Let us note that by the variation of constants formula, the solution of equation (1) satisfies the following equation:

$$\begin{aligned} x(n) ={}& x_{0}\prod _{j=n_{0}}^{n-1}b(j) + \sum_{k=n_{0}}^{n-1} \Biggl(\prod_{j=k+1}^{n-1}b(j) \Biggr)\sum _{i=n_{0}}^{k}K(k,i)x(i) \\ &{}+ \sum_{k=n_{0}}^{n-1} \Biggl(\prod _{j=k+1}^{n-1}b(j) \Biggr)a(k)\quad \mbox{for }n\geq n_{0}, \end{aligned}$$

(3)

where \(x(n_{0})=x_{0}\). Using this fact we prove the following theorems.

### Theorem 1

*Suppose that*

$$ \sum_{n=n_{0}}^{\infty} \bigl\vert a(n) \bigr\vert < \infty $$

(4)

*and there exists a nonnegative real constant*
*B*
*such that*

$$ \prod_{i=n_{0}}^{n-1} \bigl\vert b(i) \bigr\vert \leq B, \quad n\geq n_{0}. $$

(5)

*Assume also that there exists*
\(\alpha\in(0,1)\)
*such that*

$$ \sum_{k=n_{0}}^{n-1} \prod _{j=k+1}^{n-1} \bigl\vert b(j) \bigr\vert \sum _{i=n_{0}}^{k} \bigl\vert K(k,i) \bigr\vert \leq\alpha\quad \textit{for } n \geq n_{0}. $$

(6)

*Then the solutions of equation* (1) *are equi*-*bounded*.

### Proof

By (4), there exists a nonnegative real constant *A* such that \(\sum_{n=n_{0}}^{\infty} \vert a(n)\vert =A\). Let \(M_{1}>0\) be given. Choose a real number \(M_{2}\) such that \(\frac {(A+M_{1}) B}{1-\alpha} \leq M_{2}\). Let \(u_{0}\) be a real number such that \(\vert u_{0}\vert \leq M_{1}\). We define a subset *S* of *BS* by

$$S= \bigl\{ x \in BS\colon x_{0}=u_{0} \mbox{ and } \Vert x\Vert \leq M_{2} \bigr\} . $$

Then *S* is a closed subset of *BS*. Now, we define the mapping \(T\colon S\rightarrow BS\) as follows:

$$\begin{aligned} (Tx) (n)={}& u_{0}\prod_{j=n_{0}}^{n-1}b(j) + \sum_{k=n_{0}}^{n-1} \Biggl(\prod _{j=k+1}^{n-1}b(j) \Biggr)\sum _{i=n_{0}}^{k}K(k,i)x(i) \\ &{}+ \sum_{k=n_{0}}^{n-1} \Biggl(\prod _{j=k+1}^{n-1}b(j) \Biggr)a(k)\quad \mbox{for }n\geq n_{0}. \end{aligned}$$

(7)

We will prove that *T* has a fixed point in *S*.

Firstly, we show that \(T(S)\subset S\). Indeed, if \(x \in S\), then by (7), (6), (4) and (5), we get

$$\begin{aligned} \bigl\vert (Tx) (n) \bigr\vert \leq{}& M_{1} \prod_{j=n_{0}}^{n-1} \bigl\vert b(j) \bigr\vert \\ & {}+ M_{2} \sum_{k=n_{0}}^{n-1} \prod _{j=k+1}^{n-1} \bigl\vert b(j) \bigr\vert \sum_{i=n_{0}}^{k} \bigl\vert K(k,i) \bigr\vert + \prod_{j=n_{0}}^{n-1} \bigl\vert b(j) \bigr\vert \sum_{k=n_{0}}^{n-1} \bigl\vert a(k) \bigr\vert \\ \leq{}& B M_{1}+ \alpha M_{2} + B A \leq M_{2}\quad \mbox{for } n\geq n_{0}. \end{aligned} $$

Moreover, we have \((Tx)(n_{0})=u_{0}\). Hence, \(T(x) \in S\) and \(T(S)\subset S\).

