On Local and Nonlocal Discrete p-Laplacian Equations via Clark’s Theorem

Abstract

By employing Clark’s theorem we prove the existence of infinitely many homoclinic solutions to the local and nonlocal discrete p-Laplacian equations on the integers. Our results extend and correct the reasoning of some recent findings expressed in the literature.

1 Introduction and the Main Result

A wide class of difference equations represents the discrete counterpart of ordinary and partial differential equations and is usually studied in connection with numerical analysis (see, e.g., [1, 7, 8, 12, 17, 18, 21, 23] and references therein). The fractional Laplacian, understood as a positive power of the classical Laplacian, has been present in several areas of mathematics such as functional analysis, harmonic analysis, potential theory, fractional calculus, and probability [13, 25]. The fractional p-Laplacian on $\mathbb {R} ^{N},p>1,$ is defined, for $0<s<1$ and good enough functions U, as

$$\begin{aligned} \left( -\Delta \right) _{p}^{s}U(x)=\lim _{\varepsilon \rightarrow 0^{+}}\int _{ \mathbb {R} ^{n}\setminus B(x,\varepsilon )}\frac{\left| U(x)-U(y)\right| ^{p-2}\left( U(x)-U(y)\right) }{\left| x-y\right| ^{N+ps}}dy,\ \ \ \ \text {x}\in \mathbb {R} ^{N},\nonumber \\ \end{aligned}$$

(1)

where $B(x,\varepsilon )$ is the ball centered at $x\in \mathbb {R} ^{N}$ with radius $\varepsilon$. Up to some normalization constant depending on N, p, and s, this definition is consistent with the one of the fractional Laplacian $\left( -\Delta \right) ^{s}$ in the case $p=2$ (see [15]). In [15], there is the generalization of this operator, i.e.

$$\begin{aligned} \mathcal {L}_{K}u(x)=\lim _{\varepsilon \rightarrow 0^{+}}\int _{ \mathbb {R} ^{n}\setminus B(x,\varepsilon )}\left| U(x)-U(y)\right| ^{p-2}\left( U(x)-U(y)\right) K(x-y)dy,\ \ \ \ \ x\in \mathbb {R} ^{N}, \end{aligned}$$

where the kernel $K: \mathbb {R} ^{N}\smallsetminus \{0\}\rightarrow (0,+\infty )$ satisfies

$(K_{1})$:

$mK\in L^{1}(\mathbb {R}^{N}),$ where $m(x)=\min \{\left| x\right| ^{p},1\}$,

$(K_{2})$:

there exist $\theta >0$ and $s\in (0,1)$ such that $K(x)\ge \theta \left| x\right| ^{-(N+ps)}$ for all $x\in \mathbb {R}^{N}\smallsetminus \{0\}.$

The main difficulties to overcome in numerical approaches are the nonlocality and singularity of these operators. The first approach to discretization of operator (1) with $p=2$ was discussed in [4]. The authors defined the fractional powers of the discrete Laplacian $\left( -\Delta \right) ^{s},$ $0<s<1,$ where $\left( -\Delta \right) (u(k))=-(u(k+1)-2u(k)+u(k-1))$ on $\mathbb {Z},$ with the semigroup method as

$$\begin{aligned} \left( -\Delta \right) ^{s}u(k)=\frac{1}{\Gamma (-s)}\int _{0}^{\infty }\left( e^{t\Delta }u(k)-u(k)\right) \frac{dt}{t^{1+s}}, \end{aligned}$$

where $\Gamma$ denotes the Gamma function and $e^{t\Delta }u(k)=w(t,k)$ is the solution to the semidiscrete heat equation

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _{t}w(t,k)=\Delta w(t,k), &{} \text {in } \mathbb {Z} \times (0,\infty ), \\ w(0,k)=u(k) &{} \text {on } \mathbb {Z}. \end{array} \right. \end{aligned}$$

Theorem 1.1 in [4] shows that

$$\begin{aligned} \left( -\Delta \right) ^{s}u(k)=\sum _{m\in \mathbb {Z},m\ne k}(u(k)-u(m))K_{s}(k-m),\ \ \ \ \ \ \ \ k\in \mathbb {Z}, \end{aligned}$$

(2)

for

$$\begin{aligned} u\in \left\{ u: \mathbb {Z} \rightarrow \mathbb {R}:\sum _{m\in \mathbb {Z} }\frac{\left| u(m)\right| }{\left( 1+\left| m\right| \right) ^{1+2s}}<\infty \right\} , \end{aligned}$$

where the discrete kernel $K_{s}$ is given by

$$\begin{aligned} K_{s}(j)=\frac{4^{s}\Gamma (1/2+s)}{\sqrt{\pi }\left| \Gamma (-s)\right| }\cdot \frac{\Gamma (\left| j\right| -s)}{\Gamma (\left| j\right| +1+s)}, \end{aligned}$$

(3)

for any $j\in \mathbb {Z} {\setminus } \{0\},$ and $K_{s}(0)=0.$ This kernel satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} \frac{c_{s}}{\left| j\right| ^{1+2s}}\le K_{s}(j)\le \frac{C_{s}}{ \left| j\right| ^{1+2s}} &{} \text {for every }j\in \mathbb {Z} \smallsetminus \{0\}, \\ K_{s}(0)=0. &{} \end{array} \right. \end{aligned}$$

Moreover, if u is bounded then

$$\begin{aligned} \lim _{s\rightarrow 1^{-}}\left( -\Delta \right) ^{s}u=\left( -\Delta \right) u\ \text { on } \mathbb {Z}. \end{aligned}$$

(4)

We can generalize the operator given in (2) to a discrete fractional p-Laplacian in the following way: for $0<s<1,$ $p>1$ and good enough sequences $u: \mathbb {Z} \rightarrow \mathbb {R}$ we put

$$\begin{aligned} \left( -\Delta \right) _{p}^{s}u(k)=2\sum _{m\in \mathbb {Z},m\ne k}\left| u(k)-u(m)\right| ^{p-2}(u(k)-u(m))K_{s,p}(k-m),\ \ \ \ \ \ k\in \mathbb {Z},\nonumber \\ \end{aligned}$$

(5)

where the discrete kernel $K_{s,p}$ satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} \frac{c_{s,p}}{\left| j\right| ^{1+ps}}\le K_{s,p}(j)\le \frac{ C_{s,p}}{\left| j\right| ^{1+ps}} &{} \text {for every }j\in \mathbb {Z} \smallsetminus \{0\}, \\ K_{s,p}(0)=0 &{} \end{array} \right. \end{aligned}$$

(6)

and $0<c_{s,p}\le C_{s,p}$ are constants. The class of operators $\left( -\Delta \right) _{2}^{s}$ contains the fractional powers of the discrete Laplacian $\left( -\Delta \right) ^{s}$ defined in [4]. Note that showing (4) requires using the explicit form of the kernel given in (3) (see the proof of Theorem 1.1 in [4]). However, the only constraint imposed on the kernel $K_{s,p}$ is condition (6). The author does not know whether $\lim _{s\rightarrow 1^{-}}\left( -\Delta \right) _{p}^{s}u=\left( -\Delta \right) _{p}u$ on $\mathbb {Z},$ even for $p=2$, where $\left( -\Delta \right) _{p}u$ is a local discrete p-Laplacian, i.e.

$$\begin{aligned}{} & {} \left( -\Delta \right) _{p}u(k)=\left| u(k+1)-u(k)\right| ^{p-2}\left( u(k+1)-u(k)\right) \nonumber \\{} & {} -\left| u(k)-u(k-1)\right| ^{p-2}\left( u(k)-u(k-1)\right) \end{aligned}$$

(7)

for all $k\in \mathbb {Z}.$

With such discretization in hand, discrete counterparts of continuous problems were considered in [9, 10, 27, 28].

