Firstly, we give the convergence of the nonlinear term.
Lemma 2.1
Assume that
\(u_{n}\rightharpoonup u\)
in
E, then
\(\int_{\mathbb {R}^{{N}}}K(x)f(u_{n})u_{n}\rightarrow\int_{\mathbb{R}^{{N}}}K(x)f(u)u\)
and
\(\int_{\mathbb{R}^{3}}K(x)F(u_{n})\to\int_{\mathbb{R}^{3}}K(x)F(u)\).
Proof
By Proposition 1.2, \(u_{n}\to u\) strongly in \(L^{q}_{K}\) for \(q\in[2,6)\). Thus after passing to a subsequence, \(u_{n}(x)\to u(x)\) for a.e. \(x\in\mathbb{R}^{N}\). From the convergence of \(u_{n}\to u\) in \(L^{2}_{K}\) and \(L^{p+1}_{K}\), there exist two functions \(a\in L^{2}\) and \(b\in L^{p+1}\) such that \(K(x)u_{n}^{2}\leqslant a^{2}(x)\) and \(K(x)|u_{n}|^{p+1}\leqslant[b(x)]^{p+1}\) for a.e. \(x\in\mathbb{R}^{N}\). Therefore, (f4) implies that
$$\begin{aligned} \biggl\vert \int_{\mathbb{R}^{N}}K(x)f(u_{n})u_{n}\biggr\vert & \leqslant \int_{\mathbb{R}^{N}}C\bigl\vert K(x)\bigr\vert u_{n}|^{2}+CK(x)|u_{n}|^{p+1} \\ &\leqslant \int_{\mathbb{R}^{N}}\bigl[C\bigl(a(x)\bigr)^{2}+C\bigl(b(x) \bigr)^{p+1}\bigr]\,dx. \end{aligned}$$
According to the dominated convergence theorem, \(\int_{\mathbb {R}^{3}}K(x)f(u_{n})u_{n}\to\int_{\mathbb{R}^{3}}K(x)f(u)u\). A similar argument implies that \(\int_{\mathbb{R}^{3}}K(x)F(u_{n})\to\int _{\mathbb{R}^{3}}K(x)F(u)\). The proof is completed. □
The next lemma shows that \(M\neq\emptyset\).
Lemma 2.2
Assume that (f1)-(f4) hold. If
\(u\in E\)
with
\(u^{\pm}\neq 0\), then there is a unique pair
\((t,s)\)
of positive numbers such that
\(tu^{+}+su^{-}\in M\).
Proof
Fix \(u\in E\) with \(u^{\pm}\neq0\), \(\int_{\mathbb{R}^{N}}\nabla\phi _{u^{+}}\cdot\nabla\phi_{u^{-}}=\int_{\mathbb{R}^{N}}V_{2}(x)\phi _{u^{+}}(u^{-})^{2}\); since \(-\Delta\phi_{u^{-}}=V_{2}(x)(u^{-})^{2}\), we can obtain
$$ \int_{\mathbb{R}^{N}}V_{2}(x)\phi_{u^{+}}\bigl|u^{-}\bigr|^{2} \,dx= \int_{\mathbb {R}^{N}}V_{2}(x)\phi_{u^{-}} \bigl|u^{+}\bigr|^{2}\,dx= \int_{\mathbb{R}^{N}}\nabla\phi_{u^{+}}\cdot\nabla \phi_{u^{-}}. $$
