Solvability for a discrete fractional mixed type sum-difference equation boundary value problem in a Banach space

Abstract

In this paper, by means of Darbo’s fixed point theorem, we establish the existence of solutions to a nonlinear discrete fractional mixed type sum-difference equation boundary value problem in a Banach space. Additionally, as an application, we give an example to demonstrate the main result.

1 Introduction

Throughout this paper, we denote $\mathbb{N}_{a}=\{a, a+1, a+2, \ldots\}$ and $\mathbb{N}_{a}^{b}=\{a, a+1, \ldots, b\}$ for $a,b\in\mathbb{R}$ with $b-a\in\mathbb{N}_{1}$. Moreover, for any Banach-valued function u on $\mathbb{N}_{a}$, we make the convention that $\sum_{s=k_{1}}^{k_{2}}u(s)=\theta$ if $k_{1},k_{2}\in\mathbb{N}_{a}$ with $k_{1}>k_{2}$, where θ is the zero element of a given Banach space.

In this paper, we consider the following discrete fractional mixed type sum-difference equation boundary value problem in Banach space E:

$$ \textstyle\begin{cases} \Delta^{\beta}u(t) +f (t+\beta-1,u(t+\beta-1),(Tu)(t),(Su)(t) ) =\theta,\quad t\in\mathbb{N}_{0},\\ u(\beta-2)=\theta, \qquad\Delta^{\beta-1}u(\infty)=u_{\infty}, \end{cases} $$

(1.1)

where $\beta\in(1,2]$, $\Delta^{\beta}$ denotes the discrete Riemann-Liouville fractional difference of order β, $f:\mathbb{N}_{\beta-1}\times E\times E\times E\rightarrow E$ is a continuous function, θ is the zero element of E, $\Delta^{\beta-1}u(\infty) =\lim_{t\rightarrow+\infty}\Delta^{\beta-1}u(t) =u_{\infty}\in E$, and

$$(Tu) (t)=\sum_{s=0}^{t}k(t,s)u(s+\beta-1), \qquad (Su) (t)=\sum_{s=0}^{\infty}h(t,s)u(s+ \beta-1), $$

where $k:D\rightarrow\mathbb{R}$, $D=\{(t,s)\in\mathbb{N}_{0}\times\mathbb{N}_{0}:s\leq t\}$, $h:\mathbb{N}_{0}\times\mathbb{N}_{0}\rightarrow\mathbb{R}$.

In the last years, discrete fractional calculus and fractional difference equations with various boundary conditions have been studied more intensively. For details, see [1–16]; we particularly should note that the recent monograph of Goodrich and Peterson [1] is an extremely useful textbook for readers to obtain the fundamental background in discrete fractional calculus. However, as far as we know, most of the recent papers, mainly based on the Krasnosel’skii fixed point theorem, are concerning about the existence of positive solutions for discrete fractional boundary value problems on a finite interval, and few papers can be found in the literature for discrete boundary value problems on an infinite interval [16].

Lv and Feng [16] initially introduced some basic conceptions and fundamental results on discrete fractional calculus for any Banach-valued function and also, using of the contraction mapping principle, investigated the existence and uniqueness of solutions for a class of fractional mixed type sum-difference equation boundary value problems on discrete infinite intervals in Banach spaces. This is the first attempt to study the discrete fractional difference equation boundary value problems in abstract spaces.

It is well known that the measure of noncompactness is a very powerful tool to deal with differential equations [17–20], difference equations [21–23], integration equations [24], and differential inclusions [25, 26]. So, in this paper, we l employ noncompact measures and Darbo’s fixed point theorem to establish some conditions for the existence of solutions to problem (1.1). We point out that the main result is even new and efficient for integer order case of $\beta=2$.

The outline for the remainder of this paper is as follows. In Section 2, we recall some useful preliminaries. In Section 3, we establish the existence result of problem (1.1), and finally we present in Section 4 an example illustrating the abstract theory.

2 Preliminaries

In this section, we begin by presenting here some necessary definitions for discrete fractional calculus, and more preliminary facts can be found, for example, in [1, 16, 27, 28].

Definition 2.1

For any t and ν, the falling factorial function is defined as

$$t^{\underline{\nu}} =\frac{\Gamma(t+1)}{\Gamma(t+1-\nu)}, $$

provided that the right-hand side is well defined. We make the convention that if $t+1-\nu$ is a pole of the gamma function and $t+1$ is not a pole, then $t^{\underline{\nu}}=0$.

Definition 2.2

The νth discrete fractional sum of a function $f:\mathbb{N}_{a}\rightarrow E$ for $\nu>0$ is defined by

$$\Delta^{-\nu}_{a}f(t) =\frac{1}{\Gamma(\nu)}\sum _{s=a}^{t-\nu}(t-s-1)^{\underline{\nu-1}}f(s),\quad t\in \mathbb{N}_{a+\nu}. $$

Also, we define the trivial sum $\Delta_{a}^{-0}f(t)=f(t)$, $t\in \mathbb{N}_{a}$.

Definition 2.3

The νth discrete Riemann-Liouville fractional difference of a function $f:\mathbb{N}_{a}\rightarrow E$ for $\nu>0$ is defined by

$$\Delta^{\nu}_{a}f(t) =\Delta^{n} \Delta_{a}^{-(n-\nu)}f(t),\quad t\in\mathbb{N}_{a+n-\nu}, $$

where n is the smallest integer greater than or equal to ν, and $\Delta^{n}$ is the nth-order forward difference operator. If $\nu =n\in\mathbb{N}_{1}$, then $\Delta_{a}^{n}f(t)=\Delta^{n}f(t)$.

We denote by $C(\mathbb{N}_{a}^{b}, E)$ the Banach space of all functions $\varpi: \mathbb{N}_{a}^{b}\rightarrow E$ with the usual supremum norm $\|\varpi\|_{0}=\sup\{\|\varpi(t)\|: t\in\mathbb {N}_{a}^{b}\}$. Define the space

$$X= \biggl\{ u:\mathbb{N}_{\beta-2}\rightarrow E \Bigm| \sup _{t\in\mathbb{N}_{\beta-2}}\frac{\|u(t)\|}{1+t^{\underline {\beta-1}}} < +\infty \biggr\} $$

equipped with the norm

$$\|u\|_{X}=\sup_{t\in\mathbb{N}_{\beta-2}}\frac{\|u(t)\| }{1+t^{\underline{\beta-1}}}. $$

Furthermore, by means of the linear functional analysis theory we can easily verify that $(X,\|\cdot\|_{X} )$ is a Banach space. It is worth reminding that here we use α, $\alpha_{C}$, and $\alpha_{X}$ to denote the Kuratowski noncompactness measure of bounded sets in Banach spaces $E, C(\mathbb{N}_{a}^{b},E)$, and X, respectively. For more details on the Kuratowski noncompactness measure, we refer the reader to [29, 30]. We state the following properties of the Kuratowski measure of noncompactness and the Darbo’s fixed point theorem, which are needed for the sequel discussion.

