On Ulam Stability of an Operatorial Equation

Abstract

An iterative method generates a sequence associated with an equation, that, under appropriate conditions, converges to a solution of that equation. A perturbation of the equation produces also a perturbation of the sequence. In this paper, we study the Ulam stability (the behavior of the solutions of the perturbed equation with respect to the solutions of the exact equation) of an operatorial equation of the form $x_{n+1}=T_nx_n+a_n$, where $T_n:X \rightarrow X$, $n \in \mathbb {N}$, are linear and bounded operators acting on a Banach space X. As applications we obtain some stability results for the case of Volterra, Fredholm and Gram–Schmidt operators. In this way, we improve and complement some results on this topic.

1 Introduction

In what follows, by $\mathbb {N}=\{0,1,2,\ldots \}$ we denote the set of all nonnegative integers and by X a Banach space over $\mathbb {K}\in \{\mathbb {R},\mathbb {C}\}$. Let $T:X \rightarrow X$ be a bounded linear operator and consider the equation $x=Tx+y$, where $y \in X$ is a given element. By using the fixed point method to solve the equation, we get a sequence of successive approximations $(x_n)_{n \ge 0}$, satisfying the linear difference equation $x_{n+1}=Tx_{n}+y, n \ge 0, x_0 \in X,$ converging to the solution. The Ulam stability of this difference equation was studied in [5].

In this paper, we consider a generalization of the previous difference equation, more precisely the linear difference equation

$$\begin{aligned} x_{n+1}=T_nx_n+a_n,\ n \in \mathbb {N}, \end{aligned}$$

(1.1)

where $(T_n)_{n\ge 0}$ is a sequence of bounded linear operators, $T_n:X \rightarrow X$, and $(a_n)_{n \ge 0}$ a sequence in X. We study its Ulam stability, which concerns the behavior of the solutions of Eq. (1.1) under perturbations, with respect to the solutions of the unperturbed equation. A particular case of this equation was considered and studied in [12] for the case of matrix operators $(T_n)_{n \ge 0}$. For various results on difference equations we refer the reader to [14, 15].

Definition 1.1

Equation (1.1) is called Ulam stable if there exists a constant $L\ge 0$ such that for every $\varepsilon >0$ and every sequence $(x_n)_{n\ge 0}$ in X satisfying

$$\begin{aligned} \Vert x_{n+1}-T_nx_n-a_n\Vert \le \varepsilon ,\ n\in \mathbb {N}, \end{aligned}$$

(1.2)

there exists a sequence $(y_n)_{n\ge 0}$ in X such that

$$\begin{aligned} y_{n+1}=T_ny_n+a_n,\ n\in \mathbb {N} \end{aligned}$$

(1.3)

and

$$\begin{aligned} \Vert x_n-y_n\Vert \le L\varepsilon ,\ n\in \mathbb {N}. \end{aligned}$$

(1.4)

The sequence $(x_n)_{n \ge 0}$ satisfying (1.2) for some $\varepsilon >0$ is called approximate solution of Eq. (1.1).

So we can reformulate Definition 1.1 as follows: Eq. (1.1) is Ulam stable if for every approximate solution of it there exists an exact solution close to it. The number L from (1.4) is called an Ulam constant of Eq. (1.1). Further, we denote by $L_R$ the infimum of all Ulam constants of (1.1). If $L_R$ is an Ulam constant for (1.1), then we call it the best Ulam constant or the Ulam constant of Eq. (1.1). Generally, the infimum of all Ulam constants of an equation is not an Ulam constant of that equation (see [8, 27]).

The origin of stability for functional equations is a question formulated by S. M. Ulam during a talk given to Madison University, Wisconsin, and concerns the approximate homomorphism of a metric group [29]. A first partial answer to Ulam’s question was given a year later by D. H. Hyers for the Cauchy functional equation in Banach spaces [17]. The topic was intensively studied by many authors in the last 50 years; for results, various generalizations and extensions on Ulam stability we refer the reader to [2, 8, 18, 22]. We recall also the results obtained in [7, 20, 21] on Ulam stability for some second-order linear functional equations in connection with Fibonacci and Lucas sequences.

Some results on Ulam stability for the linear difference equations in Banach spaces were obtained by Brzdek et al. [9,10,11, 24]. Buse et al. [6, 13] proved that a discrete system $X_{n+1}=AX_n, n \in \mathbb {N}$, where A is a $m \times m$ complex matrix, is Ulam stable if and only if A possesses a discrete dichotomy. Recently, Baias and Popa obtained results on Ulam stability of linear difference equations of order one and two and determined the best Ulam constant in [3, 4]. Popa and Rasa obtained an explicit representation of the best Ulam constant of some classical operators in approximation theory in [25, 26].

2 Main Results

In this section, we present some results on Ulam stability for Eq. (1.1). First, we give a result which will be useful in the sequel.

Lemma 2.1

Suppose that the sequence $(x_n)_{n\ge 0}$ satisfies Eq. (1.1). Then

$$\begin{aligned} x_{n}=T_{n-1}T_{n-2} \ldots T_0 x_{0}+\sum \limits _{k=1}^{n-1} T_{n-1} \ldots T_{k}a_{k-1} + a_{n-1}, \ n\ge 2. \end{aligned}$$

If in addition $T_n$, $n \ge 0$, are invertible operators, then

$$\begin{aligned} x_{n}=T_{n-1} \ldots T_0\left( x_{0}+\sum \limits _{k=1}^{n} T_0^{-1}T_1^{-1} \ldots T_{k-1}^{-1} a_{k-1}\right) , \ n\ge 1. \end{aligned}$$

Proof

Induction. $\square$

The first result on Ulam stability for Eq. (1.1) is contained in the following theorem.

