Before giving the proof of the main theorem, we need some preliminary results.
Lemma 3.1
For every
\(\lambda\in[1,2]\), \(\Phi_{\lambda}\)
is
τ-upper semicontinuous and
\(\Phi_{\lambda}'\)
is weakly sequentially continuous. Moreover, \(\Phi_{\lambda}\)
maps bounded sets to bounded sets.
Proof
The proof was given in [33] (also see [31, 38]); here we omit the details. □
Similar to Lemma 3.8 in [38] (also see [39]), we have the following lemma, which will be helpful for our argument.
Lemma 3.2
Assume that
\(\xi(x)>0\), \(\xi(x)\in L^{\infty }(\mathbb{R}^{N})\cap L^{r}(\mathbb{R}^{N})\), and
\(\frac{1}{r}+\frac {p}{2^{*}}=1\). Then
$$\begin{aligned} \beta_{k}=\sup_{z\in E^{+}_{k},\|z\|=1} \biggl(\int _{\mathbb{R}^{N}}\xi (x)|z|^{p} \biggr)^{\frac{1}{p}} \rightarrow0, \quad\textit{as } k\rightarrow \infty. \end{aligned}$$
(3.1)
Proof
Clearly, \(0\leq\beta_{k+1}\leq\beta_{k}\), hence \(\beta _{k}\rightarrow\beta\geq0\). For every k, there exists \(z_{k}\in E^{+}_{k}\) such that \(\|z_{k}\|=1\) and
$$\begin{aligned} 0\leq\beta_{k}^{p}-\int_{\mathbb{R}^{N}} \xi(x)|z_{k}|^{p}<\frac{1}{k}. \end{aligned}$$
Up to a subsequence, we have \(z_{k}\rightharpoonup z\) in E. By the definition of \(E^{+}_{k}\) we have \(z=0\). Moreover, by Sobolev imbedding theorem we have \(|z_{k}|_{2^{*}}\leq C\) for some \(C>0\). Observe that \(\xi(x)\in L^{r}\) implies that for every \(\varepsilon>0\), there exists \(R>0\) such that
$$\begin{aligned} \biggl(\int_{B^{c}_{R}(0)}\bigl|\xi(x)\bigr|^{r} \biggr)^{\frac{1}{r}}<\varepsilon. \end{aligned}$$
Since \(E\hookrightarrow L_{\mathrm{loc}}^{p}\) compactly for \(p\in[2, 2^{*})\). Hence, there exists \(k_{0}>0\) such that
$$\begin{aligned} \int_{B_{R}(0)}\xi(x)|z_{k}|^{p}< \varepsilon, \quad\forall k\geq k_{0}. \end{aligned}$$
Now since \(\frac{1}{r}+\frac{p}{2^{*}}=1\), for k large enough, we deduce by using the Hölder inequality that
$$\begin{aligned} \int_{\mathbb{R}^{N}}\xi(x)|z_{k}|^{p} =&\int _{B_{R}(0)}\xi (x)|z_{k}|^{p}+\int _{B^{c}_{R}(0)}\xi(x)|z_{k}|^{p} \\ \leq&\int_{B_{R}(0)}\xi(x)|z_{k}|^{p}+ \biggl(\int_{B^{c}_{R}(0)}\bigl|\xi (x)\bigr|^{r} \biggr)^{1/r} \biggl(\int_{B^{c}_{R}(0)}|z_{k}|^{2^{*}} \biggr)^{p/2^{*}} \\ \leq& \varepsilon+\varepsilon|z_{k}|^{p}_{2^{*}} \\ \leq&\varepsilon \bigl(1+C^{p} \bigr). \end{aligned}$$
We see that the desired conclusion holds by taking the limit. □
Lemma 3.3
Let (V), (H1)-(H4) be satisfied. There exists
\(r_{k}>0\)
independent of
\(\lambda\in[1,2]\)
such that
$$\begin{aligned} b_{k}(\lambda):=\inf_{z\in E_{k}^{+}, \|z\|=r_{k}}\Phi_{\lambda}(z)>0, \end{aligned}$$
and
\(b_{k}(\lambda)\rightarrow\infty\)
uniformly in
λ
as
\(k\rightarrow\infty\).
