Global solutions of nonlinear Schrödinger equations

Abstract

We study the nonlinear Schrödinger equation in $\mathbb {R}^n$ without making any periodicity assumptions on the potential or on the nonlinear term. This prevents us from using concentration compactness methods. Our assumptions are such that the potential does not change the essential spectrum of the linear operator. This results in $[0, \infty )$ being the absolutely continuous part of the spectrum. If there are an infinite number of negative eigenvalues, they will converge to 0. In each case we obtain nontrivial solutions. We also obtain least energy solutions.

1 Introduction

We consider the semilinear Schrödinger equation

$$\begin{aligned} - \Delta u + V(x) u = f(x,u), \quad u \in H^1({\mathbb {R}^n}), \end{aligned}$$

(1)

where V(x) is a given potential. One wishes to find solutions and, in particular, the so called “least energy solutions.” These are solutions that minimize the corresponding energy functional. The existence of solutions depends both on the linear operator $\mathcal {A}$ and the nonlinear term f(x, u).

Many authors have studied the problem for the Schrödinger equation (1) under various stipulations (cf., e.g., [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20, 23, 25, 26, 28,29,30,31,32] and references quoted in them). In almost all cases it was required to stipulate that the spectrum of the linear operator $\mathcal {A}u = - \Delta u + V(x) u$ have a gap. This caused writers to make various assumptions on the potential V(x) to guarantee that this is the case. However, many of these assumptions caused the nature of the spectrum to be far different from that of $- \Delta u .$ Thus, any theorem proved for

$$\begin{aligned} \mathcal {A}u= - \Delta u + V(x) u \end{aligned}$$

did not hold for $\mathcal {A}u=- \Delta u.$

Some authors assumed

$$\begin{aligned} \inf _{\mathbb {R}^n} V(x) > 0 ;\;\; \lim _{|x|\rightarrow \infty } V(x) = \infty . \end{aligned}$$

Others assumed that there exists a constant B such that $V(x) \le B$ for all $x \in \mathbb {R}^n,$ $V(x) \rightarrow B \; as\; |x| \rightarrow \infty$ and $\sigma ( - \Delta + V(x) ) >0$ together with other assumptions. Another approach assumes that for every $M >0$ the set $\omega = \{x \in \mathbb {R}^n : V(x)<M \}$ has finite Lebesgue measure. Others assumed that V(x) is in some combination of $L^p(\mathbb {R}^n)$ spaces. In each case the growth of f(x, t) is controlled by the growth of V(x). In most cases the resulting spectrum of $\mathcal {A}= - \Delta + V(x)$ is discrete, consisting only of isolated eigenvalues of finite multiplicity tending to $+ \infty .$ All of these assumptions cause restrictions on the the nonlinear term depending on V. In most cases the hypothesis

$$\begin{aligned} \mu F(x,t) \le tf(x,t), \quad |x| > R, \end{aligned}$$

is used, where $\mu >2$ and

$$\begin{aligned} F(x,t)= \int _0^tf(x,s) \, ds. \end{aligned}$$

A different approach is to assume that the potential is periodic in the coordinates of $\mathbb {R}^n$ and then apply concentration compactness methods. In this case the resulting spectrum of $\mathcal {A}= - \Delta + V(x)$ is absolutely continuous and consists of a finite number of disjoint closed intervals. In order to apply this method, f(x, t) must be periodic in x as well. In the few publications where 0 is permitted to be in $\sigma ( - \Delta + V(x) ),$ an interval of the form $(-\varepsilon , 0)$ is required to be free of the spectrum.

The purpose of the present paper is solve the Eq. (1) under assumptions on V(x) such that the essential spectrum of $\mathcal {A}= - \Delta + V(x)$ is the the same as that of $- \Delta ,$ i.e., $[0,\infty ).$ The situation is different if there are no negative eigenvalues, one negative eigenvalue or two or more negative eigenvalues. If there are no negative eigenvalues, one can solve under the same hypotheses for f(x, t) that can be used for the equation

$$\begin{aligned} - \Delta u = f(x,u), \quad u \in H^1({\mathbb {R}^n}). \end{aligned}$$

(2)

Otherwise, the hypotheses on f(x, t) need only take the negative eigenvalues into consideration. We can even deal with the case where the negative eigenvalues converge to 0. We do not need an interval of the form $(-\varepsilon , 0)$ to be free of the spectrum. In each of these cases different methods must be employed, requiring different assumptions on the nonlinear term.

Concerning the function V(x) we make the following assumptions:

$(V_1)$ :

$$\begin{aligned} \sup _y \int _{|x-y| <\delta }|V(x)| \;\omega _{2}(x-y)\,dx \rightarrow 0 \; as\; \delta \rightarrow 0, \end{aligned}$$

and

$(V_2)$ :

$$\begin{aligned} \int _{|x-y| <1}|V(x)| \;\omega _{2}(x-y)\,dx \rightarrow 0 \; as\; |y| \rightarrow \infty , \end{aligned}$$

where

$$\begin{aligned} \omega _s(x) = {\left\{ \begin{array}{ll}|x|^{s-n}, \quad 0<s<n\\ 1-\ln |x|^2, \quad s=n \\ 1, \quad s>n.\end{array}\right. }\end{aligned}$$

These assumptions imply that there is a forms extension $\mathcal {A}$ of the operator

$$\begin{aligned} -\Delta u(x)+V(x)u(x) \end{aligned}$$

on the space $H=H^{1,2} (\mathbb {R}^n)$ having essential spectrum equal to $[0,\infty )$ and a (possibly empty) discrete, countable negative spectrum consisting of isolated eigenvalues of finite multiplicity with a finite lower bound $-L$

$$\begin{aligned} - \infty \,<\, -L \,\le \, \lambda _0 \,<\, \lambda _1 \,<\, \lambda _2 \,<\, \ldots \,<\, \lambda _l \,<\, \cdots <0. \end{aligned}$$

(3)

For each $l>0,$ define the subspaces $M=M_l$ and $N=N_l$ of H as

$$\begin{aligned} N \;=\; \bigoplus _{k \,<\, l} E(\lambda _k) \;,\quad M \;=\; N^{\perp } \;,\quad H \,=\, M \oplus N \;. \end{aligned}$$

For the operator $\mathcal {A}$ there are three possibilities: (a) it has no negative eigenvalues, (b) it has only one negative eigenvalue, and (c) it has two or more negative eigenvalues. What is interesting is that each of these possibilities must be dealt with differently. We shall study all of them separately.