Theorem 2.2

Suppose that $(T_n)_{n\ge 0}$ is a sequence of invertible operators such that

$$\begin{aligned} \limsup \Vert T_n^{-1}\Vert <1. \end{aligned}$$

(2.1)

Then for every $\varepsilon >0$ and every sequence $(x_n)_{n\ge 0}$ in X satisfying

$$\begin{aligned} \Vert x_{n+1}-T_nx_n-a_n\Vert \le \varepsilon ,\ n\in \mathbb {N}, \end{aligned}$$

(2.2)

there exists a unique sequence $(y_n)_{n\ge 0}$ in X such that

$$\begin{aligned} y_{n+1}=T_ny_n+a_n,\ n\in \mathbb {N} \end{aligned}$$

(2.3)

and

$$\begin{aligned} \Vert x_n-y_n\Vert \le L\varepsilon , \ n\in \mathbb {N}, \end{aligned}$$

(2.4)

where

$$\begin{aligned} L=\sup \limits _{n\in \mathbb {N}} \sum \limits _{k=0}^{\infty } \Vert T_n^{-1} \ldots T_{n+k}^{-1}\Vert < \infty . \end{aligned}$$

Proof

Existence. Suppose that $(x_n)_{n\ge 0}$ satisfies (2.2) and let

$$\begin{aligned} b_n:=x_{n+1}-T_nx_n-a_n,\ n\in \mathbb {N}. \end{aligned}$$

Then $\Vert b_n\Vert \le \varepsilon , \ n\ge 0$, and, according to Lemma 2.1

$$\begin{aligned} x_{n}=T_{n-1} \ldots T_0\left( x_{0}+\sum \limits _{k=1}^{n} T_0^{-1} \ldots T_{k-1}^{-1} (a_{k-1}+b_{k-1})\right) , \ n\ge 1. \end{aligned}$$

Remark further that the series

$$\begin{aligned} \sum \limits _{n=1}^{\infty } \Vert T_0^{-1}\Vert \cdots \Vert T_{n-1}^{-1}\Vert \end{aligned}$$

is convergent. Indeed, denoting $c_n=\Vert T_0^{-1}\Vert \cdots \Vert T_{n-1}^{-1}\Vert , \ n \ge 1$, we get

$$\begin{aligned} \limsup \frac{c_{n+1}}{c_n}=\limsup \Vert T_n^{-1}\Vert <1. \end{aligned}$$

It follows that the series

$$\begin{aligned} \sum \limits _{n=1}^{\infty } T_0^{-1} \ldots T_{n-1}^{-1} b_{n-1} \end{aligned}$$

is convergent, too. This is a simple consequence of the first comparison test, since

$$\begin{aligned} \Vert T_0^{-1} \ldots T_{n-1}^{-1} b_{n-1}\Vert&\le \Vert T_0^{-1} \ldots T_{n-1}^{-1}\Vert \cdot \Vert b_{n-1}\Vert \le \Vert T_0^{-1} \ldots T_{n-1}^{-1}\Vert \varepsilon \\&\le \Vert T_0^{-1}\Vert \cdots \Vert T_{n-1}^{-1}\Vert \varepsilon , \ n \ge 1. \end{aligned}$$

Put now

$$\begin{aligned} \sum \limits _{n=1}^{\infty } T_0^{-1} \ldots T_{n-1}^{-1} b_{n-1}=s, s \in X, \end{aligned}$$

and define $(y_n)_{n\ge 0}$ by

$$\begin{aligned} y_{n+1}=T_ny_n+a_n,\ n \ge 1, \ y_0=x_0+s. \end{aligned}$$

Then

$$\begin{aligned} y_{n}=T_{n-1} \ldots T_0\left( y_0+\sum \limits _{k=1}^{n} T_0^{-1} \ldots T_{k-1}^{-1}a_{k-1}\right) , \ n\ge 1, \end{aligned}$$

and

$$\begin{aligned} x_n-y_n&=T_{n-1} \ldots T_0\left( x_0-y_0+\sum \limits _{k=1}^{n} T_0^{-1} \ldots T_{k-1}^{-1}b_{k-1}\right) \\&=T_{n-1} \ldots T_0\left( -\sum \limits _{n=1}^{\infty } T_0^{-1} \ldots T_{n-1}^{-1}b_{n-1}+\sum \limits _{k=1}^{n} T_0^{-1} \ldots T_{k-1}^{-1}b_{k-1}\right) \\&=-T_{n-1} \ldots T_0\left( \sum \limits _{k=0}^{\infty } T_0^{-1} \ldots T_{n+k}^{-1}b_{n+k}\right) \\&=-\sum \limits _{k=0}^{\infty } T_n^{-1} \ldots T_{n+k}^{-1}b_{n+k}, \ n\ge 1. \end{aligned}$$

Hence

$$\begin{aligned} \Vert x_n-y_n\Vert&\le \sum \limits _{k=0}^{\infty } \Vert T_n^{-1} \ldots T_{n+k}^{-1}\Vert \cdot \Vert b_{n+k}\Vert \nonumber \\&\le \varepsilon \sum \limits _{k=0}^{\infty } \Vert T_n^{-1} \ldots T_{n+k}^{-1}\Vert \le L \varepsilon , \ n \in \mathbb {N}. \end{aligned}$$

(2.5)

We prove now that $L< +\infty$. Indeed, since $\limsup \Vert T_n^{-1}\Vert <1$, it follows that there exists a constant $q \in \mathbb {R}$ and $n_0 \in \mathbb {N}$ such that

$$\begin{aligned} \Vert T_n^{-1}\Vert \le q<1, \ n \ge n_0, \end{aligned}$$

which implies that

$$\begin{aligned} \sum \limits _{k=0}^{\infty } \Vert T_n^{-1} \ldots T_{n+k}^{-1}\Vert&\le \sum \limits _{k=0}^{\infty } \Vert T_n^{-1}\Vert \cdots \Vert T_{n+k}^{-1}\Vert \\&\le \sum \limits _{k=0}^{\infty } 2^{k+1}=\frac{q}{1-q}, \ n \ge n_0. \end{aligned}$$

For $n=n_0-1$, $n_0\ge 1$, we have

$$\begin{aligned} \sum \limits _{k=0}^{\infty } \Vert T_n^{-1} \ldots T_{n+k}^{-1}\Vert&= \sum \limits _{k=0}^{\infty } \Vert T_{n_0-1}^{-1} \ldots T_{n_0+k-1}^{-1}\Vert \\&= \Vert T_{n_0-1}^{-1}\Vert +\Vert T_{n_0-1}^{-1}T_{n_0}^{-1}\Vert + \cdots + \Vert \prod _{k=n_0-1}^{\infty } T_k^{-1}\Vert \\&\le \Vert T_{n_0-1}^{-1}\Vert \left( 1+\sum \limits _{k=0}^{\infty } \Vert T_{n_0}^{-1} \ldots T_{n_0+k}^{-1}\Vert \right) <\infty \end{aligned}$$

since the series

$$\begin{aligned} \sum \limits _{k=0}^{\infty } \Vert T_{n_0}^{-1} \ldots T_{n_0+k}^{-1}\Vert \end{aligned}$$

is convergent. Analogously, the above series will be convergent for all $n \in \mathbb {N}, \ n<n_0$ and the proof is done.

