The following Symmetric Mountain Pass Theorem is crucial in proving the existence of infinitely many solutions with 4-superlinear nonlinearities.
Proposition 2.1
(Symmetric Mountain Pass Theorem [21])
Let X be an infinite-dimensional Banach space, \(X=Y\oplus Z\), where Y is a finite-dimensional space. If \(J\in C^{1}(X,\mathbf{R})\) satisfies the \((Ce)\) condition, and
- (\(J_{1}\)):
-
\(J(0)=0\), \(J(-u)=J(u)\) for all \(u\in X\);
- (\(J_{2}\)):
-
there exist two constants \(\delta ,\alpha >0\) such that \(J|_{\partial B_{\delta }\cap Z}\geq \alpha \);
- (\(J_{3}\)):
-
for any finite-dimensional subspace \(\widetilde{X}\subset X\), there exists \(R=R(\widetilde{X})>0\) such that \(J(u)\leq 0\) on \(\widetilde{X}\setminus B_{R}\),
then J possesses an unbounded sequence of critical values.
In [21], the Symmetric Mountain Pass Theorem is established under the \((PS)\) condition. Since the Deformation Theorem is still valid under the \((Ce)\) condition, we see that the Symmetric Mountain Pass Theorem also holds under the \((Ce)\) condition. The following embedding lemma, which follows from [17] or [33], plays a significant role in recovering the compactness result.
Lemma 2.1
If V satisfies the condition \((V_{1})\), then the embedding map from E into \(l^{q}\) is compact for \(2\leq q\leq +\infty \).
With the help of Lemma 2.1, we have the following compactness result.
Lemma 2.2
Under the assumption of Theorem 1.1, the functional I satisfies the \((Ce)\) condition.
Proof
Suppose that \(\{u^{n}\} \) is a \((Ce)\) sequence,
$$ \bigl\vert I\bigl(u^{n}\bigr) \bigr\vert \leq M,\qquad \bigl(1+ \bigl\Vert u^{n} \bigr\Vert \bigr)I'\bigl(u^{n} \bigr)\rightarrow 0\quad \text{as } n\rightarrow \infty . $$
It suffices to prove that \(\{u^{n}\}\) has a converging subsequence in E. We first obtain that \(\{u_{n}\} \) is bounded in E. Otherwise, \(\|u^{n}\|\rightarrow +\infty \) as \(n\rightarrow \infty \). Let \(v^{n}=u^{n}/\|u^{n}\|\). Moving, if necessary, to a subsequence, we assume \(v^{n}\rightharpoonup v\) in E, by Lemma 2.1, \(v^{n}\rightarrow v\) in \(l^{q}\) and \(v_{k}^{n}\rightarrow v_{k}\) for any \(k\in \mathbf{Z}\) as \(n\rightarrow \infty \). There are only two cases \(v=0\) or \(v\neq 0\). If \(v=0\), by \((f_{3})\), we have
$$ \begin{aligned} \frac{1}{ \Vert u^{n} \Vert ^{2}}(M+1)&\geq \frac{1}{ \Vert u^{n} \Vert ^{2}} \biggl(I \bigl(u^{n}\bigr)-\frac{1}{4} \bigl\langle I' \bigl(u^{n}\bigr),u^{n} \bigr\rangle \biggr) \\ &\geq \frac{1}{4}\min \{a,1\}+\frac{1}{ \Vert u^{n} \Vert ^{2}}\sum _{k\in \mathbf{Z}}\biggl(\frac{1}{4}f_{k} \bigl(u^{n}_{k}\bigr)u_{k}^{n}-F_{k} \bigl(u^{n}_{k}\bigr)\biggr) \\ &\geq \frac{1}{4}\min \{a,1\}-\alpha \sum_{k\in \mathbf{Z}} \bigl\vert v_{k}^{n} \bigr\vert ^{2}, \end{aligned} $$