The following notation will be used throughout the paper:

$$\begin{aligned} \Vert u\Vert _q:= & {} \left( \int _{\mathbb {R}^n}|u(x)|^qdx\right) ^{1/q}, \;\; \Vert u\Vert =\Vert u\Vert _2,\\ (u,v)= & {} \int _{\mathbb {R}^n}u(x) v(x) \, dx, \;\;a(u,v) = (\mathcal {A}u,v), \;\; a(u) = a(u,u). \end{aligned}$$

Let q be any number satisfying

$$\begin{aligned} 2&< q \; \le 2n/(n-2),\quad n >2 \\ 2&< q \; < \infty ,\quad ~ n \le 2 \end{aligned}$$

(3.1.3)

and let f(x, t) be a Carathéodory function on $\mathbb {R}^n\times {\mathbb {R}}$. This means that f(x, t) is continuous in t for a.e. $x\in \mathbb {R}^n$ and measurable in x for every $t\in {\mathbb {R}}$. We make the following assumptions

(A)   The function f(x, t) satisfies

$$\begin{aligned} |f(x,t)| \le S(x)^q(|t|^{q-1} + W(x)) \end{aligned}$$

and

$$\begin{aligned} f(x,t)/S(x)^q = o\left( |t|^{q-1}\right) \;\text{ as } |t|\rightarrow \infty , \end{aligned}$$

where $S(x) > 0$ is a function in $L^q(\mathbb {R}^n)$ satisfying

$$\begin{aligned} \Vert Su\Vert _q \le C\Vert u\Vert _H,\quad u\in H, \end{aligned}$$

and W is a function in $L^{\infty }(\mathbb {R}^n)$. Here

$$\begin{aligned} \Vert u\Vert _q := \left( \int _{\mathbb {R}^n}|u(x)|^qdx \right) ^{1/q}. \end{aligned}$$

Our other hypotheses depend only on the primitive

$$\begin{aligned} F(x,t)=\int _0^t f(x,s) \, ds \end{aligned}$$

of f(x, t).

Let

$$\begin{aligned} G(u)=(\mathcal {A}u,u)-2\int _{\mathbb {R}^n} F(x,u)\,dx, \quad u \in H. \end{aligned}$$

(4)

It follows that G is a continuously differentiable functional on the whole of H (cf., e.g., [25]). It is easily checked that $u \in H$ is a (weak) solution of (1) iff it is a critical point of G(u). Our methods will make use of this fact. They involve finding linking sets A, B which separate G,  i.e., are such that

$$\begin{aligned} \sup _A G \le \inf _B G. \end{aligned}$$

If the spectrum of $\mathcal {A}$ has no gap, i.e., consists only of the interval $[0,\infty ),$ the choice of the sets A, B is very limited. If it has one gap, the subspace in the gap is very useful. If it has two gaps, we obtain two useful subspaces. In our situation, if there are no negative eigenvalues, there are no gaps. If there is one negative eigenvalue, there is one gap. If there are two or more negative eigenvalues, there are at least two gaps. Therefore we consider three different situations.

2 The space $N_1$

We let $N_1$ be the set of measurable functions h(x) on $\mathbb {R}^n$ satisfying

$$\begin{aligned} \sup _u \frac{\Vert hu\Vert }{\Vert u\Vert _H} < \infty . \end{aligned}$$

It is clear that bounded functions are in $N_1.$ It can be shown that $L^{p}(\mathbb {R}^n) \subset N_1$ for $p \ge n.$ For a general description of this space, cf. [21]. These functions will be used in our theorems because of the following properties:

Lemma 1

If $g^{-1} \in N_1,$ the following statements are true:

• 1. there is a constant C such that

  $$\begin{aligned} \Vert u\Vert _H \le C \Vert u\Vert '_H, \quad u \in H, \end{aligned}$$

  (5)

  where

  $$\begin{aligned} (\Vert u\Vert '_H)^2 = \Vert \nabla u\Vert ^2+\Vert gu\Vert ^2. \end{aligned}$$

• 2. If $\mathcal {A}$ has no negative eigenvalues, then there is an $\varepsilon >0$ such that

  $$\begin{aligned} a(u) + \Vert gu\Vert ^2 \ge \varepsilon \Vert u\Vert _H^2, \quad u \in H. \end{aligned}$$

  (6)

• 3. If $\mathcal {A}$ has only one negative eigenvalue $\lambda _0,$ then there is an $\varepsilon >0$ such that

  $$\begin{aligned} a(u) + \Vert gu\Vert ^2 \ge \varepsilon \Vert u\Vert _H^2, \quad u \in M. \end{aligned}$$

  (7)

• 4. If $\mathcal {A}$ has negative eigenvalues $\lambda _{l-1}, \lambda _l$, then there is an $\varepsilon >0$ such that

  $$\begin{aligned} a(u) -\lambda _l \Vert u\Vert ^2+ \Vert gu\Vert ^2 \ge \varepsilon \Vert u\Vert _H^2, \quad u \in M. \end{aligned}$$

  (8)

If $g \in N_1,$ then there is a constant C such that

$$\begin{aligned} \Vert u\Vert '_H \le C \Vert u\Vert _H, \quad u \in H. \end{aligned}$$

(9)

If $g, g^{-1} \in N_1,$ then the two norms are equivalent.

Lemma 1 will be proved in Sect. 8.

3 No negative eigenvalues

In this case,

$$\begin{aligned} a(u)=(\mathcal {A}u,u) \ge 0, \quad u \in H. \end{aligned}$$

There are no gaps in the spectrum of $\mathcal {A}.$

We have

Theorem 2

Assume

• 1. There is a function g(x) such that $g, g^{-1} \in N_1$ and

  $$\begin{aligned} 2F(x,u) \le -g(x)^2 |u|^2+W(x), \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \end{aligned}$$

  where $W(x) \in L^1(\mathbb {R}^n).$

Then the Eq. (1) has a solution.

Remark 3

It is clear from the equation that the solution obtained will be nontrivial if

$$\begin{aligned} f(x,0) \ne 0. \end{aligned}$$

To guarantee that a solution will be nontrivial even when $f(x,0) = 0,$ we have

Theorem 4

Assume

• 1. There are constants $0<\alpha <2,\delta >0,$ and a function g(x) such that $g, g^{-1} \in N_1$ and

  $$\begin{aligned} 2F(x,u)&\le - g(x)^2|u|^2, \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \; |u| < \delta , \\&\le -g(x)^2 |u|^2+S(x)^q |u|^\alpha , \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \; |u| > \delta . \end{aligned}$$

• 2. There is a locally bounded function h(t) such that

  $$\begin{aligned} 2F(x,u) \ge -h(u), \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \end{aligned}$$

  and

  $$\begin{aligned} c_0=\sup _\mathbb {R}h(u)/u^2 <\infty . \end{aligned}$$

Then the Eq. (1) has a nontrivial solution.

