Stable standing waves of nonlinear Schrödinger equations with potentials and general nonlinearities

Abstract

The existence and nonexistence of the minimizer of the $L^2$-constraint minimization problem $e(\alpha ):=\inf \{ E(u)|\ u\in H^1(\mathbb {R}^N),\ \left\| u\right\| _{L^2(\mathbb {R}^N)}^2=\alpha \}$ are studied. Here,

$$\begin{aligned} E(u):=\frac{1}{2}\int _{\mathbb {R}^N}|\nabla u|^2+V(x) |u|^2dx-\int _{\mathbb {R}^N}F(|u|)dx, \end{aligned}$$

$V(x)\in C(\mathbb {R}^N)$, $0\not \equiv V(x)\le 0$, $V(x)\rightarrow 0$ ($|x|\rightarrow \infty$) and $F(s) = \int _0^s f(t) dt$ is a rather general nonlinearity. We show that there exists $\alpha _0\ge 0$ such that $e(\alpha )$ is attained for $\alpha >\alpha _0$ and $e(\alpha )$ is not attained for $0<\alpha <\alpha _0$. We study differences between the cases $V(x)\not \equiv 0$ and $V(x)\equiv 0$, and obtain sufficient conditions for $\alpha _0=0$. In particular, if $N=1,2$, then $\alpha _0=0$, and hence $e(\alpha )$ is attained for all $\alpha >0$.

1 Introduction and main theorems

In this paper we are interested in the attainability of the $L^2$-constraint minimization problem

$$\begin{aligned} e(\alpha ):=\inf _{u\in M(\alpha )}E(u), \end{aligned}$$

where $\alpha >0$ is a constant,

$$\begin{aligned} \begin{aligned} E(u)&:=\frac{1}{2}\int _{\mathbb {R}^N}|\nabla u|^2+V(x)|u|^2dx -\int _{\mathbb {R}^N}F(|u|)dx, \quad F(s) := \int _0^s f(t) dt \ \ \text {for} \ s \ge 0, \\ M(\alpha )&:=\left\{ u\in H^1(\mathbb {R}^N) \ |\ \left\| u\right\| _{L^2(\mathbb {R}^N)}^2=\alpha \right\} , \quad H^1(\mathbb {R}^N) = H^1(\mathbb {R}^N,\mathbb {C}), \end{aligned} \end{aligned}$$

and f(s) and V(x) satisfy certain assumptions. This problem plays a role when we study the orbital stability of the standing wave of the nonlinear Schrödinger equation

$$\begin{aligned} iU_t=-\Delta U+V(x)U-f(U)\ \text {for}\ (t,x)\in \mathbb {R}\times \mathbb {R}^N. \end{aligned}$$

(1.1)

The standing wave is a solution of (1.1) of the special form $U(t,x)=e^{i\lambda t}u(x)$ and the orbital stability is defined in Theorem A. We impose the following assumptions (F1)–(F4) on f(s):

• (F1) $f\in C(\mathbb {C},\mathbb {C})$, $f(0)=0$.

• (F2) $f(s)\in \mathbb {R}$ for $s\in \mathbb {R}$, $f(e^{i\theta }z)=e^{i\theta }f(z)$ for $\theta \in \mathbb {R}$ and $z\in \mathbb {C}$.

• (F3) $\lim _{s\rightarrow 0}f(s)/s=0$.

• (F4) $\lim _{s\rightarrow \infty }f(s)/|s|^{{p_\mathrm{c}}}=0$, where ${p_\mathrm{c}}:=1+4/N$.

We impose the following assumption (V1) on V(x):

• (V1) $V(x)\in C(\mathbb {R}^N)$, $0\not \equiv V(x)\le 0$ and $\lim _{|x|\rightarrow \infty }V(x)=0$.

The assumptions (F1)–(F4) and (V1) are assumed throughout the present paper. In addition to (F1)–(F4) and (V1), we introduce the following conditions:

• (F5) f(s) is locally Hölder continuous with exponent $\nu \in (0,1)$ in $\mathbb {R}$, $f(s)>0$ for $s>0$ and there exists $\delta _1>0$ such that f(s)/s is nondecreasing in $(0,\delta _1)$.

• (F6) If $N\ge 5$, then $\liminf _{s\rightarrow 0}f(s)/|s|^{{p_\mathrm{sg}}}>0$, where ${p_\mathrm{sg}}:=N/(N-2)$.

• (V2) If $N\ge 5$, then

  $$\begin{aligned} V\in W^{1,\infty }(\mathbb {R}^N) \ \ \text {and}\ \ \nabla V(x)\cdot x\le \frac{(N-2)^2}{2|x|^2}\quad \text {for a.e.}\ x\in \mathbb {R}^N\setminus \{0\} . \end{aligned}$$

In order to obtain the orbital stability we further need the following:

• (F7) There exist $K>0$ and $1<p<2^*-1$ such that $|f(z_1)-f(z_2)|\le K(1+|z_1|+|z_2|)^{p-1}|z_1-z_2|$ for $z_1$, $z_2\in \mathbb {C}$. Here, $2^*=2N/(N-2)$ if $N\ge 3$, and $2^*=\infty$ if $N=1,2$.

• (F8) There exist $L>0$ and $1<p<{p_\mathrm{c}}$ such that $F(|s|)\le L(|s|^2+|s|^{p+1})$ for $s\in \mathbb {R}$.

It is known that the global well-posedness of (1.1) in $H^1(\mathbb {R}^N)$ holds if (F1), (F2), (F7) and (F8) hold and $V(x)\in L^{\infty }(\mathbb {R}^N)$. See [10, Corollary 6.1.2] for details.

To state our main theorems we recall related results. Lions [20] showed that every minimizing sequence for $e(\alpha )$ has a convergent subsequence in $H^1(\mathbb {R}^N)$ if and only if the strict subadditivity condition holds, i.e.,

$$\begin{aligned} e(\alpha )<e(\beta )+e_{\infty }(\alpha -\beta )\ \text {for all}\ \beta \in \left\{ \begin{aligned}&(0,\alpha )&\quad \text {if } V(x) \equiv 0,\\&[0,\alpha )&\quad \text {if } V(x) \not \equiv 0. \end{aligned} \right. \end{aligned}$$

(1.2)

Here, $e_{\infty }(\alpha )$ is the problem at infinity, i.e.,

$$\begin{aligned} e_{\infty }(\alpha ):=\inf _{u\in M(\alpha )}E_{\infty }(u), \end{aligned}$$

where

$$\begin{aligned} E_{\infty }(u):=\frac{1}{2}\int _{\mathbb {R}^N}|\nabla u|^2dx-\int _{\mathbb {R}^N}F(|u|)dx. \end{aligned}$$

The characterization (1.2) holds for rather wide class of functionals E(u). However, it is not easy to check (1.2) for given f and V.

First, we consider the homogeneous case $V(x)\equiv 0$. Then, $E(u)=E_{\infty }(u)$ and $e(\alpha )=e_{\infty }(\alpha )$. In the model case $f(u)=|u|^{p-1}u$$(1<p<{p_\mathrm{c}})$, Cazenave–Lions [11] showed that (1.2) holds for all $\alpha >0$ and that $e_{\infty }(\alpha )<0$ for all $\alpha >0$. In the case of a general nonlinear term f, the attainability for $e_{\infty }(\alpha )$ was mentioned in [11, Remark II.3]. However, in [11] the following condition was assumed:

$$\begin{aligned} \text {there exists }u_0\in L^2(\mathbb {R}^N)\hbox { such that } \left\| u_0\right\| _{L^2(\mathbb {R}^N)}\le \alpha \hbox { and }E_{\infty }(u_0)<0. \end{aligned}$$

(1.3)

The same attainability problem for $e_{\infty }(\alpha )$ was recently studied by [5, 13, 22]. In particular, Shibata [22] showed that there exists $\alpha _{0,\infty }\in [0,\infty )$ uniquely determined by f and N such that

$$\begin{aligned} e_{\infty }(\alpha ) {\left\{ \begin{array}{ll} =0 &{}\quad \text {if}\ 0\le \alpha \le \alpha _{0,\infty },\\ <0 &{} \quad \text {if}\ \alpha >\alpha _{0,\infty }. \end{array}\right. } \end{aligned}$$

(1.4)

Moreover, he showed that $e_{\infty }(\alpha )$ is not attained for $0<\alpha <\alpha _{0,\infty }$ and $e_{\infty }(\alpha )$ is attained for $\alpha >\alpha _{0,\infty }$. See Proposition 2.1 of the present paper for details. It was shown in [22, Lemma 2.3] that $e_{\infty }(\alpha )$ is nonincreasing. Hence the assumption (1.3) leads to $e_{\infty }(\alpha )<0$ for each $\alpha \ge \Vert u_0 \Vert _{L^2(\mathbb {R}^N)}$.

Our result is about the attainability of the inhomogeneous problem $e(\alpha )$.

Theorem A

Suppose (F1)–(F5) and (V1), and suppose (F6) or (V2). Let $\alpha _{0,\infty }$ be given in (1.4). Then there exists $\alpha _0\in [0,\alpha _{0,\infty }]$ such that the following hold:

• (i) If $\alpha >\alpha _0$, then $e(\alpha )<0$ and every minimizing sequence for $e(\alpha )$ has a strong convergent subsequence in $H^1(\mathbb {R}^N)$. Therefore, $e(\alpha )$ is attained, the set of all the minimizers, which is denoted by $S_{\alpha }$, is precompact and (1.2) holds. Moreover, if (F7) and (F8) hold, then $S_\alpha$ is orbitally stable, i.e., for any $\varepsilon >0$, there exists $\delta >0$ such that for any solution U of (1.1) with $\mathrm{dist}_{H^1}(U(0,\,\cdot \,),S_{\alpha })<\delta$ satisfies

  $$\begin{aligned} \mathrm{dist}_{H^1}(U(t,\,\cdot \,),S_{\alpha })<\varepsilon \ \text {for all}\ t\in \mathbb {R}. \end{aligned}$$

• (ii) If $0<\alpha <\alpha _0$, then $e(\alpha )=0$ and $e(\alpha )$ is not attained.

Remark 1.1

• (i) Notice that (F6) and (V2) are necessary only for $N \ge 5$. Therefore, when $1 \le N \le 4$, Theorem A holds under (F1)–(F5) and (V1) (for the orbital stability, we also need (F7) and (F8)).

• (ii) If $\alpha _{0,\infty }=0$, then $\alpha _0 = 0$ and Theorem A (i) always occurs. Remark that if $N \ge 5$, then ${p_\mathrm{sg}}< {p_\mathrm{c}}$. Hence, when $N \ge 5$ and (F6) hold, we have $\alpha _{0,\infty } = 0 = \alpha _0$ by [22, Theorem 1.3] (see also Proposition 2.2 below).

• (iii) Compared to the conditions (F1)–(F4), the conditions (F5) and (F6) seem technical. The condition (F5) is used in interaction estimates in Lemmas 2.4 and 3.3 and (F6) is used to prove the nonexistence of the minimizer in Lemma 3.1.