which implies that \(\frac{1}{4}\min \{a,1\}\leq 0\). That is impossible. If \(v\neq 0\), then \(\Omega _{1}:=\{k\in \mathbf{Z}| v_{k}\neq 0\}\neq \emptyset \). For any \(k\in \Omega _{1}\), we have \(|u_{k}^{n}|\rightarrow +\infty \) as \(n\rightarrow \infty \). By \((f_{2})\), one obtains that for any \(k\in \Omega _{1}\),
$$ \frac{F_{k}(u^{n}_{k})}{ \vert u^{n}_{k} \vert ^{4}} \bigl\vert v_{k}^{n} \bigr\vert ^{4}\rightarrow + \infty \quad \text{as } n\rightarrow \infty , $$
combined with Fatou’s Lemma, which implies that
$$ \sum_{k\in \Omega _{1}}\frac{F_{k}(u^{n}_{k})}{ \vert u^{n}_{k} \vert ^{4}} \bigl\vert v_{k}^{n} \bigr\vert ^{4} \rightarrow + \infty \quad \text{as } n\rightarrow \infty . $$
(2)
It follows from \((f_{2})\) that there exists \(L_{1}>0\) such that
$$ F_{k}(t)\geq 0 \quad\text{for any } k\in \mathbf{Z} \text{ and } \vert t \vert \geq L_{1}. $$
(3)
By \((f_{1})\), we obtain \(|F_{k}(t)|\leq Ct^{2} \quad\text{for any } k\in \mathbf{Z} \text{ and } |t| \leq L_{1}\). Combining with (3), we have
$$ F_{k}(t)\geq -Ct^{2} \quad\text{for any } k\in \mathbf{Z} \text{ and } t \in \mathbf{R}. $$
Hence, we obtain
$$ \begin{aligned} &\sum_{k\in \mathbf{Z}\setminus \Omega _{1}} \frac{F_{k}(u^{n}_{k})}{ \Vert u^{n} \Vert ^{4}} \geq -\frac{C}{ \Vert u^{n} \Vert ^{4}} \sum_{k\in \mathbf{Z}\setminus \Omega _{1}} \bigl\vert u^{n}_{k} \bigr\vert ^{2} \geq -C \frac{ \Vert u^{n} \Vert _{2}^{2}}{ \Vert u^{n} \Vert ^{4}}\geq -\frac{C}{ \Vert u^{n} \Vert ^{2}}, \end{aligned} $$
which implies that
$$ \begin{aligned} \liminf_{ n\rightarrow \infty }\sum _{k\in \mathbf{Z} \setminus \Omega _{1}}\frac{F_{k}(u^{n}_{k})}{ \Vert u^{n} \Vert ^{4}}\geq 0. \end{aligned} $$
(4)
Note that
$$ I\bigl(u^{n}\bigr)+\sum_{k\in \mathbf{Z}}F_{k} \bigl(u^{n}_{k}\bigr)\leq \frac{1}{2}\max \{a,1\} \bigl\Vert u^{n} \bigr\Vert ^{2}+\frac{b}{4} \bigl\Vert u^{n} \bigr\Vert ^{4}. $$
Dividing by \(\|u^{n}\|^{4}\) on both sides and letting \(n\rightarrow \infty \), we obtain
$$ \begin{aligned} \frac{b}{4}&\geq \limsup_{ n\rightarrow \infty } \sum_{k \in \mathbf{Z}}\frac{F_{k}(u^{n}_{k})}{ \Vert u^{n} \Vert ^{4}} \\ &\geq \limsup_{ n\rightarrow \infty } \biggl(\sum_{k\in \mathbf{Z} \setminus \Omega _{1}} \frac{F_{k}(u^{n}_{k})}{ \Vert u^{n} \Vert ^{4}}+\sum_{k \in \Omega _{1}}\frac{F_{k}(u^{n}_{k})}{ \Vert u^{n} \Vert ^{4}} \biggr) \rightarrow +\infty , \end{aligned} $$
via (2) and (4), which is impossible. In any case, we obtain a contradiction and hence \(\{u^{n}\} \) is bounded in E.