4 Only one negative eigenvalue

Let $\lambda _0 <0$ be the eigenvalue. In this case

$$\begin{aligned} a(u) \ge \lambda _0 \Vert u\Vert ^2, \quad u \in H. \end{aligned}$$

We can make use of the fact that there is a gap in the spectrum of $\mathcal {A}.$ We have

Theorem 5

Assume

• 1.
  $$\begin{aligned} 2F(x,u) \ge \lambda _0|u|^2, \quad u \in \mathbb {R},\; x \in \mathbb {R}^n. \end{aligned}$$

• 2. There are constants $0<\alpha <2,\delta >0,$ and a function g(x) such that $g, g^{-1} \in N_1$ and

  $$\begin{aligned} 2F(x,u)&\le - g(x)^2|u|^2, \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \; |u| < \delta , \\&\le -g(x)^2 |u|^2+S(x)^q |u|^\alpha , \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \; |u| > \delta . \end{aligned}$$

Then the Eq. (1) has a solution.

To obtain a nontrivial solution we have

Theorem 6

Assume

• 1.
  $$\begin{aligned} 2F(x,u) \ge \lambda _0|u|^2, \quad u \in \mathbb {R},\; x \in \mathbb {R}^n. \end{aligned}$$

• 2. There are constants $0<\alpha <2,\delta >0,$ and a function g(x) such that $g, g^{-1} \in N_1$ and

  $$\begin{aligned} 2F(x,u)&\le - g(x)^2|u|^2, \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \; |u| < \delta , \\&\le -g(x)^2 |u|^2+S(x)^q |u|^\alpha , \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \; |u| > \delta . \end{aligned}$$

• 3. There is a locally bounded function h(t) such that

  $$\begin{aligned} 2F(x,u) \ge -h(u), \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \end{aligned}$$

  and

  $$\begin{aligned} c_0=\sup _{\mathbb {R}} h(u)/u^2 < \infty . \end{aligned}$$

Then the Eq. (1) has a nontrivial solution.

5 Two or more negative eigenvalues

Here again we can make use of the fact that there is more than one gap in the spectrum. This allows us to use more complicated linking methods.

Let $\lambda _{l-1}, \; \lambda _l$ be two consecutive negative eigenvalues of $\mathcal {A}.$ We have

Theorem 7

Assume

• 1.
  $$\begin{aligned} 2F(x,u) \ge \lambda _{l-1} |u|^2, \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \end{aligned}$$

• 2. There are constants $0<\alpha <2,\delta >0,$ and a function g(x) such that $g, g^{-1} \in N_1$ and

  $$\begin{aligned} 2F(x,u)&\le - g(x)^2|u|^2 + \lambda _l |u|^2, \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \; |u| < \delta , \\&\le -g(x)^2 |u|^2+\lambda _l|u|^2 +S(x)^q |u|^\alpha , \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \; |u| > \delta . \end{aligned}$$

Then the Eq. (1) has a solution.

To obtain a nontrivial solution we have

Theorem 8

Assume

• 1.
  $$\begin{aligned} 2F(x,u) \ge \lambda _{l-1} |u|^2, \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \end{aligned}$$

• 2. There are constants $0<\alpha <2,\delta >0,$ and a function g(x) such that $g, g^{-1} \in N_1$ and

  $$\begin{aligned} 2F(x,u)&\le - g(x)^2|u|^2 + \lambda _l |u|^2, \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \; |u| < \delta , \\&\le -g(x)^2 |u|^2+\lambda _l|u|^2 +S(x)^q |u|^\alpha , \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \; |u| > \delta . \end{aligned}$$

• 3. There is a locally bounded function h(t) such that

  $$\begin{aligned} 2F(x,u) \ge -h(u), \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \end{aligned}$$

  and

  $$\begin{aligned} c_0=\sup _{\mathbb {R}} h(u)/u^2 < \infty . \end{aligned}$$

Then the Eq. (1) has a nontrivial solution.

Remark 9

Note that the hypothesis of Theorem 2 is stronger than hypotheses 1 and 2 of Theorem 5. Hypothesis 1 of Theorem 5 requires

$$\begin{aligned} 2F(x,u) \ge \lambda _{l} |u|^2, \quad u \in \mathbb {R},\; x \in \mathbb {R}^n, \end{aligned}$$

which is stronger than hypothesis 1 of Theorem 7. Consequently, the hypotheses of Theorem 4 are stronger than those of Theorem 6 which are stronger than those of Theorem 8.

6 Least energy solutions

Let $\mathcal {M}$ be the set of all solutions of (1). A solution $\tilde{u}$ is called a “least energy solution” if it minimizes the functional

$$\begin{aligned} G(u)=a(u)-2\int _{\mathbb {R}^n} F(x,u)\,dx \end{aligned}$$

(10)

over the set $\mathcal {M}.$

We have

Theorem 10

If we add the following hypothesis to Theorems 2–8, then Eq. (1) has a least energy solution: The function given by

$$\begin{aligned} H(x,u)=uf(x,u) -2 F(x,u) \end{aligned}$$

(11)

satisfies

$$\begin{aligned} H(x,u) \ge -W(x) \in L^1(\mathbb {R}^n), \quad u \in \mathbb {R}, \; x \in \mathbb {R}^n. \end{aligned}$$

(12)

We shall prove Theorems 2–10 in Sect. 8. In the next section we describe the construction of the operator $\mathcal {A}.$ We obtain the largest self-adjoint extension of $\mathcal {A}_0$ which preserves the essential spectrum.

7 The operator $\mathcal {A}$

The following was proved in [21] (Theorem 10.9, ch. 6, p. 153.)