• (iv) If we assume $0\not \equiv V(x)\ge 0$ and $\lim _{|x|\rightarrow \infty }V(x)=0$ instead of (V1), then $e(\alpha )$ is not attained for all $\alpha >0$, and $e(\alpha )=e_{\infty }(\alpha )$ for $\alpha \ge 0$. See Theorem A.1 in Appendix A.

As mentioned above, in [22, Theorem 1.3], Shibata observed whether $\alpha _{0,\infty } > 0$ or $\alpha _{0,\infty } = 0$. We also consider the same question: whether $\alpha _0>0$ or $\alpha _0=0$ under the presence of the potential term V(x).

Theorem B

Suppose (F1)–(F4) and (V1). Then the following (i) and (ii) hold:

• (i) In addition, assume that there exists an $s_0>0$ such that $f(s) \ge 0$ in $[0,s_0]$ and the following (V3) holds:

  • (V3) $\displaystyle \inf _{ \Vert \varphi \Vert _{L^2(\mathbb {R}^N)} = 1} \int _{\mathbb {R}^N} \left( |\nabla \varphi |^2 + V(x) \varphi ^2 \right) d x < 0$.

  Then $\alpha _0=0$. Moreover, when $N=1,2$, (V1) implies (V3) and $\alpha _{0} = 0$.

• (ii) Suppose $N \ge 3$ and the following condition (F9) in addition to (F1)–(F4) and (V1):

  • (F9) $\limsup _{s\downarrow 0}F(s)/s^{{p_\mathrm{c}}+1}<\infty$.

  Then there exists $\alpha _1 = \alpha _1(N, f) > 0$ satisfying the following property: for each $\alpha \in (0,\alpha _1)$ we may find a $c_\alpha >0$ such that $V(x) \ge - c_\alpha |x|^{-2}$ for $|x| > 0$ implies $\alpha _0 \ge \alpha >0$.

Remark 1.2

Notice that Theorem B (i) may be used to see a difference between the cases $V(x) \equiv 0$ and $V(x) \not \equiv 0$. Indeed, since (F6) plays a role only for $N\ge 5$, when $N=1,2$, if (V1), (F1)–(F5) and (F9) hold, then we obtain $0= \alpha _{0} < \alpha _{0,\infty }$ due to Theorems A, B (i) and [22, Theorem 1.3].

Let us mention other related results. For the homogeneous problem $e_{\infty }(\alpha )$, Bellazzini et al. [5] showed that there exists $\bar{\alpha }\ge 0$ such that $e_{\infty }(\alpha )$ is attained for $\alpha >\bar{\alpha }$ if (F5’) given in Proposition 2.1, (F8) and the following assumption are satisfied:

$$\begin{aligned} \text {there exist}\ C_1,\ C_2\ge 0,\ 1<q\le p<2^*-1\ \text {such that}\ |f(s)|\le C_1|s|^q+C_2|s|^p.\nonumber \\ \end{aligned}$$

(1.5)

Moreover, they proved that $\bar{\alpha }=0$ if

$$\begin{aligned} \text {there exists}\ 1<p<{p_\mathrm{c}}\ \text {such that}\ F(s)>s^{p+1}\ \text {for small}\ s>0. \end{aligned}$$

(1.6)

Note that (F10) in Proposition 2.2 is a generalization of (1.6). In [22] the threshold $\alpha _{0,\infty }$ was found and Proposition 2.1 was obtained. In particular, the nonexistence part (Proposition 2.1 (ii)) was proved. In Garrisi–Georgiev [13] the one-dimensional case was studied and the orbital stability of the minimizers was obtained if (1.5), (F5$'$) and the following hold:

$$\begin{aligned} \text {there exist }1<p<5(={p_\mathrm{c}})\hbox { and }s_0\ge 0\hbox { such that }\ F(s)\le C|s|^{p+1}\ \text {for}\ s\ge s_0. \end{aligned}$$

See [12] for a quasilinear homogeneous problem and [7] for a Schrödinger-Poisson problem with pure power nonlinearity. For the inhomogeneous problem $e(\alpha )$, in [6, 8, 18] the attainability was studied. In [6, 8], they deal with the rather special type of nonlinearity, that is, $f(u) = |u|^{p-1} u$ in [6] and $f(u) = Q(x) |u|^{p-1} u$ in [8]. In Jeanjean–Squassina [18, 2.4 A Stuart’s type problem] the nonlinear term is F(x, u). They showed that $e(\alpha )$ is attained if F satisfies

$$\begin{aligned} \lim _{|x|\rightarrow \infty }F(x,s)=0\ \text {uniformly in}\ s\in \mathbb {R}. \end{aligned}$$

(1.7)

Here, (1.7) leads to the weak lower semicontinuity of E(u) which our problem does not satisfy.

Let us explain technical details for the proof of Theorem A. To prove Theorem A, we try to establish (1.2) in a scheme similar to [22], and a difficulty is to exclude dichotomy since we treat $V \in L^\infty (\mathbb {R}^N)$ and E(u) is not weak lower semicontinuous. Furthermore, since our nonlinearity is general and there is a term V(x), a scaling argument in [10] or the scaled function $u( \lambda x)$ in the homogeneous case may not be useful. Therefore, we need to bring another idea to overcome this difficulty. In this paper, we perform a careful interaction estimate to exclude dichotomy in Lemma 3.3 where (F5) is used. This usage of the interaction estimate is inspired by Hirata [15] where the unconstrained problem is studied and we try to apply this estimate in the $L^2$-constraint setting. To do so, we modify any minimizing sequence to be an approximated positive solution of the Euler-Lagrange equation and prove the precompactness of the modified minimizing sequence. This reduction is done in Lemmas 2.6 and 2.8, and is also used in [16] for the homogeneous case. In addition to the reduction, to follow the scheme in [22], we also need the nonexistence result of the minimizer for which the condition $1 \le N \le 4$, (F6) or (V2) is used. See Lemma 3.1. Here we also have a difference between the cases $V(x) \equiv 0$ and $V(x) \not \equiv 0$ because the scaled function $u( \lambda x)$ may not be useful.

Finally we make a comment on the usage of the interaction estimate. Our argument is also applied to a minimizing problems with two constraint conditions and potentials. This will be discussed in [17].

This paper consists of five sections. In Sect. 2 we recall fundamental properties of the problems $e(\alpha )$ and $e_{\infty }(\alpha )$. In Sect. 3 we study the existence and nonexistence of the minimizers of $e(\alpha )$ and prove Theorem A. In Sect. 4 we prove Theorem B. In “Appendix A” we show that $e(\alpha )$ is not attained if $0\not \equiv V(x)\ge 0$ and $\lim _{|x|\rightarrow \infty }V(x)=0$.

Notations

• For $p\ge 1$, $L^p(\Omega )$ denotes the space of complex-valued measurable functions u on $\Omega \subset \mathbb {R}^N$ satisfying $\int _{\Omega }|u|^pdx<\infty$ whose norm is defined by $\left\| u\right\| _{L^p(\Omega )}:=\left( \int _{\Omega }|u|^pdx\right) ^{1/p}$. When $\Omega =\mathbb {R}^N$, write $\left\| u\right\| _p:=\left\| u\right\| _{L^p(\mathbb {R}^N)}$.

• $L^{\infty }(\Omega )$ denotes the space of complex-valued essentially bounded measurable functions u on $\Omega \subset \mathbb {R}^N$ whose norm is defined by $\left\| u\right\| _{L^{\infty }(\Omega )}:=\text {esssup}_{x\in \Omega }|u(x)|$. When $\Omega =\mathbb {R}^N$, write $\left\| u\right\| _{\infty }:=\text {esssup}_{x\in \mathbb {R}^N}|u(x)|$.

• We regard $L^2(\mathbb {R}^N)$ as a Hilbert space over $\mathbb {R}$ by the inner product $\left<u,v\right>_{L^2}:= \mathrm{Re} \int _{\mathbb {R}^N}f(x)\overline{g(x)}dx$.

• The set H stands for the space of complex-valued measurable functions u of the Sobolev space of order 1 whose norm is defined by $\left\| u\right\| _H:=\left( \int _{\mathbb {R}^N}|u|^2dx+\int _{\mathbb {R}^N}|\nabla u|^2dx\right) ^{1/2}$, i.e., $H:=H^1(\mathbb {R}^N)$. We denote its inner product by $\langle u, v \rangle _{H} := \langle \nabla u , \nabla v \rangle _{L^2} + \langle u , v \rangle _{L^2}$ and the dual space of H by $H^*$.

2 Preliminaries

We first recall known facts about the homogeneous problem $e_{\infty }(\alpha )$.

Proposition 2.1

([22, Theorems 1.1 and 1.5]) Suppose (F1)–(F4) and the following (F5$'$):

• (F5’) There exists $s_0>0$ such that $F(s_0)>0$.

Then there exists a unique $\alpha _{0,\infty }\in [0,\infty )$ such that (1.4) and the following (i) and (ii) hold:

• (i) If $\alpha >\alpha _{0,\infty }$, then every minimizing sequence for $e_{\infty }(\alpha )$ has a convergent subsequence in H up to translations. Therefore, $e_{\infty }(\alpha )$ is attained, the set of all minimizers is precompact in H up to translations and (1.2) holds. Moreover, in addition, if (F7) and (F8) hold, then the set of all minimizers is orbitally stable.

• (ii) If $0<\alpha <\alpha _{0,\infty }$, then $e_{\infty }(\alpha )$ is not attained.

Note that (F5) implies (F5$'$). The next proposition concerns when $\alpha _{0,\infty } = 0$ or $\alpha _{0,\infty } > 0$ holds.

Proposition 2.2

([22, Theorems 1.3]) Suppose (F1)–(F4) and (F5$'$). Then the following (i) and (ii) hold:

• (i) If the following (F10) holds:

  • (F10) $\liminf _{s\downarrow 0}F(s)/s^{{p_\mathrm{c}}+1}=\infty$,

  then $\alpha _{0,\infty }=0$.

• (ii) If (F9) holds, then $\alpha _{0,\infty }>0$.

Next, we collect some properties about F(s). We begin with a variant of [22, Lemma 2.2 (i)].

Lemma 2.3

Suppose (F1)–(F4), $u_0 \in H$ and that $(u_n)$ is bounded in H. If $\Vert u_n - u_0\Vert _p \rightarrow 0$ for some $p \in [2,\infty ]$, then $\lim _{n\rightarrow \infty }\int _{\mathbb {R}^N}F(|u_n|)dx=\int _{\mathbb {R}^N}F(|u_0|)dx$.

Proof

We remark that we may assume $u_n \ge 0$ without loss of generality since $\Vert |u_n| - |u_0| \Vert _p \le \Vert u_n - u_0 \Vert _p$ and $\Vert \nabla |u| \Vert _2 \le \Vert \nabla u \Vert _2$ (see [19, Theorem 6.17]). By Sobolev’s inequality and Hölder’s inequality, $\Vert u_n - u_0 \Vert _q \rightarrow 0$ for any $q \in (2,2^*)$. We also set $M_0:= \sup _{n \ge 1} \Vert u_n \Vert _H < \infty$.