Moving if necessary to a subsequence, we can assume \(u^{n}\rightharpoonup u\) in E. It follows that
$$\begin{aligned}& \bigl\langle I'\bigl(u^{n} \bigr)-I'(u),u^{n}-u\bigr\rangle \\& \quad =\bigl(a+b \bigl\Vert \Delta u^{n} \bigr\Vert _{2}^{2}\bigr)\sum_{k\in \mathbf{Z}}\Delta u^{n}_{k-1} \Delta \bigl(u^{n}_{k-1}-u_{k-1} \bigr)+\sum_{k\in \mathbf{Z}}V_{k}u^{n}_{k} \bigl(u^{n}_{k}-u_{k}\bigr) \\& \qquad {}-\bigl(a+b \Vert \Delta u \Vert _{2}^{2}\bigr)\sum _{k\in \mathbf{Z}}\Delta u_{k-1} \Delta \bigl(u^{n}_{k-1}-u_{k-1}\bigr)-\sum _{k\in \mathbf{Z}}V_{k}u_{k} \bigl(u^{n}_{k}-u_{k} \bigr) \\& \qquad {}-\sum_{k\in \mathbf{Z}}\bigl(f_{k} \bigl(u^{n}_{k}\bigr)-f_{k}(u_{k}) \bigr) \bigl(u^{n}_{k}-u_{k}\bigr) \\& \quad =\bigl(a+b \bigl\Vert \Delta u^{n} \bigr\Vert _{2}^{2}\bigr)\sum_{k\in \mathbf{Z}} \bigl\vert \Delta \bigl(u^{n}_{k-1}-u_{k-1}\bigr) \bigr\vert ^{2}+ \sum_{k\in \mathbf{Z}}V_{k} \bigl\vert u^{n}_{k}-u_{k} \bigr\vert ^{2} \\& \qquad {}-b\bigl( \Vert \Delta u \Vert _{2}^{2}- \bigl\Vert \Delta u^{n} \bigr\Vert _{2}^{2}\bigr)\sum _{k\in \mathbf{Z}}\Delta u_{k-1}\Delta \bigl(u^{n}_{k-1}-u_{k-1}\bigr) \\& \qquad {}-\sum_{k\in \mathbf{Z}}\bigl(f_{k} \bigl(u^{n}_{k}\bigr)-f_{k}(u_{k}) \bigr) \bigl(u^{n}_{k}-u_{k}\bigr) \\& \quad \geq \min \{a,1\} \bigl\Vert u^{n}-u \bigr\Vert ^{2} -b\bigl( \Vert \Delta u \Vert _{2}^{2}- \bigl\Vert \Delta u^{n} \bigr\Vert _{2}^{2}\bigr)\sum _{k\in \mathbf{Z}}\Delta u_{k-1}\Delta \bigl(u^{n}_{k-1}-u_{k-1}\bigr) \\& \qquad {}-\sum_{k\in \mathbf{Z}}\bigl(f_{k} \bigl(u^{n}_{k}\bigr)-f_{k}(u_{k}) \bigr) \bigl(u^{n}_{k}-u_{k}\bigr). \end{aligned}$$
One has
$$ \begin{aligned} \min \{a,1\} \bigl\Vert u^{n}-u \bigr\Vert ^{2} \leq{}& \bigl\langle I' \bigl(u^{n}\bigr)-I'(u),u^{n}-u \bigr\rangle \\ &{}+\sum_{k\in \mathbf{Z}}\bigl(f_{k} \bigl(u^{n}_{k}\bigr)-f_{k}(u_{k}) \bigr) \bigl(u^{n}_{k}-u_{k}\bigr) \\ &{}+b\bigl( \Vert \Delta u \Vert _{2}^{2}- \bigl\Vert \Delta u^{n} \bigr\Vert _{2}^{2}\bigr)\sum _{k\in \mathbf{Z}}\Delta u_{k-1}\Delta \bigl(u^{n}_{k-1}-u_{k-1}\bigr). \end{aligned} $$
(5)
By the boundedness of \(\{u^{n}\}\) and \(u^{n}\rightharpoonup u\) in E, it is obvious that
$$ \bigl\langle I'_{n}\bigl(u^{n} \bigr)-I'(u),u^{n}-u\bigr\rangle \rightarrow 0\quad \text{as } n\rightarrow \infty .$$
(6)
By \((f_{1})\), Lemma 2.1 and Lebesgue’s dominated convergence theorem
$$ \sum_{k\in \mathbf{Z}}\bigl(f_{k} \bigl(u^{n}_{k}\bigr)-f_{k}(u_{k}) \bigr) \bigl(u^{n}_{k}-u_{k}\bigr) \rightarrow 0 \quad \text{as } n\rightarrow \infty . $$