In the present paper we deal with the following fractional discrete p -Laplacian equations

$$\begin{aligned} \left\{ \begin{array}{ll} \left( -\Delta \right) _{p}^{s}u(k)+V(k)\left| u(k)\right| ^{p-2}u(k)=f(k,u(k)) &{} \text{ for } \text{ all } k\in \mathbb {Z} \\ u(k)\rightarrow 0 &{} \text{ as } |k|\rightarrow \infty . \end{array} \right. \end{aligned}$$

(8)

Here $s\in (0,1],p>1$ is a real number, $V:{\mathbb {Z}}\rightarrow \mathbb {(} 0,+\infty )$, while $f:{\mathbb {Z}}\times {\mathbb {R}}\rightarrow {\mathbb {R} }$ is a continuous function. For $0<s<1,$ $\left( -\Delta \right) _{p}^{s}$ is the nonlocal discrete p-Laplacian given by (5) and for $s=1$, $\left( -\Delta \right) _{p}^{1}$ is the local discrete p -Laplacian given by (7).

Recently, the existence and multiplicity of solutions to the problem of type (8) have been studied. For results on equations in $\mathbb {Z}$ driven by the local discrete p-Laplacian we refer the reader to [6, 8, 11, 12, 21,22,23, 26]. In [8], the authors studied the existence of multiple homoclinic solutions (i.e. with $\lim _{\left| k\right| \rightarrow \infty }u(k)=0$) via critical point theory. In papers [6, 11, 12, 23], the authors imposing various conditions on nonlinearity considered the existence results of a sequence of infinitely many homoclinic orbits, by using the symmetric mountain pass theorem or fountain theorem or dual fountain theorem. Infinitely many solutions have also been obtained in [22] directly using the variational method, in [26] by applying Nehari manifold methods and in [21] by employ of the Ricceri’s theorem.

For results on problems (8) in $\mathbb {Z}$ driven by the nonlocal discrete p-Laplacian we refer the reader to [9, 10, 27, 28]. In [27, 28], by using Ekeland’s variational principle, together with the mountain pass theorem, at least two homoclinic solutions were obtained. By using the Nehari manifold method, at least two homoclinic solutions to (8) were acquired in [9]. Finally, in [10], an infinite number of solutions have been obtained from different versions of Clark’s theorem.

In this paper, our goal is to apply the variational method and a variant of Clark’s theorem to find a sequence of homoclinic solutions for problem (8). To do this, we construct the appropriate space of solutions $E_{s}$ ( $s=1$ in the local case and $0<s<1$ in the nonlocal case) and the energy functional $J_{s}$ defined on it such that its critical points are solutions to our problem. By imposing appropriate conditions on the nonlinearity f, we obtain from Clark’s theorem a sequence of solutions in $E_{s}$ with norms tending to zero. Let $l^{r},1\le r\le \infty ,$ denote the sequence Lebesgue space with standard norm $\left\| \cdot \right\| _{r}$. Since $E_{s}\subset l^{p}$, we immediately get that the solutions are homoclinic.

A special case of our contributions reads as follows. We assume that potential V(k) and the nonlinearity f(k, t) satisfy the following conditions:

(V):

$V(k)\ge V_{0}>0$ for all $k\in \mathbb {Z}$, $V(k)\rightarrow +\infty$ as $\left| k\right| \rightarrow +\infty ;$

$(f_{1})$:

there exists $\varepsilon >0\$such that$\ f(k,-t)=-f(k,t)$ for all $k\in \mathbb {Z}$ and $t\in \left( -\varepsilon ,\varepsilon \right) ;$

$(f_{2})$:

there exists a positive $a\in l^{\frac{p}{p-1}}$ such that

$$\begin{aligned} \underset{t\rightarrow 0}{\lim \sup }~\frac{\left| f(k,t)\right| }{ a(k)+\left| t\right| ^{p-1}}<+\infty \end{aligned}$$

(9)

uniformly for all $k\in \mathbb {Z}$;

$(f_{3})$:

there exist $1<\sigma <p$ and an infinitesubset $Z_{0}\subset \mathbb {Z}$ such that

$$\begin{aligned} \underset{t\rightarrow 0}{\lim \inf }\frac{F(k,t)}{\left| t\right| ^{\sigma }}>0 \end{aligned}$$

(10)

for all $k\in Z_{0}$;

where F(k, t) is the primitive function of f(k, t), that is $F(k,t)=\int _{0}^{t}f(k,s)ds$ for every $t\in \mathbb {R}$ and $k\in \mathbb {Z}$.

Let us note that conditions $(f_{1})-(f_{3})$ control the behaviour of f only for a small t. Note also that since f(k, t) is odd in t, $f(k,0)=0$ for all $k\in \mathbb {Z}.$ As f is continuous and $a(k)>0,$ we always have $\lim _{t\rightarrow 0}\frac{\left| f(k,t)\right| }{a(k)+\left| t\right| ^{p-1}} =0.$ Hence the uniformity plays a key role in $(f_{2}).$ Finally, note that in (10) we do not assume uniformity in the limit. The simplest example for f satisfying $(f_{1})-(f_{3})$ is given by

$$\begin{aligned} f(k,t)=a(k)\left| t\right| ^{\sigma -2}t+\alpha \left| t\right| ^{p-2}t, \end{aligned}$$

(11)

where $1<\sigma <p,$ $\alpha >0$ and $\{a(k)\}\in l^{1}$ is positive.

Now we state our main result.

Theorem 1

Let assumptions (V) and $(f_{1})-(f_{3})$ hold. Then problem ( 8) has infinitely many non-trivial homoclinic solutions with negative energy and with their norms tending to zero.