Lemma 2.1

([22])

Let $A\subseteq C(\mathbb {N}_{a}^{b}, E)$ be bounded. Then

• (i) $\alpha_{C}(A)=\alpha (A (\mathbb{N}_{a}^{b} ) )$;

• (ii) $\alpha (A (\mathbb{N}_{a}^{b} ) ) =\sup_{t\in\mathbb{N}_{a}^{b}}\alpha (A (t ) )$,

where $A(t)=\{\varpi(t) : \varpi\in A\}$ and $A (\mathbb{N}_{a}^{b} )=\bigcup_{t\in\mathbb{N}_{a}^{b}}A(t)$.

Lemma 2.2

([30])

Let D be a bounded, closed, and convex subset of a Banach space E. If an operator $\mathcal {A}:D\rightarrow D$ is a strict set contraction, then $\mathcal{A}$ has a fixed point in D.

Remark 2.1

A bounded and continuous operator $\mathcal{A}:D\rightarrow E$ is called a strict set contraction if there is a constant $\lambda\in[0,1)$ such that $\alpha (\mathcal{A}S)\leq\lambda\alpha(S)$ for any bounded set $S\subset D$.

3 Main result

In this section, we establish the existence of solutions for problem (1.1) by using Darbo’s fixed point theorem. For convenience and shortness of our presentation, for any $u\in X$, we denote

$$g_{u}(t)=f \bigl(t+\beta-1,u(t+\beta-1),(Tu) (t),(Su) (t) \bigr),\quad t \in\mathbb{N}_{0}, $$

for further discussion and list the following conditions:

(C₁)

$$\sup_{t\in\mathbb{N}_{0}} \sum_{s=0}^{t} \bigl|k(t,s)\bigr| < +\infty,\qquad \sup_{t\in\mathbb{N}_{0}} \sum _{s=0}^{\infty} \frac{ |h(t,s)| [1+(s+\beta-1)^{\underline{\beta-1}} ]}{1+(t+\beta-1)^{\underline{\beta-1}}} < +\infty; $$

(C₂) There exist nonnegative numbers $q_{i}$, $i\in\mathbb{N}_{1}^{3}$, and functions $p_{1}, p_{2}:\mathbb{N}_{\beta -1}\rightarrow[0,\infty)$ with $p_{1}^{*}=\sum_{t=\beta-1}^{\infty} p_{1}(t)(1+t^{\underline{\beta-1}}) <\Gamma(\beta)$ and $p_{2}^{*}=\sum_{t=\beta-1}^{\infty} p_{2}(t)<+\infty$ such that

$$\bigl\| f(t,u,v,w)\bigr\| \leq p_{1}(t) \bigl(q_{1}\|u\|+q_{2} \|v\|+q_{3}\|w\|\bigr)+p_{2}(t) $$

for $(t,u,v,w)\in\mathbb{N}_{\beta-1}\times E\times E\times E$;

(C₃) For any positive number $r\in\mathbb {N}_{\beta-1}$, $f(t,u,v,w)$ is uniformly continuous on $\mathbb{N}_{\beta-1}^{r}\times B_{E}(\theta,r)\times B_{E}(\theta,r)\times B_{E}(\theta,r)$, where $B_{E}(\theta,r)=\{x\in E : \|x\|\leq r\}$;

(C₄) There exist functions $l_{i}:\mathbb {N}_{\beta-1}\rightarrow[0,+\infty)$, $i\in\mathbb{N}_{1}^{3}$, such that

$$\alpha \bigl(f(t,V_{1},V_{2},V_{3}) \bigr) \leq l_{1}(t)\alpha(V_{1})+l_{2}(t) \alpha(V_{2})+l_{3}(t)\alpha (V_{3}),\quad t\in \mathbb{N}_{\beta-1}, $$

for all bounded sets $V_{i}\subset E$, $i\in\mathbb{N}_{1}^{3}$, and

$$\sum_{t=\beta-1}^{\infty} \bigl(1+t^{\underline{\beta-1}} \bigr) \bigl[l_{1}(t)+k^{*}l_{2}(t)+h^{*}l_{3}(t) \bigr]< \Gamma(\beta). $$

Moreover, we set

$$k^{*}=\sup_{t\in\mathbb{N}_{0}} \sum_{s=0}^{t} \bigl|k(t,s)\bigr|\quad \mbox{and}\quad h^{*}=\sup_{t\in\mathbb{N}_{0}} \sum _{s=0}^{\infty} \frac{ |h(t,s)| [1+(s+\beta-1)^{\underline{\beta-1}} ]}{1+(t+\beta-1)^{\underline{\beta-1}}} $$

when (C₁) holds.

Next, we state and prove the following lemmas, which are necessary for the proof of the main result.

Lemma 3.1

Assume that (C₁) and (C₂) hold. Then, for any $u\in X$,

$$ \sum_{t=0}^{\infty} \bigl\| g_{u}(t)\bigr\| \leq p_{1}^{*}\bigl(q_{1}+q_{2}k^{*}+q_{3}h^{*} \bigr)\|u\|_{X}+p_{2}^{*}. $$

(3.1)

Proof

For any $u\in X$, $t\in\mathbb{N}_{0}$, using (C₁), (C₂), and the monotonicity of $t^{\underline{\beta-1}}$ for $t\in\mathbb{N}_{\beta-1}$ produces

$$\begin{aligned} \bigl\| g_{u}(t)\bigr\| ={}&\bigl\| f \bigl(t+\beta-1,u(t+\beta-1),(Tu) (t),(Su) (t) \bigr)\bigr\| \\ \leq{}&p_{1}(t+\beta-1) \bigl(q_{1}\bigl\| u(t+\beta-1) \bigr\| +q_{2}\bigl\| (Tu) (t)\bigr\| +q_{3}\bigl\| (Su) (t)\bigr\| \bigr) +p_{2}(t+\beta-1) \\ \leq{}&p_{1}(t+\beta-1) \bigl[1+(t+\beta-1)^{\underline{\beta-1}} \bigr] \Biggl\{ q_{1} +q_{2}\sum_{s=0}^{t} \frac{|k(t,s)|[1+(s+\beta-1)^{\underline{\beta-1}}]}{1+(t+\beta -1)^{\underline{\beta-1}}} \\ &{} +q_{3}\sum_{s=0}^{\infty} \frac{|h(t,s)|[1+(s+\beta-1)^{\underline{\beta-1}}]}{1+(t+\beta -1)^{\underline{\beta-1}}} \Biggr\} \|u\|_{X} +p_{2}(t+\beta-1) \\ \leq{}&p_{1}(t+\beta-1) \bigl[1+(t+\beta-1)^{\underline{\beta-1}} \bigr] \bigl(q_{1} +q_{2}k^{*} +q_{3}h^{*} \bigr) \|u\|_{X} +p_{2}(t+\beta-1). \end{aligned}$$

(3.2)

By summing both sides of (3.2) we get (3.1). So, this proof is completed. □

Lemma 3.2

Let $h:\mathbb{N}_{0}\rightarrow E$ be given, and $\beta\in(1,2]$. The unique solution of

$$\textstyle\begin{cases} \Delta^{\beta}u(t) +h(t) =\theta,\quad t\in\mathbb{N}_{0},\\ u(\beta-2)=\theta,\qquad \Delta^{\beta-1}u(\infty)=u_{\infty}, \end{cases} $$

is

$$u(t)=\sum_{s=0}^{\infty}G(t,s)h(s) + \frac{u_{\infty}}{\Gamma(\beta)}t^{\underline{\beta-1}}, \quad t\in\mathbb{N}_{\beta-2}, $$

where

$$G(t,s)=\frac{1}{\Gamma(\beta)} \textstyle\begin{cases} t^{\underline{\beta-1}}-(t-s-1)^{\underline{\beta-1}}, &s\in\mathbb{N}_{0}^{t-\beta},\\ t^{\underline{\beta-1}},&s\in\mathbb{N}_{t-\beta+1}. \end{cases} $$

Remark 3.1

The proof of Lemma 3.2 is similar to that of Lemma 3.2 in [16]. Hence, we omit it here. Moreover, in view of the expression of $G(t,s)$, we can easily verify that $G(t,s)\geq0$ and $\frac{G(t,s)}{1+t^{\underline{\beta-1}}}<\frac {1}{\Gamma(\beta)}$ for $(t,s)\in\mathbb{N}_{\beta-2}\times\mathbb{N}_{0}$.