Next, we prove that *T* is a contraction. Let \(y,z \in S\). Then, using (6) and (7), we get

$$\begin{aligned} \bigl\vert (Ty) (n) - (Tz) (n) \bigr\vert &\leq\sum _{k=n_{0}}^{n-1} \prod _{j=k+1}^{n-1} \bigl\vert b(j) \bigr\vert \sum _{i=n_{0}}^{k} \bigl\vert K(k,i) \bigr\vert \bigl\vert y(i)-z(i) \bigr\vert \\ & \leq\sup_{n \geq n_{0}} \bigl\vert y(n)-z(n) \bigr\vert \Biggl( \sum_{k=n_{0}}^{n-1} \prod _{j=k+1}^{n-1} \bigl\vert b(j) \bigr\vert \sum _{i=n_{0}}^{k} \bigl\vert K(k,i) \bigr\vert \Biggr) \\ & \leq\alpha\sup_{n \geq n_{0}} \bigl\vert y(n)-z(n) \bigr\vert \quad \mbox{for }n\geq n_{0}. \end{aligned} $$

Thus

$$\Vert Ty-Tz\Vert = \sup_{n\geq n_{0}} \bigl\vert (Ty) (n) - (Tz) (n) \bigr\vert \leq \alpha \Vert y-z\Vert . $$

Hence, *T* is a contraction. By the contraction mapping principle, *T* has a unique fixed point \(x^{*}\) in *S*. It means that \((Tx^{*})(n)=x^{*}(n)\). So, from (7) we have

$$x^{*}(n)= u_{0}\prod_{j=n_{0}}^{n-1}b(j) + \sum_{k=n_{0}}^{n-1} \Biggl(\prod _{j=k+1}^{n-1}b(j) \Biggr)\sum _{i=n_{0}}^{k}K(k,i)x^{*}(i) + \sum_{k=n_{0}}^{n-1} \Biggl(\prod _{j=k+1}^{n-1}b(j) \Biggr)a(k). $$

By comparing the above equation and (3), we see that the unique fixed point \(x^{*}\) is also a solution of equation (1). Hence, by the definition of the set *S*, the solutions of equation (1) are equi-bounded. This completes the proof. □

Theorem 1 extends Theorem 2.1 in [1] and Theorem 1 in [9].

The following example illustrates the result presented in Theorem 1.

### Example 1

Let us consider the linear Volterra difference equation

$$ x(n+1)=\frac{1}{n(n+1)}+ x(n)+\sum_{i=1}^{n} \frac {i}{n^{2}(n+1)^{2}}x(i),\quad n\geq1. $$

(8)

Here

$$a(n)=\frac{1}{n(n+1)}, \qquad b(n)\equiv1\quad \mbox{and}\quad K(n,i)= \frac {i}{n^{2}(n+1)^{2}}. $$

Hence \(A=\sum_{n=1}^{\infty}\frac{1}{n(n+1)}= 1\), \(B =\prod_{j=1}^{\infty}b(j)=1\),

$$\sum_{k=1}^{n-1}\prod _{j=k+1}^{n-1}\bigl\vert b(j)\bigr\vert \sum _{i=1}^{k} \bigl\vert K(k,i)\bigr\vert = \sum _{k=1}^{n-1} \frac {1}{k^{2}(k+1)^{2}}\sum _{i=1}^{k}i=\frac{1}{2}\sum _{k=1}^{n-1} \frac{1}{k(k+1)}\leq\frac{1}{2}= \alpha. $$

So, by Theorem 1 the solutions of (1) are equi-bounded. For example, for \(M_{1}=1\) we have \(M_{2}\geq\frac{(A+M_{1}) B}{1-\alpha}= 4\). Hence, all solutions with the initial condition \(\vert x(1)\vert \leq1\) have the property \(\Vert x\Vert \leq4\).

Note that for equation (8), Theorem 1 in [28] cannot be applied since the assumption given below of this theorem

$$\bigl\vert b(n) \bigr\vert + \bigl\vert K(n,n) \bigr\vert +\varphi(0) \leq1-\alpha, $$

where \(\alpha\in(0,1)\), \(\varphi(n)\geq0\), \(n\geq1\), is not satisfied.

In the next example we present a Volterra difference equation for which condition (24) from [2] does not hold. Hence, Theorem 5.1 [2] is not applicable for this example whereas our Theorem 1 is.

