Let
$$S= \bigl\{ \bigl\{ u(n)\bigr\} _{n\in \mathbb {Z}} : u(n)\in{{ \mathbb {R}}}^{\mathcal{N}}, n \in \mathbb {Z}\bigr\} . $$
As usual, for \(1\le q<\infty\), set
$$l^{q}\bigl(\mathbb {Z}, {\mathbb {R}}^{\mathcal{N}}\bigr)= \biggl\{ u\in S : \sum _{n\in \mathbb {Z}}\bigl|u(n)\bigr|^{q}< \infty \biggr\} $$
and
$$l^{\infty}\bigl(\mathbb {Z}, {\mathbb {R}}^{\mathcal{N}}\bigr)= \Bigl\{ u\in S : \sup _{n\in \mathbb {Z}}\bigl|u(n)\bigr|< \infty \Bigr\} , $$
and their norms are defined by
$$\begin{aligned}& \|u\|_{q}= \biggl(\sum_{n\in \mathbb {Z}}\bigl|u(n)\bigr|^{q} \biggr)^{1/q}, \quad\forall\ u\in l^{q}\bigl(\mathbb {Z}, {\mathbb {R}}^{\mathcal{N}} \bigr); \\& \|u\|_{\infty}=\sup_{n\in \mathbb {Z}}\bigl|u(n)\bigr|, \quad\forall u\in l^{\infty}\bigl(\mathbb {Z}, {\mathbb {R}}^{\mathcal{N}}\bigr), \end{aligned}$$
respectively. In particular, \(l^{2}(\mathbb {Z}, {\mathbb {R}}^{\mathcal{N}})\) is a Hilbert space with the following inner product:
$$(u, v)_{l^{2}}=\sum_{n\in \mathbb {Z}}u(n)\cdot v(n),\quad \forall u, v\in l^{2}\bigl(\mathbb {Z}, {\mathbb {R}}^{\mathcal{N}}\bigr). $$
Let \(\{\mathcal{E}(\lambda): -a_{0} \le\lambda\le b_{0}\}\) and \(|\mathcal{A}|\) be the spectral family and the absolute value of \(\mathcal{A}\), respectively, and \(|\mathcal {A}|^{1/2}\) be the square root of \(|\mathcal{A}|\). Set \(\mathcal{U}=\mathrm{id}-\mathcal{E}(0)-\mathcal{E}(0-)\). Then \(\mathcal {U}\) commutes with \(\mathcal{A}\), \(|\mathcal{A}|\) and \(|\mathcal{A}|^{1/2}\), and \(\mathcal{A} = \mathcal{U}|\mathcal{A}|\) is the polar decomposition of \(\mathcal{A}\) (see [19, Theorem 4.3.3]).
As in [20], let \(E=l^{2}(\mathbb {Z}, {\mathbb {R}}^{\mathcal{N}})\) and
$$E^{-}=\mathcal{E}(0)E,\qquad E^{+}=\bigl[\mathrm{id}-\mathcal{E}(0) \bigr]E. $$
For any \(u\in E\), it is easy to see that
$$ u^{-}:=\mathcal{E}(0)u\in E^{-},\qquad u^{+}:=\bigl[\mathrm{id}-\mathcal {E}(0)\bigr]u\in E^{+},\qquad u=u^{-}+u^{+}, $$
(2.1)
and
$$ \mathcal{A}u^{-}=-|\mathcal{A}|u^{-},\qquad \mathcal {A}u^{+}=|\mathcal{A}|u^{+}, \quad\forall u\in E. $$
(2.2)
Let
$$ (u, v)= \bigl(|\mathcal{A}|^{1/2}u, |\mathcal{A}|^{1/2}v \bigr)_{l^{2}}, \quad\forall u, v\in E. $$
(2.3)
Then E is a Hilbert space with the above inner product, and the corresponding norm is