Theorem 11

Let P(D) be an elliptic constant coefficient operator of order 2 on $\mathbb {R}^n,$ and let V(x) be a function satisfying $(V_1)$ and $(V_2).$ If $\rho (P_0)$ is not empty, then P(D)+V has a forms extension operator $\mathcal {A}$ such that $\sigma _e(\mathcal {A})=\sigma _e(P_0).$

Here $P_0$ is the closure of the operator P(D) restricted to $C^\infty _0(\mathbb {R}^n),$ and $\sigma _e$ is the essential spectrum. Any point not in the essential spectrum is either a point in the resolvent or an isolated eigenvalue of finite multiplicity. If $P(D) = -\Delta ,$ then $\sigma (P_0)=[0,\infty ).$ Consequently, in our case, the negative spectrum of $\mathcal {A}$ can have at most a countable number of isolated eigenvalues of finite multiplicity having a finite lower bound. The theorem is proved by showing that the bilinear form

$$\begin{aligned} a(u,v)=(P(D)u,v) +(Vu,v), \quad u,v \in H \end{aligned}$$

is bounded and closed on H. Moreover, if the coefficients of P(D) and the function V(x) are real, then the bilinear form is symmetric and $\mathcal {A}$ is selfadjoint.

8 Proof of Lemma 1.

Proof

We have

$$\begin{aligned} (u,u) =(gu,g^{-1}u) \le \Vert gu\Vert \cdot \Vert g^{-1}u\Vert \le C \Vert gu\Vert \cdot \Vert u\Vert _H \le \varepsilon \Vert u\Vert _H^2 + K_{\varepsilon } \Vert gu\Vert ^2. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert u\Vert _H^2 \le \varepsilon \Vert u\Vert _H^2 + \Vert \nabla u \Vert ^2 + K_{\varepsilon } \Vert gu\Vert ^2. \end{aligned}$$

To prove (6), assume that there is a sequence $u^{(k)} \in H$ such that $\Vert u^{(k)}\Vert ' _H =1$ and $a(u^{(k)}) + \Vert gu^{(k)}\Vert ^2 \rightarrow 0,$ where

$$\begin{aligned} (\Vert u\Vert '_H)^2 = \Vert \nabla u\Vert ^2+\Vert gu\Vert ^2. \end{aligned}$$

Since

$$\begin{aligned} \rho _k=\Vert u^{(k)}\Vert _H \le C, \end{aligned}$$

there is a renamed subsequence such that $u^{(k)}$ converges to a limit $u \in H$ weakly in H,  strongly in $L^2_{loc}(\mathbb {R}^n)$ and a.e. in $\mathbb {R}^n.$ Since $a(u^{(k)}) \ge 0,$ we have $a(u^{(k)}) \rightarrow 0$ and $\Vert gu^{(k)}\Vert \rightarrow 0.$ By our hypothesis on V(x),  there is a renamed subsequence of $u^{(k)}$ such that $b( u^{(k)}) \rightarrow b(u),$ where $b(u)=(Vu,u).$ Thus, $1+b(u)=0,$ showing that $u \ne 0.$ Since $g u^{(k)} \rightarrow gu$ a.e., we have $u =0,$ providing a contradiction. The same proof can be used to prove (7) if we keep in mind that (7) holds only on M. To prove (8), assume that there is a sequence $u^{(k)} \in M$ such that $\Vert u^{(k)}\Vert ' _H =1$ and $a(u^{(k)}) -\lambda _l\Vert u^{(k)}\Vert ^2+ \Vert gu^{(k)}\Vert ^2 \rightarrow 0.$ Since

$$\begin{aligned} \rho _k=\Vert u^{(k)}\Vert ' _H \le C, \end{aligned}$$

there is a renamed subsequence such that $u^{(k)}$ converges to a limit $u \in H$ weakly in H,  strongly in $L^2_{loc}(\mathbb {R}^n)$ and a.e. in $\mathbb {R}^n.$ Also, there is a renamed subsequence such that $\Vert \nabla u^{(k)} \Vert \rightarrow \nu$ and $\Vert g u^{(k)}\Vert \rightarrow \tau ,$ where $\nu ^2+\tau ^2=1.$ Since $a(u^{(k)}) -\lambda _l \Vert u^{(k)}\Vert ^2\ge 0$ in M,  we have $a(u^{(k)}) -\lambda _l\Vert u^{(k)}\Vert ^2\rightarrow 0$ and $\Vert gu^{(k)}\Vert \rightarrow 0.$ By our hypothesis on V(x),  there is a renamed subsequence of $u^{(k)}$ such that $b( u^{(k)}) \rightarrow b(u),$ where $b(u)=(Vu,u).$ Since $a(u^{(k)}) -\lambda _l\Vert u^{(k)}\Vert ^2 \rightarrow 0,$ and $a(u^{(k)}) \rightarrow \nu ^2 +b(u),$ we see that $\nu ^2+b(u)\le 0.$ Since $gu^{(k)} \rightarrow gu$ a.e. and $\Vert gu^{(k)}\Vert \rightarrow 0,$ we have $\tau =0$ and $u =0.$ But then, $\nu = 1$ and $1+ b(u) \le 0,$ showing that $u \ne 0,$ providing a contradiction. $\square$

9 Proofs of the Theorems

We now give the proof of Theorem 2.

Proof

We define

$$\begin{aligned} G(u)=a(u)-2\int _{\mathbb {R}^n} F(x,u(x))\,dx,\quad u \in H. \end{aligned}$$

(13)

By Lemma 1 there is an $\varepsilon >0$ such that

$$\begin{aligned} a(u) + \Vert gu\Vert ^2 \ge \varepsilon \Vert u\Vert _H^2, \quad u \in H. \end{aligned}$$

(14)

Thus

$$\begin{aligned} G(u) \rightarrow \infty , \quad \Vert u\Vert _H \rightarrow \infty , \end{aligned}$$

(15)

i.e., G(u) is coercive. Let

$$\begin{aligned} c= \inf _HG. \end{aligned}$$

By Corollary 3.22, p. 29, of [24] there is a sequence $\{u^{(k)}\} \subset H$ such that

$$\begin{aligned} G(u^{(k)})= & {} a(u^{(k)}) -2\int _{\mathbb {R}^n} F\left( x,u^{(k)}(x)\right) \,dx \rightarrow c, \end{aligned}$$

(16)

$$\begin{aligned} \left( G'(u^{(k)}),z\right) /2= & {} a\left( u^{(k)}, z\right) -\int _{\mathbb {R}^n} f\left( x,u^{(k)}\right) \cdot z(x)\,dx \rightarrow 0, \quad z \in H \end{aligned}$$

(17)

and

$$\begin{aligned} \left( G'(u^{(k)}),u^{(k)}\right) /2= a\left( u^{(k)}\right) -\int _{\mathbb {R}^n} f(x,u^{(k)})\cdot u^{(k)}\,dx \rightarrow 0. \end{aligned}$$

