For k\in \mathbb{N}, define a sequence

{b}_{k}(n):=\{\begin{array}{cc}1,\hfill & k=0,\hfill \\ {\sum}_{\ell =1}^{m}|{a}_{\ell}(n)|{b}_{k-1}({h}_{\ell}(n)-1),\hfill & k\ge 1\hfill \end{array}

(11)

for n\ge {n}_{0}+k(\tau +1).

**Theorem 1** (Correction of Theorem A)

*Suppose that there exists*
r\in \mathbb{N}
*such that*

\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}{b}_{r}(n)<1.

*Then* (1) *is exponentially stable*.

*Proof* Let us prove for all k\in \mathbb{N} that

|x(n+1)|\le {b}_{k}(n)\underset{n-k(\tau +1)\le j\le n}{max}\left\{|x(j)|\right\}\phantom{\rule{1em}{0ex}}\text{for all}n\ge {n}_{0}+k(\tau +1).

(12)

We proceed by induction in *k*. From (1), for k=1, we have

\begin{array}{rcl}|x(n+1)|& \le & \sum _{\ell =1}^{m}|{a}_{\ell}(n)|\left|x({h}_{\ell}(n))\right|\\ \le & \sum _{\ell =1}^{m}|{a}_{\ell}(n)|\underset{n-(\tau +1)\le j\le n}{max}\left\{|x(j)|\right\}\\ =& {b}_{1}(n)\underset{n-(\tau +1)\le j\le n}{max}\left\{|x(j)|\right\}\end{array}

(13)

for all n\ge {n}_{0}+\tau +1. Thus, the claim is true for k=1. Assume now that the claim is true for some k\ge 1. From (12) and (13), for all n\ge {n}_{0}+(k+1)(\tau +1), we have

\begin{array}{rcl}|x(n+1)|& \le & \sum _{\ell =1}^{m}|{a}_{\ell}(n)|{b}_{k}({h}_{\ell}(n)-1)\underset{{h}_{\ell}(n)-1-k(\tau +1)\le j\le {h}_{\ell}(n)-1}{max}\left\{|x(j)|\right\}\\ \le & \sum _{\ell =1}^{m}|{a}_{\ell}(n)|{b}_{k}({h}_{\ell}(n)-1)\underset{n-(k+1)(\tau +1)\le j\le n}{max}\left\{|x(j)|\right\}\\ =& {b}_{k+1}(n)\underset{n-(k+1)(\tau +1)\le j\le n}{max}\left\{|x(j)|\right\},\end{array}

which shows that (12) is true when *k* is replaced with (k+1). Using (12) with k=r, we see that the solution is exponentially stable by Theorem B. □

Theorem 1 with r=1 immediately yields the following result.

**Corollary 1**
*Assume that*

\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\sum _{\ell =1}^{m}|{a}_{\ell}(n)|<1.

*Then* (1) *is exponentially stable*.

**Remark 1** The claim of Theorem A for r=0 is correct.

Setting r=2 in Theorem 1, we obtain the following corollary, which is also proved in [[1], Theorem 2.17].

**Corollary 2**
*Assume that*

\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\sum _{{\ell}_{1}=1}^{m}|{a}_{{\ell}_{1}}(n)|\sum _{{\ell}_{2}=1}^{m}\left|{a}_{{\ell}_{2}}({h}_{{\ell}_{1}}(n)-1)\right|<1.

*Then* (1) *is exponentially stable*.

**Remark 2** Theorem A for the nondelay equation

x(n+1)=a(n)x(n)\phantom{\rule{1em}{0ex}}\text{for}n\ge {n}_{0}

is correct. Indeed, Theorem 1 reduces to Theorem A since for k\ge 1, we get

{b}_{k}(n)=\prod _{j=0}^{k-1}|a(n-j)|\phantom{\rule{1em}{0ex}}\text{for}n\ge {n}_{0}+k.

Setting r=3 in Theorem 1 gives us the following corollary.