Denote \(A:=b\int_{\mathbb{R}^{{N}}}|\nabla u^{+}|^{2}\,dx\int_{\mathbb {R}^{{N}}}|\nabla u^{-}|^{2}\,dx+\int_{\mathbb{R}^{{N}}}V_{2}(x)\phi _{u^{-}}|u^{+}|^{2}\,dx\) for convenience. We introduce the following notations:
$$\begin{aligned}& a_{1}=\bigl\Vert u^{+}\bigr\Vert ^{2}, \qquad a_{2}=b \biggl( \int_{\mathbb{R}^{N}}\bigl\vert \nabla u^{+}\bigr\vert ^{2} \biggr)^{2}; \end{aligned}$$
(2.1)
$$\begin{aligned}& b_{1}=\bigl\Vert u^{-}\bigr\Vert ^{2},\qquad b_{2}=b \biggl( \int_{\mathbb{R}^{N}}\bigl\vert \nabla u^{-}\bigr\vert ^{2} \biggr)^{2}. \end{aligned}$$
(2.2)
Since \(-\Delta\phi_{u}=V_{2}(x)u^{2}=V_{2}(x)(u^{+})^{2}+V_{2}(x)(u^{-})^{2}=-\Delta (\phi_{u^{+}}+\phi_{u^{-}})\), \(\phi_{u}=\phi_{u^{+}}+\phi_{u^{-}}\). Then, \(tu^{+}+su^{-}\) is contained in M if and only if
$$ \left \{ \textstyle\begin{array}{l} g_{1}(t,s):=a_{1}t^{2}+a_{2}t^{4}+At^{2}s^{2}-\int_{\mathbb {R}^{{N}}}Kf(tu^{+})(tu^{+})=0, \\ g_{2}(t,s):=b_{1}s^{2}+b_{2}s^{4}+At^{2}s^{2}-\int_{\mathbb {R}^{{N}}}Kf(su^{-})(su^{-})=0. \end{array}\displaystyle \right . $$
(2.3)
We have from (f1), (f2), and (2.3) \(g_{1}(t,t)>0\) and \(g_{2}(t,t)>0\) for \(t>0\) small and \(g_{1}(s,s)<0\), \(g_{2}(s,s)<0\) for \(s>0\) large. Thus there exist \(0<\delta<R\) such that
$$ g_{1}(\delta,\delta)>0, \qquad g_{1}(R,R)< 0; \qquad g_{2}(\delta,\delta )< 0,\qquad g_{2}(R,R)< 0. $$
(2.4)
From (2.3), it follows that \(g_{1}(\delta,s)\), \(g_{1}(R,s)\) both are increasing for \(s\geq0\) and \(g_{2}(t,R)\), \(g_{2}(t,\delta)\) both are increasing for \(t\geq0\). Therefore, by (2.4), we have
$$ g_{1}(\delta,s)>0, \qquad g_{1}(R,s)< 0, \quad s\in[\delta,R], $$
(2.5)
and
$$ g_{2}(t,\delta)>0, \qquad g_{2}(t,R)< 0, \quad t\in[\delta,R]. $$
(2.6)
By (2.5), (2.6), and Miranda’s theorem [29], there exists a point \((t_{u},s_{u})\) with \(t_{u},s_{u}\in(\delta,R)\) such that \(g_{1}(t_{u},s_{u})=g_{2}(t_{u},s_{u})=0\).