$$ \|u\|=\bigl\| |\mathcal{A}|^{1/2}u\bigr\| _{2}, \quad\forall u \in E. $$
(2.4)
By virtue of (2.1)-(2.4), one has the decomposition \(E=E^{-}\oplus E^{+}\) orthogonal with respect to both \((\cdot, \cdot)_{l^{2}}\) and \((\cdot, \cdot)\). Moreover,
$$ -\underline{\Lambda}\bigl\| u^{-}\bigr\| _{2}^{2} \le\bigl\| u^{-}\bigr\| ^{2}\le a_{0}\bigl\| u^{-} \bigr\| _{2}^{2},\qquad \bar{\Lambda}\bigl\| u^{+} \bigr\| _{2}^{2}\le\bigl\| u^{+}\bigr\| ^{2}\le b_{0}\bigl\| u^{+}\bigr\| _{2}^{2}, \quad \forall u\in E, $$
(2.5)
and
$$ \Lambda_{0}\|u\|_{2}^{2}\le\|u \|^{2}\le\max\{a_{0}, b_{0}\}\|u \|_{2}^{2}, \quad \forall u\in E. $$
(2.6)
Let X be a real Hilbert space with \(X=X^{-}\oplus X^{+}\) and \(X^{-}\bot X^{+}\). For a functional \(\varphi \in C^{1}(X, \mathbb {R})\), φ is said to be weakly sequentially lower semi-continuous if for any \(u_{k}\rightharpoonup u\) in X one has \(\varphi(u)\le\liminf_{n\to\infty}\varphi(u_{k})\), and \(\varphi'\) is said to be weakly sequentially continuous if \(\lim_{k\to\infty}\langle\varphi'(u_{k}), v\rangle= \langle\varphi'(u), v\rangle\) for each \(v\in X\).
Lemma 2.1
([21, Theorem 2.1])
Let
X
be a real Hilbert space with
\(X=X^{-}\oplus X^{+}\)
and
\(X^{-}\bot X^{+}\), and let
\(\varphi\in C^{1}(X, \mathbb {R})\)
of the form
$$\varphi(u)=\frac{1}{2} \bigl(\bigl\| u^{+}\bigr\| ^{2}- \bigl\| u^{-}\bigr\| ^{2} \bigr)-\psi (u),\quad u=u^{-}+u^{+} \in X^{-}\oplus X^{+}. $$
Suppose that the following assumptions are satisfied:
-
(LS1)
\(\psi\in C^{1}(X, \mathbb {R})\)
is bounded from below and weakly sequentially lower semi-continuous;
-
(LS2)
\(\psi'\)
is weakly sequentially continuous;
-
(LS3)
there exist
\(r>\rho>0\)
and
\(e\in X^{+}\)
with
\(\|e\|=1\)
such that
$$\kappa:=\inf\varphi\bigl(S^{+}_{\rho}\bigr) > \sup\varphi( \partial Q), $$
where
$$S^{+}_{\rho}= \bigl\{ u\in X^{+} : \|u\|=\rho \bigr\} ,\qquad Q= \bigl\{ se+v : v\in X^{-}, s\ge0, \|se+v\|\le r \bigr\} . $$
Then for some
\(c\ge\kappa\), there exists a sequence
\(\{u_{n}\} \subset X\)
satisfying
$$ \varphi(u_{n})\rightarrow c,\quad \bigl\| \varphi'(u_{n}) \bigr\| \bigl(1+\|u_{n}\| \bigr)\rightarrow0. $$
(2.7)
Such a sequence is called a Cerami sequence on the level
c, or a
\((C)_{c}\)
sequence.