Our theorem improves the results in [10]. In [10] the set of conditions imposed on the nonlinearity is inconsistent and many arguments are fallacious in the proofs. To be more precise, in [10] problem (8) with $0<s<1$ was investigated under the following set of hypotheses:

$(\hat{V})$:

$V(k)\ge V_{0}>0$ for all $k\in \mathbb {Z}$ and there is $r>0$ such that $\lim _{\left| y\right| \rightarrow \infty }\mu (\{k\in \mathbb {Z}:V(k)\le b\}\cap U_{r}(y))=0$ for any $b>0$, where $U_{r}(y)$ is the neighbourhood in $\mathbb {Z}$ centred at point y with radius r and $\mu$ is the counting measure on $\mathbb {Z}$.

$(\hat{f}_{1})$:

$\displaystyle f(k,t)$ is odd in t; 

$(\hat{f}_{2})$:

$\displaystyle \lim \nolimits _{t\rightarrow 0}\frac{\left| f(k,t)\right| }{\left| t\right| ^{p-1}}=0$ uniformly for all $k\in \mathbb {Z}$;

$(\hat{f}_{3})$:

there exist $\alpha >0$, $1<\sigma <p$ and an infinite subset $Z_{0}\subset \mathbb {Z}$ such that

$$\begin{aligned} F(k,t)\ge \alpha \left| t\right| ^{\sigma } \end{aligned}$$

(12)

for all $(k,t)\in Z_{0}\times \mathbb {R}$;

Observe that $(\hat{f}_{1}),(\hat{f}_{3})$ are global properties. Unfortunately, $(\hat{f}_{2})$ and $(\hat{f}_{3})$ are inconsistent. Indeed, from $(\hat{f}_{2})$ there exists $\varepsilon >0$ such that we have $\left| f(k,t)\right| \le \alpha p\left| t\right| ^{p-1}$ on $\mathbb {Z} \times (-\varepsilon ,\varepsilon )$ and so $F(k,t)\le \alpha \left| t\right| ^{p}$ on $\mathbb {Z} \times (-\varepsilon ,\varepsilon ).$ Since $1<\sigma <p,$ it is inconsistent with (12) for sufficiently small t. Moreover, the authors consider the nonlinearity in a special form $f(k,t)=a(k)\left| t\right| ^{q-2}t+b(k)\left| t\right| ^{r-2}t,$ where $1<q<p<r,$ with the assumption

$(\hat{f}_{4})$:

there is $a_{0}>0$ such that $0<a_{0}\le a(k)\in l^{\frac{p}{p-q}}$ and $0\le b(k)\in l^{\infty }.$

However, there is no such $\{a(k)\}.$ Let us list some false reasoning in [10]. The authors illegitimately use the norm $\left\| \cdot \right\| _{\sigma }$ on $l^{\sigma }$ for $1<\sigma <p$ to elements from the solution space (pp. 11 and 12). Furthermore, the authors incorrectly show in proofs that the energy functional related to problem ( 8) is bounded from below (p. 9) and satisfies the (PS) condition (pp. 10 and 12). To be more precise in the case of the (PS) condition, there are incorrect estimates for $1<p<2.$

Let us note that our assumption $(f_{2})$ used for obtaining the required properties of energy functional is strictly weaker than those used in the literature. To the best of our knowledge, assumptions $(\hat{f}_{2})$ and

$(\hat{f}_{5})$:

$\sup _{\left| t\right| \le T}\left| f(\cdot ,t)\right| \in l^{1}$ for all $T>0$

have been used for this purpose so far. Our toy example (11) does not satisfy $(\hat{f}_{2})$ and $(\hat{f}_{5}).$

Finally, let us make some remarks about the potential V. In the continuous setting, for problems considered on $\mathbb {R} ^{N}$, we can find in the literature some assumptions on V(x),  $x\in \mathbb {R} ^{N},$ that guarantee compactness of the embedding of the solution space into the Lebesgue space (see [2, 3, 19]):

$(V_{R})$:

$V\in C(\mathbb {R}^{N},\mathbb {R}),$ $V(x)\ge V_{0}>0$ for all $x\in \mathbb {R}^{N}$ and $V(x)\rightarrow +\infty$ as $\left| x\right| \rightarrow +\infty .$

$(V_{BW})$:

$V\in C( \mathbb {R} ^{N}, \mathbb {R} ),$ $V(x)\ge V_{0}>0$ for all $x\in \mathbb {R} ^{N}$ and for every $M>0$

$$\begin{aligned} \lambda (\{x\in \mathbb {R} ^{N}:V(x)\le M\})<+\infty , \end{aligned}$$

where $\lambda$ denotes Lebesgue measure in $\mathbb {R} ^{N}.$

$(V_{BF})$:

$V\in (L^{2}( \mathbb {R} ^{N}))_{\text {loc}}$ is positive and

$$\begin{aligned} \int _{S(x)}\frac{1}{V(y)}dy\rightarrow 0 \ \ \text {for}\left| x\right| \rightarrow +\infty , \end{aligned}$$

where $S(x)=\{y\in \mathbb {R} ^{N}:\left| y-x\right| <1\}$ is the unit sphere centred at x.

The condition $(V_{R}),$ i.e. the coerciveness of V,  is strictly weaker than $(V_{BW}),$ and $(V_{BW})$ is strictly weaker than $(V_{BF})$ (see [Proposition 3.1 and Remark 3.2][20]).

It is easy to observe that if we create the discrete counterparts of conditions $(V_{R}),(V_{BW})$ and $(V_{BF}),$ they will all be equivalent. In particular, (V) and $(\hat{V})$ are equivalent. We also note that conditions (V) and $(\hat{V})$ guarantee compactness of the embedding of the suitable solution space into $l^{p}$ (see [Lemma 2.3] [9] and [Lemma 2.3][10]).

Our main tool is a variant of Clark’s theorem from [14]. Here we state it for the reader’s convenience. For $J\in C^{1}(E, \mathbb {R} ),$ where E is a Banach space, we say J satisfies the Palais-Smale (write by (PS), for short) condition if any sequence $\{u^{(n)}\}\subset E$ for which $J(u^{(n)})$ is bounded and $J^{\prime }(u^{(n)})\rightarrow 0$ has a convergent subsequence.

Theorem 2

Let E be a Banach space, $J\in C^{1}(E, \mathbb {R} )$. Assume J satisfies the (PS) condition, is even and bounded from below, and $J(0)=0$. If for any $m\in \mathbb {N}$, there exists a m-dimensional subspace $E_{m}$ of E and $\varrho _{m}>0$ such that $\sup _{E_{m}\cap S_{\varrho _{m}}}J<0$, where $S_{\varrho }=\{u\in E:\left\| u\right\| =\varrho \}$, then there exists a sequence of critical points $\{u^{(m)}\}$ satisfying $J(u^{(m)})<0$ for all m and $\left\| u^{(m)}\right\| \rightarrow 0$ as $m\rightarrow \infty$.