Proof
Observe that, given \(\varepsilon>0\), there is \(C_{\varepsilon}>0\) such that for all \(z\in E_{k}^{+}\)
$$\begin{aligned} \bigl|H_{z}(x,z)\bigr|\leq\varepsilon|z|+C_{\varepsilon} \xi(x)|z|^{p-1} \end{aligned}$$
(3.2)
and
$$\begin{aligned} \bigl|H(x,z)\bigr|\leq\frac{1}{2}\varepsilon|z|^{2}+ \frac{1}{p}C_{\varepsilon}\xi (x)|z|^{p}. \end{aligned}$$
(3.3)
It then follows that
$$\begin{aligned} \Phi_{\lambda}(z) \geq&\frac{1}{2}\|z\|^{2}- \frac{1}{2}\lambda \varepsilon\int_{\mathbb{R}^{N}}|z|^{2}- \frac{1}{p}\lambda C_{\varepsilon}\int_{\mathbb{R}^{N}} \xi(x)|z|^{p} \\ \geq&\frac{1}{2}\|z\|^{2}-\frac{1}{2}c\lambda\varepsilon \|z\|^{2}-\frac {1}{p}\lambda C_{\varepsilon}\int _{\mathbb{R}^{N}}\xi(x)|z|^{p}. \end{aligned}$$
By choosing \(\varepsilon=\frac{1}{2\lambda c}\), we obtain
$$\begin{aligned} \Phi_{\lambda}(z)\geq\frac{1}{4}\|z\|^{2}- \frac{1}{p}\lambda C_{\varepsilon}\int_{\mathbb{R}^{N}} \xi(x)|z|^{p}. \end{aligned}$$
By (3.1), we have
$$\begin{aligned} \Phi_{\lambda}(z)\geq\frac{1}{4}\|z\|^{2}-C_{1} \beta_{k}^{p}\|z\| ^{p}=\frac{1}{4} \biggl( \|z\|^{2}-\frac{C_{1}}{4}\beta_{k}^{p} \|z \|^{p} \biggr). \end{aligned}$$
If we set \(r_{k}=(\frac{C_{1}}{2}\beta_{k}^{p})^{\frac{1}{2-p}}\), then for every \(z\in E_{k}^{+}\) such that \(\|z\|=r_{k}\), we have
$$\begin{aligned} \Phi_{\lambda}(z)\geq\tilde{b}_{k}:=\frac{1}{8} \biggl( \frac{C_{1}}{2}\beta _{k}^{p} \biggr)^{\frac{2}{2-p}}= \frac{1}{8}r_{k}^{2}>0. \end{aligned}$$
Moreover, by (3.1) again, \(\beta_{k}\rightarrow0\) as \(k\rightarrow\infty\), we have \(\tilde{b}_{k}\rightarrow\infty\), and hence \(b_{k}(\lambda)\rightarrow\infty\) uniformly in λ as \(k\rightarrow\infty\). □
Lemma 3.4
Let (V), (H1)-(H4) be satisfied. There exists
\(\rho_{k}>0\)
independent of
\(\lambda\in[1,2]\)
such that
$$\begin{aligned} a_{k}(\lambda):=\sup_{z\in E_{k}^{-},\|z\|=\rho_{k}}\Phi_{\lambda }(z) \leq0. \end{aligned}$$
Proof
Since \(\Phi_{\lambda}\leq\Phi_{1}\) for all \(\lambda\in [1,2]\), it suffices to show the conclusion holds for \(\lambda=1\). Suppose to the contrary that there exists a sequence \(z_{n}\in E_{k}^{-}\) such that \(\Phi_{1}(z_{n})>0\) for all n and \(\|z_{n}\|\rightarrow\infty\) as \(n\rightarrow\infty\). Set \(w_{n}=\frac{z_{n}}{\|z_{n}\|}=w_{n}^{+}+w_{n}^{-}\), where \(w_{n}^{-}\in E^{-}\), \(w_{n}^{+}\in\bigoplus_{j=0}^{k}\mathbb {R}e_{j}\subset E^{+}\). Then \(1=\|w_{n}\|^{2}=\|w_{n}^{+}\|^{2}+\|w_{n}^{-}\|^{2}\) and