For any $u\in X$, define the operator $\mathcal{F}$ by

$$(\mathcal{F}u) (t)=\sum_{s=0}^{\infty} G(t,s)g_{u}(s) +\frac{u_{\infty}}{\Gamma(\beta)}t^{\underline{\beta-1}},\quad t\in \mathbb{N}_{\beta-2}, $$

and due to Lemma 3.1 and Remark 3.1, we have

$$\begin{aligned} \frac{\|(\mathcal{F}u)(t)\|}{1+t^{\underline{\alpha-1}}} &\leq\sum_{s=0}^{\infty} \frac{G(t,s)}{1+t^{\underline{\beta-1}}} \bigl\| g_{u}(s) \bigr\| +\frac{\|u_{\infty}\|t^{\underline{\beta-1}}}{ \Gamma(\beta)(1+t^{\underline{\beta-1}})} \\ &\leq\frac{1}{\Gamma(\beta)} \bigl\{ p_{1}^{*} \bigl(q_{1}+q_{2}k^{*}+q_{3}h^{*} \bigr)\|u\|_{X}+p_{2}^{*}+\|u_{\infty }\| \bigr\} , \end{aligned}$$

which implies that $\mathcal{F}:X\rightarrow X$ is well defined and bounded. Furthermore, from Lemma 3.2 we know that the existence of solutions for problem (1.1) is equivalent to that of fixed points of $\mathcal{F}$ in X.

Lemma 3.3

Suppose that (C₁), (C₂), and (C₃) are satisfied. Then the operator $\mathcal{F}:X\rightarrow X$ is continuous.

Proof

Let $\{u_{n}\}_{n=1}^{\infty}\subset X$ and $u\in X$ be such that $\|u_{n}-u\|_{X}\rightarrow0$ as $n\rightarrow\infty$. Then, $\{u_{n}\}_{n=1}^{\infty}$ is a bounded set of X, that is, there exists $M>0$ such that $\|u_{{n}}\|_{X}\leq M$ for $n\in\mathbb {N}_{1}$. By taking limit we also have that $\|u\|_{X}\leq M$.

In view of (C₂), for any $\epsilon>0$, there exists a positive integer L such that

$$\begin{aligned}& \sum_{t=L+\beta}^{\infty}p_{1}(t) \bigl(1+t^{\underline{\beta-1}}\bigr) < \frac{\Gamma(\beta)}{6M(q_{1}+q_{2}k^{*}+q_{3}h^{*})}\epsilon, \end{aligned}$$

(3.3)

$$\begin{aligned}& \sum_{t=L+\beta}^{\infty}p_{2}(t) < \frac{\Gamma(\beta)}{6}\epsilon. \end{aligned}$$

(3.4)

On the other hand, condition (C₃) yields that there exists $N>0$ such that for any $n>N$ and $t\in\mathbb{N}_{0}^{L}$,

$$ \bigl\| g_{u_{n}}(t)-g_{u}(t)\bigr\| \leq\frac{\Gamma(\beta)}{3(L+1)} \epsilon. $$

(3.5)

Therefore, for $t\in\mathbb{N}_{\beta-2}^{L+\beta-1}$ and $n>N$, by (C₁), (C₂), (3.3)-(3.5), and Remark 3.1 we obtain that

$$\begin{aligned} &\frac{\|(\mathcal{F}u_{n})(t)-(\mathcal{F}u)(t)\|}{1+t^{\underline {\beta-1}}} \\ &\quad\leq\sum_{s=0}^{\infty}\frac{G(t,s)}{1+t^{\underline{\beta-1}}} \bigl\| g_{u_{n}}(s)-g_{u}(s)\bigr\| \\ &\quad< \frac{1}{\Gamma(\beta)} \Biggl\{ \sum_{s=0}^{L} \bigl\| g_{u_{n}}(s)-g_{u}(s)\bigr\| +\sum_{s=L+1}^{\infty} \bigl\| g_{u_{n}}(s)-g_{u}(s)\bigr\| \Biggr\} \\ &\quad\leq\frac{1}{\Gamma(\beta)}\sum_{s=0}^{L} \bigl\| g_{u_{n}}(s)-g_{u}(s)\bigr\| +\frac{1}{\Gamma(\beta)}\sum _{s=L+1}^{\infty} \bigl\{ \bigl\| g_{u_{n}}(s)\bigr\| +\bigl\| g_{u}(s)\bigr\| \bigr\} \\ &\quad\leq\frac{\epsilon}{3} +\frac{1}{\Gamma(\beta)} \sum _{s=L+1}^{\infty} \bigl\{ \bigl\| f \bigl(s+\beta-1,u_{n}(s+ \beta-1),(Tu_{n}) (s),(Su_{n}) (s) \bigr) \bigr\| \\ &\qquad{}+ \bigl\| f \bigl(s+\beta-1,u(s+\beta-1),(Tu) (s),(Su) (s) \bigr) \bigr\| \bigr\} \\ &\quad\leq\frac{\epsilon}{3}+ \frac{1}{\Gamma(\beta)} \sum_{s=L+1}^{\infty} \bigl\{ p_{1}(s+\beta-1) \bigl(q_{1}\bigl\| u_{n}(s+ \beta-1)\bigr\| +q_{2}\bigl\| (Tu_{n}) (s)\bigr\| +q_{3} \bigl\| (Su_{n}) (s)\bigr\| \bigr) \\ &\qquad{}+p_{1}(s+\beta-1) \bigl(q_{1}\bigl\| u(s+\beta-1) \bigr\| +q_{2}\bigl\| (Tu) (s)\bigr\| +q_{3}\bigl\| (Su) (s)\bigr\| \bigr) +2p_{2}(s+\beta-1) \bigr\} \\ &\quad\leq\frac{\epsilon}{3} +\frac{2}{\Gamma(\beta)}\sum _{s=L+\beta}^{\infty}p_{2}(s) +\frac{1}{\Gamma(\beta)} \sum _{s=L+1}^{\infty} p_{1}(s+\beta-1) \bigl[1+(s+\beta-1)^{\underline{\beta-1}} \bigr] \\ &\qquad{}\times \Biggl\{ q_{1} +q_{2}\sum _{\tau=0}^{s} \frac{|k(s,\tau)|[1+(\tau+\beta-1)^{\underline{\beta-1}}]}{1+(s+\beta -1)^{\underline{\beta-1}}} +q_{3}\sum _{\tau=0}^{\infty} \frac{|h(s,\tau)|[1+(\tau+\beta-1)^{\underline{\beta-1}}]}{1+(s+\beta -1)^{\underline{\beta-1}}} \Biggr\} \\ &\qquad{}\times\bigl(\|u_{n}\|_{X}+\|u\|_{X}\bigr) \\ &\quad< \frac{\epsilon}{3}+\frac{\epsilon}{3}+ \frac{2M(q_{1}+q_{2}k^{*}+q_{3}h^{*})}{\Gamma(\beta)} \sum _{s=L+\beta}^{\infty}p_{1}(s) \bigl(1+s^{\underline{\beta-1}} \bigr) \\ &\quad< \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon. \end{aligned}$$