Next, by (F3) and (F4), for each $\varepsilon >0$, one may find a $C_\varepsilon >0$ such that

$$\begin{aligned} |f(s)| \le \varepsilon |s| + C_\varepsilon |s|^{p_c} \quad \text {for all }s \in \mathbb {R}. \end{aligned}$$

From

$$\begin{aligned} \begin{aligned} \left| F(u_n) - F(u_0) \right|&= \left| \int _0^1 \frac{d}{d \theta } F \left( \theta u_n + (1-\theta ) u_0 \right) d \theta \right| \\&\le \int _0^1 \left| f( \theta u_n + (1-\theta ) u_0 ) \right| d \theta |u_n-u_0| \\&\le \int _0^1 \left\{ \varepsilon \left( u_n + u_0 \right) + C_\varepsilon \left( u_n + u_0 \right) ^{p_c} \right\} d \theta | u_n - u_0 | \\&= \left\{ \varepsilon \left( u_n + u_0 \right) + C_\varepsilon \left( u_n + u_0 \right) ^{p_c} \right\} | u_n - u_0 | \end{aligned} \end{aligned}$$

and Hölder’s inequality, we have

$$\begin{aligned} \begin{aligned} \left| \int _{\mathbb {R}^N} \left\{ F(u_n) - F(u_0) \right\} d x \right|&\le \int _{\mathbb {R}^N} \left\{ \varepsilon \left( u_n + u_0 \right) + C_\varepsilon \left( u_n + u_0 \right) ^{p_c} \right\} | u_n - u_0 | d x \\&\le \varepsilon \left( \Vert u_n \Vert _2 + \Vert u_0\Vert _2 \right) \Vert u_n - u_0 \Vert _2 \\&\quad + C_\varepsilon \left\| u_n + u_0 \right\| _{p_c+1}^{p_c} \Vert u_n - u_0 \Vert _{p_c+1}. \end{aligned} \end{aligned}$$

Noting $2<p_c+1 < 2^*$, we obtain

$$\begin{aligned} \limsup _{n\rightarrow \infty } \left| \int _{\mathbb {R}^N} \left\{ F(u_n) - F(u_0) \right\} d x \right| \le 4M_0^2 \varepsilon . \end{aligned}$$

Since $\varepsilon >0$ is arbitrary, $\int _{\mathbb {R}^N} F(u_n) dx \rightarrow \int _{\mathbb {R}^N} F(u_0) dx$ as $n \rightarrow \infty$. $\square$

Next, we borrow one lemma from [15], which is used for the interaction estimate in the proof of Lemma 3.3. For a proof, see [15].

Lemma 2.4

([15, Lemma 4.4]) Assume (F1) and (F5). Let $\delta _1 >0$ be as in (F5). Then the following (i) and (ii) hold:

• (i) There exists $\delta _2\in (0,\delta _1]$ such that

  $$\begin{aligned} F(u_1)+F(u_2)-F(u_1+u_2)+\frac{1}{2}(f(u_1)u_2+f(u_2)u_1)\le 0\ \ \text {for}\ \ u_1,u_2\in [0,\delta _2]. \end{aligned}$$

• (ii) For each compact set $K\subset (0,\infty )$, there exist $C_K>0$ and $\delta _K>0$ such that

  $$\begin{aligned}&F(u_1)+F(u_2)-F(u_1+u_2)+\frac{1}{2}(f(u_1)u_2+f(u_2)u_1)\\&\quad \le -C_Ku_2\ \ \text {for}\ \ u_1\in K\ \text {and}\ u_2\in [0,\delta _K]. \end{aligned}$$

In the next lemma we state fundamental properties of $e(\alpha )$ and $e_{\infty }(\alpha )$.

Lemma 2.5

Assume (F1)–(F4) and (V1). Then the following hold:

• (i) $e(\alpha )>-\infty$ for $\alpha >0$.

• (ii) For $\alpha >0$, every minimizing sequence for $e(\alpha )$ is bounded in H.

• (iii) $e(\alpha )\le e_{\infty }(\alpha )\le 0$ for $\alpha \ge 0$.

• (iv) $e(\alpha )\le e(\beta )+e_{\infty }(\alpha -\beta )$ for $0\le \beta <\alpha$.

• (v) $e(\alpha )$ is nonincreasing in $\alpha \ge 0$.

Proof

• (i) The proof is almost the same as [22, Lemma 2.2 (ii)]. By the assumptions (F1)–(F4), for $\varepsilon >0$, there exists a positive constant $C_{\varepsilon }>0$ such that

  $$\begin{aligned} F(|u|)\le C_{\varepsilon }|u|^2+\varepsilon |u|^{{p_\mathrm{c}}+1}. \end{aligned}$$

  (2.1)

  By the Gagliardo–Nirenberg inequality we have

  $$\begin{aligned} \left\| u\right\| _{{p_\mathrm{c}}+1}^{{p_\mathrm{c}}+1}\le C\left\| u\right\| _2^{4/N}\left\| \nabla u\right\| _2^2. \end{aligned}$$

  (2.2)

  Thus, (2.1) and (2.2) give

  $$\begin{aligned} \left| \int _{\mathbb {R}^N}F(|u|)dx\right| \le C_{\varepsilon }\left\| u\right\| _2^2+\varepsilon C\alpha ^{2/N}\left\| \nabla u\right\| _2^2. \end{aligned}$$

  We choose $\varepsilon >0$ such that $\varepsilon C\alpha ^{2/N}=1/4$. Then for $u\in M(\alpha )$,

  $$\begin{aligned} \int _{\mathbb {R}^N}F(|u|)dx\le C_{\varepsilon }\alpha +\frac{1}{4}\left\| \nabla u\right\| _2^2, \end{aligned}$$

  which implies

  $$\begin{aligned} E(u)\ge \frac{1}{4}\left\| \nabla u\right\| _2^2-C_\varepsilon \alpha . \end{aligned}$$

  (2.3)

  Hence, (i) holds.

• (ii) Since $u\in M(\alpha )$, the conclusion immediately follows from (2.3).

• (iii) Because $E(u)\le E_{\infty }(u)$ for each $u \in H$ due to (V1), we easily see that $e(\alpha )\le e_{\infty }(\alpha )$. For the inequality $e_{\infty }(\alpha )\le 0$, see [22, Lemma 2.3 (i)].

• (iv) For $\varepsilon >0$, we can find $\varphi _{\varepsilon }$, $\psi _{\varepsilon }\in C_0^{\infty }(\mathbb {R}^N)$ such that

  $$\begin{aligned} \varphi _{\varepsilon }\in M(\beta ),\ \psi _{\varepsilon }\in M(\alpha -\beta ),\ E(\varphi _{\varepsilon })\le e(\beta )+\varepsilon ,\ E_{\infty }(\psi _{\varepsilon })\le e_{\infty }(\alpha -\beta )+\varepsilon . \end{aligned}$$

  Let $u_{\varepsilon ,n}(x):=\varphi _{\varepsilon }(x)+\psi _{\varepsilon }(x-n\mathbf{e}_1)$. Since $\varphi _{\varepsilon }$ and $\psi _{\varepsilon }$ have compact support, we see that $u_{\varepsilon ,n}\in M(\alpha )$ for large n and that $e(\alpha )\le E(u_{\varepsilon ,n})=E(\varphi _{\varepsilon })+E(\psi _{\varepsilon }(\cdot -n\mathbf{e}_1))$. From $E(\psi _{\varepsilon } (\cdot - n \mathbf{e}_1) ) \rightarrow E_\infty (\psi _{\varepsilon })$ as $n\rightarrow \infty$ thanks to (V1), it follows that

  $$\begin{aligned} e(\alpha )\le \lim _{n\rightarrow \infty } \left( E(\varphi _{\varepsilon })+E(\psi _{\varepsilon }(\cdot -n\mathbf{e}_1)) \right) = E(\varphi _{\varepsilon })+E_{\infty }(\psi _{\varepsilon })\le e(\beta )+e_{\infty }(\alpha -\beta )+2\varepsilon . \end{aligned}$$

  Since $\varepsilon >0$ is arbitrary, (iv) holds.

• (v) By (iii) and (iv), we have

  $$\begin{aligned} e(\alpha )\le e(\beta )+e_{\infty }(\alpha -\beta )\le e(\beta )\ \text {for}\ 0\le \beta <\alpha . \end{aligned}$$

  Thus, $e(\alpha )$ is nonincreasing in $\alpha$.

$\square$

In the next two lemmas we collect some properties of a minimizing sequence for $e(\alpha )$.

Lemma 2.6

Assume (F1)–(F4) and (V1). The following hold:

• (i) Let $(u_n)\subset M(\alpha )$ be a minimizing sequence for $e(\alpha )$, and let $|u_n|(x):=|u_n(x)|$. Then $(|u_n|)$ is also a minimizing sequence.

• (ii) If $u_0 \in H$ and $(u_n)$ is a minimizing sequence for $e(\alpha )$ with $\Vert u_n - u_0 \Vert _{2} \rightarrow 0$, then $\Vert u_n - u_0 \Vert _{H} \rightarrow 0$. Furthermore, if $u_0 \in H$ and $(u_n)$ is a minimizing sequence of for $e(\alpha )$ and $\Vert |u_n| - |u_0| \Vert _2 \rightarrow 0$, then $\Vert u_n - u_0 \Vert _H \rightarrow 0$.

Proof

• (i) By $\left\| \nabla |u_n|\right\| _2^2\le \left\| \nabla u\right\| _2^2$ ([19, Theorem 6.17]) and $|u_n| \in M(\alpha )$, we see that $E(|u_n|)\le E(u_n)$ and $(|u_n|)$ is also a minimizing sequence.

• (ii) From $\Vert u_n - u_0 \Vert _2 \rightarrow 0$, it follows that

  $$\begin{aligned} u_0\in M(\alpha )\ \ \text {and}\ \ \lim _{n\rightarrow \infty }\int _{\mathbb {R}^N}V(x)u_n^2dx=\int _{\mathbb {R}^N}V(x)u_0^2dx. \end{aligned}$$

  (2.4)

  Moreover, by Lemma 2.5 (ii), $(u_n)$ is bounded in H. Thanks to $\Vert u_n - u_0 \Vert _2 \rightarrow 0$, we obtain $u_n \rightharpoonup u_0$ weakly in H. Thus, Lemma 2.3 and the weak lower semicontinuity of $\Vert \nabla \cdot \Vert _2$ yield

  $$\begin{aligned} e(\alpha ) \le E(u_0) \le \liminf _{n\rightarrow \infty } E(u_n) = \lim _{n\rightarrow \infty } E(u_n) = e(\alpha ), \end{aligned}$$

  which implies $\Vert \nabla u_n \Vert _2^2 \rightarrow \Vert \nabla u_0 \Vert _2^2$. Combining this fact with $\nabla u_n \rightharpoonup \nabla u_0$ weakly in $L^2(\mathbb {R}^N)$, we observe that $\Vert \nabla u_n - \nabla u_0 \Vert _2 \rightarrow 0$ and $\Vert u_n - u_0 \Vert _{H} \rightarrow 0$.