2 Proof of the Main Result

To use Clark’s theorem, we need to construct an appropriate solution space and define an energy functional on it such that its critical points are solutions to our problem. The conditions imposed on the nonlinearity f control its behavior only in the neighbourhood of the point $t=0$. As will be seen below, this is enough to obtain most of the needed properties of the energy functional, such as its continuous differentiability and the weak-strong continuity of it and its derivative. On the other hand, in order to obtain boundedness from below of the energy functional, we need the appropriate behavior of the nonlinearity f on $\mathbb {Z} \times \mathbb {R}$. Therefore, we will use the truncation method.

By $(f_{1})$ and $(f_{2})$ we can find $\varepsilon >0$ and $\alpha >0$ such that

$$\begin{aligned} f(k,-t)=-f(k,t) \end{aligned}$$

(13)

and

$$\begin{aligned} \left| f(k,t)\right| \le \alpha \left( a(k)+\left| t\right| ^{p-1}\right) \end{aligned}$$

(14)

for all $k\in \mathbb {Z}$ and $t\in (-2\varepsilon ,2\varepsilon ).$

Choose $g\in C_{0}^{\infty }( \mathbb {R}, \mathbb {R} )$ such that $g(-t)=g(t),$ $0\le g(t)\le 1$ for all $t\in \mathbb {R},$ $g(t)=1$ for $\left| t\right| \le \varepsilon$ and $g(t)=0$ for $\left| t\right| \ge 2\varepsilon .$ Let

$$\begin{aligned} \tilde{f}(k,t)=f(k,t)g(t) \end{aligned}$$

for all $k\in \mathbb {Z},$ $t\in \mathbb {R}.$ Then $\tilde{f}$ satisfies (13) and (14) for all $k\in \mathbb {Z},$ $t\in \mathbb {R}.$ Let $\tilde{F}(k,t)=\int _{0}^{t}\tilde{f}(k,s)ds$ for every $t\in \mathbb {R}$ and $k\in \mathbb {Z}.$ It is important for us to note that there is $M>0$ such that

$$\begin{aligned} \left| \tilde{F}(k,t)\right| \le M\ \ \ \text {for all }k\in \mathbb {Z},\ t\in \mathbb {R}. \end{aligned}$$

(15)

Indeed, it is easy to see that for all $k\in \mathbb {Z}$ we have $\left| \tilde{f}(k,t)\right| \le \alpha \left( \left\| a\right\| _{\infty }+(2\varepsilon )^{p-1}\right)$ if $\left| t\right| <2\varepsilon$ and $\tilde{f}(k,t)=0$ if $\left| t\right| \ge 2\varepsilon .$ Hence (15) holds.

We consider the auxiliary problem

$$\begin{aligned} \left\{ \begin{array}{ll} \left( -\Delta \right) _{p}^{s}u(k)+V(k)\left| u(k)\right| ^{p-2}u(k)=\tilde{f}(k,u(k)) &{} \text{ for } \text{ all } k\in \mathbb {Z} \\ u(k)\rightarrow 0 &{} \text{ as } |k|\rightarrow \infty . \end{array} \right. \end{aligned}$$

(16)

Clearly, if u is a solution of problem (16) with $\left\| u\right\| _{\infty }<\varepsilon ,$ then u is a solution of problem ( 8). So, if we find a sequence of non-trivial solutions of (16) with $\left\| \cdot \right\| _{\infty }$-norms tending to zero, we obtain infinite many non-trivial solutions of (8).

If $0<s<1,$ for the solution space, we take

$$\begin{aligned} E_{s}=\left\{ u: \mathbb {Z} \rightarrow \mathbb {R}:\sum _{j\in \mathbb {Z} }\sum _{m\in \mathbb {Z} }\left| u(j)-u(m)\right| ^{p}K_{s,p}(j-m)+\sum _{k\in \mathbb {Z} }V(k)\left| u(k)\right| ^{p}<\infty \right\} \end{aligned}$$

equipped with the norm

$$\begin{aligned} \left\| u\right\| =\left( \sum _{j\in \mathbb {Z} }\sum _{m\in \mathbb {Z} }\left| u(j)-u(m)\right| ^{p}K_{s,p}(j-m)+\sum _{k\in \mathbb {Z} }V(k)\left| u(k)\right| ^{p}\right) ^{\frac{1}{p}}. \end{aligned}$$

If $s=1,$ for the solution space, we take

$$\begin{aligned} E_{1}=\left\{ u: \mathbb {Z} \rightarrow \mathbb {R}:\sum _{k\in \mathbb {Z} }\left| u(k)-u(k-1)\right| ^{p}+\sum _{k\in \mathbb {Z} }V(k)\left| u(k)\right| ^{p}<\infty \right\} \end{aligned}$$

equipped with the norm

$$\begin{aligned} \left\| u\right\| =\left( \sum _{k\in \mathbb {Z} }\left| u(k)-u(k-1)\right| ^{p}+\sum _{k\in \mathbb {Z} }V(k)\left| u(k)\right| ^{p}\right) ^{\frac{1}{p}}. \end{aligned}$$

Then $E_{s}$ is a reflexive Banach space which compactly embeds into $l^{r}$ for all $p\le r\le \infty$ (see [10] for $0<s<1$ and [6, 8] for $s=1$). Moreover, $\left\| u\right\| _{\infty }\le V_{0}^{-1/p}\left\| u\right\|$ for all $u\in E_{s}.$

Define functionals $J_{s},I_{s},K:E_{s}\rightarrow \mathbb {R}$ as

$$\begin{aligned} J_{s}(u)= & {} I_{s}(u)-K(u), \\ I_{s}(u)= & {} \frac{1}{p}\sum _{j\in \mathbb {Z} }\sum _{m\in \mathbb {Z} }\left| u(j)-u(m)\right| ^{p}K_{s,p}(j-m)+\frac{1}{p}\sum _{k\in \mathbb {Z} }V(k)\left| u(k)\right| ^{p},\ \\ {}{} & {} \qquad \text { if }0<s<1, \\ I_{1}(u)= & {} \frac{1}{p}\sum _{k\in \mathbb {Z} }\left| u(k)-u(k-1)\right| ^{p}+\frac{1}{p}\sum _{k\in \mathbb {Z} }V(k)\left| u(k)\right| ^{p},\ \ \text {if }s=1, \\ K(u)= & {} \sum _{k\in \mathbb {Z} }\tilde{F}(k,u(k)). \end{aligned}$$

for $u\in E_{s},0<s\le 1$. By [Lemma 2.5] [9], $I_{s}\in C^{1}(E_{s}, \mathbb {R} ),0<s<1,$ with

$$\begin{aligned}{} & {} \left\langle I_{s}^{\prime }(u),v\right\rangle =\sum _{j\in \mathbb {Z} }\sum _{m\in \mathbb {Z} }\left| u(j)-u(m)\right| ^{p-2}\left( u(j)-u(m)\right) \left( v(j)-v(m)\right) K_{s,p}(j-m)\\{} & {} +\sum _{k\in \mathbb {Z} }V(k)(\left| u(k)\right| ^{p-2}u(k)v(k) \end{aligned}$$

for all $u,v\in E_{s}.$ By [8][Proposition 5], $I_{1}\in C^{1}(E_{1}, \mathbb {R} )$ with

$$\begin{aligned}{} & {} \left\langle I_{1}^{\prime }(u),v\right\rangle =\sum _{k\in \mathbb {Z} }\left| u(k)-u(k-1)\right| ^{p-2}\left( u(k)-u(k-1)\right) \left( v(k)-v(k-1)\right) \\{} & {} +\sum _{k\in \mathbb {Z} }V(k)(\left| u(k)\right| ^{p-2}u(k)v(k) \end{aligned}$$

for all $u,v\in E_{1}.$

Now, we will prove the needed properties of the functional K. First, we will recall the Vitali Theorem for sequence spaces (see [Proposition 3.128][5]).