$$ 0<\frac{\Phi(z_{n})}{\|z_{n}\|^{2}}=\frac{1}{2} \bigl( \bigl\| w_{n}^{+} \bigr\| ^{2}-\bigl\| w_{n}^{-} \bigr\| ^{2} \bigr)-\int_{\mathbb{R}^{N}}\frac{H(x,z_{n})}{|z_{n}|^{2}}|w_{n}|^{2}. $$
(3.4)
By (H1), we have
$$ \bigl\| w_{n}^{-}\bigr\| ^{2}<\bigl\| w_{n}^{+} \bigr\| ^{2}=1-\bigl\| w_{n}^{-}\bigr\| ^{2}, $$
therefore
$$ 0\leq\bigl\| w_{n}^{-}\bigr\| ^{2}\leq \frac{1}{2} \quad\mbox{and}\quad \frac{1}{2}\leq \bigl\| w_{n}^{+} \bigr\| ^{2}\leq1. $$
Since \(\bigoplus_{j=0}^{k}\mathbb{R}e_{j}\) is finite dimensional, going to a subsequence if necessary, we may assume \(w_{n}^{+}\rightarrow w^{+}\neq0\), \(w_{n}^{-}\rightharpoonup w^{-}\), and \(w_{n}(x)\rightarrow w(x)\) a.e. on \(\mathbb{R}^{N}\). Hence \(w\neq0\) and \(|z_{n}|=\|z_{n}\||w_{n}|\rightarrow\infty\). By (H2) and Fatou’s lemma, we have
$$\begin{aligned} \int_{\mathbb{R}^{N}}\frac {H(x,z_{n})}{|z_{n}|^{2}}|w_{n}|^{2} \rightarrow\infty, \quad\mbox{as } n\rightarrow\infty. \end{aligned}$$
By (3.4), we get a contradiction. □
Combining Lemmas 3.1, 3.3, 3.4, and Proposition 2.1, we have the following lemma.
Lemma 3.5
Let (V), (H1)-(H4) be satisfied, for almost every
\(\lambda\in[1,2]\), there exists a sequence
\(\{ z_{k}^{n}(\lambda)\}\)
such that
$$\begin{aligned} \sup_{n}\bigl\| z_{k}^{n}(\lambda)\bigr\| <\infty,\qquad \Phi'_{\lambda} \bigl(z_{k}^{n}(\lambda ) \bigr)\rightarrow0,\qquad \Phi_{\lambda} \bigl(z_{k}^{n}( \lambda) \bigr)\rightarrow c_{k}(\lambda), \quad\textit{as } n \rightarrow\infty. \end{aligned}$$
Lemma 3.6
Let (V), (H1)-(H4) be satisfied, for almost every
\(\lambda\in[1,2]\), there exists some
\(\{z_{k}(\lambda)\}\)
such that
$$\begin{aligned} \Phi'_{\lambda} \bigl(z_{k}(\lambda) \bigr)=0, \qquad \Phi_{\lambda} \bigl(z_{k}(\lambda ) \bigr)=c_{k}( \lambda). \end{aligned}$$
Proof
Let \(\{z_{k}^{n}(\lambda)\}\) be the sequence obtained in Lemma 3.5. Here for notational simplicity, we write \(z_{k}^{n}(\lambda )=z_{k}^{n}\). Since \(\{z_{k}^{n}\}\) is bounded, without restriction we can assume that \(\{z_{k}^{n}\}\) is either vanishing or nonvanishing. If \(\{ z_{k}^{n}\}\) is vanishing, i.e.,
$$\begin{aligned} \lim_{n\rightarrow\infty}\sup_{y\in\mathbb{R^{N}}}\int _{B(y,R)}\bigl|z_{k}^{n}\bigr|^{2}=0,\quad \forall R>0, \end{aligned}$$
by Lions’ concentration compactness principle in [40] (also see [38]), \(z_{k}^{n}\rightarrow0\) in \(L^{p}\) for all \(p\in(2,2^{*})\). By using the Hölder inequality and (3.2), we have