In the following, we prove the uniqueness of \((t,s)\). If there are two pairs \((t_{1},s_{1})\) and \((t_{2},s_{2})\) such that \(t_{1}u^{+}+su^{-}\in M\) and \(t_{2}u^{+}+su^{-}\in M\), then \((t_{1}/t_{2})t_{2}u^{+}+(s_{1}/s_{2})s_{2}u^{-}\in M\). Therefore, we may assume that \(u\in M\) and \(tu^{+}+su^{-}\in M\). We need only to prove that \(t=s=1\). We assume that \(0< t\leq s\). Then, by (2.3) and \(\theta<1\), we have \(b_{1}-\theta\int _{\mathbb{R}^{N}}V(x)(u^{-})^{2}>0\) and
$$ \biggl[b_{1}-\theta \int_{\mathbb{R}^{N}}V(x) \bigl(u^{-}\bigr)^{2}\biggr] \frac{1}{s^{2}}+b_{2}+A\geq \int_{\mathbb{R}^{N}}\frac{[Kf(su^{-})-\theta V(x)su^{-}]}{(su^{-})^{3}}\bigl(u^{-}\bigr)^{4}. $$
Since \(u\in M\),
$$\begin{aligned}& \biggl[b_{1}-\theta \int_{\mathbb{R}^{N}}V(x) \bigl(u^{-}\bigr)^{2}\biggr] \biggl[ \frac{1}{s^{2}}-1 \biggr] \\& \quad \geq \int_{\mathbb{R}^{N}} \biggl\{ \frac{[Kf(su^{-})-\theta V(x)su^{-}]}{(su^{-})^{3}}-\frac{[Kf(u^{-})-\theta V(x)u^{-}]}{(u^{-})^{3}} \biggr\} \bigl(u^{-}\bigr)^{4}. \end{aligned}$$
According to condition (f3), \(s\leq1\). On the other hand, since \(t\leq s\) and \(u\in M\), similarly, we can obtain
$$\begin{aligned}& \biggl[a_{1}-\theta \int_{\mathbb{R}^{N}}V(x) \bigl(u^{+}\bigr)^{2}\biggr] \biggl[ \frac{1}{t^{2}}-1 \biggr] \\& \quad \leq \int_{\mathbb{R}^{N}} \biggl\{ \frac{[Kf(tu^{+})-\theta V(x)tu^{+}]}{(tu^{+})^{3}}-\frac{[Kf(u^{+})-\theta V(x)u^{+}]}{(u^{+})^{3}} \biggr\} \bigl(u^{+}\bigr)^{4}. \end{aligned}$$
Therefore, \(t\geq1\). Consequently, \(t=s=1\). □
Lemma 2.3
For
\(u\in E\)
with
\(u^{\pm}\neq0\), the pair
\((t_{u},s_{u})\)
obtained in Lemma
2.2
is the maximum point of the function
\(\psi :\mathbb{R}_{+}\times\mathbb{R}_{+}\rightarrow\mathbb{R}\)
defined as
\(\psi(t,s)=J(tu^{+}+su^{-})\), where
\(\mathbb{R}_{+}=[0,\infty)\).
Proof
From the proof of Lemma 2.2, \((t_{u},s_{u})\) is a critical point of ψ in \(\mathbb{R}_{+}\times\mathbb{R}_{+}\). By assumptions (f1), (f2), and the proof of Lemma 2.2, we deduce that \(J(tu^{+}+su^{-})>0\) for \(t,s>0\) small and \(\psi(t,s)\rightarrow -\infty\) as \(|(t,s)|\rightarrow\infty\). Therefore, \((t_{u},s_{u})\) is the maximum point. In the following, we prove that \(t_{u},s_{u}>0\). It is sufficient to check that a maximum point cannot be achieved on the boundary of \(\mathbb{R}_{+}\times\mathbb{R}_{+}\). Without loss of generality, we may assume that \((0,s_{0})\) is a maximum point of ψ. Then, for \(t>0\) small, by condition (f1),