(18)

Since G(u) is coercive,

$$\begin{aligned} \rho _k=\Vert u^{(k)}\Vert ' _H \le C, \end{aligned}$$

where

$$\begin{aligned} (\Vert u\Vert '_H)^2 = \Vert \nabla u\Vert ^2+\Vert gu\Vert ^2. \end{aligned}$$

Thus there is a renamed subsequence such that $u^{(k)}$ converges to a limit $u \in H$ weakly in H,  strongly in $L^2_{loc}(\mathbb {R}^n)$ and a.e. in $\mathbb {R}^n.$ From (17) we see that

$$\begin{aligned} (G'(u),z)/2=a( u, z)-\int _{\mathbb {R}^n} f(x,u(x)) \cdot z(x)\,dx =0, \quad z \in C_0^\infty (\mathbb {R}^n), \end{aligned}$$

from which we conclude easily that u is a solution of (1). $\square$

Proof of Theorem 4

By Lemma 1, there is an $\varepsilon >0$ such that

$$\begin{aligned} a(u) + \Vert gu\Vert ^2 \ge \varepsilon \Vert u\Vert _H^2, \quad u \in H. \end{aligned}$$

(19)

Consequently,

$$\begin{aligned} G(u) \ge \varepsilon \Vert u\Vert _H^2 - \int _{|u| > \delta } S(x)^q|u(x)|^\alpha \ge \varepsilon \Vert u\Vert _H^2 - C \Vert u\Vert _H^q \end{aligned}$$

by Hypothesis (A). As a consequence, there are positive constants $\eta , \rho$ such that

$$\begin{aligned} G(u) \ge \eta , \quad \Vert u\Vert _H=\rho . \end{aligned}$$

(20)

Since $0 \in \sigma _e(\mathcal {A}),$ there is a $\varphi \in H$ such that $\Vert \varphi \Vert =1$ and $a(\varphi )=(\mathcal {A}\varphi , \varphi ) < \eta /2.$ Consequently,

$$\begin{aligned} G(s\varphi ) = s^2a(\varphi )-2\int F(x,s\varphi ) \le s^2 \eta /2 +\int h(s\varphi ). \end{aligned}$$

Thus,

$$\begin{aligned} \limsup _{s \rightarrow \infty } G(s\varphi )/s^2 \le \eta /2 + C, \end{aligned}$$

since

$$\begin{aligned} \limsup _{s \rightarrow \infty } \int \frac{h(s\varphi )}{s^2\varphi ^2} \varphi ^2 \le C\Vert \varphi \Vert ^2. \end{aligned}$$

This implies that there is a sequence $u^{(k)}$ in H such that

$$\begin{aligned} G(u^{(k)}) \rightarrow c\ge \eta /2, \quad G'(u^{(k)})/(\Vert u^{(k)}\Vert _H+1)^2 \rightarrow 0 \end{aligned}$$

(21)

(Theorem 2.7.1 of [22]).

If

$$\begin{aligned} \rho _k=\Vert u^{(k)}\Vert ' _H \le C, \end{aligned}$$

there is a renamed subsequence such that $u^{(k)}$ converges to a limit $u \in H$ weakly in H,  strongly in $L^2_{loc}(\mathbb {R}^n)$ and a.e. in $\mathbb {R}^n.$ In particular we have

$$\begin{aligned} \left( G'(u^{(k)}),z\right) /2=a\left( u^{(k)}, z\right) -\int f\left( x,u^{(k)}\right) \cdot z(x)\,dx \rightarrow 0, \quad z \in H. \end{aligned}$$

(22)

From this we see that

$$\begin{aligned} \left( G'(u),z\right) /2=a( u, z)-\int f(x,u(x)) \cdot z(x)\,dx =0, \quad z \in C_0^\infty (\mathbb {R}^n), \end{aligned}$$

from which we conclude easily that u is a solution of (1). Moreover, since $\Vert u^{(k)}\Vert '_H$ is bounded, (21) implies $G(u)=c\ge \eta /2$ (Theorem 3.4.1 of [22]). Since, $G(0)=0,$ we see that $u \ne 0.$

If

$$\begin{aligned} \rho _k=\Vert u^{(k)}\Vert '_H \rightarrow \infty , \end{aligned}$$

let $\tilde{u}^{(k)}=u^{(k)}/\rho _k.$ Then, $\Vert \tilde{u}^{(k)}\Vert '_H =1.$ There is a renamed subsequence such that $\tilde{u}^{(k)}$ converges to a function $\tilde{u}(x) \in H$ weakly in H,  strongly in $L^2_{loc}(\mathbb {R}^n),$ a.e. in $\mathbb {R}^n,$ and such that $\Vert \nabla \tilde{u}^{(k)} \Vert \rightarrow \nu$ and $\Vert g\tilde{u}^{(k)}\Vert \rightarrow \tau ,$ where $\nu ^2+\tau ^2=1.$ Let $b(u,v)=(Vu,v),\; b(u)=b(u,u).$ Since $\tilde{u}^{(k)}$ converges to $\tilde{u}$ weakly in H,  we see that $b(\tilde{u}^{(k)}) \rightarrow b(\tilde{u}).$ Hence, $\Vert \nabla \tilde{u}^{(k)} \Vert ^2 +b(\tilde{u}^{(k)}) +\Vert g\tilde{u}^{(k)}\Vert ^2 \rightarrow \nu ^2 +b(\tilde{u}) +\tau ^2.$ Note that

$$\begin{aligned} \int S^q |u|^\alpha \le \left( \int |Su|^{\alpha \cdot (q/\alpha )}\right) ^ {(\alpha /q)} \left( \int S^{(q-\alpha ) \cdot q/(q-\alpha )}\right) ^{(q-\alpha )/q} \le \Vert Su\Vert _q^\alpha \cdot \Vert S\Vert _q^{q-\alpha }. \end{aligned}$$

Thus,

$$\begin{aligned} 0 \leftarrow G(u^{(k)})/\rho _k^2 \ge \Vert \nabla \tilde{u}^{(k)} \Vert ^2 +b(\tilde{u}^{(k)}) +\Vert g\tilde{u}^{(k)}\Vert ^ 2 - O(\rho _k^{\alpha -2}) \rightarrow \nu ^2 +b(\tilde{u}) +\tau ^2. \end{aligned}$$

(23)

