1 Introduction

Let \(A\) be a bounded linear operator acting on a complex Banach space \(X\). A well known theorem of Krein [8, 10] says that the system \(\dot{x}(t)=Ax(t)\) is uniformly exponentially stable if and only if for each \(\mu \in \mathbb {R}\) and each \(y_0\in X\) the solution of the Cauchy problem

$$\begin{aligned} \left\{ \begin{aligned} \dot{y}(t)&= Ay(t)+e^{i\mu t}y_0, \\ y(0)&= 0 \end{aligned} \right. \end{aligned}$$

is bounded. The proof of this classic result can be found in [1]. This result can also be extended for strongly continuous bounded semigroups, see [3–5, 11].

Under a slightly different assumption the result on stability is also preserved for any strongly continuous semigroups acting on complex Hilbert spaces, see for example [12, 13] and references therein. See also [2, 9], for counter-examples. In [7, 14] the same result were extended for square size matrices in both continuous and discrete cases.

In [6], similar results were obtained in the following manner. Let \(P_1^0(\mathbb {R}_+, X)\) denotes the set of all continuous X-valued functions such that \(f(t + 1) = f(t)\) for all \(t\ge 0\) with \(f(0) = 0\) and let \(T = \{T(t)\}_{t\ge 0}\) be a strongly continuous semigroup on the Banach space X. If the condition

$$\begin{aligned} \sup \limits _{ t>0}\big \Vert \int _0^te^{i\mu \xi }T(t-\xi )f(\xi )d\xi \big \Vert < \infty , \end{aligned}$$

holds for all \(\mu \in \mathbb {R}\) and every \(f\in P_1^0(\mathbb {R}_+, X)\) then \(T\) is uniformly exponentially stable.

In this article we follow a similar approach of the last quoted paper and study the system \(x_{n+1}=T(1)x_n\), where \(T(1)\) is the algebraic generator of the discrete semi group \(\mathbb {T}=\{T(n):n\in \mathbb {Z_+}\}\).

2 Notations and Preliminaries

Let \(\mathcal {L}(X)\) be the Banach algebra of all linear and bounded operators acting on a Banach space \(X\). The norm on \(X\) and \(\mathcal {L}(X)\) is denoted by \(\Vert \Vert \). By \(\mathbb {R}\) we denote the set of all real numbers and \(\mathbb {Z_+}\) the set of all non-negative integers.

Let \(\mathcal {S}(\mathbb {Z_+}, X)\) be the space of all \(X\)-valued bounded sequences endowed with the supremum norm denoted by \(\Vert .\Vert _{\infty }\), and \(P_{0}^{q}(\mathbb {Z_+},X)\) be the space of \(q\)-periodic bounded sequences \(z(n)\) with \(z(0)=0\) and let \(q\ge 2\) be a given integer. Then clearly \(P_{0}^{q}(\mathbb {Z_+},X)\) is a closed subspace of \(\mathcal {S}(\mathbb {Z_+}, X).\)

Let \(A\) is a bounded linear operator on \(X\) and \(\sigma (A)\) is its spectrum. By \(r(A)\) we denote the spectral radius of \(A\), and is defined as

$$\begin{aligned} r(A):=\sup \{|\lambda |: \lambda \in \sigma (A)\}. \end{aligned}$$

It is well known that \(r(A):=\lim _{n \rightarrow \infty }\Vert A^{n}\Vert ^{\frac{1}{n}}\). The resolvent set of \(A\) is defined as \(\rho (A):=\mathbb {C}\backslash \sigma (A)\), i.e the set of all \(\lambda \in \mathbb {C}\) for which \(A-\lambda I\) is an invertible operator in \(\mathcal {L}(X)\).

We recall that \(A\) is power bounded if there exists a positive constant \(M\) such that \(\Vert A^{n}\Vert \le M\) for all \(n\in \mathbb {Z_+}\).

