a)\cap V_k)\rightarrow [0,1]$ be a continuous cut-off function satisfying

$$\begin{aligned} \Psi (t,u)=\left\{ \begin{array}{l} 1,\quad \text {if}\ \widetilde{J}(h(t,u))\ge c_k,\\ 0,\quad \text {if}\ {\widetilde{J}}(h(t,u))\le c_k-\varepsilon . \end{array}\right. \end{aligned}$$

We may assume that $\Psi (t,-u)=\Psi (t,u)$ because ${\widetilde{J}}$ is even in u. If not we replace $\Psi (t,u)$ by $\frac{1}{2}\big (\Psi (t,u)+\Psi (t,-u)\big )$. Now we use W to deform $h\in \widetilde{\Gamma }$ as follows. For $r\in [0,1/2]$ we define $g_r=(g_{r,1},g_{r,2}):[0,1]\times (S(a)\cap V_k)\rightarrow S(a)\times {\mathbb {R}}$ for $(t,u)\in S$ by

$$\begin{aligned} g_{r,1}(t,u) = \sqrt{1-\frac{r^2\Psi ^2(t,u)\Vert W_1(t,u)\Vert _{L^2}^2}{a^2}}h_1(t,u) +r\Psi (t,u)W_1(t,u), \end{aligned}$$

and

$$\begin{aligned} g_{r,2}(t,u) = h_2(t,u)+r\Psi (t,u)W_2(t,u). \end{aligned}$$

For $(t,u)\in [0,1]\times (S(a)\cap V_k){\setminus } S$ we set $g_r(t,u)=h(t,u)$. Observe that $g_r$ is continuous and equivariant: $g_r(t,-u)=\tau g_r(t,u)$. In addition, (3.2) implies

$$\begin{aligned} \max \{{\widetilde{J}}(h(0,u)),{\widetilde{J}}(h(1,u))\} =\max \{\widetilde{J}(u^{-t_k},0),{\widetilde{J}}(u^{t_k},0)\} =\max \{J(u^{-t_k}),J(u^{t_k})\}<1, \end{aligned}$$

hence $c_k-\varepsilon >1$ yields

$$\begin{aligned} g_r(0,u)=h(0,u)=(u^{-t_k},0),\quad g_r(1,u)=h(1,u)=(u^{t_k},0). \end{aligned}$$

Therefore, $g_r\in \widetilde{\Gamma }_k$ for $r\in [0,1/2]$.

The rest of the proof proceeds as the one of [12, Lemma 2.3], leading to a contradiction. $\square$

From Lemma 3.5, it is possible to find a (PS) sequence $\{v_{k,n}\}\subset S(a)$ for $J|_{S(a)}$ with the additional property $\lim _{n\rightarrow \infty }I(v_{k,n})=0$ at the level $c_k$.

Lemma 3.6

There exists a sequence $\{v_{k,n}\}\subset S(a)$, also denoted by $\{v_n\}$ for simplicity of notation, such that, as $n\rightarrow +\infty$,

$$\begin{aligned} J(v_n) \rightarrow c_k, \quad \big (J|_{S(a)}\big )^{\prime }(v_n) \rightarrow 0, \quad I(v_n) \rightarrow 0. \end{aligned}$$

Proof

The proof is essentially the same as the proof of [12, Lemma 2.4]. We only sketch it and refer to [12] for more details. For n large such that $\frac{1}{n}<c_k-1$, let $g_n\in \widetilde{\Gamma }_k$ be such that

$$\begin{aligned} \underset{t\in [0,1],\ u\in S(a)\cap V_k}{\max }\widetilde{J}(g_n(t,u))\le c_k+\frac{1}{n}. \end{aligned}$$

We may assume that $g_n(t,u)=(g_{n,1}(t,u),0)$ with $g_{n,1}\in \Gamma _k$. By Lemma 3.5 there exists $(w_n,s_n)\in S(a)\times {\mathbb {R}}$ such that:

• (i) ${\widetilde{J}}(w_n,s_n)\in [c_k-\frac{1}{n}, c_k+\frac{1}{n}]$,

• (ii) $\min _{t\in [0,1],\ u\in S(a)\cap V_k} \Vert (w_n,s_n)-g_n(t,u)\Vert _{\mathbf {E}}\le \frac{1}{\sqrt{n}}$,

• (iii) ${\Vert \big ({\widetilde{J}}|_{S(a)\times {\mathbb {R}}}\big )'(w_n,s_n)\Vert} \le {\frac{2}{\sqrt{n}}}$.