Thus, $1+b(\tilde{u})= \nu ^2+b(\tilde{u}) + \tau ^2\le 0.$ This implies that $b(\tilde{u}) \ne 0,$ and consequently $\tilde{u}(x) \not \equiv 0.$ But $g \tilde{u}^{(k)} \rightarrow g\tilde{u}$ a.e. This means that $\tilde{u} =0,$ a contradiction. Hence, the $\rho _k$ are bounded, and the proof is complete. $\square$

Proof of Theorem 5

We follow the proof of Theorem 4. By Lemma 1 there is an $\varepsilon >0$ such that (7) holds. This implies that there are positive constants $\eta , \rho$ such that

$$\begin{aligned} G(u) \ge \eta , \quad \Vert u\Vert _H=\rho , \; u \in M \end{aligned}$$

(24)

by the argument given there. Note that

$$\begin{aligned} G(v) =a(v)-2\int F(x,v) = \lambda _0 \Vert v\Vert ^2- 2\int F(x,v) \le 0, \quad v \in N \end{aligned}$$

by Hypothesis 1. Define $A=M \cap {\mathbf{B}}_\rho , \;\; B=N,$ where

$$\begin{aligned} {\mathbf{B}}_\rho =\{u \in H: \Vert u\Vert _H <\rho \} . \end{aligned}$$

Then A links B (Example 2, p. 38 of [22]). If $G_1=-G,$ then

$$\begin{aligned} \sup _A G_1 \le \inf _B G_1. \end{aligned}$$

By Theorem 13.7, p. 259 of [22] or Corollary 3.22, p. 29 of [24] there is a sequence $\{u^{(k)}\} \subset H$ such that

$$\begin{aligned} G\left( u^{(k)}\right)= & {} a\left( u^{(k)}\right) -2\int _{\mathbb {R}^n} F\left( x,u^{(k)}(x)\right) \,dx \rightarrow c, \end{aligned}$$

(25)

$$\begin{aligned} \left( G'(u^{(k)}),z\right) /2= & {} a\left( u^{(k)}, z\right) -\int _{\mathbb {R}^n} f\left( x,u^{(k)}\right) \cdot z(x)\,dx \rightarrow 0, \quad z \in H \end{aligned}$$

(26)

and

$$\begin{aligned} \left( G'(u^{(k)}),u^{(k)}\right) /2= a\left( u^{(k)}\right) -\int _{\mathbb {R}^n} f\left( x,u^{(k)}\right) \cdot u^{(k)}\,dx \rightarrow 0. \end{aligned}$$

(27)

If

$$\begin{aligned} \rho _k=\Vert u^{(k)}\Vert ' _H \le C, \end{aligned}$$

where

$$\begin{aligned} \left( \Vert u\Vert '_H\right) ^2 = \Vert \nabla u\Vert ^2+\Vert gu\Vert ^2, \end{aligned}$$

then there is a renamed subsequence such that $u^{(k)}$ converges to a limit $u \in H$ weakly in H,  strongly in $L^2_{loc}(\mathbb {R}^n)$ and a.e. in $\mathbb {R}^n.$ From (26) we see that

$$\begin{aligned} \left( G'(u),z\right) /2=a( u, z)-\int _{\mathbb {R}^n} f(x,u(x)) \cdot z(x)\,dx =0, \quad z \in C_0^\infty (\mathbb {R}^n), \end{aligned}$$

from which we conclude easily that u is a solution of (1).

If

$$\begin{aligned} \rho _k=\Vert u^{(k)}\Vert '_H \rightarrow \infty , \end{aligned}$$

let $\tilde{u}^{(k)}=u^{(k)}/\rho _k.$ Then, $\Vert \tilde{u}^{(k)}\Vert '_H =1.$ There is a renamed subsequence such that $\tilde{u}^{(k)}$ converges to a function $\tilde{u}(x) \in H$ weakly in H,  strongly in $L^2_{loc}(\mathbb {R}^n),$ a.e. in $\mathbb {R}^n,$ and such that $\Vert \nabla \tilde{u}^{(k)} \Vert \rightarrow \nu$ and $\Vert g\tilde{u}^{(k)}\Vert \rightarrow \tau ,$ where $\nu ^2+\tau ^2=1.$ Let $b(u,v)=(Vu,v),\; b(u)=b(u,u).$ Since $\tilde{u}^{(k)}$ converges to $\tilde{u}$ weakly in H,  we see that $b(\tilde{u}^{(k)}) \rightarrow b(\tilde{u}).$ Hence, $0 \le \Vert \nabla \tilde{u}^{(k)} \Vert ^2 +b(\tilde{u}^{(k)}) +\Vert g\tilde{u}^{(k)}\Vert ^2 \rightarrow \nu ^2 +b(\tilde{u}) +\tau ^2.$ Since $a(\tilde{u}^{(k)}) \ge 0,$ we have $\nu ^2 +b(\tilde{u}) \ge 0.$ Note that

$$\begin{aligned} \int S^q |u|^\alpha \le \left( \int |Su|^{\alpha \cdot (q/\alpha )}\right) ^ {(\alpha /q)} \left( \int S^{(q-\alpha ) \cdot q/(q-\alpha )}\right) ^{(q-\alpha )/q} \le \Vert Su\Vert _q^\alpha \cdot \Vert S\Vert _q^{q-\alpha }. \end{aligned}$$

Moreover,

$$\begin{aligned} G\left( u^{(k)}\right) /\rho _k^2 =\Vert \nabla \tilde{u}^{(k)} \Vert ^2 +b\left( \tilde{u}^{(k)}\right) +\Vert g\tilde{u}^{(k)}\Vert ^ 2 + O\left( \rho _k^{\alpha -2}\right) \rightarrow \nu ^2 +b(\tilde{u}) +\tau ^2=0. \end{aligned}$$

This implies $\nu ^2+b(\tilde{u}) =0$ and $\tau ^2=0.$ But $g \tilde{u}^{(k)} \rightarrow g\tilde{u}$ a.e. This means that $\tilde{u} =0$ and $\nu =1,$ so that

$$\begin{aligned} 1+b(\tilde{u})= 0. \end{aligned}$$

But this implies that $b(\tilde{u}) \ne 0,$ and consequently $\tilde{u}(x) \not \equiv 0,$ a contradiction. Hence, the $\rho _k$ are bounded, and the proof is complete. $\square$