3 Exponential Stability of Discrete Semigroups

We recall that a discrete semigroup is a family \(\mathbb {T}=\{T(n):n\in \mathbb {Z_+}\}\) of bounded linear operators acting on \(X\) which satisfies the following conditions

  1. (1)

    \(T(0)=I\), the identity operator on \(X\),

  2. (2)

    \(T(n+m)=T(n)T(m)\) for all \(n,m \in \mathbb {Z_+}\).

Let \(T(1)\) denote the algebraic generator of the semigroup \(\mathbb {T}\). Then it is clear that \(T(n)=T^{n}(1)\) for all \(n\in \mathbb {Z_+}\). The growth bound of \(\mathbb {T}\) is denoted by \(\omega _{0}(\mathbb {T})\) and is defined as

$$\begin{aligned} \omega _{0}(\mathbb {T})&:= \inf \{\omega \in \mathbb {R}: \hbox { there exists} \ M_{\omega }> 0\\&\quad \hbox {such that} \ \Vert T(n)\Vert \le M_{\omega }e^{\omega n}\,\hbox {for all}\,n\in \mathbb {Z_+}\}. \end{aligned}$$

The family \(\mathbb {T}\) is uniformly exponentially stable if \(\omega _{0}(\mathbb {T})\) is negative, or equivalently, if there exist two positive constants \(M\) and \(\omega \) such that \(\Vert T(n)\Vert \le M e^{-\omega n}\) for all \(n \in \mathbb {Z_+}\).

We recall the following lemma without proof from [6], so that the paper will be self contained

Lemma 3.1

Let \(A\in \mathcal {L}(X)\). If the inequality

$$\begin{aligned} \sup _{n\in \mathbb {Z_+}}\big |\big |\sum _{k=0}^{n}e^{i\mu k}A^{k}\big |\big |=M_{\mu }<\infty ,\ \hbox {for all} \ \mu \in \mathbb {R}, \end{aligned}$$
(3.1)

then \(r(A)< 1\).

We denote by \(\mathcal {A}\) the set of all X-valued, q-periodic sequences \(z\) such that

$$\begin{aligned} z(n)=n(n-q)T(n)x, \ \ n\in \{0,1,\dots ,q-1\}, \ \ x\in X. \end{aligned}$$

Now we state and prove our main result.

Theorem 3.2

Let \(T(1)\) is the algebraic generator of the discrete semigroup \(\mathbb {T}=\{T(n):n\in \mathbb {Z_+}\}\) on \(X\) and \(\mu \in \mathbb {R}\). Then the following holds true.

  1. (1)

    If the system \(x_{n+1}=T(1)x_n\) is uniformly exponentially stable then for each real number \(\mu \) and each \(q\)-periodic sequence \((z(n))\) with \(z(0)=0\) the solution of the Cauchy Problem

    $$\begin{aligned} \left\{ \begin{aligned} y_{n+1}&= T(1)y_{n}+e^{i\mu (n+1)}z(n+1), \\ y(0)&= 0 \end{aligned} \right. \quad {(T(1), \mu ,0)} \end{aligned}$$

    is bounded.

  2. (2)

    If for each real number \(\mu \) and each \(q\)-periodic sequence \((z(n))\) in \(\mathcal {A}\) the solution of the Cauchy Problem \((T(1), \mu ,0)\) is bounded, with the assumption that the operator \(e^{i\mu n}\sum \nolimits _{\nu =1}^{q-1}e^{i\mu \nu }\nu (q-\nu )x\) is non zero for each non zero \(x\) in \(X\). Then \(\mathbb {T}\) is uniformly exponentially stable.

Proof

(1) Here we show that if \(\mathbb {T}\) is uniformly exponentially stable then the solution (\(y_n)\) of \((T(1), \mu ,0)\) is bounded.