  Assume that $(u_n)$ is a minimizing sequence for $e(\alpha )$ with $\Vert |u_n| - |u_0| \Vert _2 \rightarrow 0$. By Lemma 2.5 (ii), $(u_n)$ is bounded in H, hence, choosing a subsequence if necessary, we may assume $u_n \rightarrow u_0$ in $L^2_\mathrm{loc} (\mathbb {R}^N)$ without loss of generality. Since $\Vert |u_{n}| - |u_0|\Vert _2 \rightarrow 0$ and $u_{n} \rightarrow u_0$ in $L^2_\mathrm{loc}(\mathbb {R}^N)$, we may find a $w_0 \in L^2(\mathbb {R}^N)$ and a subsequence $(u_{n_k})$ such that $|u_{n_k}(x)| \le w_0(x)$ and $u_{n_k}(x) \rightarrow u_0(x)$ a.e. $\mathbb {R}^N$. The dominated convergence theorem gives $\Vert u_{n_k} - u_0 \Vert _{2} \rightarrow 0$ and the former assertion gives $\Vert u_{n_k} - u_0 \Vert _{H} \rightarrow 0$ due to the fact that $(u_{n_k})$ is a minimizing sequence for $e(\alpha )$. Since the limit is independent of subsequences, we have $\Vert u_n - u_0 \Vert _H \rightarrow 0$ and the proof is completed.

$\square$

Remark 2.7

A similar argument to the proof of Lemma 2.6 shows that if $u_0 \in M(\alpha )$ is a minimizer, then so is $|u_0(x)|$. Hence, when $e(\alpha )$ is attained, we may always find a nonnegative minimizer.

Lemma 2.8

Let $(u_n)\subset M({\alpha })$ be a minimizing sequence for $e(\alpha )$. Then there exist $(v_n)\subset M(\alpha )$ and $(\lambda _n)\subset \mathbb {R}$ such that $(\lambda _n)$ is bounded and

$$\begin{aligned} \left\| u_n-v_n\right\| _H\rightarrow 0,\quad E'(v_n)+\lambda _nQ'(v_n)\rightarrow 0\ \text {strongly in}\ H^*, \end{aligned}$$

(2.5)

where $Q(u) := \Vert u \Vert _2^2$. Furthermore, if $(u_n)$ is real-valued, then we may choose $v_n$ as real-valued function.

Remark 2.9

We notice that if $(v_n)$ in Lemma 2.8 has a strongly convergent subsequence in H, then so is $(u_n)$.

Proof of Lemma 2.8

We first remark that Q is smooth and $Q'(u) u = 2 Q(u)$. By $M(\alpha ) = Q^{-1} (\alpha )$, we notice that $M(\alpha )$ is closed and a Hilbert manifold with codimension 1. Moreover, the tangent space of $M(\alpha )$ at u and the tangent derivative $D_{T_uM(\alpha )} E$ of E at u are given by

$$\begin{aligned} \begin{aligned}&T_u M(\alpha ) = \left\{ v \in H \ |\ \langle \nabla Q(u) , v \rangle _{H} = 0 \right\} , \\&D_{T_uM(\alpha )} E (u) = E'(u) - \frac{ E'(u) \nabla Q(u)}{\Vert \nabla Q(u) \Vert _{H}^2} Q'(u), \end{aligned} \end{aligned}$$

(2.6)

where $\nabla Q(u) \in H$ is the unique element satisfying $\langle \nabla Q(u) , v \rangle _{H} = Q'(u) v$ for every $v \in H$.

We now apply Ekeland’s variational principle for E(u) and $(u_n)$ on $M(\alpha )$ to get $v_n \in M(\alpha )$ satisfying

$$\begin{aligned} \Vert u_n - v_n \Vert _{H} \le \sqrt{\varepsilon _n}, \quad E(v_n) \le E(w) + \sqrt{\varepsilon _n} \Vert v_n - w \Vert _{H} \quad \text {for each }w \in M(\alpha ), \end{aligned}$$

(2.7)

where $\varepsilon _n := E(u_n) - e(\alpha ) \ge 0$. Putting $w=u_n$ in (2.7) and the fact $v_n \in M(\alpha )$ assert that $(v_n)$ is also a minimizing sequence. In addition, (2.6) and (2.7) imply that

$$\begin{aligned} \left\| D_{T_{v_n}M(\alpha )} E(v_n) \right\| _{ (T_{u_n} M(\alpha ))^*} := \sup \left\{ D_{T_{v_n}M(\alpha )} E(v_n) \varphi \ |\ \Vert \varphi \Vert _{H}=1, \ \varphi \in T_{v_n}M(\alpha ) \right\} \rightarrow 0.\nonumber \\ \end{aligned}$$

(2.8)

Since $(v_n)$ is bounded in H, $E'$ maps bounded sets into bounded sets and $\Vert \nabla Q(v_n) \Vert _{H} \ge 2\alpha / \Vert v_n \Vert _H$ for any $n \ge 1$ due to $Q'(v_n) v_n = 2 Q(v_n) = 2\alpha$, setting $\lambda _n := - E'(v_n) \nabla Q(v_n) / \Vert \nabla Q(v_n) \Vert _{H}^2$, from (2.6) and (2.8), we see that (2.5) holds.

If $(u_n)$ is real-valued, then we restrict ourselves into $H_{\mathbb {R}} := \{ u \in H \ |\$u$~\text {is real-valued} \}$ and $M_\mathbb {R}(\alpha ) := M(\alpha ) \cap H_{\mathbb {R}}$. Since $e(\alpha ) = \inf _{u \in M_{\mathbb {R}}(\alpha )} E(u)$ holds, we may use the above argument on $M_{\mathbb {R}}(\alpha )$ to obtain real-valued functions $(v_n)$ satisfying (2.5). Thus we complete the proof. $\square$

3 Proof of Theorem A

We first observe the case when $e(\alpha )$ is not attained.

Lemma 3.1

Assume (F1)–(F5) and (V1) and assume (F6) or (V2). If there are $\alpha >0$ and $\beta >0$ such that $e(\alpha )=e(\beta )$ and $\alpha >\beta$, then $e(\beta )$ is not attained.

Proof

We first prove the following:

$$\begin{aligned} \text {If }e(\,\cdot \,)\hbox { is constant in }[\beta ,{\beta }+\varepsilon )\hbox { for small }\varepsilon >0, \hbox {then }e(\beta )\hbox { is not attained.} \end{aligned}$$

(3.1)

Remark that (3.1) implies our conclusion. Indeed, we see by Lemma 2.5 (v) that $e(\,\cdot \,)$ is nonincreasing. Since $e(\alpha )=e(\beta )$, we observe that $e(\,\cdot \,)$ is constant in the interval $[\beta ,\alpha ]$. Then by (3.1), $e(\beta )$ is not attained.

Now we prove (3.1) by contradiction and let $u_0\in M(\beta )$ be a minimizer for $e(\beta )$. Thanks to Remark 2.7, we may assume $u_0 \ge 0$. Notice that $u_0$ is a (classical) solution of

$$\begin{aligned} -\Delta u_0+V(x)u_0-f(u_0)=-2\lambda u_0\ \ \text {in}\ \ \mathbb {R}^N\end{aligned}$$

(3.2)

for some $\lambda \in \mathbb {R}$. Next, we show by contradiction that $\lambda \le 0$. If $\lambda >0$, then

$$\begin{aligned} \left. \frac{d}{dt}E(tu_0)\right| _{t=1} =\int _{\mathbb {R}^N}|\nabla u_0|^2+V(x)u_0^2-f(u_0)u_0dx=-2\lambda \int _{\mathbb {R}^N}|u_0|^2dx=-2\lambda \beta <0. \end{aligned}$$

Hence, for sufficiently small $\eta >0$, the monotonicity of $e(\alpha )$ yields

$$\begin{aligned} e(\beta +\varepsilon )\le e((1+\eta )^2\beta )\le E((1+\eta )u_0)<E(u_0)=e(\beta ), \end{aligned}$$

which is a contradiction. Thus, $\lambda \le 0$.

We prove (3.1). Since $V(x)\le 0\le u_0(x)$ and $\lambda \le 0$, by (3.2) and $f(s) \ge 0$ ($s \ge 0$) due to (F5), we have

$$\begin{aligned} -\Delta u_0\ge f(u_0) \ge 0 \ \ \text {in}\ \ \mathbb {R}^N\ \ \text {and}\ \ u_0\ge 0\ \ \text {in}\ \ \mathbb {R}^N. \end{aligned}$$

(3.3)

Hence, the strong maximum principle and $u_0 \in M(\beta )$ give $u_0>0$ in $\mathbb {R}^N$.

If $N=1,2$, then $-\Delta u_0\ge 0$ in $\mathbb {R}^N$. Since $u_0$ is a positive super-harmonic function in $\mathbb {R}$ or $\mathbb {R}^2$, we see that $u_0$ is constant (see [21, Chapter 2, Theorem 29] for $N=2$). However, this contradicts $u_0 \in L^2(\mathbb {R}^N)$ and $e(\beta )$ is not attained.

If $N=3,4$, then we show that (3.2) has no solution in H. This claim is proved in [16, Lemma A.2], however, we give another simple proof which is similar to [4, Lemma 3.12]. Let $c_1>0$ and $w(x):=u_0(x)-c_1|x|^{2-N}$. Here $c_1>0$ can be chosen so that $w(x)\ge 0$ for all $|x|=1$ due to $u_0 > 0$ in $\mathbb {R}^N$. From $-\Delta w=-\Delta u_0\ge 0$ for $|x|>1$ and $w(x)\rightarrow 0$ as $|x|\rightarrow \infty$, the weak maximum principle asserts that $w\ge 0$ in $|x|\ge 1$, which implies $u_0(x)\ge c_1|x|^{2-N}$ for $|x|\ge 1$. However, this contradicts $u_0\in L^2(\mathbb {R}^N)$ when $N=3,4$. Hence, $e(\beta )$ is not attained.

We consider the case $N\ge 5$. In this case we assume (F6) or (V2). If (F6) holds, then it follows from the result of [1] that (3.3) has no solution. Hence, $e(\beta )$ is not attained.