Meanwhile, for $t\in\mathbb{N}_{L+\beta}$ and $n>N$, applying (3.3)-(3.5) again, we can easily verify that

$$\frac{\|(\mathcal{F}u_{n})(t)-(\mathcal{F}u)(t)\|}{1+t^{\underline {\beta-1}}}< \epsilon. $$

Then, we conclude that $\|\mathcal{F}u_{n}-\mathcal{F}u\|_{X}\leq \epsilon$ for $n>N$. So $\mathcal{F}$ is continuous, and the proof is completed. □

Lemma 3.4

Let B be a bounded subset of X. If (C₁) and (C₂) hold, then for any $\epsilon>0$, there exists a positive number $N\in\mathbb{N}_{\beta-2}$ such that $\|\frac{(\mathcal{F}u)(t_{2})}{1+t_{2}^{\underline{\beta-1}}} -\frac{(\mathcal{F}u)(t_{1})}{1+t_{1}^{\underline{\beta-1}}} \| <\epsilon$ for each $u\in B$ and any $t_{1}, t_{2}\in\mathbb{N}_{N}$.

Proof

In view of Lemma 3.1 and the boundedness of B, there exists $M>0$ such that

$$ \sum_{t=0}^{\infty}\bigl\| g_{u}(t)\bigr\| \leq M \quad\mbox{for any } u\in B. $$

(3.6)

Note that

$$\begin{aligned} & \biggl\| \frac{(\mathcal{F}u)(t_{2})}{1+t_{2}^{\underline{\beta-1}}} - \frac{(\mathcal{F}u)(t_{1})}{1+t_{1}^{\underline{\beta-1}}} \biggr\| \\ &\quad\leq \frac{1}{\Gamma(\beta)} \Biggl\| u_{\infty}+\sum _{s=0}^{\infty}g_{u}(s) \Biggr\| \biggl|\frac{t_{2}^{\underline{\beta-1}}}{1+t_{2}^{\underline{\beta-1}}} - \frac{t_{1}^{\underline{\beta-1}}}{1+t_{1}^{\underline{\beta-1}}} \biggr| \\ &\qquad{}+ \frac{1}{\Gamma(\beta)} \Biggl\| \sum_{s=0}^{t_{2}-\beta} \frac{(t_{2}-s-1)^{\underline{\beta -1}}}{1+t_{2}^{\underline{\beta-1}}} g_{u}(s) -\sum_{s=0}^{t_{1}-\beta} \frac{(t_{1}-s-1)^{\underline{\beta -1}}}{1+t_{1}^{\underline{\beta-1}}} g_{u}(s) \Biggr\| . \end{aligned}$$

(3.7)

Observing (3.7) together with $\lim_{t\rightarrow+\infty}\frac {t^{\underline{\beta-1}}}{1+t^{\underline{\beta-1}}}=1$, we only need to show that for any $\epsilon>0$, there exists sufficiently large positive number $N\in\mathbb{N}_{\beta-2}$ such that, for any $t_{1},t_{2}\in\mathbb{N}_{N}$,

$$\Biggl\| \sum_{s=0}^{t_{2}-\beta}\frac{(t_{2}-s-1)^{\underline{\beta -1}}}{1+t_{2}^{\underline{\beta-1}}} g_{u}(s) -\sum_{s=0}^{t_{1}-\beta} \frac{(t_{1}-s-1)^{\underline{\beta -1}}}{1+t_{1}^{\underline{\beta-1}}} g_{u}(s) \Biggr\| < \epsilon. $$

Relation (3.6) yields that there exists a positive number $L\in\mathbb {N}_{0}$ such that

$$ \sum_{t=L+1}^{\infty}\bigl\| g_{u}(t)\bigr\| \leq \frac{\epsilon}{3}\quad \mbox{uniformly with respect to } u\in B. $$

(3.8)

On the other hand, from the monotonicity of $\iota^{\underline{\beta -1}}$ we can declare that $\lim_{t\rightarrow+\infty}\frac {(t-L-1)^{\underline{\beta-1}}}{1+t^{\underline{\beta-1}}}=1$.

In fact, for any $t\in\mathbb{N}_{L+\beta+1}$, $L\in\mathbb {N}_{0}$, we have

$$\begin{aligned} \frac{(t-L-1)^{\underline{\beta-1}}}{t^{\underline{\beta-1}}} &=\frac{\Gamma(t-L)\Gamma(t-\beta+2)}{\Gamma(t+1)\Gamma(t-\beta +1-L)} \\ &=\frac{(t+1-\beta)^{\underline{L}}}{t^{\underline{L}}} \\ &=\frac{\prod_{j=0}^{L-1}(t-j-\beta+1)}{\prod_{j=0}^{L-1}(t-j)} \\ &=\prod_{j=0}^{L-1} \biggl(1+ \frac{1-\beta}{t-j} \biggr), \end{aligned}$$

which implies that

$$\lim_{t\rightarrow+\infty}\frac{(t-L-1)^{\underline{\beta -1}}}{1+t^{\underline{\beta-1}}} =\lim_{t\rightarrow+\infty} \frac{\frac{(t-L-1)^{\underline{\beta -1}}}{t^{\underline{\beta-1}}}}{ \frac{1}{t^{\underline{\beta-1}}}+1}=\frac{1}{0+1}=1. $$

So, there exists $N\in\mathbb{N}_{L+\beta+1}$ such that for any $t_{1},t_{2}\in\mathbb{N}_{N}$ and $s\in\mathbb{N}_{0}^{L}$,

$$\begin{aligned} & \biggl|\frac{(t_{2}-s-1)^{\underline{\beta-1}}}{1+t_{2}^{\underline {\beta-1}}} -\frac{(t_{1}-s-1)^{\underline{\beta-1}}}{1+t_{1}^{\underline{\beta-1}}} \biggr| \\ &\quad\leq \biggl(1-\frac{(t_{2}-s-1)^{\underline{\beta -1}}}{1+t_{2}^{\underline{\beta-1}}} \biggr) + \biggl(1-\frac{(t_{1}-s-1)^{\underline{\beta-1}}}{1+t_{1}^{\underline {\beta-1}}} \biggr) \\ &\quad\leq \biggl(1-\frac{(t_{2}-L-1)^{\underline{\beta -1}}}{1+t_{2}^{\underline{\beta-1}}} \biggr) + \biggl(1-\frac{(t_{1}-L-1)^{\underline{\beta-1}}}{1+t_{1}^{\underline {\beta-1}}} \biggr) \\ &\quad< \frac{\epsilon}{3M}. \end{aligned}$$