Proof of Theorem 6

We follow the proof of Theorem 4. By Lemma 1 there is an $\varepsilon >0$ such that (7) holds. This implies that there are positive constants $\eta , \rho$ such that

$$\begin{aligned} G(u) \ge \eta , \quad \Vert u\Vert _H=\rho , \; u \in M \end{aligned}$$

(28)

as in the proof of Theorem 4. Note that

$$\begin{aligned} G(v) =a(v)-2\int F(x,v) = \lambda _0 \Vert v\Vert ^2- 2\int F(x,v) \le 0, \quad v \in N \end{aligned}$$

by Hypothesis 1. Since $\mathcal {A}$ is not invertible on M, there is a $\varphi \in M$ such that $\Vert \varphi \Vert =1$ and $a(\varphi )=(\mathcal {A}\varphi , \varphi ) < \eta .$ For $R > \rho ,$ let

$$\begin{aligned} A_R= & {} [N \cap {\mathbf{B}}_R] \cup \{ v +s \varphi : s \ge 0,\; \Vert v+s\varphi \Vert =R\}\\ B= & {} M \cap \partial \mathbf{B}_\rho , \end{aligned}$$

where

$$\begin{aligned} {\mathbf{B}}_\rho =\{u \in H: \Vert u\Vert _H <\rho \} . \end{aligned}$$

By Example 3, p.38 of [22], $A_R, \; B$ link each other. Now

$$\begin{aligned} G(v+s \varphi )&\le a(v) +s^2 \eta + \int \frac{h(v+s \varphi )}{(v+s \varphi )^2} (v+s \varphi )^2 \\&\le \lambda _0 \Vert v\Vert ^2 +s^2 \eta +c_0 \Vert v+s \varphi \Vert ^2 \\&\le (c_0 +\lambda _0) \Vert v\Vert ^2 + (c_0+ \eta ) s^2 \\&\le (2c_0 +\lambda _0 +\eta ) R^2. \end{aligned}$$

We can now apply Theorem 2.7.3 of [22] to conclude that there is a sequence $u^{(k)}$ in H such that

$$\begin{aligned} G\left( u^{(k)}\right) \rightarrow c\ge \eta , \quad G'\left( u^{(k)}\right) /\left( \Vert u^{(k)}\Vert _H+1\right) ^2 \rightarrow 0. \end{aligned}$$

(29)

We can now follow the proofs of Theorem 5 to reach the desired conclusion. $\square$

Proof of Theorem 7

Define the subspaces M and N of H as before:

$$\begin{aligned} N = \bigoplus _{k \,<\, l} E(\lambda _k) \;,\quad M \;=\; N^{\perp } \;,\quad H \,=\, M \oplus N \;. \end{aligned}$$

Let

$$\begin{aligned} G(u)=a(u)-2\int _{\mathbb {R}^n} F(x,u)\,dx. \end{aligned}$$

(30)

We note that Hypothesis 1 implies

$$\begin{aligned} G(v) \le 0,\quad v \in N. \end{aligned}$$

(31)

In fact, we have

$$\begin{aligned} G(u)=a(u)-2\int _{\mathbb {R}^n} F(x,u)\,dx \le \int _{\mathbb {R}^n}\left[ \lambda _{l-1}u^2-2f(x,u)\right] \, dx \le 0, \quad u \in N. \end{aligned}$$

In view of inequality (8), we see that there are positive constants $\eta , \rho$ such that

$$\begin{aligned} G(u) \ge \eta , \quad \Vert u\Vert _H=\rho , \; u \in M. \end{aligned}$$

(32)

Take

$$\begin{aligned} A= & {} \partial {\mathbf{B}}_{\rho } \cap M,\\ B= & {} N, \end{aligned}$$

where

$$\begin{aligned} {\mathbf{B}}_\rho =\{u \in H: \Vert u\Vert _H <\rho \} . \end{aligned}$$

By Example 8, p. 22 of [24], A links B. Moreover,

$$\begin{aligned} \sup _A[-G] \le 0 \le \inf _B[-G]. \end{aligned}$$

(33)

Hence, we may apply Corollary 2.8.2 of [22] to conclude that that there is a sequence $\{u^{(k)}\} \subset H$ such that

$$\begin{aligned} G\left( u^{(k)}\right)= & {} a\left( u^{(k)}\right) -2\int _{\mathbb {R}^n} F\left( x,u^{(k)}(x)\right) \,dx \rightarrow c\le 0, \end{aligned}$$

(34)

$$\begin{aligned} \left( G'(u^{(k)}),z\right) /2= & {} a\left( u^{(k)}, z\right) -\int _{\mathbb {R}^n} f\left( x,u^{(k)}\right) \cdot z(x)\,dx \rightarrow 0, \quad z \in H \end{aligned}$$

(35)

and

$$\begin{aligned} \left( G'(u^{(k)}),u^{(k)}\right) /2= a\left( u^{(k)}\right) -\int _{\mathbb {R}^n} f\left( x,u^{(k)}\right) \cdot u^{(k)}\,dx \rightarrow 0. \end{aligned}$$

(36)

If

$$\begin{aligned} \rho _k=\Vert u^{(k)}\Vert ' _H \le C, \end{aligned}$$

there is a renamed subsequence such that $u^{(k)}$ converges to a limit $u \in H$ weakly in H strongly in $L^2_{loc}(\mathbb {R}^n)$ and a.e. in $\mathbb {R}^n.$ From (35) we see that

$$\begin{aligned} \left( G'(u),z\right) /2=a(u, z)-\int _{\mathbb {R}^n}f(x,u(x))\cdot z(x)\,dx =0, \quad z \in C_0^\infty (\mathbb {R}^n), \end{aligned}$$

from which we conclude easily that u is a solution of (1). We can now follow the proof of Theorem 4 until we reach (23). In our case this becomes

$$\begin{aligned} G\left( u^{(k)}\right) /\rho _k^2 \ge \Vert \nabla \tilde{u}^{(k)} \Vert ^2 +b\left( \tilde{u}^{(k)}\right) +\Vert g\tilde{u}^{(k)}\Vert ^ 2 -\lambda _l \Vert \tilde{u}^{(k)}\Vert ^ 2- O\left( \rho _k^{\alpha -2}\right) . \end{aligned}$$

(37)

Since $\lambda _l \le 0,$

$$\begin{aligned} G\left( u^{(k)}\right) /\rho _k^2 \rightarrow 0, \end{aligned}$$

and

$$\begin{aligned} \Vert \nabla \tilde{u}^{(k)} \Vert ^2 +b(\tilde{u}^{(k)}) +\Vert g\tilde{u}^{(k)}\Vert ^ 2 \rightarrow \nu ^2 +b(\tilde{u}) +\tau ^2, \end{aligned}$$

we again have $1+b(\tilde{u})\le 0,$ which leads to the desired conclusion. $\square$