Uniqueness. Suppose that for a sequence $(x_n)_{n\ge 0}$ satisfying the relation (2.2) there exist two sequences $(y_n)_{n\ge 0}$ and $(z_n)_{n\ge 0}$ satisfying the relations (2.3) and (2.4). Then, in view of (2.4)

$$\begin{aligned} \Vert x_n-y_n\Vert \le L \varepsilon , \ \Vert x_n-z_n\Vert \le L \varepsilon , \ n \ge 0, \end{aligned}$$

hence

$$\begin{aligned} \Vert y_n-z_n\Vert \le \Vert y_n-x_n\Vert +\Vert x_n-z_n\Vert \le 2L \varepsilon , \ n \ge 0. \end{aligned}$$

On the other hand, taking account of Lemma 2.1, one gets

$$\begin{aligned} y_n-z_n=T_{n-1} \ldots T_0(y_0-z_0), \ n \ge 1, \end{aligned}$$

or equivalently

$$\begin{aligned} y_0-z_0=T_0^{-1} \ldots T_{n-1}^{-1}(y_n-z_n). \end{aligned}$$

Thus

$$\begin{aligned} \Vert y_0-z_0\Vert&\le \Vert T_0^{-1} \ldots T_{n-1}^{-1}\Vert \cdot \Vert y_n-z_n\Vert \nonumber \\&\le 2L\varepsilon \cdot \Vert T_0^{-1} \ldots T_{n-1}^{-1}\Vert , n\ge 1. \end{aligned}$$

(2.6)

Now, since the series $\sum \nolimits _{n=1}^{\infty } \Vert T_0^{-1}\Vert \cdots \Vert T_{n-1}^{-1}\Vert$ is convergent, we get that the series

$$\begin{aligned} \sum \limits _{n=1}^{\infty } \Vert T_0^{-1} \ldots T_{n-1}^{-1}\Vert \end{aligned}$$

is convergent too, and consequently

$$\begin{aligned} \lim _{n \rightarrow \infty } \Vert T_0^{-1} \ldots T_{n-1}^{-1}\Vert =0. \end{aligned}$$

(2.7)

Finally, from (2.6) and (2.7), we get $y_0=z_0$, so $y_n=z_n$, for all $n \in \mathbb {N}$. $\square$

Theorem 2.3

Suppose that $(T_n)_{n \ge 0}$ is a sequence of nonzero operators with

$$\begin{aligned} \liminf \frac{1}{\Vert T_n\Vert }>1. \end{aligned}$$

(2.8)

Then there exists a constant $L\ge 0$ such that for every $\varepsilon >0$ and every sequence $(x_m)_{m \ge 0}$ in X satisfying

$$\begin{aligned} \Vert x_{n+1}-T_nx_n-a_n\Vert \le \varepsilon ,\ n\in \mathbb {N}, \end{aligned}$$

(2.9)

there exists a sequence $(y_n)_{n\ge 0}$ in X with the properties

$$\begin{aligned} y_{n+1}=T_ny_n+a_n,\ n\in \mathbb {N}, \end{aligned}$$

(2.10)

$$\begin{aligned} \Vert x_n-y_n\Vert \le L\varepsilon , \ n\ge 0. \end{aligned}$$

(2.11)

Proof

Suppose that $(x_n)_{n\ge 0}$ satisfies (2.9) and let $b_n:=x_{n+1}-T_nx_n-a_n, \ n \in \mathbb {N}$. Then $\Vert b_n\Vert \le \varepsilon , \ n \in \mathbb {N}$, and

$$\begin{aligned} x_n=T_{n-1} \ldots T_0 x_{0}+\sum \limits _{k=1}^{n-1} T_{n-1} \ldots T_{k}(a_{k-1}+b_{k-1}) + a_{n-1}+b_{n-1}, \ n\ge 2, \end{aligned}$$

according to Lemma 2.1.

Consider now $(y_n)_{n \ge 0}$ given by (2.10) with $y_0=x_0$. Then

$$\begin{aligned} y_n=T_{n-1} \ldots T_0 y_{0}+\sum \limits _{k=1}^{n-1} T_{n-1} \ldots T_{k}a_{k-1} + a_{n-1}, \ n\ge 2. \end{aligned}$$

Hence

$$\begin{aligned} \Vert x_n-y_n\Vert&=\Vert \sum \limits _{k=1}^{n-1} T_{n-1} \ldots T_{k}b_{k-1} + b_{n-1}\Vert \nonumber \\&\le \varepsilon + \varepsilon \sum \limits _{k=1}^{n-1} \Vert T_{n-1} \ldots T_{k}\Vert \le \varepsilon \left( 1+\sum \limits _{k=1}^{n-1} \Vert T_{n-1}\Vert \cdots \Vert T_{k}\Vert \right) .\nonumber \\ \end{aligned}$$

(2.12)

Taking account of (2.8), we find $q \in \mathbb {R}$ and $n_0 \in \mathbb {N}$ such that

$$\begin{aligned} \frac{1}{\Vert T_n\Vert }\ge q>1, \ n\ge n_0. \end{aligned}$$

Thus for every $n\ge n_0+1$ we have

$$\begin{aligned} 1+\sum \limits _{k=1}^{n-1} \Vert T_{n-1}\Vert \cdots \Vert T_{k}\Vert&=1+\sum \limits _{k=1}^{n_0-1} \Vert T_{n-1}\Vert \cdots \Vert T_{k}\Vert +\sum \limits _{k=n_0}^{n-1} \Vert T_{n-1}\Vert \cdots \Vert T_{k}\Vert \\&\le 1+\sum \limits _{k=1}^{n_0-1} \Vert T_{n-1}\Vert \cdots \Vert T_{k}\Vert + \sum \limits _{k=n_0}^{n-1} \frac{1}{q^{n+k}} \\&< 1+\frac{1}{q-1}+\sum \limits _{k=1}^{n_0-1} \Vert T_{n-1}\Vert \cdots \Vert T_{k}\Vert < \infty . \end{aligned}$$

Taking

$$\begin{aligned} L_1:=1+\frac{1}{q-1}+\sum \limits _{k=1}^{n_0-1} \Vert T_{n-1}\Vert \cdots \Vert T_{k}\Vert \end{aligned}$$

we get

$$\begin{aligned} \Vert x_n-y_n\Vert \le L_1\varepsilon , \ \text {for all} \ n \in \mathbb {N}^*, \ n\ge n_0. \end{aligned}$$

Finally, for

$$\begin{aligned} L:=\max _{n \le n_0} \left\{ L_1, 1+\sum _{k=1}^{n-1} \Vert T_{n-1}\Vert \cdots \Vert T_k\Vert \right\} \end{aligned}$$

we obtain

$$\begin{aligned} \Vert x_n-y_n\Vert \le L\varepsilon , \ n\ge n_0. \end{aligned}$$

$\square$

Remark 2.4

The condition (2.8) in Theorem 2.3 can be replaced by the following: there exists $q \in (0,1)$ such that

$$\begin{aligned} \Vert T_n\Vert \le q, \ \text {for all} \ n \in \mathbb {N}. \end{aligned}$$

Thus, following the lines of the above proof, the Ulam constant can be chosen $L=\frac{1}{1-q}$.