(3.9)

Now taking $t_{1},t_{2}\in\mathbb{N}_{N}$, by (3.8) and (3.9) we get

$$\begin{aligned} & \Biggl\| \sum_{s=0}^{t_{2}-\beta} \frac{(t_{2}-s-1)^{\underline{\beta-1}}}{1+t_{2}^{\underline{\beta-1}}} g_{u}(s) -\sum_{s=0}^{t_{1}-\beta} \frac{(t_{1}-s-1)^{\underline{\beta -1}}}{1+t_{1}^{\underline{\beta-1}}} g_{u}(s) \Biggr\| \\ &\quad\leq\sum_{s=0}^{L} \biggl| \frac{(t_{2}-s-1)^{\underline{\beta-1}}}{1+t_{2}^{\underline {\beta-1}}} -\frac{(t_{1}-s-1)^{\underline{\beta-1}}}{1+t_{1}^{\underline{\beta-1}}} \biggr| \bigl\| g_{u}(s)\bigr\| \\ &\qquad{}+\sum_{s=L+1}^{t_{2}-\beta} \frac{(t_{2}-s-1)^{\underline{\beta-1}}}{1+t_{2}^{\underline{\beta-1}}} \bigl\| g_{u}(s)\bigr\| +\sum_{s=L+1}^{t_{1}-\beta} \frac{(t_{1}-s-1)^{\underline{\beta-1}}}{1+t_{1}^{\underline{\beta-1}}} \bigl\| g_{u}(s)\bigr\| \\ &\quad< \frac{\epsilon}{3M}\sum_{s=0}^{\infty} \bigl\| g_{u}(s)\bigr\| +2\sum_{s=L+1}^{\infty} \bigl\| g_{u}(s)\bigr\| \\ &\quad\leq \epsilon. \end{aligned}$$

Therefore, the proof is completed. □

Lemma 3.5

Let B be a bounded subset of X. If (C₁) and (C₂) hold, then

$$\alpha_{X}(\mathcal{F}B) =\sup_{t\in\mathbb{N}_{\beta-2}} \alpha \biggl( \frac{(\mathcal{F}B)(t)}{1+t^{\underline{\beta-1}}} \biggr). $$

Proof

First, we claim that $\alpha_{X}(\mathcal{F}B) \leq\sup_{t\in\mathbb{N}_{\beta-2}} \alpha (\frac{(\mathcal{F}B)(t)}{1+t^{\underline{\beta-1}}} )$.

In view of Lemma 3.4, we know that for any $\epsilon>0$, there exists a positive number $N\in\mathbb{N}_{\beta-2}$ such that, for $t_{1}, t_{2}\in\mathbb{N}_{N}$,

$$ \biggl\| \frac{(\mathcal{F}u)(t_{2})}{1+t_{2}^{\underline{\beta-1}}} -\frac{(\mathcal{F}u)(t_{1})}{1+t_{1}^{\underline{\beta-1}}} \biggr\| < \epsilon \quad \mbox{uniformly with respect to } u\in B. $$

(3.10)

Denote by $\mathcal{F}B|_{\mathbb{N}_{\beta-2}^{N}}$ the restriction of $\mathcal{F}B$ on $\mathbb{N}_{\beta-2}^{N}$. By Lemma 2.1 we obtain

$$\alpha_{X} (\mathcal{F}B|_{\mathbb{N}_{\beta-2}^{N}} ) =\sup_{t\in\mathbb{N}_{\beta-2}^{N}} \alpha \biggl(\frac{(\mathcal{F}B)(t)}{1+t^{\underline{\beta-1}}} \biggr) \leq\sup_{t\in\mathbb{N}_{\beta-2}}\alpha \biggl(\frac{(\mathcal{F}B)(t)}{1+t^{\underline{\beta-1}}} \biggr). $$

So, there exists a partition of B such that $B=\bigcup_{i=1}^{n}B_{i}$, $\mathcal{F}B|_{\mathbb{N}_{\beta-2}^{N}}=\bigcup_{i=1}^{n} \mathcal{F}B_{i}|_{\mathbb{N}_{\beta-2}^{N}}$ and

$$ \operatorname{diam}_{X} (\mathcal{F}B_{i}|_{\mathbb{N}_{\beta-2}^{N}} ) < \sup_{t\in\mathbb{N}_{\beta-2}}\alpha \biggl(\frac{(\mathcal{F}B)(t)}{1+t^{\underline{\beta-1}}} \biggr) + \epsilon,\quad i\in\mathbb{N}_{1}^{n}, $$

(3.11)

where $\operatorname{diam}_{X}(\cdot)$ denotes the diameter of a bounded subset of X. Moreover, for all $\mathcal{F}u_{1}, \mathcal{F}u_{2}\in \mathcal{F}B_{i}$, $i\in\mathbb{N}_{1}^{n}$, and $t\in\mathbb{N}_{N}$, (3.10) and (3.11) imply that

$$\begin{aligned} & \biggl\| \frac{(\mathcal{F}u_{1})(t)}{1+t^{\underline{\beta-1}}} -\frac{(\mathcal{F}u_{2})(t)}{1+t^{\underline{\beta-1}}} \biggr\| \\ &\quad\leq \biggl\| \frac{(\mathcal{F}u_{1})(t)}{1+t^{\underline{\beta-1}}} -\frac{(\mathcal{F}u_{1})(N)}{1+N^{\underline{\beta-1}}} \biggr\| + \biggl\| \frac{(\mathcal{F}u_{1})(N)}{1+N^{\underline{\beta-1}}} - \frac{(\mathcal{F}u_{2})(N)}{1+N^{\underline{\beta-1}}} \biggr\| + \biggl\| \frac{(\mathcal{F}u_{2})(N)}{1+N^{\underline{\beta-1}}} -\frac{(\mathcal{F}u_{2})(t)}{1+t^{\underline{\beta-1}}} \biggr\| \\ &\quad< \epsilon +\sup_{t\in\mathbb{N}_{\beta-2}} \alpha \biggl(\frac{(\mathcal{F}B)(t)}{1+t^{\underline{\beta-1}}} \biggr) +\epsilon +\epsilon \\ &\quad=\sup_{t\in\mathbb{N}_{\beta-2}} \alpha \biggl(\frac{(\mathcal{F}B)(t)}{1+t^{\underline{\beta-1}}} \biggr) +3 \epsilon. \end{aligned}$$

(3.12)

Hence, it follows from (3.11) and (3.12) that

$$\operatorname{diam}_{X}(\mathcal{F}B_{i}) \leq\sup _{t\in\mathbb{N}_{\beta-2}} \alpha \biggl(\frac{(\mathcal{F}B)(t)}{1+t^{\underline{\beta-1}}} \biggr) +3\epsilon. $$