(7)
Let us consider the functional \(P: E\rightarrow \mathbf{R}\),
$$ P(w)=\sum_{k\in \mathbf{Z}}\Delta u_{k-1}\Delta w_{k-1}. $$
Since \(|P(w)|\leq \|u\|\|w\|\), we can deduce that P is a continuous linear functional on E. By \(u^{n}\rightharpoonup u\) in E, we obtain
$$ P\bigl(u^{n}-u\bigr)=\sum_{k\in \mathbf{Z}}\Delta u_{k-1}\Delta \bigl(u^{n}_{k-1}-u_{k-1} \bigr) \rightarrow 0 \quad \text{as } n\rightarrow \infty . $$
By the boundedness of \(\{u^{n}\}\) in E, we have
$$ b\bigl( \Vert \Delta u \Vert _{2}^{2}- \bigl\Vert \Delta u^{n} \bigr\Vert _{2}^{2} \bigr)\sum_{k\in \mathbf{Z}} \Delta u_{k-1}\Delta \bigl(u^{n}_{k-1}-u_{k-1}\bigr)\rightarrow 0 \quad \text{as } n\rightarrow \infty . $$
(8)
It follows from (5)–(8) that \(\|u^{n}-u\|\rightarrow 0\) as \(n\rightarrow \infty \). Thus, \(u^{n}\rightarrow u\) strongly in E as \(n\rightarrow \infty \). □
Let \(\{e^{j}\}\) be an orthonormal basis of E and define \(X_{j}=\operatorname{span}\{e_{j}\}\), \(Y_{m}=\bigoplus _{j=1}^{m}X_{j}\) and \(Z_{m}=\overline{\bigoplus _{j=m+1}^{\infty }X_{j}}\) for any \(m\in \mathbf{N}\).
Lemma 2.3
Under the assumption \((V_{1})\), for any \(2\leq q\leq +\infty \),
$$ \beta _{m}(q):=\sup_{u\in Z_{m}, \Vert u \Vert =1} \Vert u \Vert _{q}\rightarrow 0 \quad \textit{as } m\rightarrow \infty . $$
(9)
Proof
It is obvious that \(0<\beta _{m+1}(q)\leq \beta _{m}(q)\), so that \(\beta _{m}(q)\rightarrow \beta (q)\geq 0\) as \(m\rightarrow \infty \). For every \(m\in \mathbf{N}\), there exists \(u^{m}\in \mathbf{Z}_{m}\) with \(\|u^{m}\|=1\) such that
$$ \bigl\Vert u^{m} \bigr\Vert _{q}> \frac{\beta _{m}(q)}{2}. $$
(10)
For any \(w\in E\), \(w=\sum_{j=1}^{\infty } c_{j}e_{j}\), by the Cauchy–Schwarz inequality, we have
$$ \begin{aligned} \bigl\vert \bigl\langle u^{m},w\bigr\rangle \bigr\vert &= \Biggl\vert \Biggl\langle u^{m},\sum _{j=1}^{ \infty } c_{j}e_{j}\Biggr\rangle \Biggr\vert = \Biggl\vert \Biggl\langle u^{m},\sum _{j=m}^{\infty } c_{j}e_{j} \Biggr\rangle \Biggr\vert \\ &\leq \bigl\Vert u^{m} \bigr\Vert \Biggl\Vert \sum _{j=m}^{\infty } c_{j}e_{j} \Biggr\Vert = \Biggl\Vert \sum_{j=m}^{ \infty } c_{j}e_{j} \Biggr\Vert \rightarrow 0 \end{aligned} $$
as \(m\rightarrow \infty \), which implies that \(u^{m}\rightharpoonup 0\) in E. The compact embedding of \(E\hookrightarrow l^{q}\), \(q\in [2,\infty ]\), implies that \(u^{m}\rightarrow 0\) in \(l^{q}\). Let \(m\rightarrow \infty \) in (10) and we obtain \(\beta _{m}(q)\rightarrow 0\) as \(m\rightarrow \infty \). □