Finally, we present a nonstability result for Eq. (1.1). Taking into account that the stability results presented above hold for $\Vert T_n\Vert <1$ or $\Vert T_n^{-1}\Vert <1,$ $n \ge n_0$, we will consider for nonstability results the case $\Vert T_n\Vert =1$, $n \in \mathbb {N}$.

Theorem 2.5

Suppose that $\Vert T_n\Vert =1$, for all $n \in \mathbb {N}$ and there exists $u_0 \in \overline{B}(0_X,1)$ such that

$$\begin{aligned} \lim _{n \rightarrow \infty } n\Vert T_{n-1} \ldots T_0u_0\Vert =+\infty . \end{aligned}$$

(2.13)

Then for every $\varepsilon >0$ there exists a sequence $(x_n)_{n \ge 0}$ in X satisfying

$$\begin{aligned} \Vert x_{n+1}-T_nx_n-a_n\Vert \le \varepsilon ,\ n\in \mathbb {N}, \end{aligned}$$

such that for every sequence $(y_n)_{n\ge 0}$ given by the recurrence

$$\begin{aligned} y_{n+1}=T_ny_n+a_n,\ n\in \mathbb {N}, \ y_0 \in X, \end{aligned}$$

we have

$$\begin{aligned} \sup _{n \in \mathbb {N}}\Vert x_n-y_n\Vert =+\infty , \end{aligned}$$

i.e., Eq. (1.1) is not Ulam stable.

Proof

Let $\varepsilon >0$ and consider the sequence $(x_n)_{n\ge 0}$ defined by the relation

$$\begin{aligned} x_{n+1}=T_nx_n+a_n+\varepsilon T_n\ldots T_0 u_0, \ n \in \mathbb {N}. \end{aligned}$$

Then, according to Lemma 2.1, we get

$$\begin{aligned} x_n&=T_{n-1} \ldots T_0x_0 + \sum \limits _{k=1}^{n-1} T_{n-1} \ldots T_{k}(a_{k-1}+\varepsilon T_{k-1} \ldots T_0u_0) \\&\quad + a_{n-1} + \varepsilon T_{n-1} \ldots T_0u_0, \ n\ge 2. \end{aligned}$$

On the other hand,

$$\begin{aligned} \Vert x_{n+1}-T_nx_n-a_n\Vert =\varepsilon \Vert T_n\ldots T_0u_0\Vert \le \varepsilon , \ n \in \mathbb {N}, \end{aligned}$$

hence $(x_n)_{n\ge 0}$ is an approximate solution of Eq. (1.1). Let $(y_n)_{n \ge 0}$ be an arbitrary sequence in X, $y_{n+1}=T_ny_n+a_n,\ n\in \mathbb {N}, \ y_0 \in X$. Then

$$\begin{aligned} y_n=T_{n-1}\ldots T_0y_0 + \sum \limits _{k=1}^{n-1} T_{n-1} \ldots T_{k}a_{k-1} + a_{n-1}, \ n\ge 1, \end{aligned}$$

therefore

$$\begin{aligned} x_n-y_n&=T_{n-1}\ldots T_0(x_0-y_0) + \varepsilon \sum \limits _{k=1}^{n} T_{n-1} \ldots T_0 u_0, \\&=T_{n-1}\ldots T_0(x_0-y_0) + n\varepsilon T_{n-1} \ldots T_0 u_0, \ n \ge 1. \end{aligned}$$

It follows

$$\begin{aligned} \Vert x_n-y_n\Vert&= \Vert T_{n-1}\ldots T_0(x_0-y_0) + \varepsilon nT_{n-1} \ldots T_0 u_0\Vert \\&\ge \left| \Vert T_{n-1}\ldots T_0(x_0-y_0)\Vert - \varepsilon n\Vert T_{n-1} \ldots T_0 u_0\Vert \right| , \ n \ge 1. \end{aligned}$$

The sequence $(T_{n-1}\ldots T_0(x_0-y_0))_{n\ge 1}$ is bounded, since

$$\begin{aligned} \Vert T_{n-1}\ldots T_0(x_0-y_0)\Vert \le \Vert T_{n-1}\Vert \cdots \Vert T_0\Vert \Vert x_0-y_0\Vert =\Vert x_0-y_0\Vert , \ n \ge 1, \end{aligned}$$

therefore $\lim _{m \rightarrow \infty } n\Vert \mathrm{x_{m}-y_{m}}\Vert =+\infty .$ $\square$

Similar results can be obtained when we replace $(x_n)_{n\ge 0}$, $(T_n)_{n\ge 0}$ and $(a_n)_{n\ge 0}$ by $X_n:=(x_1(n),x_2(n),\ldots ,x_p(n))^T \in X^p$, $A_n \in \mathbb {K}^{p \times p}$ and $B_n \in X^p$, respectively. In this way we get Ulam stability results for systems of linear difference equations (see [30]). Note that on $X^p$, the following norm $\Vert Y\Vert _{\infty }:=\max _{1\le i \le p} \Vert y_i\Vert$ ($Y=(y_1,y_2,\ldots ,y_p)^T$) is considered, alongside the matrix norm $\Vert A\Vert _{\infty }=\max _{1\le i \le p} \sum _{j=1}^{p} |a_{ij}|$ of $A \in \mathbb {K}^{p \times p}$ (which is simply the maximum absolute row sum of the matrix), induced by the vector norm $\Vert \cdot \Vert _{\infty }$ on $\mathbb {K}^p$. Also, one can easily verify that $\Vert AY\Vert _{\infty }\le \Vert A\Vert _{\infty } \Vert Y\Vert _{\infty }$ and $\Vert AB\Vert _{\infty }\le \Vert A\Vert _{\infty } \Vert B\Vert _{\infty }$, for any $A, B \in \mathbb {K}^{p \times p}$ and $Y \in X^p$. Finally, it is worth mentioning here that one can replace the above norms with some submultiplicative ones in order to obtain similar stability results for the equation

$$\begin{aligned} X_{n+1}=A_nX_n+B_n,\ n\ge 0. \end{aligned}$$

(2.14)

Corollary 2.6

Suppose that $(A_n)_{n\ge 0}$ is a sequence of invertible matrices in $\mathbb {K}^{p \times p}$ with

$$\begin{aligned} \limsup \Vert A_n^{-1}\Vert _{\infty }<1. \end{aligned}$$

(2.15)