Since $\mathcal{F}B=\bigcup_{i=1}^{n}\mathcal{F}B_{i}$, we get that $\alpha_{X}(\mathcal{F}B) \leq\sup_{t\in\mathbb{N}_{\beta-2}} \alpha (\frac{(\mathcal{F}B)(t)}{1+t^{\underline{\beta-1}}} ) +3\epsilon$, which by the arbitrariness of ϵ implies that

$$\alpha_{X}(\mathcal{F}B) \leq\sup_{t\in\mathbb{N}_{\beta-2}} \alpha \biggl(\frac{(\mathcal{F}B)(t)}{1+t^{\underline{\beta-1}}} \biggr). $$

Next, we show that $\sup_{t\in\mathbb{N}_{\beta-2}} \alpha (\frac{(\mathcal{F}B)(t)}{1+t^{\underline{\alpha-1}}} ) \leq\alpha_{X}(\mathcal{F}B)$. For any given $\epsilon>0$, there exists a partition $\mathcal {F}B=\bigcup_{i=1}^{n}\mathcal{F}B_{i}$ such that $\mathrm {diam}_{X}(\mathcal{F}B_{i})<\alpha_{X}(\mathcal{F}B)+\epsilon$. Therefore, for any $t\in\mathbb{N}_{\beta-2}$ and $u_{1}, u_{2}\in B_{i}$, we obtain

$$\biggl\| \frac{(\mathcal{F}u_{2})(t)}{1+t^{\underline{\beta-1}}} -\frac{(\mathcal{F}u_{1})(t)}{1+t^{\underline{\beta-1}}} \biggr\| \leq\biggl\| \mathcal{F}u_{2}- \mathcal{F}u_{1}\biggr\| _{X}< \alpha_{X}(\mathcal {F}B)+ \epsilon. $$

In accordance with $(\mathcal{F}B)(t)=\bigcup_{i=1}^{n}\mathcal {F}B_{i}(t)$, we get $\alpha (\frac{(\mathcal{F}B)(t)}{1+t^{\underline{\beta-1}}} ) \leq\alpha_{X}(\mathcal{F}B)+\epsilon$. Because ϵ is arbitrary, we have

$$\sup_{t\in\mathbb{N}_{\beta-2}}\alpha \biggl(\frac{(\mathcal{F}B)(t)}{1+t^{\underline{\beta-1}}} \biggr) \leq \alpha_{X}(\mathcal{F}B). $$

Consequently, the proof of this lemma is complete. □

With all auxiliary results in hand, now we state the main result.

Theorem 3.1

If (C₁)-(C₄) hold, then problem (1.1) has at least one solution u in X.

Proof

Choose

$$R>\frac{\|u_{\infty}\|+p_{2}^{*}}{\Gamma(\beta )-p_{1}^{*}(q_{1}+q_{2}k^{*}+q_{3}h^{*})} $$

and let

$$B=\bigl\{ u\in X : \|u\|_{X}\leq R\bigr\} . $$

Then, for any $u\in B$, by Lemma 3.1 and Remark 3.1 we have

$$\begin{aligned} \frac{\|(\mathcal{F}u)(t)\|}{1+t^{\underline{\alpha-1}}} &\leq\sum_{s=0}^{\infty} \frac{G(t,s)}{1+t^{\underline{\beta-1}}} \bigl\| g_{u}(s) \bigr\| +\frac{\|u_{\infty}\|t^{\underline{\beta-1}}}{ \Gamma(\beta)(1+t^{\underline{\beta-1}})} \\ &\leq\frac{1}{\Gamma(\beta)} \bigl\{ p_{1}^{*} \bigl(q_{1}+q_{2}k^{*}+q_{3}h^{*} \bigr)\|u\|_{X}+p_{2}^{*}+\|u_{\infty }\| \bigr\} \leq R , \end{aligned}$$

which implies that $\mathcal{F}: B\rightarrow B$.

Set $D=\overline{\mathrm{co}}(\mathcal{F}B)$. Obviously, D is a bounded, convex, and closed subset of B. In the sequel, we show that the operator $\mathcal{F}: D\rightarrow D$ is a strict contraction.

Observing that $\mathcal{F}D\subset\mathcal{F}B\subset D$, together with Lemma 3.3, we know that $\mathcal{F}: D\rightarrow D$ is bounded and continuous. Finally, we show that there exists a constant $\lambda \in[0,1)$ such that $\alpha_{X}(\mathcal{F}V)\leq\lambda\alpha_{X}(V)$ for any $V\subset D$. Moreover, in view of Lemma 3.5, we only need to verify

$$ \sup_{t\in\mathbb{N}_{\beta-2}}\alpha \biggl(\frac{(\mathcal{F}V)(t)}{1+t^{\underline{\beta-1}}} \biggr) \leq\lambda \alpha_{X}(V). $$

(3.13)

For a positive integer $n>t-\beta$, define

$$\begin{aligned} (\mathcal{F}_{n}u) (t) =\frac{1}{\Gamma(\beta)} \sum _{s=0}^{t-\beta} \bigl[t^{\underline{\beta-1}}-(t-s-1)^{\underline{\beta-1}} \bigr] g_{u}(s) +\frac{1}{\Gamma(\beta)}\sum_{s=t-\beta+1}^{n}t^{\underline{\beta-1}} g_{u}(s) +\frac{u_{\infty}}{\Gamma(\beta)}t^{\underline{\beta-1}}. \end{aligned}$$

Then from (C₁) and (C₂), for any $u\in V$, we have

$$\begin{aligned} & \biggl\| \frac{(\mathcal{F}_{n}u)(t)}{1+t^{\underline{\beta-1}}} -\frac{(\mathcal{F}u)(t)}{1+t^{\underline{\beta-1}}} \biggr\| \\ &\quad\leq\frac{1}{\Gamma(\beta)}\sum_{s=n+1}^{\infty} \bigl\| g_{u}(s)\bigr\| \\ &\quad\leq\frac{1}{\Gamma(\beta)} \sum_{s=n+1}^{\infty} \bigl\{ p_{1}(s+\beta+1) \bigl[1+(s+\beta-1)^{\underline{\beta-1}} \bigr] \bigl(q_{1}+q_{2}k^{*}+q_{3}h^{*} \bigr)R+p_{2}(s+\beta-1) \bigr\} , \end{aligned}$$

which implies that

$$d_{\mathrm{H}} \biggl(\frac{(\mathcal{F}_{n}V)(t)}{1+t^{\underline{\beta-1}}}, \frac{(\mathcal{F}V)(t)}{1+t^{\underline{\beta-1}}} \biggr) \rightarrow0 \quad \mbox{as } n\rightarrow\infty, t\in\mathbb{N}_{\beta-2}, $$

where $d_{\mathrm{H}}(\cdot,\cdot)$ denotes the Hausdorff metric in space E. So, by the properties of noncompactness measure we obtain

$$ \lim_{n\rightarrow\infty}\alpha \biggl(\frac{(\mathcal {F}_{n}V)(t)}{1+t^{\underline{\beta-1}}} \biggr) =\alpha \biggl(\frac{(\mathcal{F}V)(t)}{1+t^{\underline{\beta-1}}} \biggr), \quad t\in\mathbb{N}_{\beta-2}. $$

(3.14)

Now we estimate

$$\alpha \biggl(\frac{(\mathcal{F}_{n}V)(t)}{1+t^{\underline{\beta-1}}} \biggr) \quad (n>t-\beta). $$