Now we define a functional Φ on E by
$$ \Phi(u)=\frac{1}{2}\sum_{n\in \mathbb {Z}} \bigl[p(n+1)\triangle u\cdot \triangle u+L(n)u\cdot u) \bigr]-\sum _{n\in \mathbb {Z}}W(n, u). $$
(2.8)
For any \(u\in E\), there exists an \(n_{0}\in \mathbb {N}\) such that \(|u(n)|\le1\) for \(|n|\ge n_{0}\). Hence, under assumptions (PL), (W1), and (W2), the functional Φ is of class \(C^{1}(E, \mathbb {R})\). Moreover,
$$\begin{aligned} \bigl\langle \Phi'(u), v \bigr\rangle = \sum _{n\in \mathbb {Z}} \bigl[p(n+1)\triangle u\cdot\triangle v+L(n)u\cdot v \bigr] -\sum_{n\in \mathbb {Z}}\nabla W(n, u)\cdot v, \quad\forall u, v \in E. \end{aligned}$$
(2.9)
By virtue of (2.1), (2.2), (2.3), and (2.4), one has
$$ \Phi(u)=\frac{1}{2} \bigl(\bigl\| u^{+} \bigr\| ^{2}-\bigl\| u^{-}\bigr\| ^{2} \bigr)-\sum _{n\in \mathbb {Z}}W(n, u), \quad\forall u\in E, $$
(2.10)
and
$$ \bigl\langle \Phi'(u), v \bigr\rangle = \bigl(u^{+}, v\bigr)-\bigl(u^{-}, v\bigr)-\sum _{n\in \mathbb {Z}}\nabla W(n, u)\cdot v,\quad \forall u, v\in E. $$
(2.11)
Furthermore, the critical points of Φ in E are solutions of system (1.1) with \(u(\pm\infty)=0\); see [6, 10].
Let
$$ \Psi(u)=\sum_{n\in \mathbb {Z}}W(n, u),\quad \forall u \in E. $$
(2.12)
Then, by standard arguments, we can prove the following two lemmas.
Lemma 2.2
Suppose that (PL), (W1), and (W2) are satisfied. Then Ψ is nonnegative, weakly sequentially lower semi-continuous, and
\(\Psi'\)
is weakly sequentially continuous.
Lemma 2.3
Suppose that (PL), (W1), and (W2) are satisfied. Then there is a
\(\rho>0\)
such that
\(\kappa:=\inf\Phi(S_{\rho}^{+})>0\), where
\(S_{\rho }^{+}=\partial B_{\rho}\cap E^{+}\).
Let \(m_{0}:=\inf_{n\in \mathbb {Z}, |x|=1}M(n)x\cdot x\). Then (W3) implies that \(m_{0}> \bar{\Lambda}\). Since \(\sigma(\mathcal{A})\) is a union of closed intervals, we can choose \(e\in[\mathcal {E}(m_{1})-\mathcal{E}(\bar{\Lambda})]E\subseteq E^{+}\), where \(\bar{\Lambda}< m_{1}<m_{0}\). Thus,
$$ \bar{\Lambda}\|e\|_{2}^{2}\le\|e \|^{2}\le m_{1}\|e\|_{2}^{2}< m_{0}\|e\|_{2}^{2}\le \sum _{n\in \mathbb {Z}}M(n)e\cdot e. $$
(2.13)
Lemma 2.4
Suppose that (PL), (W1), (W2), and (W3) are satisfied. Then there is a
\(r_{0}>0\)
such that
\(\sup\Phi(\partial Q)\le0\), where
$$ Q= \bigl\{ w+se : w\in E^{-}, s\ge0, \|w+se\|\le r_{0} \bigr\} . $$
(2.14)
Proof