$$\begin{aligned} \psi(t,s_{0}) =&J\bigl(tu^{+}+s_{0}u^{-} \bigr) \\ =&\frac{t^{2}}{2}\bigl\Vert u^{+}\bigr\Vert ^{2}+ \frac{t^{4}}{4}b\bigl\vert \nabla u^{+}\bigr\vert ^{4}_{2}+ \frac {t^{4}}{4} \int_{\mathbb{R}^{{N}}}V_{2}(x)\phi_{u^{+}} \bigl(u^{+}\bigr)^{2}- \int _{\mathbb{R}^{{N}}}K(x)F\bigl(tu^{+}\bigr)\,dx \\ &{}+\frac{t^{2}s_{0}^{2}}{2}b\bigl\vert \nabla u^{+}\bigr\vert _{2}^{2} \bigl\vert \nabla u^{-}\bigr\vert ^{2}_{2}+ \frac{t^{2}s_{0}^{2}}{2} \int_{\mathbb{R}^{{N}}}V_{2}(x)\phi _{u^{+}} \bigl(u^{-}\bigr)^{2}\,dx \\ &{}+\frac{s_{0}^{2}}{2}\bigl\Vert u^{-}\bigr\Vert ^{2}+\frac{s_{0}^{4}}{4}b\bigl\vert \nabla u^{-}\bigr\vert _{2}^{4}+\frac{s_{0}^{4}}{4} \int_{\mathbb{R}^{{N}}}V_{2}(x)\phi _{u^{-}} \bigl(u^{-}\bigr)^{2}- \int_{\mathbb{R}^{{N}}}K(x)F\bigl(s_{0}u^{-}\bigr)\,dx \\ >&\frac{s_{0}^{2}}{2}\bigl\Vert u^{-}\bigr\Vert ^{2}+ \frac{s_{0}^{4}}{4}b\bigl\vert \nabla u^{-}\bigr\vert _{2}^{4}+ \frac{s_{0}^{4}}{4} \int_{\mathbb{R}^{{N}}}V_{2}(x)\phi _{u^{-}} \bigl(u^{-}\bigr)^{2}- \int_{\mathbb{R}^{{N}}}K(x)F\bigl(s_{0}u^{-}\bigr)\,dx \\ =&\psi(0,s_{0}). \end{aligned}$$
The pair \((0,s_{0})\) is not a maximum point of ψ in \(\mathbb {R}_{+}\times\mathbb{R}_{+}\). □
Lemma 2.4
-
(i)
If
\(\langle J'(u),u^{\pm}\rangle<0\), then
\((t_{u},s_{u})\in(0,1)^{2}\).
-
(ii)
If
\(u^{\pm}\ne0\)
is such that
\(\langle J'(u),u^{\pm}\rangle \leqslant0\), then
\((t_{u},s_{u})\in(0,1]^{2}\).
-
(iii)
If
\(\langle J'(u),u^{\pm}\rangle>0\), then
\((t_{u},s_{u})\in(1,\infty)^{2}\).
Proof
Under the assumptions of (i), (ii), and (iii), we have \(u^{\pm}\neq0\). Let \(c_{1}=a_{1}-\theta \int_{\mathbb{R}^{N}}V(x)(u^{+})^{2}\) and \(d_{1}=b_{1}-\theta\int_{\mathbb{R}^{N}}V(x)(u^{-})^{2}\), then \(c_{1},d_{1}>0\). Since \(t_{u}u^{+}+s_{u}u^{-}\in M\), we have from (2.3)
$$\begin{aligned} \bigl\langle J'(u),u^{+}\bigr\rangle =&c_{1} \biggl[1-\frac{1}{t^{2}_{u}} \biggr]+A \biggl[1-\frac{s^{2}_{u}}{t^{2}_{u}} \biggr] \\ &{}- \int_{\mathbb{R}^{N}} \biggl[\frac{K(x)f(u^{+})-\theta V(x)(u^{+})}{(u^{+})^{3}} -\frac{K(x)f(t_{u}u^{+})-\theta V(x)(t_{u}u^{+})}{(t_{u}u^{+})^{3}} \biggr] \bigl(u^{+}\bigr)^{4} \end{aligned}$$
(2.7)
and
$$\begin{aligned} \bigl\langle J'(u),u^{-}\bigr\rangle =&d_{1} \biggl[1-\frac{1}{s^{2}_{u}} \biggr]+A \biggl[1-\frac{t^{2}_{u}}{s^{2}_{u}} \biggr] \\ &{}- \int_{\mathbb{R}^{N}} \biggl[\frac{K(x)f(u^{-})-\theta V(x)(u^{-})}{(u^{-})^{3}} -\frac{K(x)f(s_{u}u^{-})-\theta V(x)(s_{u}u^{-})}{(s_{u}u^{-})^{3}} \biggr] \bigl(u^{-}\bigr)^{4}. \end{aligned}$$
(2.8)
For convenience, we introduce some notations as follows:
$$\begin{aligned}& D_{11}=c_{1} \biggl[1-\frac{1}{t^{2}_{u}} \biggr]+A \biggl[1- \frac {s^{2}_{u}}{t^{2}_{u}} \biggr], \\& D_{12}= \int_{\mathbb{R}^{N}} \biggl[\frac{K(x)f(u^{+})-\theta V(x)(u^{+})}{(u^{+})^{3}} -\frac{K(x)f(t_{u}u^{+})-\theta V(x)(t_{u}u^{+})}{(t_{u}u^{+})^{3}} \biggr] \bigl(u^{+}\bigr)^{4}, \\& D_{21}=d_{1} \biggl[1-\frac{1}{s^{2}_{u}} \biggr]+A \biggl[1- \frac {t^{2}_{u}}{s^{2}_{u}} \biggr], \\& D_{22}= \int_{\mathbb{R}^{N}} \biggl[\frac{K(x)f(u^{-})-\theta V(x)(u^{-})}{(u^{-})^{3}} -\frac{K(x)f(s_{u}u^{-})-\theta V(x)(s_{u}u^{-})}{(s_{u}u^{-})^{3}} \biggr] \bigl(u^{-}\bigr)^{4}. \end{aligned}$$
Without loss of generality, we may assume that \(t_{u}\geqslant s_{u}\).
(i) If \(t_{u}\geqslant1\), then \(D_{11}\geqslant0\) and by the condition (f3), we have \(D_{12}\leqslant0\). It follows from (2.7) that \(\langle J'(u),u^{+}\rangle\geqslant0\). This is a contradiction. Thus, \((t_{u},s_{u})\in(0,1)^{2}\).
(ii) If \(t_{u}>1\), then \(D_{11}>0\) and by the condition (f3), we have \(D_{12}\leqslant0\). Therefore, \(\langle J'(u),u^{+}\rangle>0\). This is a contradiction. Thus, \((t_{u},s_{u})\in(0,1]^{2}\).
(iii) If \(s_{u}\leqslant1\), similarly \(D_{21}\leqslant0\) and \(D_{22}\geqslant0\). It follows from (2.8) that \(\langle J'(u),u^{-}\rangle\leqslant0\), which is a contradiction. Thus, \((t_{u},s_{u})\in(1,\infty)^{2}\). □
By Lemma 2.3, we may define the minimization problem
$$ m=\inf\bigl\{ J(u):u\in M\bigr\} . $$
(2.9)
Lemma 2.5
Suppose that conditions (f1)-(f4) hold, then
\(m>0\)
can be achieved.
Proof
For any \(u\in M\), we have \(\langle J'(u),u\rangle=0\). Then we get from conditions (f1) and (f4)
$$\begin{aligned} \|u\|^{2}&\leq \int_{\mathbb{R}^{{N}}}\bigl(|\nabla u|^{2}+V_{1}(x)u^{2}+V_{2}(x) \phi_{u}u^{2}\bigr)\,dx+b|\nabla u|_{2}^{4} \\ &= \int_{\mathbb{R}^{{N}}}K(x)f(u)u \\ &\leq\varepsilon \int_{\mathbb{R}^{N}} K(x)u^{2}+C_{\varepsilon} \int _{\mathbb{R}^{N}} K(x)u^{p+1}. \end{aligned}$$
Then by Proposition 1.2, we can get
$$ \|u\|^{2}\leq C_{1}\varepsilon\|u\|^{2}+C_{2} \|u\|^{p+1}, $$
where \(C_{1}, C_{2}>0\) are constants. So there exists a constant \(\alpha>0\) such that \(\|u\|\geq\alpha\). Moreover, by (f3), we have
$$ \bigl[K(x)f(t)t-\theta V(x)t^{2}\bigr]-4\biggl[K(x)F(t)-\theta \frac{1}{2}V(x)t^{2}\biggr]\geq0,\quad x\in\mathbb{R}^{N},t \in\mathbb{R}. $$
Hence, we have
$$\begin{aligned} J(u) =&J(u)-\frac{1}{4}\bigl\langle J'(u),u\bigr\rangle \\ =&\frac{1}{4}\|u\|^{2}+\frac{1}{4} \int_{\mathbb {R}^{{N}}}K(x) \bigl(f(u)u-4F(u)\bigr) \\ \geq&\frac{1}{4}\|u\|^{2}-\frac{1}{4}\theta \int_{\mathbb{R}^{N}}V(x)u^{2} \\ \geq&\frac{1}{4}(1-\theta)\alpha^{2}. \end{aligned}$$
This implies that \(m\geq\frac{1}{4}(1-\theta)\alpha^{2}>0\).