Clearly (ii) implies

$$\begin{aligned} \lim _{n\rightarrow \infty }s_n=0. \end{aligned}$$

(3.6)

Set $v_n=w_n^{s_n}$. Then by (i)

$$\begin{aligned} \lim _{n\rightarrow \infty }J(v_n)=\lim _{n\rightarrow \infty }{\widetilde{J}}(w_n,s_n)=c_k. \end{aligned}$$

Since $I(v_n)=\partial _s{\widetilde{J}}(w_n,s_n)$, by (iii) we have

$$\begin{aligned} \lim _{n\rightarrow \infty }I(v_n)=0. \end{aligned}$$

Moreover, since $\langle J'(v_n),\varphi \rangle _{E^*\times E}=\langle \partial _u{\widetilde{J}}(w_n,s_n), \varphi ^{-s_n}\rangle _{E^*\times E}$ for any $\varphi \in T_{S(a)}v_n$, by (iii) and (3.6), for n sufficiently large,

$$\begin{aligned} \Vert \big (J|_{S(a)}\big )'(v_n)\Vert&=\sup _{\varphi \in T_{S(a)}v_n,\ \Vert \varphi \Vert _E=1}\langle J'(v_n),\varphi \rangle _{E^*\times E}\\&=\sup _{\varphi \in T_{S(a)}v_n,\ \Vert \varphi \Vert _E=1}\langle \partial _u{\widetilde{J}}(w_n,s_n), \varphi ^{-s_n}\rangle _{E^*\times E}\\&\le \frac{2}{\sqrt{n}}\sup _{\varphi \in T_{S(a)}v_n,\ \Vert \varphi \Vert _E=1}\Vert \varphi ^{-s_n}\Vert _E\le \frac{4}{\sqrt{n}}. \end{aligned}$$

This completes the proof. $\square$

With (PS) sequences on hand, we are in a position to study their compactness. Since we are working in the space E consisting of radially symmetric functions so that E is imbedded compactly in $L^q({\mathbb {R}}^N)$ for any $q\in (2,2^*)$, it is easier to show the compactness of (PS) sequences in the present case compared with the argument in Sect. 2.

Lemma 3.7

Let $\{v_{k,n}\}_{n=1}^\infty \subset S(a)$ be the sequence obtained in Lemma  3.6. Then up to a subsequence $\{v_{k,n}\}_{n=1}^\infty$ converges strongly in E to some $u_k\in S(a)$ as $n\rightarrow \infty$. Moreover, there exists $\lambda _k<0$ such that $(\lambda _k,u_k)$ is a solution of (1.2) and $J(u_k)=c_k$.

Proof

In view of the fact that $v_{k,n}$ satisfies (2.10) and (2.11), we see that $\{v_{k,n}\}_{n=1}^\infty$ is bounded in E. Therefore, we may assume that there exists $u_k\in E$ such that up to a subsequence $v_{k,n}\rightarrow u_k$ weakly in E. Since $N\ge 2$, E is imbedded compactly in $L^q({\mathbb {R}}^N)$ for any $q\in (2,2^*)$. Then $v_{k,n}\rightarrow u_k$ strongly in $L^q({\mathbb {R}}^N)$ for any $q\in (2,2^*)$.

Since

$$\begin{aligned} |F(s)|\le C(|s|^r+|s|^p)\quad \text {and}\quad 2<\frac{2Nr}{N+\alpha }\le \frac{2Np}{N+\alpha }<2^*, \end{aligned}$$

we have $F(v_{k,n})\rightarrow F(u_k)$ strongly in $L^\frac{2N}{N+\alpha }({\mathbb {R}}^N)$ (see [29, Theorem A.4]). Then $I_\alpha *F(v_{k,n})\rightarrow I_\alpha *F(u_k)$ strongly in $L^\frac{2N}{N-\alpha }({\mathbb {R}}^N)$ as $I_\alpha *:L^\frac{2N}{N+\alpha }({\mathbb {R}}^N)\rightarrow L^\frac{2N}{N-\alpha }({\mathbb {R}}^N)$ is a bounded linear operator. Since

$$\begin{aligned} 2<\frac{2N(r-1)}{\alpha +2}\le \frac{2N(p-1)}{\alpha +2}<2^*, \end{aligned}$$

we can choose a number q such that

$$\begin{aligned} \frac{2N}{\alpha +2}<q<\frac{2N}{\alpha } \end{aligned}$$

(3.7)

and

$$\begin{aligned} 2<q(r-1)\le q(p-1)<2^*. \end{aligned}$$

(3.8)