### Example 2

Let us consider the linear Volterra difference equation

$$ x(n+1)= \biggl\lfloor \frac{n+1}{2} \biggr\rfloor ^{(-1)^{n}} x(n)+\sum_{i=1}^{n} \frac{1}{2^{k+i}}x(i), \quad n \geq1. $$

(9)

Here

$$b(n)= \biggl\lfloor \frac{n+1}{2} \biggr\rfloor ^{(-1)^{n}}= \biggl(1,1, \frac{1}{2},2, \frac {1}{3},3,\frac{1}{4},4, \dots \biggr) \quad \mbox{and}\quad K(n,i)=\frac{1}{2^{n+i}}. $$

It is easy to verify that conditions (5) and (6) are satisfied. So, by Theorem 1 the solutions of (9) are equi-bounded.

### Remark 1

From Theorem 3.1 in [7] it follows that the boundedness of equation (2) is equivalent to the stability of its zero solution. Hence, and by Theorem 1, conditions (5) and (6) ensure the stability of the zero solution of equation (2).

Now, we provide conditions for the asymptotic stability of the zero solution of equation (2).

### Theorem 2

*If condition* (6) *holds and*

$$ \lim_{n\to\infty}\prod_{i=n_{0}}^{n-1}b(i)=0, $$

(10)

*then the zero solution of equation* (2) *is asymptotically stable*.

### Proof

By (10) there exists a constant \(B_{1}\) such that \(\vert \prod_{i=n_{0}}^{n-1}b(i)\vert \leq B_{1}\) for \(n\geq n_{0}\). Let \(M >0\) be given. Choose \(\mu>0\) such that

$$ \mu\leq\frac{(1-\alpha)M }{B_{1}}. $$

(11)

Let \(u_{0}\) be a real number such that \(\vert u_{0} \vert \leq\mu\). We define a subset \(S_{1}\) of *BS* by

$$S_{1}= \Bigl\{ x \in BS\colon x_{0}=u_{0}, \Vert x\Vert \leq M \mbox{ and } \lim_{n\to\infty} x(n)=0 \Bigr\} . $$

Then \(S_{1}\) is a closed subset of *BS*. We define the mapping \(T_{1}\) for \(x\in S_{1}\) as follows:

$$ (T_{1}x) (n)= u_{0}\prod _{j=n_{0}}^{n-1}b(j) + \sum_{k=n_{0}}^{n-1} \Biggl(\prod_{j=k+1}^{n-1}b(j) \Biggr)\sum _{i=n_{0}}^{k}K(k,i)x(i)\quad \mbox{for }n\geq n_{0}. $$

(12)

We show that \(T_{1}(S_{1})\subset S_{1}\). Indeed, if \(x \in S_{1}\), then by (6) and (11) we have

$$\begin{aligned} \bigl\vert (T_{1}x) (n) \bigr\vert &\leq \mu\prod_{j=n_{0}}^{n-1} \bigl\vert b(j) \bigr\vert + M \sum_{k=n_{0}}^{n-1} \prod _{j=k+1}^{n-1} \bigl\vert b(j) \bigr\vert \sum_{i=n_{0}}^{k} \bigl\vert K(k,i) \bigr\vert \\ & \leq\mu B_{1} + M \alpha \leq M \quad \mbox{for } n\geq n_{0}. \end{aligned} $$

Moreover, from (10) we immediately get that the first term on the right-hand side of (12) tends to zero. We show that

$$\Biggl\vert \sum_{k=n_{0}}^{n-1} \Biggl(\prod _{j=k+1}^{n-1}b(j) \Biggr)\sum _{i=n_{0}}^{k}K(k,i)x(i) \Biggr\vert \rightarrow0 \quad \mbox{as } n\rightarrow\infty. $$

Let \(\varepsilon>0\). Since \(\lim_{n\to\infty} x(n)=0\), there exists \(n_{1}\geq n_{0}\) such that

$$ \sup_{n\geq n_{1}} \bigl\vert x(n) \bigr\vert \leq \frac{\varepsilon}{2\alpha}. $$

(13)

By (10) there exists \(n_{2} \geq n_{1}\geq n_{0}\) such that

$$ \prod_{j=n_{2}}^{n-1} \bigl\vert b(j) \bigr\vert \leq\frac{\varepsilon }{2\alpha M} \quad \mbox{for }n\geq n_{2}. $$

(14)