Then for every $\varepsilon >0$ and every sequence $(X_n)_{n\ge 0}$ in $X^p$ with

$$\begin{aligned} \Vert X_{n+1}-A_nX_n-B_n\Vert _{\infty }\le \varepsilon ,\ n\in \mathbb {N}, \end{aligned}$$

(2.16)

there exists a unique sequence $(Y_n)_{n\ge 0}$ in $X^p$ such that

$$\begin{aligned} Y_{n+1}=A_nY_n+B_n,\ n\in \mathbb {N} \end{aligned}$$

(2.17)

and

$$\begin{aligned} \Vert X_n-Y_n\Vert _{\infty }\le \sup _{n \in \mathbb {N}} \sum \limits _{k=0}^{\infty } \Vert A_n^{-1}\Vert _{\infty } \cdots \Vert A_{n+k}^{-1}\Vert _{\infty } \varepsilon , \ n\in \mathbb {N}. \end{aligned}$$

(2.18)

Corollary 2.7

Suppose that $(A_n)_{n \ge 0}$ is a sequence of nonzero matrices in $\mathbb {K}^{p \times p}$ with

$$\begin{aligned} \liminf \frac{1}{\Vert A_n\Vert _{\infty }}>1. \end{aligned}$$

(2.19)

Then, there exists a constant $L\ge 0$ such that for every $\varepsilon >0$ and every sequence $(X_n)_n \ge 0$ in $X^p$ satisfying

$$\begin{aligned} \Vert X_{n+1}-A_nX_n-B_n\Vert _{\infty }\le \varepsilon ,\ n\in \mathbb {N}, \end{aligned}$$

there exists a sequence $(Y_n)_{n\ge 0}$ in $X^p$ with the property

$$\begin{aligned} Y_{n+1}=A_nY_n+B_n,\ n\in \mathbb {N} \end{aligned}$$

(2.20)

such that

$$\begin{aligned} \Vert X_n-Y_n\Vert _{\infty }\le L\varepsilon , \ n\ge 1. \end{aligned}$$

(2.21)

The next nonstability result for Eq. (2.14) is a simple consequence of Theorem 2.5.

Corollary 2.8

Suppose that $(A_n)_{n \ge 0}$ is a sequence of matrices in $\mathbb {K}^{p \times p}$ such that $\Vert A_n\Vert _{\infty }=1$, for all $n \in \mathbb {N}$ and there exists $U_0 \in X^p$, $\Vert U_0\Vert _{\infty }=1$ such that

$$\begin{aligned} \lim _{n \rightarrow \infty } n\Vert A_{n-1} \ldots A_0U_0\Vert _{\infty }=+\infty . \end{aligned}$$

(2.22)

Then for every $\varepsilon >0$ there exists a sequence $(X_n)_n \ge 0$ in $X^p$ satisfying

$$\begin{aligned} \Vert X_{n+1}-A_nX_n-B_n\Vert _{\infty }\le \varepsilon ,\ n\in \mathbb {N}, \end{aligned}$$

such that for every sequence $(Y_n)_{n\ge 0}$ given by the recurrence

$$\begin{aligned} Y_{n+1}=A_nY_n+B_n,\ n\in \mathbb {N}, \ y_0 \in X^p, \end{aligned}$$

we have

$$\begin{aligned} \sup _{n \in \mathbb {N}}\Vert X_n-Y_n\Vert _{\infty }=+\infty , \end{aligned}$$

i.e., Eq. (2.14) is not Ulam stable.

3 The Ulam Stability of a p-Order Linear Difference Equation with Variable Coefficients

In the sequel, we will investigate the Ulam stability of the following p-order linear recurrence with variable coefficients

$$\begin{aligned} x_{n+p}=a_{p-1}(n)x_{n+p-1}+ \cdots + a_0(n)x_n+b_n, \end{aligned}$$

(3.1)

where $(a_k(n))_{n\ge 0}$, $0\le k\le p-1$ are sequences in K and $(b_n)_{n \ge 0}$ is a sequence in X. If the recurrence has constant coefficients, we have a characterization of its Ulam stability. Namely, the equation is Ulam stable if and only if the characteristic equation has no roots on the unit circle (see [8]). Moreover, for $p=1,2,3,$ the best Ulam constant was obtained. But for equations with variable coefficients, there are few results on Ulam stability (see e.g., [23, 30]). Let us remark that Eq. (3.1) can be rewritten as

$$\begin{aligned} X_{n+1}=A_nX_n+B_n, \ n \in \mathbb {N}, \end{aligned}$$

where, for some $k \in (0,\infty )$,

$$\begin{aligned}&A_n=\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} a_{p-1}(n) &{} ka_{p-2}(n) &{} \ldots &{} k^{p-2}a_1(n) &{} k^{p-1}a_0(n) \\ \frac{1}{k} &{} 0 &{} \cdots &{} 0 &{} 0\\ 0 &{} \frac{1}{k} &{} \cdots &{} 0 &{} 0\\ \cdots &{} \cdots &{} \cdots &{} \cdots &{} \cdots \\ 0 &{} 0 &{} \cdots &{} \frac{1}{k} &{} 0\\ \end{array}\right) \in \mathbb {K}^{p \times p}, \nonumber \\&X_n=(k^{p-1}x_{n+p-1}, \ldots , kx_{n+1}, x_n)^T \in X^p \end{aligned}$$

(3.2)

and

$$\begin{aligned} B_n=(k^{p-1}b_n, 0, \ldots , 0)^T \in X^p. \end{aligned}$$

Remark that the usual matriceal form for Eq. (3.1) is obtained for $k=1$ (see [30]). The form considered in this paper for arbitrary $k>0$ is more convenient to obtain good conditions under which the stability of Eq. (3.1) holds.