On the other hand, when (V2) holds, we first observe from (3.2) that $u_0$ satisfies the Pohozaev identity:

$$\begin{aligned} 0=\frac{N-2}{2}\left\| \nabla u_0\right\| ^2_2-N \int _{\mathbb {R}^N}F(u_0)-{\lambda }u_0^2-\frac{V(x)}{2}u_0^2dx +\frac{1}{2}\int _{\mathbb {R}^N}(x\cdot \nabla V(x)) u_0^2 dx. \end{aligned}$$

Then we have

$$\begin{aligned} 0&\ge e(\beta )\\&= E(u_0)\\&=\frac{1}{2}\left\| \nabla u_0\right\| _2^2+\frac{1}{2}\int _{\mathbb {R}^N}V(x)u_0^2dx -\int _{\mathbb {R}^N}F(u_0)dx\\&=\frac{1}{N}\left\| \nabla u_0\right\| _2^2-\lambda \left\| u_0\right\| _2^2 -\frac{1}{2N}\int _{\mathbb {R}^N}x\cdot \nabla V(x)u_0^2dx\\&\ge \frac{1}{N}\left( \left\| \nabla u_0\right\| _2^2-\frac{1}{2}\int _{\mathbb {R}^N}x\cdot \nabla V(x)u_0^2dx \right) , \end{aligned}$$

where we used $\lambda \le 0$. Since $\nabla V(x)\in L^{\infty }(\mathbb {R}^N)$, the strict inequality in (V2) holds on $A\subset \mathbb {R}^N$, where the Lebesgue measure of A is strictly positive. Since $u_0 > 0$ in $\mathbb {R}^N$, we get

$$\begin{aligned} \frac{1}{2}\int _{\mathbb {R}^N}x\cdot \nabla V(x)u_0^2dx< \frac{(N-2)^2}{4}\int _{\mathbb {R}^N}\frac{u_0^2}{|x|^2}dx. \end{aligned}$$

From Hardy’s inequality, it follows that

$$\begin{aligned} 0\ge Ne(\beta )\ge \left\| \nabla u_0\right\| _2^2 -\frac{1}{2}\int _{\mathbb {R}^N}x\cdot \nabla V(x)u_0^2dx >\left\| \nabla u_0\right\| _2^2-\frac{(N-2)^2}{4}\int _{\mathbb {R}^N}\frac{u_0^2}{|x|^2}dx\ge 0. \end{aligned}$$

This is a contradiction and $e(\beta )$ is not attained. Thus (3.1) holds. $\square$

Next we observe a behavior of minimizing sequence when the compactness does not hold.

Lemma 3.2

Assume (F1)–(F5) and (V1) and assume (F6) or (V2). Let $(u_n)\subset M(\alpha )$ be a minimizing sequence for $e(\alpha )$ such that $u_n\rightharpoonup u_0$ weakly in H and let $\beta :=\left\| u_0\right\| _2^2$. If either $0<\beta <\alpha$ or both $\beta =0$ and $e(\alpha )<0$, then there exist $(y_n)\subset \mathbb {R}^N$ and $w_0\in H\backslash \{0\}$ such that

$$\begin{aligned}&|y_n|\rightarrow \infty ,\ u_n(\,\cdot \,+y_n)\rightharpoonup w_0 \hbox { weakly in }H, \end{aligned}$$

(3.4)

$$\begin{aligned}&\lim _{n\rightarrow \infty }\left\| u_n-u_0-w_0(\,\cdot \,-y_n)\right\| _2=0\ \text {and}\ \alpha =\beta +\gamma , \end{aligned}$$

(3.5)

where $\gamma :=\left\| w_0\right\| _2^2$. Moreover, the following hold:

$$\begin{aligned} E(u_0)=e(\beta ),\ E_{\infty }(w_0)=e_{\infty }(\gamma )\ \text {and}\ e(\alpha )=e(\beta )+e_{\infty }(\gamma ). \end{aligned}$$

(3.6)

Proof

We divide the proof into three steps.

Step 1:We find $(y_n)\subset \mathbb {R}^N$and$w_0\in H\backslash \{0\}$such that (3.4) holds.

First, we show by contradiction that

$$\begin{aligned} \liminf _{n\rightarrow \infty }\sup _{z\in \mathbb {Z}^N}\left\| u_n-u_0\right\| _{L^2(Q^N+z)}>0 \quad \text {where }Q^N:=[0,1]^N. \end{aligned}$$

(3.7)

Suppose on the contrary that $\sup _{z\in \mathbb {Z}^N}\left\| u_n-u_0\right\| _{L^2(Q^N+z)}\rightarrow 0$. Then, $u_n\rightarrow u_0$ strongly in $L^q(\mathbb {R}^N)$ for $2<q<2^*$ (See [23]). By Lemmas 2.3 and 2.5, we have

$$\begin{aligned} e(\alpha ) \le e(\beta )\le E(u_0)\le \lim _{n\rightarrow \infty }E(u_n)=e(\alpha ). \end{aligned}$$

(3.8)

When $\beta =0$ and $e(\alpha )<0$, we get a contradiction. Hence (3.7) holds provided $\beta = 0$ and $e(\alpha ) < 0$.

Next, let us consider the case $0<\beta <\alpha$. In this case, (3.8) asserts $e(\alpha ) = E(u_0) = e(\beta )$ and $u_0$ is a minimizer due to $\Vert u_0 \Vert _2^2 = \beta$. However, this contradicts Lemma 3.1. Therefore, (3.7) holds.

From (3.7) and $u_n \rightarrow u_0$ in $L^2_\mathrm{loc} (\mathbb {R}^N)$, we can find $(y_n)\subset \mathbb {R}^N$ such that $\left\| u_n\right\| _{L^2(Q^N+y_n)}\rightarrow c_0>0$ and $|y_n|\rightarrow \infty$. Let

$$\begin{aligned} u_n(\cdot +y_n)\rightharpoonup w_0\ \text {weekly in}\ H. \end{aligned}$$

Note that $w_0\not \equiv 0$ because $c_0>0$. Therefore, $(y_n)$ and $w_0$ satisfy (3.4). The proof of Step 1 is complete.

Since $|y_n|\rightarrow \infty$ ($n\rightarrow \infty$), we have

$$\begin{aligned} \left\| u_n-u_0-w_0(\cdot -y_n)\right\| _2^2&=\left\| u_n\right\| _2^2+\left\| u_0\right\| _2^2+\left\| w_0\right\| _2^2\nonumber \\&\quad -2\left<u_n,u_0\right>_{L^2}-2\left<u_n(\cdot +y_n),w_0\right>_{L^2}+o(1)\nonumber \\&=\left\| u_n\right\| _2^2-\left\| u_0\right\| _2^2-\left\| w_0\right\| _2^2+o(1). \end{aligned}$$

(3.9)

In particular,

$$\begin{aligned} \gamma :=\left\| w_0\right\| _2^2\le \liminf _{n\rightarrow \infty }(\left\| u_n\right\| _2^2-\left\| u_0\right\| _2^2)=\alpha -\beta . \end{aligned}$$

Note that $\gamma >0$ because $w_0\ne 0$.

Step 2:We show that$(y_n)$and$w_0$satisfy (3.5).

Let $\delta :=\lim _{n\rightarrow \infty }\left\| u_n-u_0-w_0(\cdot -y_n)\right\| _2^2$. Then, we see by (3.9) that $\delta =\alpha -\beta -\gamma$. Our aim is to show that $\delta =0$. Suppose on the contrary that

$$\begin{aligned} \delta >0. \end{aligned}$$

(3.10)

By direct calculation we have

$$\begin{aligned}&\frac{1}{2}\left( \left\| \nabla u_n\right\| _2^2-\left\| \nabla u_0\right\| _2^2-\left\| \nabla w_0(\cdot -y_n)\right\| _2^2-\left\| \nabla (u_n-u_0-w_0(\cdot -y_n))\right\| _2^2\right) \nonumber \\&\quad =-\left\| \nabla u_0\right\| _2^2+\left<\nabla u_n,\nabla u_0\right>_{L^2}-\left\| \nabla w_0(\cdot -y_n)\right\| _2^2\nonumber \\&\qquad -\left<\nabla u_0,\nabla w_0(\cdot -y_n)\right>_{L^2}+\left<\nabla u_n(\cdot +y_n),\nabla w_0\right>_{L^2}\nonumber \\&\quad =o(1). \end{aligned}$$

(3.11)

Similarly,

$$\begin{aligned} \frac{1}{2}\int _{\mathbb {R}^N}V(x)\left( |u_n|^2-|u_0|^2-|w_0(\cdot -y_n)|^2-|u_n-u_0-w_0(\cdot -y_n)|^2\right) dx=o(1).\nonumber \\ \end{aligned}$$

(3.12)

By the Brezis–Lieb lemma [9, Theorem 2], we have

$$\begin{aligned} \begin{aligned}&\int _{\mathbb {R}^N}F(|u_n|)dx = \int _{\mathbb {R}^N}F(|u_0|)dx+\int _{\mathbb {R}^N}F(|u_n-u_0|)dx+o(1),\\&\quad \int _{\mathbb {R}^N}F(|u_n(\cdot +y_n)-u_0(\cdot +y_n)|)dx = \int _{\mathbb {R}^N}F(|w_0|)dx\\&\quad +\int _{\mathbb {R}^N}F(|u_n(\cdot +y_n)-u_0(\cdot +y_n)-w_0|)dx+o(1). \end{aligned} \end{aligned}$$

Thus,

$$\begin{aligned}&\int _{\mathbb {R}^N}F(|u_n|)dx-\int _{\mathbb {R}^N}F(|u_0|)dx\nonumber \\&\quad -\int _{\mathbb {R}^N}F(|w_0(\cdot -y_n)|)dx-\int _{\mathbb {R}^N}F(|u_n-u_0-w_0(\cdot -y_n)|)dx=o(1). \end{aligned}$$

(3.13)

Combining (3.11)–(3.13), we have

$$\begin{aligned} E(u_n)-E(u_0)-E(w_0(\cdot -y_n))-E(u_n-u_0-w_0(\cdot -y_n))=o(1). \end{aligned}$$

(3.14)

Since $V(x)\rightarrow 0$ as $|x|\rightarrow \infty$, $u_n\rightharpoonup u_0$ weakly in H and $|y_n|\rightarrow \infty$, we have

$$\begin{aligned} \int _{\mathbb {R}^N}V(x)|u_n(x)-u_0(x)-w_0(x-y_n)|^2dx\rightarrow 0. \end{aligned}$$

(3.15)

Noting

$$\begin{aligned}&E(u_n-u_0-w_0(\cdot -y_n))=E_{\infty }(u_n-u_0-w_0(\cdot -y_n))\\&\quad +\frac{1}{2}\int _{\mathbb {R}^N}V(x)|u_n(x)-u_0(x)-w_0(x-y_n)|^2dx, \end{aligned}$$

we have

$$\begin{aligned} \liminf _{n\rightarrow \infty }E(u_n-u_0-w_0(\cdot -y_n))\ge e_{\infty }(\delta )\quad \text {and}\quad \liminf _{n\rightarrow \infty }E(w_0(\cdot -y_n))\ge e_{\infty }(\gamma ).\nonumber \\ \end{aligned}$$

(3.16)

Hence, by (3.14)–(3.16) we have

$$\begin{aligned} e(\alpha )\ge e(\beta )+e_{\infty }(\gamma )+e_{\infty }(\delta ). \end{aligned}$$

(3.17)

By (3.17) and Lemma 2.5 (iv), we have

$$\begin{aligned} e(\alpha )\ge e(\beta )+e_{\infty }(\gamma )+e_{\infty }(\delta )\ge e(\beta +\gamma )+e_{\infty }(\delta )\ge e(\beta +\gamma +\delta )=e(\alpha ).\nonumber \\ \end{aligned}$$

(3.18)

Hence, $e(\alpha )=e(\beta )+e_{\infty }(\gamma )+e_{\infty }(\delta )$. Since $\delta >0$, by Proposition 2.1 (i), we see that if $\gamma +\delta >\alpha _{0,\infty }$, then $e_{\infty }(\gamma )+e_{\infty }(\delta )>e_{\infty }(\gamma +\delta )$. This gives a contradiction because

$$\begin{aligned} e(\alpha )=e(\beta )+e_{\infty }(\gamma )+e_{\infty }(\delta )>e(\beta )+e_{\infty }(\gamma +\delta )\ge e(\beta +\gamma +\delta )=e(\alpha ). \end{aligned}$$

Thus, $\gamma +\delta \le \alpha _{0,\infty }$ and $e_{\infty }(\gamma )=e_{\infty }(\delta )=0$ thanks to Proposition 2.1. By (3.18) we have $e(\alpha )=e(\beta )$. Thus, when $\beta =0$ and $e(\alpha )<0$, we obtain a contradiction and (3.10) does not hold, which gives $\delta =0$.