In view of (C₁) and (C₂), we know that $\{g_{u}(t) : u\in V \}$ is bounded. Relying on Lemma 2.1, Lemma 3.5, and (C₄), we have

$$\begin{aligned} \alpha \biggl(\frac{(\mathcal{F}_{n}V)(t)}{1+t^{\underline{\beta -1}}} \biggr) \leq{}&\frac{1}{\Gamma(\beta)} \alpha \Biggl( \Biggl\{ \sum_{s=0}^{t-\beta}g_{u}(s):u \in V \Biggr\} \Biggr) +\frac{1}{\Gamma(\beta)} \alpha \Biggl( \Biggl\{ \sum _{s=t-\beta+1}^{n}g_{u}(s):u\in V \Biggr\} \Biggr) \\ \leq{}&\frac{1}{\Gamma(\beta)}\sum_{s=0}^{n} \alpha \bigl( f\bigl(s+\beta-1,V(s+\beta-1),(TV) (s),(SV) (s)\bigr) \bigr) \\ \leq{}&\frac{1}{\Gamma(\beta)}\sum_{s=0}^{n} \bigl[l_{1}(s+\beta-1)\alpha \bigl(V(s+\beta-1) \bigr) +l_{2}(s+\beta-1)\alpha \bigl((TV) (s) \bigr) \\ &{}+l_{3}(s+\beta-1)\alpha \bigl((SV) (s) \bigr) \bigr] \\ \leq&\frac{1}{\Gamma(\beta)}\sum_{s=0}^{n} \bigl[1+(s+\beta-1)^{\underline {\beta-1}}\bigr] \Biggl\{ l_{1}(s+\beta-1) \\ &{}+l_{2}(s+\beta-1) \sum_{\tau=0}^{s} \frac{|k(s,\tau)|[1+(\tau+\beta-1)^{\underline{\beta-1}}]}{ [1+(s+\beta-1)^{\underline{\beta-1}}]} \\ &{}+l_{3}(s+\beta-1) \sum_{\tau=0}^{\infty} \frac{|h(s,\tau)|[1+(\tau+\beta-1)^{\underline {\beta-1}}]}{ [1+(s+\beta-1)^{\underline{\beta-1}}]} \Biggr\} \alpha_{X}(V) \\ \leq{}&\frac{1}{\Gamma(\beta)}\sum_{s=\beta-1}^{n+\beta-1} \bigl[1+s^{\underline {\beta-1}}\bigr] \bigl[l_{1}(s) +l_{2}(s)k^{*} +l_{3}(s)h^{*}\bigr] \alpha_{X}(V). \end{aligned}$$

By (3.14) we immediately get

$$\alpha \biggl(\frac{(\mathcal{F}V)(t)}{1+t^{\underline{\beta-1}}} \biggr)\leq\lambda\alpha_{X}(V) \quad \mbox{with } \lambda=\frac{1}{\Gamma(\beta)}\sum_{s=\beta-1}^{n+\beta -1} \bigl[1+s^{\underline{\beta-1}}\bigr] \bigl[l_{1}(s) +l_{2}(s)k^{*} +l_{3}(s)h^{*}\bigr]. $$

So (3.13) holds with $\lambda\in[0,1)$, and from Lemma 2.2 we immediately obtain that problem (1.1) has at least one solution in D. Hence, the proof is completed. □

4 An example

Example 4.1

Consider the following infinite system of scalar discrete fractional difference equations:

$$ \textstyle\begin{cases} \Delta^{3/2}u_{n}(t) +\frac{3^{-(t+1)}}{n [1+(t+1/2)^{\underline{1/2}} ]^{2}} \sin[u_{n}(t+1/2)] +\frac{2^{-t}}{2n^{2} [1+(t+1/2)^{\underline{1/2}} ]}\\ \quad{}\times \{ 1+u_{2n+1}(t+1/2)+\sum_{s=0}^{t} \frac{1}{(t+s+2)^{\underline{2}}} u_{3n}(s+1/2)\\ \quad{}+\sum_{s=0}^{\infty} \frac{\cos(t^{2}s)}{(s+2)^{\underline{2}} [1+(s+1/2)^{\underline{1/2}} ]} u_{n+1}(s+1/2) \}^{1/2}=0,\quad t\in\mathbb{N}_{0},\\ u_{n}(-1/2)=0, \qquad\Delta^{1/2}u_{n}(\infty)=\frac{1}{n!}, \quad n\in \mathbb{N}_{1}. \end{cases} $$

(4.1)

Conclusion

System (4.1) has at least one solution $\{u_{n}(t)\}_{n=1}^{\infty}$ such that $u_{n}(t)\rightarrow0$ as $n\rightarrow\infty$ for $t\in\mathbb{N}_{-1/2}$.

Proof

Let $E=c_{0}=\{u=(u_{1},u_{2},\ldots,u_{n},\ldots):u_{n}\rightarrow0\}$. Evidently, $(E,\|\cdot\|)$ is a Banach space with the norm $\|u\| =\sup_{n}|u_{n}|$ for any $u\in E$. Then infinite system (4.1) can be regarded as a boundary value problem of the form (1.1) in the Banach space E. In this case, $\beta=3/2$, $\theta=(0,0,\ldots,0,\ldots)\in E$, $u_{\infty}=(1,1/2!,\ldots,1/n!,\ldots)\in E$,

$$k(t,s)=\frac{1}{(t+s+2)^{\underline{2}}},\qquad h(t,s)=\frac{\cos(t^{2}s)}{ (s+2)^{\underline{2}} [1+(s+1/2)^{\underline{1/2}} ]}, $$

and $f=(f_{1},f_{2},\ldots,f_{n},\ldots)$ with

$$ f_{n}(t,u,v,w) =\frac{3^{-t-1/2}}{n (1+t^{\underline{1/2}} )^{2}}\sin u_{n} + \frac{2^{-t+1/2}}{2n^{2} (1+t^{\underline{1/2}} )} ( 1+u_{2n+1}+v_{3n}+w_{n+1} )^{1/2}, $$

(4.2)

where $t\in\mathbb{N}_{1/2}$ and $u=(u_{1},u_{2},\ldots,u_{n},\ldots), v=(v_{1},v_{2},\ldots,v_{n},\ldots), w=(w_{1},w_{2},\ldots ,w_{n},\ldots)\in E$. From the expression of $f_{n}$ we easily to see that $f:\mathbb{N}_{1/2}\times E\times E\times E \rightarrow E$ is continuous. Furthermore, we can easily verify that

$$\begin{aligned}& k^{*}=\sup_{t\in\mathbb{N}_{0}} \sum_{s=0}^{t} \frac{1}{(t+s+2)^{\underline{2}}} =\sup_{t\in\mathbb{N}_{0}} \frac{1}{2(t+1)} = \frac{1}{2} < \infty, \\& \begin{aligned}[b] h^{*}&=\sup_{t\in\mathbb{N}_{0}} \frac{1}{1+(t+1/2)^{\underline{1/2}}} \sum_{s=0}^{\infty} \frac{|\cos(t^{2}s)|[1+(s+1/2)^{\underline{1/2}}]}{ [1+(s+1/2)^{\underline{1/2}}](s+2)^{\underline{2}}} \\ &\leq\sup_{t\in\mathbb{N}_{0}} \frac{1}{1+(t+1/2)^{\underline{1/2}}} \sum _{s=0}^{\infty} \frac{1}{(s+2)^{\underline{2}}} \\ &\leq\frac{1}{1+\Gamma(3/2)}< \frac{3}{4} < \infty. \end{aligned} \end{aligned}$$