Suppose that $a_0(n)\ne 0, \ n \in \mathbb {N}$, and let

$$\begin{aligned} e_n=\frac{1}{k^{p-1}|a_0(n)|} + \frac{|a_{p-1}(n)|}{k^{p-2}|a_0(n)|} + \cdots + \frac{|a_1(n)|}{|a_0(n)|} \end{aligned}$$

and

$$\begin{aligned} f_n=|a_{p-1}(n)|+ \cdots + k^{p-1}|a_0(n)|. \end{aligned}$$

Corollary 3.1

Suppose that there exists $k \in (0,1)$ such that

$$\begin{aligned} \limsup e_n<1. \end{aligned}$$

Then for every $\varepsilon >0$ and every sequence $(x_n)_{n \ge 0}$ in X satisfying

$$\begin{aligned} \Vert x_{n+p}-(a_{p-1}(n)x_{n+p-1}+ \cdots + a_0(n)x_n+b_n)\Vert \le \varepsilon , \ n \in \mathbb {N}, \end{aligned}$$

there exists a unique sequence $(y_n)_{n \ge 0}$ in X such that

$$\begin{aligned} y_{n+p}=a_{p-1}(n)y_{n+p-1}+ \cdots + a_0(n)y_n+b_n, \ n \in \mathbb {N} \end{aligned}$$

and

$$\begin{aligned} \Vert x_n-y_n\Vert \le L\varepsilon , \ n\ge 0, \end{aligned}$$

where

$$\begin{aligned} L=\sup _{n \in \mathbb {N}} \sum _{j=0}^{\infty } e_n e_{n+1} \ldots e_{n+j}. \end{aligned}$$

Proof

Consider the induced submultiplicative matrix norm $\Vert \cdot \Vert _{\infty }$, $A_n$, $B_n$ and $X_n$ as in relation (3.2) and observe that, since

$$\begin{aligned} A_n^{-1}=\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} 0 &{} k &{} 0 &{} \cdots &{} 0 \\ 0 &{} 0 &{} k &{} \cdots &{} 0\\ \cdots &{} \cdots &{} \cdots &{} \cdots &{} \cdots \\ 0 &{} 0 &{} 0 &{} \cdots &{} k\\ \frac{1}{k^{p-1}a_0(n)} &{} -\frac{a_{p-1}(n)}{k^{p-2}a_0(n)} &{} -\frac{a_{p-2}(n)}{k^{p-3}a_0(n)} &{} \cdots &{} -\frac{a_1(n)}{a_0(n)}\\ \end{array}\right) , \end{aligned}$$

the property $\limsup \Vert A_n^{-1}\Vert _{\infty }<1$ is equivalent to $\limsup e_n<1$. Further, take an arbitrary $\varepsilon >0$ and consider $\varepsilon _1:=k^{p-1}\varepsilon$. Then

$$\begin{aligned} \Vert X_{n+1}-A_nX_n-B_n\Vert _{\infty }&= \Vert (k^{p-1}(x_{n+p}-a_{p-1}(n)x_{n+p-1}-\cdots -b_n),\ldots , 0)^T\Vert _{\infty }\\&= k^{p-1}\Vert x_{n+p}-(a_{p-1}(n)x_{n+p-1}+ \cdots +a_0(n)x_n+b_n)\Vert \\&\le \varepsilon _1, \end{aligned}$$

i.e.,

$$\begin{aligned} \Vert x_{n+p}-(a_{p-1}(n)x_{n+p-1}+ \cdots +a_0(n)x_n+b_n)\Vert \le \varepsilon . \end{aligned}$$

Then, in view of Corollary 2.6, there exists a unique sequence $Y_n \in X^p$, $Y_{n+1}=A_nY_n+B_n$, such that

$$\begin{aligned} \Vert X_n-Y_n\Vert _{\infty } \le L \varepsilon _1\le L \varepsilon \end{aligned}$$

with $L=\sup _{n \in \mathbb {N}} \sum _{j=0}^{\infty } e_n e_{n+1} \ldots e_{n+j}$. Finally, if we take $y_n:=p_1(Y_n)$, where $p_1:X^p \rightarrow X$ is given by $p_1(z_1,z_2,\ldots ,z_p)=z_p$, one can easily check that $y_n$ is a solution of (3.1) and that $\Vert x_n-y_n\Vert \le L \varepsilon$, $n \in \mathbb {N}$. $\square$

Remark 3.2

In particular, if $e_n=k \in (0,1)$ for every $n \in \mathbb {N}$, then $L=\frac{1}{1-k}$ in Corollary 3.1.

Example 3.3

Consider the recurrence

$$\begin{aligned} x_{n+2}=\frac{3n}{2n+1}x_{n+1}+\frac{7n+9}{n+1}x_n+b_n. \end{aligned}$$

(3.3)

Then for every $\varepsilon >0$ and every sequence $(x_n)_{n \ge 0}$ in X satisfying

$$\begin{aligned} \Vert x_{n+2}-\frac{3n}{2n+1}x_{n+1}-\frac{7n+9}{n+1}x_n-b_n\Vert \le \varepsilon , \ n \in \mathbb {N} \end{aligned}$$

there exists a solution $(y_n)_{n \ge 0}$ of (3.3) such that

$$\begin{aligned} \Vert x_n-y_n\Vert \le 2\varepsilon , \ n\in \mathbb {N}. \end{aligned}$$

Proof

If we choose $k=\frac{1}{2}$, then

$$\begin{aligned} e_n=\max \left\{ \frac{1}{2}, \frac{2(n+1)}{7n+9} + \frac{3n(n+1)}{(2n+1)(7n+9)} \right\} , \end{aligned}$$

which implies that $e_n=\frac{1}{2}$, for every $n \in \mathbb {N}$. Further, taking into account Remark 3.2, we get the desired conclusion, i.e.

$$\begin{aligned} \Vert x_n-y_n\Vert \le 2\varepsilon , \ n \in \mathbb {N}. \end{aligned}$$

$\square$

Corollary 3.4

Suppose that there exists $k>1$ such that

$$\begin{aligned} \liminf \frac{1}{f_n}>1. \end{aligned}$$

Then there exists a positive constant L such that for every $\varepsilon >0$ and every sequence $(x_n)_{n \ge 0}$ in X satisfying

$$\begin{aligned} \Vert x_{n+p}-(a_{p-1}(n)x_{n+p-1}+ \cdots + a_0(n)x_n+b_n)\Vert \le \varepsilon , \ n \in \mathbb {N} \end{aligned}$$

there exists a sequence $(y_n)_{n \ge 0}$ in X satisfying (3.1) such that

$$\begin{aligned} \Vert x_n-y_n\Vert \le L\varepsilon . \end{aligned}$$

Proof

Consider the induced submultiplicative matrix norm $\Vert \cdot \Vert _{\infty }$ and $A_n$, $B_n$ and $X_n$ given by (3.2) with k replaced by $k_1:=\frac{1}{k}$. Observe also that the property $\liminf \frac{1}{f_n}>1$ is equivalent to the condition $\liminf \frac{1}{\Vert A_n\Vert _{\infty }}>1$. Then applying Corollary 2.7 we get the desired conclusion. $\square$

4 Other Applications

Consider first the Volterra operator, which is a bounded linear operator on the space $L^2[0,1]$ of complex-valued square-integrable functions on the interval [0,1]. The Volterra operator V is defined for a function $f \in L^2[0,1]$ by

$$\begin{aligned} V(f)(t)=\int _{0}^{t}f(s)\,\mathrm{{d}}s, \ t \in [0,1]. \end{aligned}$$

It is worth mentioning here that V is a quasinilpotent operator (that is, the spectral radius $\rho (V)$, is zero), but it is not nilpotent and the operator norm of V is exactly $\Vert V\Vert =\frac{2}{\pi }$ (see [16]).