Obviously, \(\Phi(w)\le0\) for \(w\in E^{-}\). It is sufficient to show that \(\Phi(w+te)\le0\) for \(t\ge0\), \(w\in E^{-}\) and \(\|w+te\|\ge r\) for large \(r>0\). Arguing indirectly, assume that for some sequence \(\{w_{k}+t_{k}e\}\subset E^{-}\oplus \mathbb {R}^{+} e\) with \(\|w_{k}+t_{k}e\| \rightarrow\infty\), \(\Phi(w_{k}+t_{k}e)\ge0\) for all \(k\in \mathbb {N}\). Set \(v_{k}=(w_{k}+t_{k}e)/\|w_{k}+t_{k}e\|=v_{k}^{-}+s_{k}e\), then \(\| v_{k}^{-}+s_{k}e\|=1\). Passing to a subsequence, we may assume that \(v_{k}\rightharpoonup v\) in E, then \(v_{k}(n)\rightarrow v(n)\) for all \(n\in \mathbb {Z}\), \(v_{k}^{-}\rightharpoonup v^{-}\) in E, \(s_{k}\rightarrow s\), and
$$ \frac{\Phi(w_{k}+t_{k}e)}{\|w_{k}+t_{k}e\|^{2}} =\frac{s_{k}^{2}}{2}\|e\|^{2}- \frac{1}{2}\bigl\| v_{k}^{-}\bigr\| ^{2} -\sum _{n\in \mathbb {Z}}\frac{W(n, w_{k}+t_{k}e)}{\|w_{k}+t_{k}e\|^{2}}\ge0. $$
(2.15)
Clearly, (2.15) yields \(s>0\). By virtue of (2.13), there exists a finite set \(\Pi\subset \mathbb {Z}\) such that
$$ s^{2}\|e\|^{2}-\bigl\| v^{-} \bigr\| ^{2}-\sum_{n\in\Pi}M(n) \bigl(se+v^{-} \bigr)\cdot\bigl(se+v^{-}\bigr)< 0. $$
(2.16)
From (W3) and (2.15), one has
$$\begin{aligned} 0 \le& \frac{s_{k}^{2}}{2}\|e\|^{2}-\frac{1}{2} \bigl\| v_{k}^{-}\bigr\| ^{2}-\sum _{n\in \Pi}\frac{W(n, w_{k}+t_{k}e)}{\|w_{k}+t_{k}e\|^{2}} \\ = & \frac{s_{k}^{2}}{2}\|e\|^{2}-\frac{1}{2} \bigl\| v_{k}^{-}\bigr\| ^{2}-\frac {1}{2}\sum _{n\in\Pi}M(n)v_{k}\cdot v_{k} -\sum _{n\in\Pi}\frac{W_{\infty}(n, w_{k}+t_{k}e)}{\|w_{k}+t_{k}e\|^{2}}. \end{aligned}$$
Clearly, \(|W_{\infty}(n, x)|\le c_{0}|x|^{2}\) for some \(c_{0}>0\) and \(W_{\infty}(n, x)/|x|^{2} \rightarrow0\) as \(|x|\to\infty\). Since \(v_{k} \rightharpoonup v\) in E, then \(v_{k}(n) \rightarrow v(n)\) for \(n\in\Pi\). Hence, one has
$$\begin{aligned} \sum_{n\in\Pi}\frac{W_{\infty}(n, w_{k}+t_{k}e)}{\|w_{k}+t_{k}e\|^{2}} = & \sum _{n\in\Pi}\frac{W_{\infty}(n, w_{k}+t_{k}e)}{|w_{k}+t_{k}e|^{2}}|v_{k}| \\ \le& c_{0}\sum_{n\in\Pi, |v(n)|=0}|v_{k}| +\sum_{n\in\Pi, |v(n)|\ne0}\frac{W_{\infty}(n, w_{k}+t_{k}e)}{|w_{k}+t_{k}e|^{2}}|v_{k}|\\ = & o(1). \end{aligned}$$
Hence
$$0\le s^{2}\|e\|^{2}-\bigl\| v^{-}\bigr\| ^{2}- \sum_{n\in\Pi}M(n) \bigl(s e+v^{-}\bigr) \cdot\bigl(s e+v^{-}\bigr), $$
a contradiction to (2.16). □
Lemma 2.5
Suppose that (PL), (W1), (W2) and (W3) are satisfied. Then there exist a constant
\(c>0\)
and a sequence
\(\{u_{k}\}\subset E\)
satisfying
$$ \Phi(u_{k})\rightarrow c,\quad \bigl\| \Phi'(u_{k}) \bigr\| \bigl(1+\|u_{k}\|\bigr)\rightarrow0. $$
(2.17)
Proof
Lemma 2.5 is a direct corollary of Lemmas 2.1, 2.2, 2.3, and 2.4. □
Lemma 2.6
Suppose that (PL), (W1), (W2), (W3), and (W4) are satisfied. Then any sequence
\(\{u_{k}\}\subset E\)
satisfying (2.17) is bounded in
E.