$$\begin{aligned} \biggl\vert \int_{\mathbb{R}^{N}}H_{z} \bigl(x,z_{k}^{n} \bigr)\cdot z_{k}^{n} \biggr\vert \leq \varepsilon\int_{\mathbb{R}^{N}}\bigl|z_{k}^{n}\bigr|^{2}+C_{\varepsilon} \int_{\mathbb{R}^{N}}\bigl|z_{k}^{n}\bigr|^{p} \rightarrow0 \end{aligned}$$
as \(n\rightarrow\infty\). Therefore, we have
$$\begin{aligned} 0< b_{k}(\lambda)\leq\Phi_{\lambda} \bigl(z_{k}^{n} \bigr)\leq\bigl\| \bigl(z_{k}^{n} \bigr)^{+}\bigr\| ^{2}-\bigl\| \bigl(z_{k}^{n} \bigr)^{-} \bigr\| ^{2} =\Phi'_{\lambda} \bigl(z_{k}^{n} \bigr)z_{k}^{n}+\lambda\int_{\mathbb {R}^{N}}H_{z} \bigl(x,z_{k}^{n} \bigr)\cdot z_{k}^{n} \rightarrow0 \end{aligned}$$
as \(n\rightarrow\infty\). This implies that \(\{z_{k}^{n}\}\) is nonvanishing, i.e., there exist \(R, \delta>0\), and a sequence \(\{y_{n}\}\subset\mathbb{R}^{N}\) such that
$$\begin{aligned} \liminf_{n\rightarrow\infty}\int_{B(y_{n}, R)}\bigl|z_{k}^{n}\bigr|^{2} \geq\delta. \end{aligned}$$
Taking a subsequence if necessary we may suppose that
$$\begin{aligned} \int_{B(y_{n}, R)}\bigl|z_{k}^{n}\bigr|^{2}\geq \frac{\delta}{2} \end{aligned}$$
for all \(n\in\mathbb{N}\). Since \(\Phi_{\lambda}\) is invariant under translation, by a standard argument, we have
$$\begin{aligned} \int_{B(0, R+\frac{\sqrt{N}}{2})}\bigl|w_{k}^{n}\bigr|^{2} \geq\frac{\delta}{2}, \end{aligned}$$
(3.5)
where \(w_{k}^{n}=z_{k}^{n}(\cdot-k_{n})\). Observe that \(\|z_{k}^{n}\|=\| w_{k}^{n}\|\). Hence \(\{w_{k}^{n}\}\) is bounded, up to a subsequence we may assume
$$\begin{aligned} w_{k}^{n}\rightharpoonup z_{k} \quad \mbox{in } E,\qquad w_{k}^{n}\rightarrow z_{k} \quad\mbox{in } L^{2}_{\mathrm{loc}} \quad\mbox{and} \quad w_{k}^{n} \rightarrow z_{k} \quad\mbox{a.e. in } \mathbb{R}^{N}. \end{aligned}$$
(3.6)
In view of (3.5), we know \(z_{k}\neq0\). By Lemma 3.1, it is easy to see that \(\Phi '_{\lambda}(z_{k})=0\).
Now, let us show that
$$\begin{aligned} \int_{\mathbb{R}^{N}}H \bigl(x,w_{k}^{n}-z_{k} \bigr)\rightarrow0, \quad\mbox{as } n\rightarrow\infty. \end{aligned}$$
(3.7)
Indeed, in virtue of assumption (H1), we know that for any \(\varepsilon>0\), there exists \(R>0\) such that
$$\begin{aligned} \biggl(\int_{B^{c}_{R}(0)}\bigl|\xi(x)\bigr|^{r} \biggr)^{\frac{1}{r}}<\varepsilon. \end{aligned}$$
Thus, by the Hölder inequality and the Sobolev embedding theorem, we have