In the case $0<\beta <\alpha$, by (3.16), $e_\infty (\delta ) = 0 = e_\infty (\gamma )$ and (3.14), we have

$$\begin{aligned} e(\beta )\le & {} E(u_0)+E(w_0(\cdot -y_n))+E(u_n-u_0-w_0(\cdot -y_n))+ o(1) \nonumber \\= & {} E(u_n)+o(1)\rightarrow e(\alpha ). \end{aligned}$$

(3.19)

Since $\left\| u_0\right\| _2^2=\beta$, by (3.19), we see that $e(\beta )$ is attained by $u_0$ as well as $e(\beta ) = e(\alpha )$. However, by Lemma 3.1, $e(\beta )$ is not attained and we obtain a contradiction. Hence, $\delta =0$ and Step 2 is proved.

Step 3:We show that $(y_n)$and$w_0$satisfy (3.6).

In Step 2 we saw that (3.14)–(3.16) hold when $\delta >0$ is assumed. However, (3.14)–(3.16) hold even in the case $\delta =0$, since (3.10) is not used in deriving (3.14)–(3.16). By (3.14)–(3.16) we have

$$\begin{aligned} \begin{aligned} e(\alpha )&=\liminf _{n\rightarrow \infty }E(u_n) \\&=\liminf _{n\rightarrow \infty }\left( E(u_0)+E(w_0(\cdot -y_n))+E(u_n-u_0-w_0(\cdot -y_n))\right) \\&\ge E(u_0) + E_\infty (w_0) + \liminf _{n\rightarrow \infty } E \left( u_n - u_0 - w_0 ( \cdot - y_n ) \right) \\&\ge e(\beta )+e_{\infty }(\gamma )+e_{\infty }(\delta ), \end{aligned} \end{aligned}$$

(3.20)

where $\delta =\lim _{n\rightarrow \infty }\left\| u_n-u_0-w_0(\cdot -y_n)\right\| _2^2$. In Step 2 we have shown that $\delta =0$, and hence $\alpha =\beta +\gamma$. Since $\gamma >0$ and $e_{\infty }(\delta )=0$, by Lemma 2.5 (iv), we have

$$\begin{aligned} e(\beta )+e_{\infty }(\gamma )+e_{\infty }(\delta )=e(\beta )+e_{\infty }(\gamma )\ge e(\alpha ). \end{aligned}$$

(3.21)

By (3.21) and (3.20) we see that $e(\alpha )=e(\beta )+e_{\infty }(\gamma )$. Hence, by (3.20), $E(u_0)=e(\beta )$ and $E_{\infty }(w_0)=e_{\infty }(\gamma )$. Thus, Step 3 is proved, and the proof of Lemma 3.2 is completed. $\square$

Now we prove the precompactness of minimizing sequence.

Lemma 3.3

Assume (F1)–(F5) and (V1) and assume (F6) or (V2). Let $\alpha >0$. If $e(\alpha )<0$, then every minimizing sequence for $e(\alpha )$ has a strong convergent subsequence in H.

Proof

Let $(u_n)\subset M(\alpha )$ be a minimizing sequence for $e(\alpha )$. By Lemma 2.6, it suffices to show that $(|u_n|)$ has a strongly convergent subsequence in $L^2(\mathbb {R}^N)$. Moreover, from Lemma 2.8 and Remark 2.9, we may assume that $(u_n)$ satisfies

$$\begin{aligned} E'(u_n)+\lambda _nQ'(u_n)&\rightarrow 0\ \text {strongly in}\ H^*\quad \text {and}\quad (u_n)_-:=\max \{-u_n(x),0\} \rightarrow 0\ \text {strongly in}\ L^2(\mathbb {R}^N)\nonumber \\ \end{aligned}$$

(3.22)

for some bounded sequence $(\lambda _n) \subset \mathbb {R}$. We may also suppose

$$\begin{aligned} u_n \rightharpoonup u_0\ \text {weakly in}\ H\ \ \text {and}\ \ \lambda _n\rightarrow \lambda \ \text {in}\ \mathbb {R}. \end{aligned}$$

Let $\beta :=\left\| u_0\right\| _2^2$. Then, $\beta \le \alpha$.

If $\beta =\alpha$, then $u_n\rightarrow u_0$ strongly in $L^2(\mathbb {R}^N)$ and Lemma 2.6 asserts that $(u_n)$ has a strongly convergent subsequence in H. Hence, the conclusion holds.

When $0 \le \beta < \alpha$, by Lemma 3.2, there exist $(y_n)\subset \mathbb {R}$ and $w_0\in H\backslash \{0\}$ such that (3.4)–(3.6) hold. From (3.22) and the definition of $w_0$ in Step 1 of Lemma 3.2, it follows that

$$\begin{aligned} -\Delta w_0 + 2 \lambda w_0 = f(w_0) \quad \mathrm{in} \ \mathbb {R}^N, \quad w_0 \ge 0 \quad \mathrm{in} \ \mathbb {R}^N. \end{aligned}$$

(3.23)

Since $f(s) \ge 0$ for $s \ge 0$ by (F5) and $-\Delta w_0 + (2\lambda )_+ w_0 \ge -\Delta w_0 + 2 \lambda w_0 = f(w_0) \ge 0$ in $\mathbb {R}^N$, the strong maximum principle and $\Vert w_0 \Vert _2^2 = \alpha - \beta > 0$ give

$$\begin{aligned} w_0 > 0 \quad \mathrm{in} \ \mathbb {R}^N. \end{aligned}$$

(3.24)

Now we may exclude the case $\beta = 0$. In this case, we have $e(\alpha ) = e_\infty (\alpha ) = E_\infty (w_0)$ and $w_0$ is a minimizer for $e_\infty (\alpha )$. However, (V1) and (3.24) give a contradiction:

$$\begin{aligned} e(\alpha ) \le E(w_0) < E_\infty (w_0) = e_\infty (\alpha ) = e(\alpha ). \end{aligned}$$

Hence, the case $\beta = 0$ does not occur.

Hereafter we prove that the case

$$\begin{aligned} 0< \beta <\alpha \end{aligned}$$

(3.25)

does not occur. Suppose on the contrary that (3.25) holds.

We divide the proof into two steps.

Step 1We show that $\lambda >0$.

By (3.23), we observe that $w_0$ satisfies the Pohozaev identity

$$\begin{aligned} 0=\frac{N-2}{2}\Vert \nabla w_0\Vert ^2_2-N\int _{\mathbb {R}^N}F(w_0)-{\lambda }w_0^2dx. \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} 0\ge e_{\infty }(\alpha -\beta )=E_{\infty }(w_0)=\frac{1}{N}\Vert \nabla w_0\Vert ^2_2-{\lambda }\Vert w_0\Vert ^2_2. \end{aligned}$$

Now we infer from (3.24) that $\lambda \ge \frac{1}{N(\alpha -\beta )}\Vert \nabla w_0\Vert ^2_2>0$.

Step 2 Conclusion.

In this step, we borrow the idea from [15]. Set

$$\begin{aligned} w_n(x):=w_0(x-n\mathbf{e}_1),\quad \tau _n:=\frac{\sqrt{\alpha }}{\Vert u_0+w_n\Vert _2}\quad \text {and}\quad \kappa _n:=\left<u_0,w_n\right>_{L^2}. \end{aligned}$$

Remark that $\tau _n (u_0 + w_n ) \in M(\alpha )$, $\kappa _n\rightarrow 0$ as $n\rightarrow \infty$ and

$$\begin{aligned} \tau _n^2=\frac{\alpha }{\alpha +2\kappa _n}=1-\frac{2\kappa _n}{\alpha }+O(\kappa _n^2) \quad \text {and} \quad \tau _n=1-\frac{\kappa _n}{\alpha }+O(\kappa _n^2). \end{aligned}$$

Since it follows from (3.22), $\Vert u_0 \Vert ^2_2 = \beta > 0$ and a similar argument to $w_0$ that

$$\begin{aligned} -\Delta u_0 + V(x) u_0 + 2 \lambda u_0 = f(u_0) \quad \mathrm{in} \ \mathbb {R}^N, \quad u_0 > 0 \quad \mathrm{in} \ \mathbb {R}^N, \end{aligned}$$

(3.26)

combining this fact with (3.23) and (3.26), we have

$$\begin{aligned} \begin{aligned}&\frac{\tau _n^2}{2}\int _{\mathbb {R}^N}|\nabla (u_0+w_n)|^2+V(x)(u_0+w_n)^2dx \\&\quad = \, \frac{1}{2}\left( 1-\frac{2\kappa _n}{\alpha }+O(\kappa _n^2)\right) \int _{\mathbb {R}^N} |\nabla u_0|^2+V(x)u_0^2\\&\qquad +|\nabla w_n|^2+V(x)w_n^2 +2\nabla u_0\cdot \nabla w_n+2V(x)u_0w_n dx \\&\quad = \, \frac{1}{2}\left( 1-\frac{2\kappa _n}{\alpha }\right) \int _{\mathbb {R}^N}|\nabla u_0|^2+V(x)u_0^2+|\nabla w_0|^2+V(x)w_n^2dx \\&\qquad +\left( 1-\frac{2\kappa _n}{\alpha }\right) \int _{\mathbb {R}^N}\frac{1}{2}(\nabla u_0\cdot \nabla w_n+V(x)u_0w_n) \\&\qquad +\frac{1}{2}(\nabla u_0\cdot \nabla w_n+V(x)u_0w_n)dx +O(\kappa _n^2) \\&\quad = \, \frac{1}{2}\left( 1-\frac{2\kappa _n}{\alpha }\right) \int _{\mathbb {R}^N}|\nabla u_0|^2+V(x)u_0^2+|\nabla w_0|^2+V(x)w_n^2dx \\&\qquad +\left( 1-\frac{2\kappa _n}{\alpha }\right) \int _{\mathbb {R}^N} \frac{1}{2}(-2\lambda u_0w_n+f(u_0)w_n) \\&\qquad +\frac{1}{2}(-2\lambda u_0w_n+f(w_n)u_0) +\frac{1}{2}V(x)u_0w_ndx +O(\kappa _n^2) \\&\quad = \, \frac{1}{2}\left( 1-\frac{2\kappa _n}{\alpha }\right) \int _{\mathbb {R}^N}|\nabla u_0|^2+V(x)u_0^2+|\nabla w_0|^2+V(x)w_n^2dx \\&\qquad +\left( 1-\frac{2\kappa _n}{\alpha }\right) \left\{ -2\lambda \kappa _n+\frac{1}{2}\int _{\mathbb {R}^N}f(u_0)w_n+f(w_n)u_0dx\right\} \\&\qquad +\frac{1}{2}\left( 1-\frac{2\kappa _n}{\alpha }\right) \int _{\mathbb {R}^N}V(x)u_0w_ndx+O(\kappa _n^2). \end{aligned} \end{aligned}$$