Theorem 3

Let $\{w^{(n)}\}_{n\in \mathbb {N} }\subset l^{1}$ be a sequence such that $w_{k}^{(n)}\rightarrow w_{k}$ for all $k\in \mathbb {Z}.$ Then the following assertions are equivalent:

• (a) $w^{(n)}\rightarrow w$ in $l^{1}.$

• (b) The family $\{\left| w^{(n)}\right| :n\in \mathbb {N} \}$ is uniformly summable, that is, for every $\varepsilon >0,$ we can find $h\in \mathbb {N}$ such that

  $$\begin{aligned} \sup _{n\in \mathbb {N} }\sum _{\left| k\right| \ge h}\left| w_{k}^{(n)}\right| <\varepsilon . \end{aligned}$$

In our case, the function $\tilde{f}$ satisfies condition (14) globally, i.e. on $\mathbb {Z} \times \mathbb {R}.$ In order to obtain the needed properties of the functional K,  it is enough to control $\tilde{f}$ in a neighbour of $t=0.$ Therefore, we formulate and prove the result with a weaker assumption. Here we follow [24].

Lemma 4

Assume (V) and let $f\in C( \mathbb {Z} \times \mathbb {R}, \mathbb {R} )$ satisfy the following condition: there exist $\varepsilon ,C>0$ and a positive $a\in l^{p^{\prime }},$ $p^{\prime }=\frac{p}{p-1},$ such that

$$\begin{aligned} \left| f(k,t)\right| \le C\left( a(k)+\left| t\right| ^{p-1}\right) \end{aligned}$$

(17)

for all $k\in \mathbb {Z}$ and $\left| t\right| <\varepsilon .$ Then the functional $K:E_{s}\rightarrow \mathbb {R},$ $K(u)=\sum _{k\in \mathbb {Z} }F(k,u(k)),$ where $F(k,t)=\int _{0}^{t}f(k,s)ds$ for every $t\in \mathbb {R}$ and $k\in \mathbb {Z},$ has the following properties:

• (a) K is sequentially weakly-strongly continuous, that is, if $u^{(n)}\rightharpoonup u$, then $K(u^{(n)})\rightarrow K(u)$.

• (b) K is of class $C^{1}$ with

  $$\begin{aligned} \left\langle K^{\prime }(u),v\right\rangle =\sum _{k\in \mathbb {Z} }f(k,u(k))v(k) \end{aligned}$$

  (18)

  for all $u,v\in E_{s}.$

• (c) $K^{\prime }$ is sequentially weakly-strongly continuous, i.e., if $u^{(n)}\rightharpoonup u$, then $K^{\prime }(u^{(n)})\rightarrow K^{\prime }(u)$ in $E_{s}^{*}.$

Proof

By (17), there exists $C_{1}>0$ such that

$$\begin{aligned} \left| F(k,t)\right| \le C_{1}(a(k)\left| t\right| +\left| t\right| ^{p}) \ \ \ \ \ \text {for all }k\in \mathbb {Z} \text { and }\left| t\right| <\varepsilon . \end{aligned}$$

(19)

For any $u\in E_{s}$ there exists $h\in \mathbb {N}$ such that $\left| u(k)\right| <\varepsilon$ for all $\left| k\right| \ge h$ and so, using (19) and Hölder’s inequality, we deduce that $F(\cdot ,u(\cdot ))$ is summable and K(u) is well defined.

(a) Assume that $u^{(n)}\rightharpoonup u$ in E. Since $E_{s}$ compactly embeds into $l^{p}$, up to a subsequence, $\{\left| u^{(n)}\right| ^{p}\}$ converges in $l^{1}.$ Thus

$$\begin{aligned} \left\{ a^{p^{\prime }}+\left| u^{(n)}\right| ^{p}+\left| u\right| ^{p}:n\in \mathbb {N} \right\} \subset l^{1}\ \end{aligned}$$

is uniformly summable by Theorem 3. In particular, there exists $h_{1}\in \mathbb {N}$ such that $\left| u^{(n)}(k)\right| ,\left| u(k)\right| <\varepsilon$ for all $\left| k\right| >h_{1}$ and $n\in \mathbb {N}.$ Now, from (19) and Young’s inequality, there exists $C_{2}>0,$ such that

$$\begin{aligned} \left| F(k,u^{(n)}(k))-F(k,u(k))\right| \le C_{2}\left( (a(k))^{p^{\prime }}+\left| u^{(n)}(k)\right| ^{p}+\left| u(k)\right| ^{p}\right) \end{aligned}$$

for all $\left| k\right| >h_{1}$ and $n\in \mathbb {N}$. Hence $\left\{ F(\cdot ,u^{(n)})-F(\cdot ,u):n\in \mathbb {N} \right\}$ is uniformly summable. As F is continuous, we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\sum _{k\in \mathbb {Z} }\left| F(k,u^{(n)}(k))-F(k,u(k))\right| =0, \end{aligned}$$

by Theorem 3. This shows $K(u^{(n)})\rightarrow K(u).$

(b) To see formula (18) take $u,v\in E_{s}$. For any $k\in \mathbb {Z}$ and $0<\left| t\right| <1$ there exists $\lambda _{k,t}\in [0,1]$ such that

$$\begin{aligned} \frac{F(k,u(k)+tv(k))-F(k,u(k))}{t}=f(k,u(k)+\lambda _{k,t}tv(k))v(k), \end{aligned}$$

by the Mean Value Theorem. We claim that

$$\begin{aligned} \lim _{t\rightarrow 0}\sum _{k\in \mathbb {Z} }\left( f(k,u(k)+\lambda _{k,t}tv(k))-f(k,u(k))\right) v(k)=0. \end{aligned}$$