The proof of Theorem 1.1
We will make use of the Symmetric Mountain Pass Theorem and Proposition 2.1 to prove Theorem 1.1. It is easy to see that \((J_{1})\) follows from the condition \((f_{4})\). It follows from \((f_{1})\) that
$$ F_{k}(t)\leq c_{1}t^{2}+c_{2} \vert t \vert ^{p}\quad \text{for any } k\in \mathbf{Z} \text{ and } t\in \mathbf{R}. $$
By Lemma 2.3, there exists \(m\in \mathbf{N}\) large enough such that
$$ \Vert u \Vert ^{2}_{2}\leq \frac{\min \{a,1\}}{4c_{1}} \Vert u \Vert ^{2} \quad \text{and}\quad \Vert u \Vert ^{p}_{p} \leq \frac{\min \{a,1\}}{4c_{2}} \Vert u \Vert ^{p} \quad\text{for any } u\in Z_{m}. $$
It follows from the above three inequalities that
$$ \begin{aligned} I(u)&=\frac{1}{2}\min \{a,1\} \Vert u \Vert ^{2} -\sum_{k\in \mathbf{Z}}F_{k} \bigl(u^{n}_{k}\bigr) \\ &\geq \frac{1}{2}\min \{a,1\} \Vert u \Vert ^{2}-c_{1} \Vert u \Vert _{2}^{2}-c_{2} \Vert u \Vert _{p}^{p} \\ &\geq \frac{1}{4}\min \{a,1\} \Vert u \Vert ^{2}\bigl(1- \Vert u \Vert ^{p-2}\bigr). \end{aligned} $$
It follows from \(p>2\) that there exist \(\delta _{1}\), \(\alpha >0\) such that \(I|_{\partial B_{\delta _{1}}\cap Z_{m}}\geq \alpha \). Thus, \((J_{2})\) holds.
It remains to prove \((J_{3})\). Since all norms are equivalent in a finite-dimensional space, there exists \(c_{4}\) such that \(\|u\|^{4}_{4}\geq c_{4}\|u\|^{4}\) for any \(u\in \widetilde{E}\).
By \((f_{2})\), for any \(M>\frac{b}{4c_{4}}\), there exists \(L_{1}>0\) such that \(F_{k}(t)\geq Mt^{4}\) for \(|t|\geq L_{1}\). It follows from \((f_{1})\) that there exists \(C_{1}>0\) such that \(F_{k}(t)\geq -C_{1}t^{2}\) for \(|t|\leq L_{1}\). From the above two inequalities, it follows that for any \(k\in \mathbf{Z}\),
$$ F_{k}(t)\geq Mt^{4}-C_{M}t^{2} \quad\text{for any } k\in \mathbf{Z} { \text{ and }} t\in \mathbf{R}, $$
where \(C_{M}=C_{1}+ML_{1}^{2}\). It follows that
$$ \begin{aligned} I(u)&\leq \frac{1}{2}\max \{a,1\} \Vert u \Vert ^{2}+\frac{b}{4} \Vert u \Vert ^{4}-\sum _{k\in \mathbf{Z}}F_{k}(u_{k}) \\ &\leq \frac{1}{2}\max \{a,1\} \Vert u \Vert ^{2}+ \frac{b}{4} \Vert u \Vert ^{4}-M \Vert u \Vert _{4}^{4}+C_{M} \Vert u \Vert _{2}^{2} \\ &\leq C \Vert u \Vert ^{2}-\biggl(Mc_{4}- \frac{b}{4}\biggr) \Vert u \Vert ^{4}, \end{aligned} $$
for all \(u\in \widetilde{E}\). Hence, there exists a large \(R=R(\widetilde{E})\) such that \(I(u)\leq 0\) on \(\widetilde{E}\setminus B_{R}\). This completes the proof. □