Let \(\{u_{n}\}\subset M\) be such that \(J(u_{n})\rightarrow m\). Then \(\{ u_{n}\}\) is bounded in E, so there exists a subsequence of \(\{u_{n}\}\) and \(u\in E \), we may still denote the subsequence by \(\{u_{n}\}\), such that \(u_{n}\rightharpoonup u\) and \(u_{n}^{\pm}\rightharpoonup u^{\pm}\) weakly in E and \(u_{n}(x)\to u(x)\) a.e. \(x\in\mathbb{R}^{N}\). Since \(\{ u_{n}\}\in M\), we have \(\langle J'(u_{n}),u_{n}^{+}\rangle=0\). That is,
$$ \bigl\Vert u_{n}^{+}\bigr\Vert ^{2}+b\vert \nabla u_{n}\vert ^{2}_{2} \bigl\vert \nabla u_{n}^{+}\bigr\vert _{2}^{2}+ \int_{\mathbb {R}^{{N}}}V_{2}(x)\phi_{u_{n}} \bigl(u_{n}^{+}\bigr)^{2}= \int_{\mathbb {R}^{{N}}}Kf\bigl(u_{n}^{+} \bigr)u_{n}^{+}. $$
(2.10)
This implies from conditions (f1) and (f4) that there exists a constant \(\rho>0\) such that \(\|u_{n}^{+}\|\geq\rho\) for all \(n\in \mathbb{N}\). By (2.10) and Lemma 2.2, we get
$$ 0< \rho^{2}\leq\bigl\Vert u_{n}^{+}\bigr\Vert ^{2}\leq \int_{\mathbb {R}^{{N}}}Kf\bigl(u_{n}^{+} \bigr)u_{n}^{+}= \int_{\mathbb{R}^{{N}}}Kf\bigl(u^{+}\bigr)u^{+}+o(1), $$
where \(o(1)\) denotes a quantity tending to zero as \(n\rightarrow +\infty\). Thus, \(u^{+}\neq0\). Similarly, we have \(u^{-}\neq0\).
Since \(u_{n}\rightharpoonup u\) in E, we have \(\phi _{u_{n}}\rightharpoonup\phi_{u}\) in \(D^{1,2}(\mathbb{R}^{{N}})\) if \(V_{2}\in L^{\infty}(\mathbb{R}^{N})\) or \(\phi_{u_{n}}\to\phi_{u}\) in \(D^{1,2}(\mathbb{R}^{{N}})\) if \(V_{2}\in L^{(6-N)/2N}(\mathbb{R}^{N})\). By the weak lower semicontinuity of the norm and Fatou’s lemma, we have
$$\begin{aligned}& \bigl\Vert u^{+}\bigr\Vert ^{2}+b\vert \nabla u \vert ^{2}_{2}\bigl\vert \nabla u^{+}\bigr\vert _{2}^{2}+ \int_{\mathbb {R}^{{N}}}V_{2}(x)\phi_{u}\bigl\vert u^{+}\bigr\vert ^{2}\,dx \\& \quad \leq \liminf_{n\rightarrow+\infty}\biggl(\bigl\Vert u_{n}^{+}\bigr\Vert ^{2}+b\vert \nabla u_{n}\vert ^{2}_{2}\bigl\vert \nabla u_{n}^{-}\bigr\vert ^{2}_{2}+ \int_{\mathbb {R}^{{N}}}V_{2}(x)\phi_{u_{n}}\bigl\vert u_{n}^{+}\bigr\vert ^{2}\,dx\biggr). \end{aligned}$$
Then from (2.10) and Lemma 2.1 we get