**Corollary 3**
*Assume that*

\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\sum _{{\ell}_{1}=1}^{m}|{a}_{{\ell}_{1}}(n)|\sum _{{\ell}_{2}=1}^{m}\left|{a}_{{\ell}_{2}}({h}_{{\ell}_{1}}(n)-1)\right|\sum _{{\ell}_{3}=1}^{m}\left|{a}_{{\ell}_{3}}({h}_{{\ell}_{2}}({h}_{{\ell}_{1}}(n)-1)-1)\right|<1.

*Then* (1) *is exponentially stable*.

**Example 3** Consider the delay difference equation (3) with (4), where p,q\in \mathbb{R}, which can be written in the two equivalent forms:

x(n+1)=-{a}_{1}(n)x(n)-{a}_{2}(n)x(n-1)\phantom{\rule{1em}{0ex}}\text{for}n\ge 0

(14)

and

x(n+1)-x(n)=-({a}_{1}(n)+1)x(n)-{a}_{2}(n)x(n-1)\phantom{\rule{1em}{0ex}}\text{for}n\ge 0,

(15)

where

{a}_{1}(n)=\{\begin{array}{cc}p,\hfill & n=2l,\hfill \\ 0,\hfill & n=2l+1\hfill \end{array}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{a}_{2}(n)=\{\begin{array}{cc}0,\hfill & n=2l,\hfill \\ q,\hfill & n=2l+1.\hfill \end{array}

Computing \{{b}_{k}(n)\} defined by (11), we see that

{b}_{k}(n)=\{\begin{array}{cc}|p{q}^{k-1}|,\hfill & n=2l,\hfill \\ |{q}^{k}|,\hfill & n=2l+1\hfill \end{array}\phantom{\rule{1em}{0ex}}\text{for}k\in \mathbb{N},n\ge 2k.

Equation (3) is exponentially stable by Theorem 1 if |q|<1 because there always exists r\in \mathbb{N} such that |p{q}^{r-1}|<1 and |{q}^{r}|<1. From (5), we see that |q|<1 is the best possible condition for the global exponential stability of (3) with (4).

Application of a recent result [[12], Theorem 6] to (14) gives us (1+|p|)|q|<1, which implies |q|<1.

The so-called ‘3/2-test’ (see [13] and [[1], Theorem A]) can be applied to (15) if p>-1 and q>0, and ensures global exponential stability when

p+q+2<\frac{3}{2}+\frac{1}{2\cdot 2}=\frac{7}{4}\phantom{\rule{1em}{0ex}}\text{or equivalently}\phantom{\rule{1em}{0ex}}p+q<-\frac{1}{4}

for which 0<q<3/4 is necessary.

It is obvious that these two results and Corollary 1 cannot deliver any answer for the exponential stability when p=1 and q=1/2.

As mentioned in Remark 2, Theorem A is valid for a nondelay scalar equation. Next, any higher-order (of the order not exceeding *d*) equation (1), with n-d<{h}_{\ell}(n)\le n, can be rewritten as the first-order system

X(n+1)=A(n)X(n),\phantom{\rule{1em}{0ex}}n=0,1,2,\dots ,

(16)

where X(n)\in {\mathbb{R}}^{d}, A(n) are d\times d matrices. Indeed, denote X(0)={(x(-d+1),x(-d+2),\dots ,x(0))}^{T}, X(n)={(x(nd-d+1),x(nd-d+2),\dots ,x(nd))}^{T} and rewrite (1) as

x(n)=\sum _{j=1}^{d}b(n,j)x(n-j),\phantom{\rule{1em}{0ex}}n\in \mathbb{N},

(17)

where

b(n,j)=\sum _{l\in \{1,\dots ,m|{h}_{l}(n)=n-j\}}{a}_{l}(n).

Then we can define the matrix A(0)={(c(i,j))}_{i=1}^{d} as follows:

and X(1)=A(0)X(0). Similarly, we construct A(n), n\in \mathbb{N} and obtain system (16). Since \parallel X(n)\parallel \ge |x(j)|, j=nd-d+1,nd-d+2,\dots ,nd, exponential (asymptotical) stability of (16) implies the relevant stability of (1). We recall that (16) is exponentially stable if there exist {n}_{0}\in \mathbb{N}, L>0, and \mu \in (0,1) such that \parallel X(n)\parallel \le L{\mu}^{n}\parallel X(0)\parallel, n\ge {n}_{0}.