Then, for \(n\geq n_{2}\), we have

$$\begin{aligned} & \Biggl\vert \sum_{k=n_{0}}^{n-1} \Biggl(\prod_{j=k+1}^{n-1}b(j) \Biggr)\sum _{i=n_{0}}^{k}K(k,i)x(i) \Biggr\vert \\ &\quad \leq\sum_{k=n_{0}}^{n-1} \prod _{j=k+1}^{n-1} \bigl\vert b(j) \bigr\vert \sum _{i=n_{0}}^{k} \bigl\vert K(k,i) \bigr\vert \bigl\vert x(i) \bigr\vert \\ &\quad \leq \sum_{k=n_{0}}^{n_{2}-1} \prod _{j=k+1}^{n-1} \bigl\vert b(j) \bigr\vert \sum _{i=n_{0}}^{k} \bigl\vert K(k,i) \bigr\vert \bigl\vert x(i) \bigr\vert \\ &\qquad {}+ \sum_{k=n_{2}}^{n-1} \prod _{j=k+1}^{n-1} \bigl\vert b(j) \bigr\vert \sum _{i=n_{0}}^{k} \bigl\vert K(k,i) \bigr\vert \bigl\vert x(i) \bigr\vert \\ &\quad \leq \sup_{n\geq n_{0}} \bigl\vert x(n) \bigr\vert \sum _{k=n_{0}}^{n_{2}-1} \prod_{j=k+1}^{n_{2}-1} \bigl\vert b(j) \bigr\vert \prod_{j=n_{2}}^{n-1} \bigl\vert b(j) \bigr\vert \sum_{i=n_{0}}^{k} \bigl\vert K(k,i) \bigr\vert \\ &\qquad {}+ \sup_{n\geq n_{2}} \bigl\vert x(n) \bigr\vert \sum _{k=n_{2}}^{n-1} \prod_{j=k+1}^{n-1} \bigl\vert b(j) \bigr\vert \sum_{i=n_{0}}^{k} \bigl\vert K(k,i) \bigr\vert . \end{aligned} $$

From the above, the definition of the set \(S_{1}\), (6), (13) and (14), we obtain

$$\Biggl\vert \sum_{k=n_{0}}^{n-1} \Biggl(\prod _{j=k+1}^{n-1}b(j) \Biggr)\sum _{i=n_{0}}^{k}K(k,i)x(i) \Biggr\vert \leq M \prod _{j=n_{2}}^{n-1} \bigl\vert b(j) \bigr\vert \alpha+ \frac{\varepsilon }{2 \alpha} \alpha\leq\frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. $$

Thus, the second term of formula (12) tends to zero, too. So, we have \(\lim_{n\to\infty}(Tx)(n)=0\). Hence, \(T_{1}(x) \in S_{1}\) and \(T_{1}(S_{1})\subset S_{1}\).

Similarly as in the proof of Theorem 1, we get that *T* has a fixed point. This fixed point solves equation (1) and tends to zero. Hence, the zero solution of equation (2) is asymptotically stable. □

### Example 3

Let us consider the linear Volterra difference equation

$$ x(n+1)=\frac{1}{n} x(n)+\sum_{i=1}^{n} \frac{1}{2n}x(i),\quad n\geq1. $$

(15)

Here

$$b(n)=\frac{1}{n}\quad \mbox{and}\quad K(n,i)= \frac{1}{2n}. $$

It is easy to verify that conditions (5) and (6) are satisfied. So, by Theorem 1 the solutions of (15) are equi-bounded. Moreover, since condition (10) is also satisfied, by Theorem 2 the zero solution of equation (15) is asymptotically stable.

Note that the result about asymptotic stability of the zero solution obtained in the section Scalar Equation (see Example 3.1) of [4] is not applicable here because assumption vi) \(\sum_{n=1}^{\infty} \vert K(n,i)\vert \leq C\), where *C* is a positive constant, for (15) is not satisfied. Similarly, Theorem 2.7 of [7] could not be applied for (15) since the kernel \(K(n,i)= \frac{1}{2n}\) does not satisfy condition (2.28).

In Proposition 3.2 of [3], the necessary and sufficient conditions for boundedness of all solutions of equations of type (2) are given. It is easy to see that for (15), assumption (17) of [3] does not hold. But, as it was shown above, each solution of this equation is bounded. Corollary 3.7 of [3] is not applicable here, too.