Proof
In view of (2.17), there exists a constant \(C_{0}>0\) such that
$$ C_{0}\ge\Phi(u_{k})-\frac{1}{2}\bigl\langle \Phi'(u_{k}), u_{k}\bigr\rangle =\sum _{n\in \mathbb {Z}}\widetilde{W}(n, u_{k}). $$
(2.18)
To prove the boundedness of \(\{u_{k}\}\), arguing by contradiction, suppose that \(\|u_{k}\| \to\infty\). Let \(v_{k}=u_{k}/\|u_{k}\|\). Then \(\|v_{k}\|=1\). Passing to a subsequence, we may assume that \(v_{k}\rightharpoonup v\) in E, then \(v_{k}(n)\rightarrow v(n)\) for all \(n\in \mathbb {Z}\). Let
$$\Pi_{k}= \biggl\{ n\in \mathbb {Z}: \frac{|\nabla W(n, u_{k})|}{|u_{k}|}\le\Lambda _{0}-\delta_{0} \biggr\} . $$
Then by using \(\Lambda_{0}\|v_{k}\|_{2}^{2}\le\|v_{k}\|^{2}\), one has
$$\begin{aligned} \sum_{n\in\Pi_{k}} \frac{|\nabla W(n, u_{k})|}{|u_{k}|}|v_{k}|\bigl|v_{k}^{+}-v_{k}^{-}\bigr| \le& (\Lambda_{0}-\delta_{0})\sum _{n\in\Pi _{k}}|v_{k}|\bigl|v_{k}^{+}-v_{k}^{-}\bigr| \\ \le& (\Lambda_{0}-\delta_{0})\|v_{k} \|_{2}^{2}\le1-\frac{\delta _{0}}{\Lambda_{0}}. \end{aligned}$$
(2.19)
If \(\delta:=\limsup_{k\to\infty}\|v_{k}\|_{\infty}=0\), then it follows from (W3), (W4), and (2.18) that
$$\begin{aligned} &\sum_{n\in \mathbb {Z}\setminus\Pi_{k}}\frac{|\nabla W(n, u_{k})|}{|u_{k}|}|v_{k}|\bigl|v_{k}^{+}-v_{k}^{-}\bigr| \\ &\quad \le \|v_{k}\|_{\infty}\bigl\| v_{k}^{+}-v_{k}^{-} \bigr\| _{\infty}\sum_{n\in \mathbb {Z}\setminus\Pi_{k}}\frac{|\nabla W(n, u_{k})|}{|u_{k}|} \\ & \quad\le C_{1}\|v_{k}\|_{\infty}\sum _{n\in \mathbb {Z}\setminus\Pi_{k}}\widetilde {W}(n, u_{k}) \\ & \quad\le C_{2}\|v_{k}\|_{\infty} = o(1). \end{aligned}$$
(2.20)
From (2.10), (2.11), (2.19), and (2.20), one gets
$$\begin{aligned} 1 +o(1) = & \frac{\|u_{k}\|^{2}-\langle\Phi'(u_{k}), u_{k}^{+}-u_{k}^{-}\rangle}{\|u_{k}\|^{2}} \\ \le& \sum_{n\in \mathbb {Z}}\frac{|\nabla W(n, u_{k})|}{|u_{k}|}|v_{k}|\bigl|v_{k}^{+}-v_{k}^{-}\bigr| \\ = & \sum_{n\in\Pi_{k}} \frac{|\nabla W(n, u_{k})|}{|u_{k}|}|v_{k}|\bigl|v_{k}^{+}-v_{k}^{-}\bigr| +\sum_{n\in \mathbb {Z}\setminus\Pi_{k}}\frac{|\nabla W(n, u_{k})|}{|u_{k}|}|v_{k}|\bigl|v_{k}^{+}-v_{k}^{-}\bigr| \\ \le& 1-\frac{\delta_{0}}{\Lambda_{0}}+o(1), \end{aligned}$$
(2.21)
a contradiction. Thus \(\delta>0\).