Proof of Theorem 8

We follow the proof of Theorem 7. We note that Hypothesis 1 implies

$$\begin{aligned} G(v) \le 0,\quad v \in N. \end{aligned}$$

(38)

In fact, we have

$$\begin{aligned} G(u)=a(u)-2\int _{\mathbb {R}^n} F(x,u)\,dx \le \int _{\mathbb {R}^n}\left[ \lambda _{l-1}u^2-2f(x,u)\right] \, dx \le 0, \quad u \in N. \end{aligned}$$

In view of inequality (8), we see that there are positive constants $\eta , \rho$ such that

$$\begin{aligned} G(u) \ge \eta , \quad \Vert u\Vert _H=\rho , \; u \in M. \end{aligned}$$

(39)

Let $\varphi$ be an eigenfunction of $\mathcal {A}$ corresponding to $\lambda _l$ and satisfying $\Vert \varphi \Vert =1.$ For $R > \rho ,$ let

$$\begin{aligned} A_R= & {} [N \cap {\mathbf{B}}_R] \cup \{ v +s \varphi : s \ge 0,\; \Vert v+s\varphi \Vert =R\}\\ B= & {} M \cap \partial \mathbf{B}_\rho , \end{aligned}$$

where

$$\begin{aligned} {\mathbf{B}}_\rho =\{u \in H: \Vert u\Vert _H <\rho \} . \end{aligned}$$

By Example 3, p.38 of [22], $A_R, \; B$ link each other. Now

$$\begin{aligned} G(v+s \varphi )&\le a(v) +s^2 \lambda _l + \int \frac{h(v+s \varphi )}{(v+s \varphi )^2} (v+s \varphi )^2 \\&\le \lambda _{l-1} \Vert v\Vert ^2 +s^2 \lambda _l +c_0 \Vert v+s \varphi \Vert ^2 \\&\le (c_0 +\lambda _{l-1}) \Vert v\Vert ^2 + (c_0+ \lambda _l) s^2 \\&\le (2c_0 +\lambda _{l-1} +\lambda _l) R^2. \end{aligned}$$

We can now apply Theorem 2.7.3 of [22] to conclude that there is a sequence $u^{(k)}$ in H such that

$$\begin{aligned} G\left( u^{(k)}\right) \rightarrow c\ge \eta /2, \quad G'\left( u^{(k)}\right) /\left( \Vert u^{(k)}\Vert _H+1\right) ^2 \rightarrow 0. \end{aligned}$$

(40)

We can now follow the proof of Theorem 4 until we reach (23). In our case this becomes

$$\begin{aligned} G\left( u^{(k)}\right) /\rho _k^2 \ge \Vert \nabla \tilde{u}^{(k)} \Vert ^2 +b\left( \tilde{u}^{(k)}\right) +\Vert g\tilde{u}^{(k)}\Vert ^ 2 -\lambda _l \Vert \tilde{u}^{(k)}\Vert ^ 2- O\left( \rho _k^{\alpha -2}\right) . \end{aligned}$$

(41)

Since $\lambda _l \le 0,$

$$\begin{aligned} G\left( u^{(k)}\right) /\rho _k^2 \rightarrow 0, \end{aligned}$$

and

$$\begin{aligned} \Vert \nabla \tilde{u}^{(k)} \Vert ^2 +b\left( \tilde{u}^{(k)}\right) +\Vert g\tilde{u}^{(k)}\Vert ^ 2 \rightarrow \nu ^2 +b(\tilde{u}) +\tau ^2, \end{aligned}$$

we again have $1+b(\tilde{u})\le 0,$ which leads to the desired conclusion. $\square$

Proof of Theorem 10. Let

$$\begin{aligned} \alpha = \inf _\mathcal {M}G(u). \end{aligned}$$

There is a sequence $\{u^{(k)}\} \in \mathcal {M}$ such that

$$\begin{aligned} G\left( u^{(k)}\right)= & {} a\left( u^{(k)}\right) -2\int _{\mathbb {R}^n} F\left( x,u^{(k)}(x)\right) \,dx \rightarrow \alpha , \end{aligned}$$

(42)

$$\begin{aligned} \left( G'(u^{(k)}),z\right) /2= & {} a\left( u^{(k)}, z\right) -\int _{\mathbb {R}^n} f\left( x,u^{(k)}\right) \cdot z(x)\,dx = 0, \quad z \in H \end{aligned}$$

(43)

and

$$\begin{aligned} \left( G'(u^{(k)}),u^{(k)}\right) /2= a\left( u^{(k)}\right) -\int _{\mathbb {R}^n}f\left( x,u^{(k)}\right) \cdot u^{(k)}\,dx = 0. \end{aligned}$$

(44)

Thus,

$$\begin{aligned} \int _{\mathbb {R}^n} H\left( x,u^{(k)}(x)\right) \,dx =G\left( u^{(k)}\right) \rightarrow \alpha . \end{aligned}$$

In view of assumption (11), we see that $\alpha \ge -\int W > - \infty .$ In view of the arguments given in the proofs of Theorems 2–8, we see that

$$\begin{aligned} \rho _k=\Vert u^{(k)}\Vert _H \le C. \end{aligned}$$

Hence, there is a renamed subsequence such that $u^{(k)}$ converges to a limit $u \in H,$ weakly in H,  strongly in $L^2_{loc}(\mathbb {R}^n)$ and a.e. in $\mathbb {R}^n.$ From (43) we see that

$$\begin{aligned} \left( G'(u),z\right) /2=a(u, z)-\int _{\mathbb {R}^n}f(x,u(x))\cdot z(x)\,dx =0, \quad z \in C_0^\infty (\mathbb {R}^n), \end{aligned}$$

from which we conclude easily that u is a solution of (1). Hence, $u \in \mathcal {M}.$ Moreover,

$$\begin{aligned} G(u)=\int _{\mathbb {R}^n} H(x, u(x))\, dx \le \liminf \int _{\mathbb {R}^n} H(x,u^{(k)}(x))\,dx =\liminf G(u^{(k)}) =\alpha . \end{aligned}$$

Thus, $G(u)=\alpha .$ If $\alpha \ne 0,$ then $u \ne 0.$ This completes the proof.