$$\begin{aligned} &\int_{\mathbb{R}^{N}}H \bigl(x,w_{k}^{n}-z_{k} \bigr)\\ &\quad\leq\int_{\mathbb {R}^{N}}\varepsilon\bigl|w_{k}^{n}-z_{k}\bigr|^{2}+C_{\varepsilon} \xi (x)\bigl|w_{k}^{n}-z_{k}\bigr|^{p} \\ &\quad=\int_{\mathbb{R}^{N}}\varepsilon\bigl|w_{k}^{n}-z_{k}\bigr|^{2}+C_{\varepsilon } \int_{B_{R}(0)}\xi(x)\bigl|w_{k}^{n}-z_{k}\bigr|^{p} +C_{\varepsilon}\int_{B^{c}_{R}(0)}\xi(x)\bigl|w_{k}^{n}-z_{k}\bigr|^{p} \\ &\quad\leq\varepsilon\bigl|w_{k}^{n}-z_{k}\bigr|_{2}^{2}+C_{\varepsilon}| \xi|_{\infty }\int_{B_{R}(0)}\bigl|w_{k}^{n}-z_{k}\bigr|^{p} +C_{\varepsilon} \biggl(\int_{B^{c}_{R}(0)}\bigl|\xi(x)\bigr|^{r} \biggr)^{\frac {1}{r}}\bigl|w_{k}^{n}-z_{k}\bigr|_{2^{*}}^{p} \\ &\quad\leq\varepsilon(C_{2}+C_{3}+C_{4}). \end{aligned}$$
Hence (3.7) holds. Similarly, we also have
$$\begin{aligned} \int_{\mathbb {R}^{N}}H_{z} \bigl(x,w_{k}^{n}-z_{k} \bigr) \bigl(w_{k}^{n}-z_{k} \bigr)\rightarrow0, \quad \mbox{as } n\rightarrow\infty. \end{aligned}$$
(3.8)
Moreover, by (3.7) and (3.8), it is easy to show that
$$\begin{aligned} \int_{\mathbb{R}^{N}}H \bigl(x,w_{k}^{n} \bigr)\rightarrow\int_{\mathbb {R}^{N}}H(x,z_{k}), \quad \mbox{as } n \rightarrow\infty \end{aligned}$$
(3.9)
and
$$\begin{aligned} \int_{\mathbb{R}^{N}}H_{z} \bigl(x,w_{k}^{n} \bigr)w_{k}^{n} \rightarrow\int_{\mathbb {R}^{N}}H_{z}(x,z_{k})z_{k}, \quad\mbox{as } n\rightarrow\infty. \end{aligned}$$
(3.10)
Therefore, by (3.9) and (3.10) we obtain
$$\begin{aligned} \Phi_{\lambda} \bigl(w_{k}^{n} \bigr) =& \frac{1}{2} \bigl\langle \Phi'_{\lambda } \bigl(w_{k}^{n} \bigr),w_{k}^{n} \bigr\rangle +\frac{\lambda}{2}\int_{\mathbb {R}^{N}}H_{z} \bigl(x,w_{k}^{n} \bigr)w_{k}^{n} - \lambda\int_{\mathbb{R}^{N}}H \bigl(x,w_{k}^{n} \bigr) \\ \rightarrow&\frac{1}{2} \bigl\langle \Phi'_{\lambda}(z_{k}),z_{k} \bigr\rangle +\frac {\lambda}{2}\int_{\mathbb{R}^{N}}H_{z}(x,z_{k})z_{k} -\lambda\int_{\mathbb{R}^{N}}H(x,z_{k}) \\ =&\Phi_{\lambda}(z_{k}), \end{aligned}$$
which implies \(\Phi_{\lambda}(z_{k})=c_{k}(\lambda)\). The proof is complete. □
By the preceding lemma, we directly obtain the following lemma.
Lemma 3.7
Let (V), (H1)-(H4) be satisfied, there exist a sequence
\(\{\lambda_{n}\}\)
and a sequence
\(\{z_{k}(\lambda _{n})\}\)
such that
$$\begin{aligned} \lambda_{n}\rightarrow1, \quad\Phi_{\lambda_{n}} \bigl(z_{k}( \lambda _{n}) \bigr)=c_{k}(\lambda_{n}),\qquad \Phi'_{\lambda_{n}} \bigl(z_{k}(\lambda_{n}) \bigr)=0. \end{aligned}$$
Lemma 3.8
Let (H4) be satisfied, then
$$\begin{aligned} \int_{\mathbb{R}^{N}} \biggl(H(x,z)-H(x,r\phi)+r^{2} \bigl(H_{z}(x,z),\phi \bigr)-\frac{1+r^{2}}{2} \bigl(H_{z}(x,z),z \bigr) \biggr)\leq C, \end{aligned}$$
where
\(z\in E\), \(\phi\in E^{+}\), \(0\leq r\leq1\), and the constant
\(C:=\int_{\mathbb{R}^{N}}|W(x)|\)
does not depend on
z, ϕ, r.