Taking into account Remark 2.4 we get the following stability result for the linear difference equation

$$\begin{aligned} x_{n+1}=Vx_n+a_n, \ x_0 \in L^2[0,1], \ n \in \mathbb {N}, \end{aligned}$$

where $(a_n)_{n \ge 0}$ is a sequence in $L^2[0,1]$.

Corollary 4.1

For every $\varepsilon >0$ and every sequence $(x_n)_n \ge 0$ in $L^2[0,1]$ satisfying

$$\begin{aligned} \Vert x_{n+1}-Vx_n-a_n\Vert \le \varepsilon ,\ n\in \mathbb {N}, \end{aligned}$$

there exists a sequence $(y_n)_{n\ge 0}$ in $L^2[0,1]$ such that

$$\begin{aligned} y_{n+1}=Vy_n+a_n,\ y_0 \in L^2[0,1], \ n\in \mathbb {N} \end{aligned}$$

and

$$\begin{aligned} \Vert x_n-y_n\Vert \le \frac{\pi \varepsilon }{\pi -2}, \ n\ge 0. \end{aligned}$$

(4.1)

Further, given a domain $\Omega$ in $R^m$, we consider a sequence of Hilbert–Schmidt kernels, that is, a sequence of functions $(k_n)_{n \ge 0}$, $k_n: \Omega \times \Omega \rightarrow \mathbb {C}$ with

$$\begin{aligned} \int _{{\Omega }}\int _{{\Omega }}|k_n(x,y)|^{{2}}\,\mathrm{{d}}x\,\mathrm{{d}}y<\infty , \ n \ge 0, \end{aligned}$$

which means that the $L^2(\Omega \times \Omega ; \mathbb {C})$ norm of each $k_n$ is finite. Further, we associated the following sequence of Hilbert–Schmidt integral operators $(K_n)_{n \ge 0}$, $K_n:L^2(\Omega ; \mathbb {C}) \rightarrow L^2(\Omega ; \mathbb {C})$ defined by

$$\begin{aligned} (K_nu)(x)=\int _{{\Omega }}k_n(x,y)u(y) \,\mathrm{{d}}y, \ u \in L^2(\Omega ; \mathbb {C}). \end{aligned}$$

Then $K_n$ is a Hilbert–Schmidt operator for every $n \in \mathbb {N}$ and its Hilbert–Schmidt norm is

$$\begin{aligned} \Vert K_n\Vert _{HS}=\Vert k_n\Vert _{L^2}. \end{aligned}$$

Hilbert–Schmidt integral operators are continuous (and hence bounded) and compact (see [28]).

Taking into account Theorem 2.3, we get the following stability result for the linear difference equation

$$\begin{aligned} x_{n+1}=K_nx_n+a_n, \ x_0 \in L^2(\Omega ; \mathbb {C}), \ n \in \mathbb {N}, \end{aligned}$$

where $(a_n)_{n \ge 0}$ is a sequence in $L^2(\Omega ; \mathbb {C})$.

Corollary 4.2

Suppose that $(K_n)_{n \ge 0}$ is a sequence of nonzero Hilbert–Schmidt integral operators and

$$\begin{aligned} \liminf \frac{1}{\Vert k_n\Vert _{L^2}}>1. \end{aligned}$$

Then, there exists a constant $L\ge 0$, such that for every $\varepsilon >0$ and every sequence $(x_n)_n \ge 0$ in $L^2(\Omega ; \mathbb {C})$ satisfying

$$\begin{aligned} \Vert x_{n+1}-K_nx_n-a_n\Vert \le \varepsilon ,\ n\in \mathbb {N}, \end{aligned}$$

there exists a sequence $(y_n)_{n\ge 0}$ in $L^2(\Omega ; \mathbb {C})$ with the property

$$\begin{aligned} y_{n+1}=K_ny_n+a_n, \ y_0 \in L^2(\Omega ; \mathbb {C}), \ n\in \mathbb {N} \end{aligned}$$

and

$$\begin{aligned} \Vert x_n-y_n\Vert \le L\varepsilon , \ n\ge 0. \end{aligned}$$

(4.2)

Remark 4.3

An example of nonstable equation is given below. Take now the Bernstein operator (see [1]) which assigns to each continuous, real-valued function $f \in C[0,1]$ (where C[0, 1] is endowed with the supremum norm) the polynomial function $B_n f$ defined by

$$\begin{aligned} B_n f(t)=\sum _{k=0}^{n} \left( {\begin{array}{c}n\\ k\end{array}}\right) t^k(1-t)^{n-k} f\left( \frac{k}{n}\right) , \ n \in \mathbb {N}. \end{aligned}$$

It is well known that $B_n$ preserves affine functions.

Further, we consider the sequence $(x_n)_{n \ge 0}$ defined by the recurrence

$$\begin{aligned} x_{n+1}=B_nx_n+a_n, \ n \in \mathbb {N}, \end{aligned}$$

(4.3)

and we show that Eq. (4.3) is not Ulam stable. Indeed, choosing $u_0(t)=1, \ t \in [0,1]$, one can easily observe that $\Vert u_0\Vert =1$ and $\Vert B_{n-1}\ldots B_0 u_0\Vert =1, \ \forall n \ge 1$. Using now Theorem 2.5, one gets the desired conclusion.

Another example concerns the Ulam stability of Eq. (1.1) for operators $T_n$, $n\ge 0$, acting on a finite dimensional Banach space X.

Suppose in what follows that a vector norm $\Vert \cdot \Vert$ on $X=\mathbb {K}^p$ is given. Then any square matrix A of order p with entries in $\mathbb {K}$ induces a linear operator $T:\mathbb {K}^p \rightarrow \mathbb {K}^p$, $Tx=Ax$, with respect to the standard basis, and one defines the corresponding induced norm on the space $\mathbb {K}^{p \times p}$ of all $p \times p$ matrices as follows:

$$\begin{aligned} \Vert A\Vert&=\sup \{\Vert Ax\Vert :x\in \mathbb {K}^p \ \text {with} \ \Vert x\Vert =1\}\\&=\left\{ \frac{\Vert Ax\Vert }{\Vert x\Vert }:x\in \mathbb {K}^p \ \text {with} \ x \not =0 \right\} . \end{aligned}$$

If we consider $\mathbb {K}^p$ to be endowed with the Euclidean norm (i.e. $\Vert x\Vert =\sqrt{|x_1|^2+\cdots +|x_n|^2}$, where $x=(x_1,\ldots ,x_n) \in \mathbb {K}^p$), then the induced matrix norm is the spectral norm. Let us recall also here that the spectral norm of a matrix A is the largest singular value of A, i.e., the square root of the largest eigenvalue of the matrix $A^*A$, where $A^*$ denotes the conjugate transpose of A.