**Theorem 2** *If there exist* \lambda \in (0,1), M>0, *and* {n}_{0}\in \mathbb{N} *such that* \parallel A(n)\parallel \le M *and* \parallel {\prod}_{j=n}^{n+k-1}A(j)\parallel \le \lambda *for every* n\ge {n}_{0} *and for some positive integer* *k*, *then* (16) *is exponentially stable*.

*Proof* Without loss of generality, we can assume M>1 and {\prod}_{j=0}^{{n}_{0}-1}\parallel A(j)\parallel \le M. Further, for any n\ge {n}_{0} denote m=[\frac{n-{n}_{0}-1}{k}], where [t] is the integer part of *t*, and obtain the estimate

\begin{array}{rcl}\parallel X(n)\parallel & =& \parallel A(n-1)\cdots A(0)X(0)\parallel \le \parallel A(n-1)\cdots A(0)\parallel \parallel X(0)\parallel \\ =& \parallel \prod _{j={n}_{0}+km}^{n-1}A(j)\cdots \prod _{j={n}_{0}}^{{n}_{0}+k-1}A(j)\prod _{j=0}^{{n}_{0}-1}A(j)\parallel \parallel X(0)\parallel \\ \le & \parallel \prod _{j={n}_{0}+km}^{n-1}A(j)\parallel \cdots \parallel \prod _{j={n}_{0}}^{{n}_{0}+k-1}A(j)\parallel \parallel \prod _{j=0}^{{n}_{0}-1}A(j)\parallel \parallel X(0)\parallel \\ \le & M{\lambda}^{m}{M}^{k}\le {M}^{k+1}{\left(\frac{1}{\lambda}\right)}^{1+{n}_{0}/k}{\left({\lambda}^{1/k}\right)}^{n}\parallel X(0)\parallel =L{\mu}^{n}\parallel X(0)\parallel \end{array}

for n\ge {n}_{0}, where L={M}^{k+1}{\lambda}^{-1-{n}_{0}/k}, \mu ={\lambda}^{1/k}. □

**Example 4** (Example 6 in [2])

If in (16)

{A}_{2m}=\frac{1}{2}\left(\begin{array}{cc}5.1& 4.9\\ 4.9& 5.1\end{array}\right),\phantom{\rule{2em}{0ex}}{A}_{2m+1}=\frac{1}{2}\left(\begin{array}{cc}7.1& -6.9\\ -6.9& 7.1\end{array}\right),\phantom{\rule{1em}{0ex}}m=0,1,2,\dots

then both {A}_{2m} and {A}_{2m+1} have the norms exceeding one (they have eigenvalues of 5 and 7, respectively), but the product

{A}_{2m}{A}_{2m+1}={A}_{2m+1}{A}_{2m}=\left(\begin{array}{cc}0.6& -0.1\\ -0.1& 0.6\end{array}\right)

has the norm \parallel {A}_{2m}{A}_{2m+1}\parallel =\parallel {A}_{2m+1}{A}_{2m}\parallel =0.7<1, thus (16) is exponentially stable.

**Example 5** Consider (16) with a 4-periodic matrix A(n), where {A}_{4m+j}=\left(\begin{array}{cc}0& -1.1\\ 1.1& 0\end{array}\right), m=0,1,2,\dots , j=1,2,3, {A}_{4m}=\left(\begin{array}{cc}0& -0.7\\ 0.7& 0\end{array}\right). Then \parallel {A}_{4m+j}\parallel =1.1>1, j=1,2,3, but (16) is exponentially stable since {A}_{n+3}{A}_{n+2}{A}_{n+1}{A}_{n}=\left(\begin{array}{cc}0.9317& 0\\ 0& 0.9317\end{array}\right) for any n=0,1,2,\dots , and \lambda =\parallel {A}_{n+3}{A}_{n+2}{A}_{n+1}{A}_{n}\parallel =0.9317<1.