Suppose, by contradiction, that there exist $\delta >0$ and a null sequence $\{t_{n}\}_{n\in \mathbb {N} }\subset [-1,1]$ such that

$$\begin{aligned} \sum _{k\in \mathbb {Z} }\left( f(k,u(k)+\lambda _{k,t_{n}}t_{n}v(k))-f(k,u(k))\right) v(k)>\delta \ \ \text { for all }n\in \mathbb {N}. \end{aligned}$$

(20)

Since $u,v\in l^{p},$ there is $h_{2}\in \mathbb {N}$ such that $\left| u(k)\right| +\left| v(k)\right| <\varepsilon$ for all $\left| k\right| >h_{2}$. By (17) and Young’s inequality, we have for all $\left| t\right| <1,$ $\lambda \in [0,1],$ $k\in \mathbb {Z}$ and some constant $C_{3}>0$

$$\begin{aligned} \left| f(k,u(k)+\lambda tv(k))v(k)\right| \le C_{3}\left( (a(k))^{p^{\prime }}+\left| u(k)\right| ^{p}+\left| v(k)\right| ^{p}\right) . \end{aligned}$$

Since $a^{p^{\prime }},\left| u\right| ^{p},\left| v\right| ^{p}\in l^{1},$ the family $\left\{ \{f(k,u(k)+\lambda _{k,t_{n}}t_{n}v(k))\}_{k\in \mathbb {Z} }:n\in \mathbb {N} \right\} \subset l^{1}$ is uniformly summable. This and the continuity of f give us

$$\begin{aligned} \lim _{n\rightarrow \infty }\sum _{k\in \mathbb {Z} }\left( f(k,u(k)+\lambda _{k,t_{n}}t_{n}v(k))-f(k,u(k))\right) v(k)=0, \end{aligned}$$

by Theorem 3, which contradicts (20). Hence K is Gateaux differentiable and formula (18) holds. Clearly, after proving (c) we will have continuous Frechet differentiability of K.

(c) Assume that $u^{(n)}\rightharpoonup u$ in E. We have

$$\begin{aligned} \left\| K^{\prime }(u^{(n)})-K^{\prime }(u)\right\| _{E_{s}^{*}}= & {} \sup _{v\in E_{s},\left\| v\right\| =1}\left| \sum _{k\in \mathbb {Z} }(f(k,u^{(n)}(k))-f(k,u(k)))v(k)\right| \\\le & {} \sup _{v\in E_{s},\left\| v\right\| =1}\sum _{k\in \mathbb {Z} }\left| f(k,u^{(n)}(k))-f(k,u(k))\right| \left| v(k)\right| . \end{aligned}$$

Set

$$\begin{aligned} \alpha =\underset{n\rightarrow \infty }{\lim \sup }\sup _{v\in E_{s},\left\| v\right\| =1}\sum _{k\in \mathbb {Z} }\left| f(k,u^{(n)}(k))-f(k,u(k))\right| \left| v(k)\right| . \end{aligned}$$

We assert that $\alpha =0$. Suppose, by contradiction, that $\alpha >0$. Hence, there exists a sequence $\{v^{(n)}\}\subset E$ with $\left\| v^{(n)}\right\| =1$ such that

$$\begin{aligned} \sum _{k\in \mathbb {Z} }\left| f(k,u^{(n)}(k))-f(k,u(k))\right| \left| v^{(n)}(k)\right| >\frac{\alpha }{2} \end{aligned}$$

(21)

for sufficiently large n. Since $E_{s}$ compactly embeds into $l^{p}$, up to subsequences, $\{\left| u^{(n)}\right| ^{p}\}$ and $\{\left| v^{(n)}\right| ^{p}\}$ converge in $l^{1}.$ Thus

$$\begin{aligned} \left\{ \left| a\right| ^{p^{\prime }}+\left| u\right| ^{p}+\left| u^{(n)}\right| ^{p}+\left| v^{(n)}\right| ^{p}:n\in \mathbb {N} \right\} \subset l^{1} \end{aligned}$$

is uniformly summable by Theorem 3. In particular, there exists $h_{3}\in \mathbb {N}$ such that $\left| u^{(n)}(k)\right| ,\left| u(k)\right| <\varepsilon$ for all $\left| k\right| >h_{3}$ and $n\in \mathbb {N}.$ From (17) and Young’s inequality, we have

$$\begin{aligned} \left| f(k,u^{(n)}(k))-f(k,u(k))\right| \left| v^{(n)}(k)\right| \le C_{4}\left( \left| a(k)\right| ^{p^{\prime }}+\left| u(k)\right| ^{p}+\left| u^{(n)}(k)\right| ^{p}+\left| v^{(n)}(k)\right| ^{p}\right) \end{aligned}$$

for some constant $C_{4}>0$ and all $\left| k\right| >h_{3},n\in \mathbb {N}.$ Hence

$$\begin{aligned} \left\{ \left| f(\cdot ,u^{(n)})-f(\cdot ,u)\right| \left| v^{(n)}\right| :n\in \mathbb {N} \right\} \subset l^{1} \end{aligned}$$

is uniformly summable. Since f is continuous,

$$\begin{aligned} \lim _{n\rightarrow +\infty }\sum _{k\in \mathbb {Z} }\left| f(k,u^{(n)}(k))-f(k,u(k))\right| \left| v^{(n)}(k)\right| =0, \end{aligned}$$

by Vitali Theorem. This contradicts (21) if $\alpha >0$. Hence, $\left\| K^{\prime }(u^{(n)})-K^{\prime }(u)\right\| _{E_{s}^{*}}\rightarrow 0$ as $n\rightarrow +\infty$. $\square$

All the above establish $J_{s}\in C^{1}(E_{s}, \mathbb {R} )$ with

$$\begin{aligned} \left\langle J_{s}^{\prime }(u),v\right\rangle= & {} \sum _{j\in \mathbb {Z} }\sum _{m\in \mathbb {Z} }\left| u(j)-u(m)\right| ^{p-2}\left( u(j)-u(m)\right) \left( v(j)-v(m)\right) K_{s,p}(j-m) \nonumber \\{} & {} +\sum _{k\in \mathbb {Z} }V(k)(\left| u(k)\right| ^{p-2}u(k)v(k)-\sum _{k\in \mathbb {Z} }\tilde{f}(k,u(k))v(k) \nonumber \\ \end{aligned}$$

(22)

if $0<s<1$ and

$$\begin{aligned} \left\langle J_{1}^{\prime }(u),v\right\rangle= & {} \sum _{k\in \mathbb {Z} }\left| u(k)-u(k-1)\right| ^{p-2}\left( u(k)-u(k-1)\right) \left( v(k)-v(k-1)\right) \\{} & {} +\sum _{k\in \mathbb {Z} }V(k)(\left| u(k)\right| ^{p-2}u(k)v(k)-\sum _{k\in \mathbb {Z} }\tilde{f}(k,u(k))v(k) \end{aligned}$$

if $s=1$, for all $u,v\in E_{s}.$ For every $k\in \mathbb {Z},$ taking in the above formulas $v=e_{k}=\{\delta _{kj}\}_{j\in \mathbb {Z} },$ where $\delta _{kj}$ is the Kronecker delta, we conclude that every critical point of $J_{s}$ is a homoclinic solution of problem (16).