From $u_0,w_0 \in L^\infty (\mathbb {R})$ with $u_0,w_0 \ge 0$, (F3) and (F5), it follows that

$$\begin{aligned} 0 \le \int _{\mathbb {R}^N} f(u_0)w_n + f(w_n) u_0 d x \le \int _{\mathbb {R}^N} C_0 \left( u_0w_n + w_nu_0 \right) d x = 2 C_0 \kappa _n . \end{aligned}$$

(3.27)

Since $V(x) \le 0$ and we may assume $1 - 2\kappa _n / \alpha \ge 0$, we have

$$\begin{aligned}&E(\tau _n(u_0+w_n))\nonumber \\&\quad \le \frac{1}{2}\left( 1-\frac{2\kappa _n}{\alpha }\right) \int _{\mathbb {R}^N} |\nabla u_0|^2+V(x)u_0^2 +|\nabla w_n|^2+V(x)w_n^2dx-2\lambda \kappa _n \nonumber \\&\qquad +\int _{\mathbb {R}^N}\frac{1}{2}(f(u_0)w_n+f(w_n)u_0)dx-\int _{\mathbb {R}^N}F(\tau _n(u_0+w_n))dx+O(\kappa _n^2)\nonumber \\&\quad \le E(u_0)+E_{\infty }(w_n)-\frac{\kappa _n}{\alpha } \int _{\mathbb {R}^N}|\nabla u_0|^2+V(x) u_0^2+|\nabla w_0|^2dx -2 \lambda \kappa _n \nonumber \\&\qquad +\int _{\mathbb {R}^N}\frac{1}{2}(f(u_0)w_n+f(w_n)u_0)dx \nonumber \\&\qquad +\int _{\mathbb {R}^N}F(u_0)+F(w_n)-F(\tau _n(u_0+w_n))dx+O(\kappa _n^2). \end{aligned}$$

(3.28)

Noting $f\in C^{\nu }_\mathrm{loc}(\mathbb {R})$ due to (F5), we have

$$\begin{aligned} \int _{\mathbb {R}^N}F(\tau _n(u_0+w_n))dx&=\int _{\mathbb {R}^N}F\left( \left( 1-\frac{\kappa _n}{\alpha }+O(\kappa _n^2)\right) (u_0+w_n)\right) dx\nonumber \\&=\int _{\mathbb {R}^N}F(u_0+w_n)+f(u_0+w_n) \left( -\frac{\kappa _n}{\alpha }\right) (u_0+w_n)dx+O(\kappa _n^{1+\nu }). \end{aligned}$$

(3.29)

By (3.28), (3.29) and $\alpha = \Vert u_0 \Vert _2^2 + \Vert w_0 \Vert ^2_2$, we have

$$\begin{aligned}&E(\tau _n(u_0+w_n))\nonumber \\&\quad \le E(u_0)+E_{\infty }(w_0) -\frac{\kappa _n}{\alpha }\int _{\mathbb {R}^N}|\nabla u_0|^2+V(x)u_0^2+|\nabla w_0|^2dx -2 \frac{\lambda }{\alpha } \left( \Vert u_0 \Vert ^2_2 + \Vert w_0 \Vert ^2_2 \right) \kappa _n \nonumber \\&\qquad +\int _{\mathbb {R}^N}\frac{1}{2}(f(u_0)w_n+f(w_n)u_0)dx\nonumber \\&\qquad +\int _{\mathbb {R}^N}F(u_0)+F(w_n)-F(u_0+w_n) +\frac{\kappa _n}{\alpha }f(u_0+w_n)(u_0+w_n)dx+O(\kappa _n^{1+\nu })\nonumber \\&\quad =E(u_0)+E_{\infty }(w_0) -\frac{\kappa _n}{\alpha }\int _{\mathbb {R}^N}f(u_0)u_0+f(w_n)w_ndx \nonumber \\&\qquad +\int _{\mathbb {R}^N}F(u_0)+F(w_n)-F(u_0+w_n) +\frac{1}{2}(f(u_0)w_n+f(w_n)u_0)dx\nonumber \\&\qquad +\frac{\kappa _n}{\alpha }\int _{\mathbb {R}^N}f(u_0+w_n)(u_0+w_n)dx+O(\kappa _n^{1+\nu })\nonumber \\&\quad =E(u_0)+E_{\infty }(w_0) +\int _{\mathbb {R}^N}F(u_0)+F(w_n)-F(u_0+w_n)+\frac{1}{2}(f(u_0)w_n+f(w_n)u_0)dx\nonumber \\&\qquad +\frac{\kappa _n}{\alpha }\int _{\mathbb {R}^N}(f(u_0+w_n)-f(u_0))u_0+(f(u_0+w_n)-f(w_n))w_ndx +O(\kappa _n^{1+\nu }). \end{aligned}$$

(3.30)

From (3.23), (3.26), $V(x) \rightarrow 0$ as $|x| \rightarrow \infty$ and $\lambda > 0$ due to Step 1, it follows that $u_0$ and $w_0$ decay exponentially as $|x| \rightarrow \infty$. In fact, we may prove that if $0<\eta _1<2\lambda <\eta _2$, then there exist $C_{\eta _1}>0$ and $C_{\eta _2}>0$ such that

$$\begin{aligned} C_{\eta _2}e^{-\sqrt{\eta _2}|x|}\le u_0(x)\le C_{\eta _1}e^{-\sqrt{\eta _1}|x|}\quad \text {and}\quad C_{\eta _2}e^{-\sqrt{\eta _2}|x|}\le w_0(x)\le C_{\eta _1}e^{-\sqrt{\eta _1}|x|}. \end{aligned}$$

(3.31)

Noting $|f(u_0+w_n)-f(u_0)|\le Cw_n^{\nu }$, we see that

$$\begin{aligned}&\int _{\mathbb {R}^N}|f(u_0+w_n)-f(u_0)||u_0|dx \le C \int _{\mathbb {R}^N} w_n^\nu u_0 dx = C\int _{\mathbb {R}^N}(w_nu_0)^{\nu }u_0^{1-\nu }dx\\&\quad \le C\left( \int _{\mathbb {R}^N}w_nu_0dx\right) ^{\nu }\left( \int _{\mathbb {R}^N}u_0dx\right) ^{1-\nu } =O(\kappa _n^{\nu }). \end{aligned}$$

By a similar argument, we have

$$\begin{aligned} \int _{\mathbb {R}^N}|f(u_0+w_n)-f(w_n)||w_n|dx&\le C\int _{\mathbb {R}^N}(u_0w_n)^{\nu }w_n^{1-\nu }dx\\&\le C\left( \int _{\mathbb {R}^N}u_0w_ndx\right) ^{\nu }\left( \int _{\mathbb {R}^N}w_ndx\right) ^{1-\nu } =O(\kappa _n^{\nu }). \end{aligned}$$

Using two inequalities, by (3.30) we have

$$\begin{aligned}&E(\tau _n(u_0+w_n))\nonumber \\&\quad \le E(u_0)+E_{\infty }(w_0) +\int _{\mathbb {R}^N}F(u_0)\nonumber \\&\qquad +F(w_n)-F(u_0+w_n)+\frac{1}{2}(f(u_0)w_n+f(w_n)u_0)dx +O(\kappa _n^{1+\nu }). \end{aligned}$$

(3.32)

Let $\delta _2>0$ be given in Lemma 2.4 (i). We can choose an $R_0>0$ such that if $n\ge 2R_0$, then

$$\begin{aligned} \max _{x\in \mathbb {R}^N\backslash (B_{R_0}(O)\cup B_{R_0}(n\mathbf{e}_1))}u_0(x)\le \delta _2\quad \text {and}\quad \max _{x\in \mathbb {R}^N\backslash (B_{R_0}(O)\cup B_{R_0}(n\mathbf{e}_1))}w_n(x)\le \delta _2. \end{aligned}$$

By Lemma 2.4 (i) we see that if $n\ge 2R_0$, then

$$\begin{aligned} \int _{\mathbb {R}^N\backslash (B_{R_0}(O)\cup B_{R_0}(n\mathbf{e}_1))} F(u_0)+F(w_n)-F(u_0+w_n)+\frac{1}{2}(f(u_0)w_n+f(w_n)u_0)dx\le 0.\nonumber \\ \end{aligned}$$

(3.33)

Next, set

$$\begin{aligned} K:=\left\{ u_0(x)|\ x\in \overline{B_{R_0}(O)}\right\} \cup \left\{ w_n(x)|\ x\in \overline{B_{R_0}(n\mathbf{e}_1)}\right\} . \end{aligned}$$

Then $K\subset (0,\infty )$ and K is compact. Let $\delta _K$ be given in Lemma 2.4. We can choose $n_{R_0}\ge 2R_0$ such that if $n\ge n_{R_0}$, then

$$\begin{aligned} \max _{x\in \overline{B_{R_0}(n\mathbf{e}_1)}}u_0(x)\le \delta _K\quad \text {and}\quad \max _{x\in \overline{B_{R_0}(O)}}w_n(x)\le \delta _K. \end{aligned}$$

By Lemma 2.4 (ii) we see that if $n\ge n_{R_0}$, then

$$\begin{aligned}&\int _{\overline{B_{R_0}(O)}\cup \overline{B_{R_0}(n\mathbf{e}_1)}} F(u_0)+F(w_n)-F(u_0+w_n)+\frac{1}{2}(f(u_0)w_n+f(w_n)u_0)dx\nonumber \\&\quad \le -C_K\left( \int _{\overline{B_{R_0}(O)}}w_n(x)dx+\int _{\overline{B_{R_0}(n\mathbf{e}_1)}}u_0(x)dx\right) . \end{aligned}$$

(3.34)

Thus, from (3.32)–(3.34), we see that if $n\ge n_{R_0}$, then

$$\begin{aligned}&E(\tau _n(u_0+w_n)) \le E(u_0)+E_{\infty }(w_0)\nonumber \\&\quad -C_K\left( \int _{\overline{B_{R_0}(O)}}w_n(x)dx+\int _{\overline{B_{R_0}(n\mathbf{e}_1)}}u_0(x)dx\right) + O(\kappa _n^{1+\nu }). \end{aligned}$$

(3.35)

Now recalling (3.31), we obtain

$$\begin{aligned} \int _{\overline{B_{R_0}(O)}}w_n(x)dx +\int _{\overline{B_{R_0}(n\mathbf{e}_1)}}u_0(x)dx\ge C_{\eta _2}e^{-\sqrt{\eta _2}n} \quad \text {for} \ \eta _2 > 2 \lambda . \end{aligned}$$

Remark also that for each $\eta _1 \in (0,2\lambda )$, it is possible to prove

$$\begin{aligned} \kappa _n\le C_{\eta _1}e^{-\sqrt{\eta _1}n}. \end{aligned}$$

For instance, see [2, Proposition 1.2], [3, Lemma II.2] and [17].