Proof
This follows from (H4) if we take \(z=z\) and \(w=r\phi-z\). □
Lemma 3.9
Let (V), (H1)-(H4) be satisfied. The sequences
\(\{z_{k}(\lambda_{n})\}\)
given in Lemma
3.7
are bounded.
Proof
For notational simplicity, we write \(z_{k}^{n}:=z_{k}(\lambda_{n})\). First, we claim that there is a constant C independent of \(z_{k}^{n}\) and \(\lambda_{n}\) such that
$$\begin{aligned} \Phi_{\lambda_{n}} \bigl(r \bigl(z_{k}^{n} \bigr)^{+} \bigr)-\Phi_{\lambda_{n}} \bigl(z_{k}^{n} \bigr)\leq C,\quad \forall r\in[0,1]. \end{aligned}$$
(3.11)
Since
$$\begin{aligned} \bigl\langle \Phi'_{\lambda_{n}} \bigl(z_{k}^{n} \bigr),\varphi \bigr\rangle = \bigl( \bigl(z_{k}^{n} \bigr)^{+},\varphi^{+} \bigr)-\lambda_{n} \bigl( \bigl(z_{k}^{n} \bigr)^{-},\varphi^{-} \bigr) -\lambda_{n}\int_{\mathbb{R}^{N}}H_{z} \bigl(x,z_{k}^{n} \bigr)\varphi=0,\quad \forall \varphi\in E, \end{aligned}$$
it follows from the definition of \(\Phi_{\lambda}\) that
$$\begin{aligned} \Phi_{\lambda_{n}} \bigl(r \bigl(z_{k}^{n} \bigr)^{+} \bigr)-\Phi_{\lambda _{n}} \bigl(z_{k}^{n} \bigr) =&\frac{1}{2} \bigl(r^{2}-1 \bigr)\bigl\| \bigl(z_{k}^{n} \bigr)^{+}\bigr\| ^{2}+\frac {1}{2}\lambda_{n}\bigl\| \bigl(z_{k}^{n} \bigr)^{-}\bigr\| ^{2} \\ &{}+\lambda_{n}\int_{\mathbb{R}^{N}}H \bigl(x,z_{k}^{n} \bigr)-\lambda_{n}\int_{\mathbb{R}^{N}}H \bigl(x,r \bigl(z_{k}^{n} \bigr)^{+} \bigr) \\ &{}+\frac{1}{2} \bigl( \bigl(z_{k}^{n} \bigr)^{+},\varphi^{+} \bigr)-\frac{1}{2}\lambda _{n} \bigl( \bigl(z_{k}^{n} \bigr)^{-}, \varphi^{-} \bigr) -\frac{1}{2} \lambda_{n}\int_{\mathbb{R}^{N}}H_{z} \bigl(x,z_{k}^{n} \bigr)\varphi. \end{aligned}$$
Take
$$\begin{aligned} \varphi = \bigl(r^{2}+1 \bigr) \bigl(z_{k}^{n} \bigr)^{-}- \bigl(r^{2}-1 \bigr) \bigl(z_{k}^{n} \bigr)^{+}= \bigl(r^{2}+1 \bigr)z_{k}^{n}-2r^{2} \bigl(z_{k}^{n} \bigr)^{+}, \end{aligned}$$
which together with Lemma 3.8 implies that
$$\begin{aligned} \Phi_{\lambda_{n}} \bigl(r \bigl(z_{k}^{n} \bigr)^{+} \bigr)-\Phi_{\lambda _{n}} \bigl(z_{k}^{n} \bigr) =&-\frac{r^{2}}{2}\lambda_{n}\bigl\| \bigl(z_{k}^{n} \bigr)^{-}\bigr\| ^{2} +\lambda_{n}\int _{\mathbb{R}^{N}} \bigl(H \bigl(x,z_{k}^{n} \bigr)-H \bigl(x,r \bigl(z_{k}^{n} \bigr)^{+} \bigr) \bigr) \\ &{}+\lambda_{n}\int_{\mathbb{R}^{N}} \biggl(r^{2} \bigl(H_{z} \bigl(x,z_{k}^{n} \bigr), \bigl(z_{k}^{n} \bigr)^{+} \bigr)- \frac {1+r^{2}}{2} \bigl(H_{z} \bigl(x,z_{k}^{n} \bigr),z_{k}^{n} \bigr) \biggr) \\ \leq& C. \end{aligned}$$
Hence (3.11) holds.