Now, we show that $J_{s}$ satisfies all assumptions in Theorem 2. Clearly, $J_{s}$ is even and $J_{s}(0)=0.$

Claim 5

$J_{s}$ is bounded from below.

First, we observe that there exists a constant $C>0$ such that

$$\begin{aligned} K(u)\le \frac{1}{2}I_{s}(u)+C \end{aligned}$$

for all $u\in E_{s}$. Indeed, applying (13)-(15) gives

$$\begin{aligned} \tilde{F}(k,t)\le & {} \alpha \left( a(k)\left| t\right| +\frac{1}{p} \left| t\right| ^{p}\right) , \\ \tilde{F}(k,t)\le & {} M \end{aligned}$$

for all $k\in \mathbb {Z},t\in \mathbb {R}.$ Moreover, by (V), there exists $h\in \mathbb {N}$ such that

$$\begin{aligned} V(k)\ge 4\alpha \end{aligned}$$

for all $\left| k\right| \ge h.$ Hence

, all the above and Young’s inequality give us

$$\begin{aligned} K(u)= & {} \sum _{k<h}\tilde{F}(k,u(k))+\sum _{\left| k\right| \ge h} \tilde{F}(k,u(k)) \\\le & {} 2hM+\alpha \sum _{\left| k\right| \ge h}a(k)\left| u(k)\right| +\frac{\alpha }{p}\sum _{\left| k\right| \ge h}\left| u(k)\right| ^{p} \\\le & {} 2hM+\frac{\alpha }{p^{\prime }}\left\| a\right\| _{p^{\prime }}+ \frac{2\alpha }{p}\sum _{\left| k\right| \ge h}\left| u(k)\right| ^{p} \\\le & {} C+\frac{1}{2p}\sum _{\left| k\right| \ge h}V(k)\left| u(k)\right| ^{p} \\\le & {} C+\frac{1}{2}I_{s}(u) \end{aligned}$$

for all $u\in E_{s}$, where $C=2hM+\frac{\alpha }{p^{\prime }}\left\| a\right\| _{p^{\prime }}$. Therefore,

$$\begin{aligned} J_{s}(u)=I_{s}(u)-K(u)\ge \frac{1}{2}I_{s}(u)-C=\frac{1}{2p}\left\| u\right\| ^{p}-C \end{aligned}$$

(23)

for all $u\in E_{s}.$ Hence, $J_{s}$ is bounded from below. This proves Claim 5.

Claim 6

$J_{s}$ satisfies the (PS) condition.

Let $\{u^{(n)}\}$ be a sequence in $E_{s}$ such that $\{J_{s}(u^{(n)})\}$ is bounded and $J_{s}^{\prime }(u^{(n)})\rightarrow 0.$ By (23), $\{u^{(n)}\}$ is bounded. Then, passing to a subsequence if necessary, it can be assumed that there exists $u\in E_{s}$ such that $u^{(n)}\rightharpoonup u$ weakly in $E_{s}$ and $u^{(n)}\rightarrow u$ strongly in $l^{p}.$ All this and Lemma 4 give us

$$\begin{aligned} \left\langle J_{s}^{\prime }(u^{(n)})-J_{s}^{\prime }(u),u^{(n)}-u\right\rangle\rightarrow & {} 0, \\ \left\langle K^{\prime }(u^{(n)})-K^{\prime }(u),u^{(n)}-u\right\rangle\rightarrow & {} 0 \end{aligned}$$

and so

$$\begin{aligned} \left\langle I_{s}^{\prime }(u^{(n)})-I_{s}^{\prime }(u),u^{(n)}-u\right\rangle \rightarrow 0. \end{aligned}$$

(24)

Now, if $s=1,$ we argue as in [12][Lemma 3.4] and we obtain $u^{(n)}\rightarrow u$ in $E_{1}.$ If $0<s<1,$we argue similarly. Let $\varphi _{p}(t)=\left| t\right| ^{p-2}t,t\in \mathbb {R}.$ Recall the following inequalities (see [16][p. 25])

$$\begin{aligned} \left( \varphi _{p}(s)-\varphi _{p}(t)\right) (s-t)\ge \left\{ \begin{array}{ll} c_{1}(p)\left| s-t\right| ^{p}, &{} \text {if }p\ge 2, \\ c_{2}(p)\frac{\left| s-t\right| ^{2}}{\left( \left| s\right| +\left| t\right| \right) ^{2-p}}, &{} \text {if }1<p<2, \end{array} \right. \end{aligned}$$

(25)

for all $s,t\in \mathbb {R},$ where $c_{1}(p),c_{2}(p)>0$ are constants.

If $p\ge 2,$ then

$$\begin{aligned}{} & {} \left\langle I_{s}^{\prime }(u^{(n)})-I_{s}^{\prime }(u),u^{(n)}-u\right\rangle \\= & {} \sum _{j\in \mathbb {Z} }\sum _{m\in \mathbb {Z} }\left( \varphi _{p}\left( u^{(n)}(j)-u^{(n)}(m)\right) -\varphi _{p}\left( u(j)-u(m)\right) \right) \\{} & {} \left( u^{(n)}(j)-u^{(n)}(m)-(u(j)-u(m))\right) K_{s,p}(j-m) \\{} & {} +\sum _{k\in \mathbb {Z} }V(k)(\varphi _{p}(u^{(n)}(k))-\varphi _{p}(u(k)))(u^{(n)}(k)-u(k)) \\\ge & {} c_{1}(p)\left\| u^{(n)}-u\right\| ^{p}. \end{aligned}$$

Hence $u^{(n)}\rightarrow u$ in $E_{s},$ by (24).