$$ \bigl\Vert u^{+}\bigr\Vert ^{2}+b\vert \nabla u\vert ^{2}_{2}\bigl\vert \nabla u^{+}\bigr\vert _{2}^{2}+ \int_{\mathbb {R}^{{N}}}V_{2}(x)\phi_{u}\bigl\vert u^{+}\bigr\vert ^{2}\,dx\leq \int_{\mathbb{R}^{{N}}}Kf\bigl(u^{+}\bigr)u^{+}. $$
(2.11)
Similarly, we have
$$ \bigl\Vert u^{-}\bigr\Vert ^{2}+b\vert \nabla u\vert _{2}^{2}\bigl\vert \nabla u^{-}\bigr\vert _{2}^{2}+ \int_{\mathbb {R}^{{N}}}V_{2}(x)\phi_{u}\bigl\vert u^{-}\bigr\vert ^{2}\,dx\leq \int_{\mathbb{R}^{{N}}}Kf\bigl(u^{-}\bigr)u^{-}. $$
(2.12)
From (2.11), (2.12), and the proof of Lemma 2.4, there exists \((\overline{t},\overline{s})\in(0,1]\times (0,1]\) such that \(\overline{u}=\overline{t}u^{+}+\overline {s}u^{-}\in M\). Moreover, according to condition (f3) we have \([K(x)f(t)t-\theta V(x)t^{2}]-4[K(x)F(t)-\frac{1}{2}\theta V(x)t^{2}]\geq0\) is nondecreasing on \((-\infty,0)\) and \((0,\infty)\), respectively. Thus,
$$\begin{aligned} m&\leq J(\overline{u})=J(\overline{u})-\frac{1}{4}\bigl\langle J'(\overline{u}),\overline{u}\bigr\rangle \\ &=\frac{1}{4}\|\overline{u}\|^{2}+\frac{1}{4} \int_{\mathbb {R}^{{N}}}K\bigl(f(\overline{u})\overline{u}-4F(\overline{u}) \bigr)\,dx \\ &=\frac{1}{4}\bigl(\bigl\| \overline{t}u^{+}\bigr\| ^{2}+\bigl\| \overline{s}u^{-}\bigr\| ^{2}\bigr)+\frac{1}{4} \int_{\mathbb{R}^{{N}}}K\bigl(f(\overline{u})\overline {u}-4F(\overline{u}) \bigr)\,dx \\ &\leq\frac{1}{4}|\nabla u|_{2}^{2}+\frac{1}{4} \int_{\mathbb {R}^{N}}V(x)\overline{u}^{2}+\frac{1}{4} \int_{\mathbb {R}^{{N}}}K\bigl(f(u)u-4F(u)\bigr)\,dx+\frac{1}{4}\theta \int_{\mathbb {R}^{N}}V(x)\bigl[u^{2}-\overline{u}^{2} \bigr] \\ &\leq\frac{1}{4}|\nabla u|_{2}^{2}+ \frac{1}{4}(1-\theta) \int_{\mathbb {R}^{N}}V(x)u^{2}+\frac{1}{4} \int_{\mathbb{R}^{{N}}}\bigl[K\bigl(f(u)u-4F(u)\bigr)+\theta V(x)u^{2}\bigr] \\ &\leq \liminf_{n\to\infty} \biggl[J(u_{n})-\frac {1}{4} \bigl\langle J'(u_{n}),u_{n}\bigr\rangle \biggr]=m. \end{aligned}$$
By the above inequality and Lemma 2.3 we deduce that \(\overline{t}=\overline{s}=1\). Thus \(\overline{u}=u\) and \(J(u)=m\). □