The condition

$$\begin{aligned} |f(s)|\le C(|s|^{r-1}+|s|^{p-1}) \end{aligned}$$

together with (3.8) implies $f(v_{k,n})\rightarrow f(u_k)$ strongly in $L^q({\mathbb {R}}^N)$ (see [29, Theorem A.4]). Let $\mu$ be the number defined by

$$\begin{aligned} \frac{N-\alpha }{2N}+\frac{1}{q}=\frac{1}{\mu }. \end{aligned}$$

(3.9)

Then $(I_\alpha *F(v_{k,n}))f(v_{k,n})\rightarrow (I_\alpha *F(u_k))f(u_k)$ strongly in $L^\mu ({\mathbb {R}}^N)$. By (3.7) and (3.9), we have

$$\begin{aligned} \frac{2N}{N+2}<\mu <2. \end{aligned}$$

(3.10)

Note that (see [8, Lemma 3]), for $v\in H^1({\mathbb {R}}^N)$,

$$\begin{aligned} \int _{{\mathbb {R}}^{N}}\nabla v_{k,n}\cdot \nabla vdx-\mu _{k,n}\int _{{\mathbb {R}}^{N}}v_{k,n}vdx-\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(v_{k,n}))f(v_{k,n})vdx=o(1)\Vert v\Vert \end{aligned}$$

(3.11)

as $n\rightarrow \infty$, where

$$\begin{aligned} \mu _{k,n}:=\frac{1}{a^2}\langle J^{\prime }(v_{k,n}),v_{k,n}\rangle . \end{aligned}$$

Since we already proved the convergence of $(I_\alpha *F(v_{k,n}))f(v_{k,n})$ to $(I_\alpha *F(u_k))f(u_k)$ in $L^\mu ({\mathbb {R}}^N)$ and since $\mu$ is in the range of (3.10), we can argue in the same way as (2.15)–(2.16) to see that, for $v\in H^1({\mathbb {R}}^N)$,

$$\begin{aligned} \int _{{\mathbb {R}}^{N}}\nabla u_k\cdot \nabla vdx-\lambda _k\int _{{\mathbb {R}}^{N}}u_kvdx-\int _{{\mathbb {R}}^{N}}(I_{\alpha }*F(u_k))f(u_k)vdx=0, \end{aligned}$$

(3.12)

where $\lambda _k=\lim _{n\rightarrow \infty }\mu _{k,n}<0$. Taking the difference between (3.11) and (3.12) with $v=v_{k,n}-u_k$, we see for $n\rightarrow \infty$:

$$\begin{aligned} \begin{aligned}& {\int_{{\mathbb {R}}^N}|\nabla (v_{k,n}-u_k)|^{2} dx} -\lambda _k {\int_{{\mathbb {R}}^{N}}(v_{k,n}-u_k)^{2} dx} \\&\quad = {{\int_{{\mathbb {R}}^{N}}} \big [(I_{\alpha }*F(v_{k,n}))f(v_{k,n})-(I_{\alpha }*F(u_n))f(u_n)\big ](v_{k,n}-u_n)dx+o(1) \rightarrow 0}. \end{aligned} \end{aligned}$$

Therefore, $\{v_{k,n}\}_{n=1}^\infty$ converges strongly in E to $u_k\in S(a)$ as $n\rightarrow \infty$ and $(\lambda _k,u_k)$ is a solution of (1.1) with $J(u_k)=c_k$. $\square$

Proof of Theorem 1.3

By Lemma 3.7, $(\lambda _k,\pm u_k)$ are radial solutions of (1.2) with $\lambda _k<0$ and $J(u_k)=c_k$. This proves the result. $\square$