So, condition (C₁) is satisfied. On the other hand, using the simple inequality

$$(1+z)^{\gamma}\leq1+\gamma z \quad\mbox{for } z\in[0,+\infty), \gamma \in(0,1), $$

we see from (4.2) that

$$\begin{aligned} \bigl|f_{n}(t,u,v,w)\bigr| &\leq\frac{3^{-t-1/2}}{n (1+t^{\underline{1/2}} )^{2}}|\sin u_{n}| + \frac{2^{-t+1/2}}{2n^{2} (1+t^{\underline{1/2}} )} \bigl(1+\|u\|+\|v\|+\|w\| \bigr)^{1/2} \\ &\leq\frac{2^{-t-1/2}}{2n^{2} (1+t^{\underline{1/2}} )} \bigl(\| u\|+\|v\|+\|w\| \bigr) +\frac{2^{-t+1/2}}{2n^{2} (1+t^{\underline{1/2}} )} + \frac{3^{-t-1/2}}{n (1+t^{\underline{1/2}} )^{2}}, \quad n\in\mathbb{N}_{1}, \end{aligned}$$

and, therefore,

$$\bigl\| f(t,u,v,w)\bigr\| \leq p_{1}(t) \bigl(\|u\|+\|v\|+\|w\| \bigr)+p_{2}(t), $$

where

$$p_{1}(t)=\frac{2^{-t-1/2}}{2(1+t^{\underline{1/2}})},\qquad p_{2}(t)=\frac{2^{-t+1/2}}{2(1+t^{\underline{1/2}})} + \frac{3^{-t-1/2}}{ (1+t^{\underline{1/2}} )^{2}},\qquad q_{1}=q_{2}=q_{3}=1, $$

which implies

$$p_{1}^{*}=1/2< \Gamma(3/2),\qquad p_{2}^{*}< 3/2. $$

So condition (C₂) is satisfied. We can also verify that (C₃) holds by (4.2). Finally, we check condition (C₄). Let $f=f^{(1)}+f^{(2)}$, where

$$\begin{aligned} &f_{n}^{(1)}(t,u,v,w) =\frac{3^{-t-1/2}}{n (1+t^{\underline{1/2}} )^{2}}\sin u_{n},\\ &f_{n}^{(2)}(t,u,v,w) =\frac{2^{-t+1/2}}{2n^{2} (1+t^{\underline{1/2}} )} ( 1+u_{2n+1}+v_{3n}+w_{n+1} )^{1/2}. \end{aligned}$$

Then we obtain that for any bounded sets $V_{i}\subset E$, $i\in\mathbb {N}_{1}^{3}$, $\alpha (f^{2}(t,V_{1},V_{2},V_{3}) )=0$, $t\in\mathbb{N}_{1/2}$. In fact, since $V_{i}$, $i\in\mathbb{N}_{1}^{3}$, are bounded, there exists $r>0$ such that $V_{i}\subset B_{E}(\theta, r)$, $i\in\mathbb{N}_{1}^{3}$. Let $\{ u^{(m)}\}_{m=1}^{\infty}\in V_{1}$, $\{v^{(m)}\}_{m=1}^{\infty}\in V_{2}$, $\{w^{(m)}\}_{m=1}^{\infty}\in V_{3}$. Then for any fixed $t\in\mathbb{N}_{1/2}$, we have

$$ \bigl|f^{(2)}_{n} \bigl(t,u^{(m)},v^{(m)},w^{(m)} \bigr) \bigr| \leq\frac{3r}{4n^{2} [1+\Gamma(3/2) ]} +\frac{1}{2n^{2} [1+\Gamma(3/2) ]}, \quad n,m\in \mathbb{N}_{1}. $$

(4.3)

Therefore, $\{f^{(2)}_{n} (t,u^{(m)},v^{(m)},w^{(m)} ) \}$ is bounded, and so, by the diagonal method we can choose a subsequence $\{m_{k}\}\subset\{m\}$ such that

$$ f^{(2)}_{n} \bigl(t,u^{(m_{k})},v^{(m_{k})},w^{(m_{k})} \bigr) \rightarrow z_{n} \quad\mbox{as } k\rightarrow\infty, n\in \mathbb{N}_{1}. $$

(4.4)

Now, (4.3) and (4.4) imply

$$ |z_{n}| \leq\frac{3r}{4n^{2} [1+\Gamma(3/2) ]} +\frac{1}{2n^{2} [1+\Gamma(3/2) ]}, \quad n,m\in\mathbb{N}_{1}. $$

(4.5)

So $z=(z_{1},z_{2},\ldots,z_{n},\ldots)\in c_{0}=E$, and it is easy to see from (4.3)-(4.5) that

$$\bigl\| f^{(2)} \bigl(t,u^{(m_{k})},v^{(m_{k})},w^{(m_{k})} \bigr)-z \bigr\| =\sup_{n} \bigl|f_{n}^{(2)} \bigl(t,u^{(m_{k})},v^{(m_{k})},w^{(m_{k})} \bigr)-z_{n} \bigr| \rightarrow0 \quad\mbox{as } k\rightarrow\infty. $$

Thus, we have proved that $f^{2}(t,V_{1},V_{2},V_{3})$ is relatively compact in E and

$$ \alpha \bigl(f^{2}(t,V_{1},V_{2},V_{3}) \bigr)=0. $$

(4.6)

On the other hand, for any $t\in\mathbb{N}_{1/2}$, $u,\overline{u}\in V_{1}$, $v,\overline{v}\in V_{2}$, $w,\overline{w}\in V_{3}$, we have

$$\bigl|f_{n}^{(1)}(t,u,v,w)-f_{n}^{(1)}(t, \overline{u},\overline{v},\overline{w})\bigr| \leq\frac{3^{-t-1/2}}{ (1+t^{\underline{1/2}} )^{2}}|u_{n}- \overline{u}_{n}|, $$

and, therefore,

$$ \alpha \bigl(f^{(1)}(t,V_{1},V_{2},V_{3}) \bigr)\leq\frac {3^{-t-1/2}}{(1+t^{\underline{1/2}})^{2}}\alpha(V_{1}). $$

(4.7)

From (4.6) and (4.7) we have

$$\alpha \bigl(f(t,V_{1},V_{2},V_{3}) \bigr) \leq \alpha \bigl(f^{(1)}(t,V_{1},V_{2},V_{3}) \bigr) \leq\frac{3^{-t-1/2}}{(1+t^{\underline{1/2}})^{2}}\alpha(V_{1}). $$

In view of $\sum_{t=1/2}^{\infty}\frac {3^{-t-1/2}}{(1+t^{\underline{1/2}})}<1/2<\Gamma(3/2)$, we get that condition (C₄) holds with $l_{1}(t)=\frac {3^{-t-1/2}}{(1+t^{\underline{1/2}})^{2}}$ and $l_{2}(t)=l_{3}(t)=0$. So by Theorem 3.1 our conclusion follows. □