Put $\eta _1:=(\sqrt{2\lambda }-\varepsilon )^2$ and $\eta _2:=(\sqrt{2\lambda }+\varepsilon )^2$. If $\varepsilon >0$ is sufficiently small, then

$$\begin{aligned} \sqrt{\eta _2}-(1+\nu )\sqrt{\eta _1}=-\nu \sqrt{2\lambda }+(2+\nu )\varepsilon <0. \end{aligned}$$

Thus,

$$\begin{aligned} \kappa _n^{1+\nu }e^{\sqrt{\eta _2}n}\le C_{\eta _1}^{1+\nu }e^{(\sqrt{\eta _2}-(1+\nu )\sqrt{\eta _1})n}\rightarrow 0 \ \text {as }n \rightarrow \infty . \end{aligned}$$

Therefore, $O(\kappa _n^{1+\nu })=o(e^{-\sqrt{\eta _2}n})$. By (3.35) we see that if n is large, then

$$\begin{aligned} e(\alpha ) \le E(\tau _n(u_0+w_n))&\le E(u_0)+E_{\infty }(w_0)-c_{\eta _2}e^{-\sqrt{\eta _2}n}+o(e^{-\sqrt{\eta _2}n})\nonumber \\&<E(u_0)+E_{\infty }(w_0)=e(\alpha ), \end{aligned}$$

which is a contradiction. Hence, (3.25) does not occur and the proof is completed. $\square$

Proof of Theorem A

Let $\alpha _0:=\inf \{\alpha \ge 0|\ e(\alpha )<0\}$. It is clear that $\alpha _0\le \alpha _{0,\infty }$. Since $\alpha _{0,\infty }$ exists and $\alpha _{0,\infty }<\infty$ thanks to Proposition 2.1, we see that $\alpha _0$ exists and $\alpha _0<\infty$. By Lemma 2.5 (v), $e(\alpha )$ is nonincreasing. Since $e(0)=0$, we easily see that $e(\alpha )=0$ for $0<\alpha <\alpha _0$ and that $e(\alpha )<0$ for $\alpha >\alpha _0$. It follows from Lemma 3.3 that if $\alpha >\alpha _0$, then every minimizing sequence has a strong convergent subsequence in H. It is well known that the orbital stability of $S_\alpha$ follows from the precompactness of every minimizing sequence for $e(\alpha )$. Moreover, Lemma 3.1 and the definition of $\alpha _0$ imply Theorem A (ii). Therefore, Theorem A holds. $\square$

4 Proof of Theorem B

Proof of Theorem B

• (i) We first prove $\alpha _0 = 0$ when (V3) holds. By (V3), there is a $\varphi \in C^\infty _0(\mathbb {R}^N)$ such that $\Vert \varphi \Vert _2 = 1$ and

  $$\begin{aligned} \frac{1}{2}\int _{\mathbb {R}^N}|\nabla \varphi |^2+V(x)\varphi ^2dx < 0. \end{aligned}$$

  Replacing $|\varphi |$ if necessary, we may suppose $\varphi \ge 0$. Let $\alpha \in (0, s_0^2 / \Vert \varphi \Vert _\infty ^2 )$. Since $\sqrt{\alpha }\varphi \in M(\alpha )$ and $F(\sqrt{\alpha } \varphi ) \ge 0$, we get

  $$\begin{aligned} e(\alpha ) \le E(\sqrt{\alpha }\varphi ) \le \frac{\alpha }{2} \int _{\mathbb {R}^N} |\nabla \varphi |^2 + V(x) \varphi ^2 d x <0. \end{aligned}$$

  By the monotonicity of $e(\alpha )$ in Lemma 2.5, we see that $\alpha _0=0$ holds.

  Next, we show that $N=1,2$ and (V1) imply (V3). Let V(x) satisfy

• (V1) and $\varphi \in C^\infty _0(\mathbb {R}^N)$. Put $\varphi _t(x) := t^{N/2} \varphi (tx)$ for $t > 0$. Choose also an $R_0>0$ so that $\int _{|x| \le R_0} V(x) dx < 0$. Then we have

  $$\begin{aligned} \begin{aligned} \int _{\mathbb {R}^N} |\nabla \varphi _t |^2 + V(x) |\varphi _t|^2 d x&= t^2 \Vert \nabla \varphi \Vert _2^2 + t^N \int _{\mathbb {R}^N} V(x) | \varphi (tx) |^2 d x \\&\le t^2 \left( \Vert \nabla \varphi \Vert ^2_2 + t^{N-2} \int _{|x| \le R_0 } V(x) |\varphi (t x)|^2 d x \right) . \end{aligned} \end{aligned}$$

  (4.1)

  Remark that

  $$\begin{aligned} \lim _{t\rightarrow 0} \int _{|x| \le R_0} V(x) |\varphi (tx)|^2 dx = |\varphi (0)|^2 \int _{|x| \le R_0} V(x) d x. \end{aligned}$$

  Hence, when $N=1$, by selecting $\varphi \in C^\infty _0(\mathbb {R})$ so that $\varphi (0) \ne 0$, if $t>0$ is sufficiently small, then (4.1) and the choice of $R_0$ imply $\int _{\mathbb {R}} |\nabla \varphi _t|^2 + V(x) |\varphi _t|^2 d x < 0$.

  When $N=2$, from $(-\log |x|)_+^\alpha \in H^1(\mathbb {R}^2)$ for $0<\alpha < 1/2$, we may find a $\psi _k \in C^\infty _0(\mathbb {R}^2)$ so that $\Vert \nabla \psi _k \Vert _2=1$, $\psi _k \ge 0$ and $\psi _k(0) \rightarrow \infty$ as $k \rightarrow \infty$. Setting $\varphi = \psi _k$ and selecting a sufficiently large $k_0$, we obtain

  $$\begin{aligned} \Vert \nabla \psi _{k_0} \Vert ^2_2 + | \psi _{k_0}(0)|^2 \int _{|x| \le R_0} V(x) d x < 0. \end{aligned}$$

  Thus, if $t>0$ is sufficiently small, then (4.1) gives $\int _{\mathbb {R}^2} |\nabla (\psi _{k_0})_t |^2 +V(x) |(\psi _{k_0})_t|^2 d x < 0$. Therefore, when $N=2$, (V3) holds.

• (ii) We show that there exists V(x) such that $\alpha _0>0$. Let $b:=\sup _{s>0}F(s)/s^{{p_\mathrm{c}}+1}$. By (F4) and (F9) we see that $b<\infty$. Let $C_0$ denote the best constant of the inequality $\left\| u\right\| _{{p_\mathrm{c}}+1}^{{p_\mathrm{c}}+1}\le C_0\left\| u\right\| _2^{4/N}\left\| \nabla u\right\| _2^2$ and define $\alpha _1 = \alpha _1(N,f) > 0$ by $\alpha _1 := (2b C_0)^{-N/2}$. For $\alpha \in (0,\alpha _1)$, we also set $c_\alpha := (N-2)^2 ( 1 - 2 b C_0 \alpha ^{2/N} ) / 4 > 0$ and suppose that $V(x) \ge - c_\alpha |x|^{-2}$ for $|x|>0$. Then by Hardy’s inequality and the definition of b, $C_0$ and $c_\alpha$, we obtain

  $$\begin{aligned} E(u)&\ge \frac{1}{2}\int _{\mathbb {R}^N}|\nabla u|^2dx-\frac{c_\alpha }{2}\int _{\mathbb {R}^N}\frac{u^2}{|x|^2}dx - b \Vert u \Vert _{{p_\mathrm{c}}+1}^{{p_\mathrm{c}}+ 1}\\&\ge \left( \frac{1}{2}-\frac{1}{2}+bC_0\alpha ^{2/N}-bC_0\alpha ^{2/N}\right) \int _{\mathbb {R}^N}|\nabla u|^2dx =0. \end{aligned}$$

  This inequality indicates that $e(\alpha )=0$ and $\alpha _0 \ge \alpha > 0$ follows from from the monotonicity of $e(\alpha )$.

Appendix A: Nonexistence of minimizer

We consider the following case:

• (V4) $0\not \equiv V(x)\ge 0$ and $\lim _{|x|\rightarrow \infty }V(x)=0$.

Theorem A.1

Suppose (V4) and the following (F11):

• (F11) $f(s)\le f(|s|)$ for $s\in \mathbb {R}$, $f(s)\ge 0$ for $s\ge 0$, $|f(s)|\le C(|s|+|s|^{{p_\mathrm{c}}})$, $\lim _{s\rightarrow \infty }f(s)/s^{{p_\mathrm{c}}}=0$.

Then $e(\alpha )=e_{\infty }(\alpha )$ for $\alpha \ge 0$ and $e(\alpha )$ is not attained for $\alpha >0$.

The assumption (F11) is weaker than (F1)–(F5).

Proof

First, we show that $e(\alpha )=e_{\infty }(\alpha )$. Since $V(x)\ge 0$, we see that $e(\alpha )\ge e_{\infty }(\alpha )$. On the other hand, for any $u\in M(\alpha )$ and $n\in \mathbb {N}$, we obtain

$$\begin{aligned} e(\alpha )\le E(u(\,\cdot \,-n\mathbf{e}_1))=E_{\infty }(u)+\frac{1}{2}\int _{\mathbb {R}^N}V(x+n\mathbf{e}_1)|u|^2dx. \end{aligned}$$

Letting $n\rightarrow \infty$, we obtain $e(\alpha )\le E_{\infty }(u)$. Since u is arbitrary, we see that $e(\alpha )\le e_{\infty }(\alpha )$. Thus, $e(\alpha )=e_{\infty }(\alpha )$.

Second, we show by contradiction that $e(\alpha )$ is not attained. Suppose on the contrary that $e(\alpha )$ is attained by $u_0\in H\cap M(\alpha )$. By Remark 2.7, we may assume $u_0 \ge 0$. Since $E \in C^1( H_\mathbb {R},\mathbb {R})$ due to (F11), there exists a $\lambda \in \mathbb {R}$ such that $-\Delta u_0 + (V(x) + 2 \lambda )_+ u_0 \ge -\Delta u_0+(V(x)+2\lambda )u_0=f(u_0)\ge 0$ in $\mathbb {R}^N$. Thus, the weak Harnack inequality [14, Theorem 8.18] yields $u_0>0$ in $\mathbb {R}^N$. Using this fact and $0\not \equiv V(x)\ge 0$, we obtain

$$\begin{aligned} e(\alpha )=E(u_0)=E_{\infty }(u_0)+\frac{1}{2}\int _{\mathbb {R}^N}V(x)u_0^2dx>E_{\infty }(u_0)\ge e_{\infty }(\alpha ). \end{aligned}$$

This is a contradiction, because $e(\alpha )=e_{\infty }(\alpha )$. Therefore, $e(\alpha )$ has no minimizer. $\square$