Going if necessary to a subsequence, we may assume the existence of \(n_{k}\in \mathbb {Z}\) such that
$$\bigl|v_{k}(n_{k})\bigr|=\|v_{k}\|_{\infty}> \frac{\delta}{2}. $$
Choose integers \(i_{k}\) and \(m_{k}\) with \(0\le m_{k}\le N-1\) such that \(n_{k}=i_{k}N+m_{k}\). Let \(\tilde{v}_{k}(n) =v_{k}(n+i_{k}N)\), then
$$ \bigl|\tilde{v}_{k}(m_{k})\bigr|> \frac{\delta}{2},\quad \forall k\in N. $$
(2.22)
Now we define \(\tilde{u}_{k}(n)=u_{k}(n+i_{k}N)\). Since \(p(n)\), \(L(n)\), and \(W(n, x)\) are N-periodic in n, then \(\tilde{u}_{k}/\|u_{k}\|=\tilde{v}_{k}\) and \(\|\tilde{u}_{k}\|=\|u_{k}\|\). Passing to a subsequence, we have \(\tilde{v}_{k}\rightharpoonup\tilde {v}\) in E, then \(\tilde{v}_{k}(n)\rightarrow \tilde{v}(n)\) for all \(n\in \mathbb {Z}\). Obviously, (2.22) implies that \(\tilde{v}(n)\ne0\) for some \(n\in\{0, 1, \ldots, N-1\}\). Let
$$E_{0}= \bigl\{ u\in E : \bigl\{ n\in \mathbb {Z}: \bigl|u(n)\bigr|>0\bigr\} \mbox{ is finite set} \bigr\} . $$
For any \(\phi\in E_{0}\), there exists an \(n_{0}\in \mathbb {N}\) such that \(\phi (n)=0\) for all \(|n|>n_{0}\). Setting \(\phi_{k}(n)=\phi(n-i_{k}N)\), then it follows from (W3) and (2.9) that
$$\begin{aligned} &\frac{\langle\Phi'(u_{k}), \phi_{k}\rangle}{\|u_{k}\|} \\ &\quad = \sum_{n\in \mathbb {Z}} \biggl[p(n+1)\triangle v_{k}\cdot\triangle\phi _{k}+L(n)v_{k} \cdot\phi_{k} -\frac{\nabla W(n, u_{k})\cdot\phi_{k}}{\|u_{k}\|} \biggr] \\ &\quad = \sum_{n\in \mathbb {Z}} \biggl[p(n+1)\triangle v_{k}\cdot\triangle\phi _{k}+\bigl(L(n)-M(n) \bigr)v_{k}\cdot\phi_{k} -\frac{\nabla W_{\infty}(n, u_{k})\cdot\phi_{k}}{\|u_{k}\|} \biggr] \\ &\quad = \sum_{n\in \mathbb {Z}} \biggl[p(n+1)\triangle \tilde{v}_{k}\cdot \triangle\phi+\bigl(L(n)-M(n)\bigr) \tilde{v}_{k}\cdot\phi -\frac{\nabla W_{\infty}(n, \tilde{u}_{k})\cdot\phi}{\|\tilde{u}_{k}\| } \biggr]. \end{aligned}$$
(2.23)
Note that
$$\begin{aligned} \biggl\vert \sum_{n\in \mathbb {Z}}\frac{\nabla W_{\infty}(n, \tilde{u}_{k})\cdot \phi}{\|\tilde{u}_{k}\|}\biggr\vert = \biggl\vert \sum_{|n|\le n_{0}}\frac{\nabla W_{\infty}(n, \tilde {u}_{k})\cdot\phi}{\|\tilde{u}_{k}\|} \biggr\vert \le\sum_{|n|\le n_{0}}\frac{|\nabla W_{\infty}(n, \tilde {u}_{k})|}{|\tilde{u}_{k}|}| \tilde{v}_{k}||\phi|=o(1). \end{aligned}$$
Hence, it follows from (2.17) and (2.23) that
$$\sum_{n\in \mathbb {Z}} \bigl[p(n+1)\triangle \tilde{v}_{k}\cdot\triangle\phi +\bigl(L(n)-M(n)\bigr) \tilde{v}_{k}\cdot\phi \bigr]=o(1), $$
(2.24)
which yields