To show the boundness of \(\{z_{k}^{n}\}\), we argue by contradiction that \(\|z_{k}^{n}\|\rightarrow\infty\) as \(n\rightarrow\infty\). Since \(\Phi_{\lambda}\geq0\), we know \(\|(z_{k}^{n})^{+}\|^{2}\geq\| (z_{k}^{n})^{-}\|^{2}\). Let \(w_{k}^{n}=z_{k}^{n}/\|z_{k}^{n}\|\), then \(w_{k}^{n}=(w_{k}^{n})^{-}+(w_{k}^{n})^{+}\) and \(\frac{1}{2}\leq\| (w_{k}^{n})^{+}\|^{2}\leq1\). Passing to a subsequence, \((w_{k}^{n})^{+}\rightharpoonup w_{k}^{+}\) in E, \((w_{k}^{n})^{+}\rightarrow w_{k}^{+}\) in \(L^{2}_{\mathrm{loc}}\), and \((w_{k}^{n})^{+}\rightarrow w_{k}^{+}\) a.e. on \(\mathbb{R}^{N}\). We see that either \(\{(w_{k}^{n})^{+}\}\) is vanishing, i.e.,
$$\begin{aligned} \lim_{n\rightarrow\infty}\sup_{y\in\mathbb{R}^{N}}\int _{B(y,R)}\bigl| \bigl(w_{k}^{n} \bigr)^{+}\bigr|^{2}=0, \quad\forall R>0, \end{aligned}$$
or nonvanishing, i.e., there exist \(R, \delta>0\) and a sequence \(\{y_{n}\}\subset\mathbb{R}^{N}\) such that
$$\begin{aligned} \liminf_{n\rightarrow\infty}\int_{B(y_{n},R)}\bigl| \bigl(w_{k}^{n} \bigr)^{+}\bigr|^{2}\geq \delta. \end{aligned}$$
If \((w_{k}^{n})^{+}\) is vanishing, Lions’ concentration compactness principle implies \((w_{k}^{n})^{+}\rightarrow0\) in \(L^{p}\) for \(p\in (2,2^{*})\). Lebesgue’s dominated convergence theorem and (3.3) imply that
$$\begin{aligned} \int_{\mathbb{R}^{N}}H \bigl(x,s \bigl(w_{k}^{n} \bigr)^{+} \bigr)\rightarrow0, \quad\mbox{for any } s\in\mathbb{R}. \end{aligned}$$
(3.12)
Let \(r_{n}=\frac{s}{\|z_{k}^{n}\|}\rightarrow0\) as \(n\rightarrow\infty \). Therefore, by (3.11) and (3.12), we have
$$\begin{aligned} c_{k}(\lambda_{n})+C =&C+\Phi_{\lambda_{n}} \bigl(z_{k}^{n} \bigr)\geq\Phi_{\lambda _{n}} \bigl(r_{n} \bigl(z_{k}^{n} \bigr)^{+} \bigr) \\ =&\frac{s^{2}}{2}\bigl\| \bigl(w_{k}^{n} \bigr)^{+}\bigr\| ^{2}-\lambda_{n}\int _{\mathbb {R}^{N}}H \bigl(x,s \bigl(w_{k}^{n} \bigr)^{+} \bigr) \\ \geq&\frac{s^{2}}{4}+o(1), \end{aligned}$$
which implies a contradiction if s is large enough. Hence \((w_{k}^{n})^{+}\) is nonvanishing, i.e., there exist \(R, \delta >0\) and a sequence \(\{y_{n}\}\subset\mathbb{R}^{N}\) such that
$$\begin{aligned} \liminf_{n\rightarrow\infty}\int_{B(y_{n},R)}\bigl| \bigl(w_{k}^{n} \bigr)^{+}\bigr|^{2}\geq \delta. \end{aligned}$$
Going to a subsequence if necessary, we have
$$\begin{aligned} \int_{B(y_{n},R)}\bigl| \bigl(w_{k}^{n} \bigr)^{+}\bigr|^{2}\geq\frac{\delta}{2}, \quad\mbox{for all } n \in \mathbb{N}. \end{aligned}$$