If $1<p<2$, then, by Hölder’s inequality, we obtain

$$\begin{aligned}{} & {} \sum _{j\in \mathbb {Z} }\sum _{m\in \mathbb {Z} }\left| u^{(n)}(j)-u(j)-(u^{(n)}(m)-u(m))\right| ^{p}K_{s,p}(j-m) \nonumber \\\le & {} \left( \sum _{j\in \mathbb {Z} }\sum _{m\in \mathbb {Z} }\frac{\left| u^{(n)}(j)-u^{(n)}(m)-(u(j)-u(m))\right| ^{2}K_{s,p}(j-m)}{\left( \left| u^{(n)}(j)-u^{(n)}(m)\right| +\left| u(j)-u(m)\right| \right) ^{2-p}}\right) ^{\frac{p}{2}}\times \nonumber \\{} & {} \left( \sum _{j\in \mathbb {Z} }\sum _{m\in \mathbb {Z} }\left( \left| u^{(n)}(j)-u^{(n)}(m)\right| +\left| u(j)-u(m)\right| \right) ^{p}K_{s,p}(j-m)\right) ^{\frac{2-p}{2}} \nonumber \\\le & {} C_{1}\left( \sum _{j\in \mathbb {Z} }\sum _{m\in \mathbb {Z} }\frac{\left| u^{(n)}(j)-u^{(n)}(m)-(u(j)-u(m))\right| ^{2}K_{s,p}(j-m)}{\left( \left| u^{(n)}(j)-u^{(n)}(m)\right| +\left| u(j)-u(m)\right| \right) ^{2-p}}\right) ^{\frac{p}{2}}\left( \left\| u^{(n)}\right\| +\left\| u\right\| \right) ^{\frac{p(2-p) }{2}} \nonumber \\ \end{aligned}$$

(26)

and similarly,

$$\begin{aligned}{} & {} \sum _{k\in \mathbb {Z} }V(k)\left| u^{(n)}(k)-u(k)\right| ^{p}\le C_{1}\left( \sum _{k\in \mathbb {Z} }V(k)\frac{\left| u^{(n)}(k)-u(k)\right| ^{2}}{\left( \left| u^{(n)}(k)\right| +\left| u(k)\right| \right) ^{2-p}}\right) ^{ \frac{p}{2}}\nonumber \\{} & {} \left( \left\| u^{(n)}\right\| +\left\| u\right\| \right) ^{\frac{p(2-p)}{2}}. \end{aligned}$$

(27)

Thus, using (25)-(27), we have

$$\begin{aligned}{} & {} \left\langle I_{s}^{\prime }(u^{(n)})-I_{s}^{\prime }(u),u^{(n)}-u\right\rangle \\\ge & {} c_{2}(p)\left( \sum _{j\in \mathbb {Z} }\sum _{m\in \mathbb {Z} }\frac{\left| u^{(n)}(j)-u^{(n)}(m)-(u(j)-u(m))\right| ^{2}K_{s,p}(j-m)}{\left( \left| u^{(n)}(j)-u^{(n)}(m)\right| +\left| u(j)-u(m)\right| \right) ^{2-p}} +\sum _{k\in \mathbb {Z}}V(k)\frac{\left| u^{(n)}(k)-u(k)\right| ^{2}}{\left( \left| u^{(n)}(k)\right| +\left| u(k)\right| \right) ^{2-p}}\right) \\\ge & {} C_{2}\frac{\left( \sum _{j\in \mathbb {Z} }\sum _{m\in \mathbb {Z} }\left| u^{(n)}(j)-u(j)-(u^{(n)}(m)-u(m))\right| ^{p}K_{s,p}(j-m)\right) ^{\frac{2}{p}}+\left( \sum _{k\in \mathbb {Z} }V(k)\left| u^{(n)}(k)-u(k)\right| ^{p}\right) ^{\frac{2}{p}}}{ \left( \left\| u^{(n)}\right\| +\left\| u\right\| \right) ^{(2-p)}} \\\ge & {} C_{3}\frac{\left\| u^{(n)}-u\right\| ^{2}}{\left( \left\| u^{(n)}\right\| +\left\| u\right\| \right) ^{(2-p)}} \end{aligned}$$

and so $u^{(n)}\rightarrow u$ in $E_{s},$ by (24). This proves Claim 6.

Claim 7

For any $m\in \mathbb {N}$, there exists a m-dimensional subspace $E_{m}$ of $E_{s}$ and $\varrho _{m}>0$ such that $\sup _{E_{m}\cap S_{\varrho _{m}}}J_{s}<0.$

Let $Z_{0}=\{k_{i}:i\in \mathbb {N} \}\subset \mathbb {Z}$ be the set with the properties listed in $(f_{3})$. Define $e^{(i)}:=e_{k_{i}},$ that is $e^{(i)}=\{e^{(i)}(k)\}_{k\in \mathbb {Z} },i\in \mathbb {N},$ with

$$\begin{aligned} e^{(i)}(k)=\left\{ \begin{array}{ll} 1, &{} \text {if }k=k_{i}, \\ 0, &{} \text {if }k\ne k_{i}. \end{array} \right. \end{aligned}$$

Define $E_{m}:=$span$\{e^{(i)}:i=1,...,m\},$ $m\in \mathbb {N}.$ Clearly, $E_{m}\subset E_{s}$ and $\dim E_{m}=m.$ Now, fix $m\in \mathbb {N}$ and write $Z_{m}:=\{k_{1},...,k_{m}\}\subset Z_{0}.$ Since $Z_{m}$ is a finite set, there exist $C_{m}>0$ and $0<\varepsilon _{m}<\varepsilon$ such that

$$\begin{aligned} \tilde{F}(k,t)\ge C_{m}\left| t\right| ^{\sigma } \end{aligned}$$

for all $k\in Z_{m}$ and $\left| t\right| <\varepsilon _{m},$ by $(f_{3})$ and the fact that $\tilde{F}=F$ on $\mathbb {Z} \times (-\varepsilon ,\varepsilon ).$ Since all norms are equivalent on the finite dimensional space $E_{m}$, we have

$$\begin{aligned} J_{s}(u)= & {} \frac{1}{p}\left\| u\right\| ^{p}-\sum _{k\in \mathbb {Z} }\tilde{F}(k,u(k))=\frac{1}{p}\left\| u\right\| ^{p}-\sum _{k\in Z_{m}} \tilde{F}(k,u(k)) \\\le & {} \frac{1}{p}\left\| u\right\| ^{p}-C_{m}\sum _{k\in Z_{m}}\left| u(k)\right| ^{\sigma }=\frac{1}{p}\left\| u\right\| ^{p}-C_{m}\left\| u\right\| _{\sigma }^{\sigma } \\\le & {} \frac{1}{p}\left\| u\right\| ^{p}-\tilde{C}_{m}\left\| u\right\| ^{\sigma } \end{aligned}$$

for all $u\in E_{m}$ with $\left\| u\right\| _{\infty }<\varepsilon _{m}.$ From this and $1<\sigma <p,$ we conclude that there exists $0<\varrho _{m}<\varepsilon _{m}$ such that $\sup _{E_{m}\cap S_{\varrho _{m}}}J_{s}<0.$ This proves Claim 7.

We have verified all the conditions of Theorem 2. Consequently, problem (16) has a sequence $\{u^{(m)}\}$ of non-trivial solutions with $\left\| u^{(m)}\right\| \rightarrow 0$ as $m\rightarrow \infty$. Since also $\left\| u^{(m)}\right\| _{\infty }\rightarrow 0$ as $m\rightarrow \infty ,$ there exists $m_{0}$ such that for every $m\ge m_{0},$ $u^{(m)}$ is the solution of problem (8). This proves Theorem 1.