$$ \sum_{n\in \mathbb {Z}} \bigl[p(n+1)\triangle \tilde{v}\cdot\triangle\phi +\bigl(L(n)-M(n)\bigr)\tilde{v}\cdot\phi \bigr]=0. $$
(2.25)
This shows that \(\tilde{v}\) is an eigenfunction of the operator ℬ, where
$$(\mathcal{B}u) (n)=\triangle\bigl[p(n)\triangle u(n-1)\bigr]-\bigl(L(n)-M(n) \bigr)u(n), \quad \forall n\in \mathbb {Z}. $$
But ℬ has only continuous spectrum in E. This contradiction shows that \(\{u_{n}\}\) is bounded. □
Proof of Theorem 1.1
In view of Lemmas 2.5 and 2.6, there exists a bounded sequence \(\{u_{k}\}\subset E\) satisfying (2.17). Thus there exists a constant \(C_{3}>0\) such that
$$ \sqrt{\Lambda_{0}}\|u_{k}\|_{\infty}\le \sqrt{\Lambda_{0}}\|u_{k}\|_{2} \le \|u_{k}\|\le C_{3},\quad \forall k\in \mathbb {N}. $$
(2.26)
Hence, by (W1) and (W2), there exists a constant \(C_{4}>0\) such that
$$ \bigl|\widetilde{W}(n, x)\bigr|\le\frac{c\Lambda_{0}}{2C_{3}^{2}}|x|^{2}+C_{4}|x|^{3}, \quad \forall (n, x)\in \mathbb {Z}\times \mathbb {R}^{\mathcal{N}}, |x|\le\frac{C_{3}}{\sqrt{\lambda_{0}}}. $$
(2.27)
If \(\delta:=\limsup_{k\to\infty}\|u_{k}\|_{\infty}=0\), then
$$ \sum_{n\in \mathbb {Z}}\bigl|u_{k}(n)\bigr|^{3} \le\|u_{k}\|_{\infty}\sum_{n\in \mathbb {Z}}\bigl|u_{k}(n)\bigr|^{2} \le\frac{C_{3}^{2}}{\Lambda_{0}}\|u_{k}\|_{\infty}=o(1). $$
(2.28)
From (2.10), (2.11), (2.17), (2.26), (2.27), and (2.28), one has
$$\begin{aligned} c = & \Phi(u_{k})-\frac{1}{2}\bigl\langle \Phi'(u_{k}), u_{k} \bigr\rangle +o(1) \\ = & \sum_{n\in \mathbb {Z}}\widetilde{W}(n, u_{k})+o(1) \\ \le& \frac{c\Lambda_{0}}{2C_{3}^{2}}\sum_{n\in \mathbb {Z}}\bigl|u_{k}(n)\bigr|^{2}+C_{4} \sum_{n\in \mathbb {Z}}\bigl|u_{k}(n)\bigr|^{3}+o(1) \\ \le& \frac{c}{2}+o(1). \end{aligned}$$
This contradiction shows that \(\delta>0\).
Going if necessary to a subsequence, we may assume the existence of \(n_{k}\in \mathbb {Z}\) such that
$$\bigl|u_{k}(n_{k})\bigr|=\|u_{k}\|_{\infty}> \frac{\delta}{2}. $$
Choose integers \(i_{k}\) and \(m_{k}\) with \(0\le m_{k}\le N-1\) such that \(n_{k}=i_{k}N+m_{k}\). Let \(v_{k}(n)=u_{k}(n+i_{k}N)\), then
$$ \bigl|v_{k}(m_{k})\bigr|> \frac{\delta}{2}, \quad\forall k\in \mathbb {N}. $$
(2.29)
Since \(p(n)\), \(L(n)\), and \(W(n, x)\) are N-periodic in n, we have \(\| v_{k}\|=\|u_{k}\|\) and
$$ \Phi(v_{k})\rightarrow c,\quad \bigl\| \Phi'(v_{k}) \bigr\| \bigl(1+\|v_{k}\|\bigr)\rightarrow0. $$
(2.30)
Passing to a subsequence, we have \(v_{k}\rightharpoonup v\) in E, \(v_{k}(n)\rightarrow v(n)\) for all \(n\in \mathbb {Z}\). Obviously, (2.29) implies that \(v\ne0\). It is easy to show that \(\Phi'(v)=0\). □