By a standard argument, we have
$$\begin{aligned} \int_{B(0, R+\frac{\sqrt{N}}{2})}\bigl| \bigl(u_{k}^{n} \bigr)^{+}\bigr|^{2}\geq\frac{\delta}{2}, \end{aligned}$$
(3.13)
where \((u_{k}^{n})^{+}=(w_{k}^{n})^{+}(\cdot-k_{n})\). Thus (3.13) implies that \(u_{k}^{+}\neq0\) and \(|z_{k}^{n}|\rightarrow\infty\). It follows from (H2) and Fatou’s lemma that
$$\begin{aligned} \int_{\mathbb{R}^{N}}\frac {H(x,z_{k}^{n})}{|z_{k}^{n}|^{2}}\bigl|w_{k}^{n}\bigr|^{2} \rightarrow\infty, \quad\mbox{as } n\rightarrow\infty, \end{aligned}$$
and
$$\begin{aligned} 0\leq\frac{\Phi_{\lambda_{n}}(z_{k}^{n})}{\|z_{k}^{n}\|^{2}}=\frac {1}{2}\bigl\| \bigl(w_{k}^{n} \bigr)^{+}\bigr\| ^{2}-\lambda_{n} \biggl( \frac{1}{2}\bigl\| \bigl(w_{k}^{n} \bigr)^{-} \bigr\| ^{2} +\int_{\mathbb{R}^{N}}\frac {H(x,z_{k}^{n})}{|z_{k}^{n}|^{2}}\bigl|w_{k}^{n}\bigr|^{2} \biggr)\rightarrow -\infty, \end{aligned}$$
as \(n\rightarrow\infty\), a contradiction. Therefore, \(\{z_{k}^{n}\}\) is bounded. □
Proof of Theorem 1.1
Clearly, the condition (F2) holds and \(\Phi_{\lambda}(-z)=\Phi_{\lambda}(z)\) for all \((\lambda,z)\in [1,2]\times E\). Lemma 3.1 implies that the conditions (F1) and (F3) hold. Lemmas 3.3 and 3.4 show that \(\Phi_{\lambda}\) possesses the geometric property of Proposition 2.1. Therefore, we can obtain the sequence \(\{z_{k}^{n}\} \) given in Lemma 3.7 by Proposition 2.1. From the relations
$$\begin{aligned} \Phi_{1} \bigl(z_{k}^{n} \bigr)= \Phi_{\lambda_{n}} \bigl(z_{k}^{n} \bigr)+( \lambda_{n}-1) \biggl(\frac{1}{2}\bigl\| \bigl(z_{k}^{n} \bigr)^{-}\bigr\| ^{2}+\int_{\mathbb {R}^{N}}H \bigl(x,z_{k}^{n} \bigr) \biggr) \end{aligned}$$
and
$$\begin{aligned} \bigl\langle \Phi'_{1} \bigl(z_{k}^{n} \bigr)-\Phi'_{\lambda_{n}} \bigl(z_{k}^{n} \bigr),\varphi \bigr\rangle = (\lambda_{n}-1) \biggl( \bigl( \bigl(z_{k}^{n} \bigr)^{-},\varphi \bigr)+\int _{\mathbb {R}^{N}}H_{z} \bigl(x,z_{k}^{n} \bigr)\varphi \biggr), \quad\forall\varphi\in E, \end{aligned}$$
we deduce, since the sequence \(\{c_{k}(\lambda_{n})\}\) is nondecreasing and bounded from above, that \(\{z_{k}^{n}\}\) is a \((PS)\)-sequence for \(\Phi_{1}\) at level \(c_{k}(1)\). By repeating the argument of Lemma 3.6 we obtain the existence of \(z_{k}\in E\) such that \(\Phi'_{1}(z_{k})=0\) and \(\Phi_{1}(z_{k})\geq b_{k}\), moreover, \(b_{k}\rightarrow\infty\) as \(k\rightarrow\infty\). This ends the proof of Theorem 1.1. □