Take now a sequence $(A_n)_{n \ge 0}$ of $p \times p$ matrices, $T_n:\mathbb {K}^p \rightarrow \mathbb {K}^p$, $T_nx=A_nx$ and denote by $\lambda _1^{(n)}, \ldots , \lambda _p^{(n)}$ and $\Lambda _1^{(n)}, \ldots , \Lambda _p^{(n)}$ the eigenvalues of $A_n$ and $A_n^*A_n$, respectively. If we suppose that

$$\begin{aligned} |\lambda _1^{(n)}|\le |\lambda _2^{(n)}| \le \cdots \le |\lambda _p^{(n)}| \ \text {and} \ \Lambda _1^{(n)}\le \Lambda _2^{(n)} \le \cdots \le \Lambda _p^{(n)}, \end{aligned}$$

then (see [19]) $\Vert T_n\Vert =\sqrt{\Lambda _p^{(n)}}$ and, if $A_n$ are additionally invertible matrices, then $\Vert T_n^{-1}\Vert =\frac{1}{\sqrt{\Lambda _1^{(n)}}}$. Moreover, if $A_n$ are normal and invertible matrices, then $\Vert T_m\Vert =|\lambda _p^{(m)}|$ and $\Vert T_m^{-1}\Vert =\frac{1}{|\lambda _1^{(m)}|}$.

The following results on Ulam stability follow easily, if we take also into account Theorems 2.2 and 2.3.

Theorem 4.4

Let $(a_n)_{n\ge 0}$ be a sequence in X and suppose that $A_n$ are invertible matrices and that there exist $q>1$ and $n_0 \in \mathbb {N}$ such that

$$\begin{aligned} \Lambda _1^{(n)}\ge q>1, \ n\ge n_0. \end{aligned}$$

(4.4)

Then for every $\varepsilon >0$ and every sequence $(x_n)_{n\ge 0}$ in $\mathbb {K}^p$ with

$$\begin{aligned} \Vert x_{n+1}-A_nx_n-a_n\Vert \le \varepsilon ,\ n\in \mathbb {N}, \end{aligned}$$

there exists a unique sequence $(y_n)_{n\ge 0}$ in $\mathbb {K}^p$ such that

$$\begin{aligned} y_{n+1}=A_ny_n+a_n,\ n\in \mathbb {N} \end{aligned}$$

and

$$\begin{aligned} \Vert x_n-y_n\Vert \le \sup _{n \in \mathbb {N}} \sum \limits _{k=0}^{\infty } \frac{1}{\sqrt{\Lambda _1^{(n)} \ldots \Lambda _1^{(n+k)}}} \varepsilon , \ n\in \mathbb {N}. \end{aligned}$$

Theorem 4.5

Suppose that $A_n$ are nonzero matrices and that there exist $q<1$ and $n_0 \in \mathbb {N}$ such that

$$\begin{aligned} \Lambda _p^{(n)}\le q<1, \ n\ge n_0. \end{aligned}$$

(4.5)

Then, there exists $L\ge 0$ such that for every $\varepsilon >0$ and every sequence $(x_n)_n \ge 0$ in $\mathbb {K}^p$ satisfying

$$\begin{aligned} \Vert x_{n+1}-A_nx_n-a_n\Vert \le \varepsilon ,\ n\in \mathbb {N}, \end{aligned}$$

there exists a sequence $(y_n)_{n\ge 0}$ in $\mathbb {K}^p$ with the property

$$\begin{aligned} y_{n+1}=A_ny_n+a_n,\ n\in \mathbb {N} \end{aligned}$$

and

$$\begin{aligned} \Vert x_n-y_n\Vert \le L\varepsilon , \ n\ge 1. \end{aligned}$$

Remark 4.6

In case $A_n$ are normal and invertible matrices, the above results hold with $|\lambda _1^{(n)}|$ and $|\lambda _p^{(n)}|$ instead of $\sqrt{\Lambda _1^{(n)}}$ and $\sqrt{\Lambda _p^{(n)}}$, respectively.

Remark 4.7

If we consider now $\mathbb {K}^p$ to be endowed with the Taxicab norm (i.e. $\Vert x\Vert _1=\sum _{i=1}^{p}|x_i|$, where $x=(x_1,\ldots ,x_p) \in \mathbb {K}^p$) or with the maximum norm (i.e. $\Vert x\Vert _{\infty }=\max \limits _{1\le i \le p} |x_i|$) then the induced matrix norms are $\Vert A\Vert _1=\max _{1\le j \le p} \sum _{i=1}^{p}|a_{ij}|$ and $\Vert A\Vert _{\infty }=\max _{1\le i \le p} \sum _{j=1}^{p}|a_{ij}|$, respectively. Further, take $A_n=(a_{ij}^{(n)})\in \mathbb {K}^{p \times p}$, for all $n \in \mathbb {N}$. Then the condition (4.5) above can be replaced by

$$\begin{aligned} \max _{1\le j \le p} \sum _{i=1}^{p}|a_{ij}^{(n)}| \le q<1, \ \text {for all} \ n\ge n_0, \end{aligned}$$

or by

$$\begin{aligned} \max _{1\le i \le p} \sum _{j=1}^{p}|a_{ij}^{(n)}| \le q<1, \ \text {for all} \ n\ge n_0, \end{aligned}$$

in order to obtain the desired conclusion, since the following inequalities

$$\begin{aligned} \frac{1}{\sqrt{p}}\Vert A\Vert _1 \le \Vert A\Vert _2 \le \sqrt{p}\Vert A\Vert _1 \end{aligned}$$

and

$$\begin{aligned} \frac{1}{\sqrt{p}}\Vert A\Vert _{\infty } \le \Vert A\Vert _2 \le \sqrt{p}\Vert A\Vert _{\infty } \end{aligned}$$

hold always true for any square matrix A of order p. It is worth mentioning also that neither the following implications

$$\begin{aligned} \liminf \frac{1}{\Vert A\Vert _1}>1&\Longrightarrow \liminf \frac{1}{\Vert A\Vert _2}>1 \\ \text {and} \\ \liminf \frac{1}{\Vert A\Vert _{\infty }}>1&\Longrightarrow \liminf \frac{1}{\Vert A\Vert _2}>1 \end{aligned}$$

nor the opposite ones seam to be valid.

The results obtained in this section generalize the results obtained in [5, 12].