Let \(M\) and \(\nu \) be positive constants such that \(\Vert T(n)\Vert \le Me^{-\nu n}\), for all \(n\in \mathbb {Z_+}\). The solution of the Cauchy Problem \((T(1), \mu ,0)\) is given by

$$\begin{aligned} y_n=\sum _{k=0}^ne^{iuk}T(n-k)z(k). \end{aligned}$$
(3.2)

Taking norm of both sides

$$\begin{aligned} \Vert y_n\Vert&= \Vert \sum _{k=0}^ne^{iuk}T(n-k)z(k)\Vert \\&\le \sum _{k=0}^n\Vert e^{iuk}T(n-k)z(k)\Vert \\&= \sum _{k=0}^n\Vert e^{iuk}\Vert \Vert T(n-k)\Vert \Vert z(k)\Vert \\&\le \sum _{k=0}^n\Vert T(n-k)\Vert \Vert z(k)\Vert \\&\le \sum _{k=0}^nMe^{-\nu (n-k)}M', \ \hbox {where} \ M'=\sup \limits _{k\in \mathbb {Z}_+}\Vert z(k)\Vert \\&= M''e^{-\nu n}\sum _{k=0}^ne^{\nu k},\ \hbox {where} \ M''=MM'\\&= M''e^{-\nu n}\left( \frac{1-e^{(n+1)\nu }}{1-e^\nu }\right) , \hbox {where} \ \nu >0\\&\le M''e^{-\nu n}\\&\le \frac{M''}{e^{\nu }}. \end{aligned}$$

Thus the solution of the Cauchy Problem \((T(1), \mu , 0)\) is bounded.

(2) The proof of the second part is not so easy. Let us divide \(n\) by \(q\) i.e. \(n=Nq+r\) for some \(N\in \mathbb {Z_+}\), where \(r \in \{0,1, \dots , q-1\}\).

For each \(j\in \mathbb {Z_+}\), we consider the set \(A_{j}:=\{1+jq,2+jq,\dots , q-1+jq\}\), also let \(B_{N}:=\{Nq+1,\dots , Nq+r\}\) if \(r\ge 1\) and \(B:=\{0,q,2q,\dots , Nq\}\) then clearly

$$\begin{aligned} \cup _{j=0}^{N-1}A_{j}\cup B_{N}\cup B=\{0,1,2,\dots ,n\}. \end{aligned}$$

From (3.2) we know that the solution of the Cauchy Problem \((T(1), \mu , 0)\) is

$$\begin{aligned} y_n=\sum _{k=0}^{n}e^{i\mu k}T(n-k)z(k). \end{aligned}$$

The sequence \(z(.)\) belongs to the set \(\mathcal {A}\) if and only if there exists \(x\in X\) such that

$$\begin{aligned} z(k)=\left\{ \begin{array}{l@{\quad }l} (k-jq)[(1+j)q-k]T(k-jq)x, &{} \hbox { if} \quad k \in A_{j}, \\ k(q-k)T(k)x, &{} \hbox { if }\quad k \in B_N,\\ 0, &{} \hbox { if }\quad k \in C. \end{array} \right. \end{aligned}$$

Then clearly \(\big (z(k)\big )\in \mathcal {A}\). Thus

$$\begin{aligned} y_n&= \sum _{k=0}^{n}e^{i\mu k}T(n-k)z(k)\\&= \sum _{k\in \cup _{j=0}^{N-1}A_{j}}e^{i\mu k}T(Nq+r-k)z(k)\\&\quad + \sum _{k\in B_{N}}e^{i\mu k}T(Nq+r-k)z(k)\\&\quad + \sum _{k\in C}e^{i\mu k}T(Nq+r-k)z(k)\\&= \sum _{j=0}^{N-1}\sum _{k=1+jq}^{(q-1+jq)}e^{i\mu k}T(Nq+r-k)z(k) \\&\quad + \sum _{k=Nq+1}^{Nq+r}e^{i\mu k}T(Nq+r-k)z(k)\\&= J_{1}+J_{2}. \end{aligned}$$

where

$$\begin{aligned} J_{1}&= \sum _{j=0}^{N-1} \sum _{k=1+jq}^{(q-1+jq)}e^{i\mu k}T(Nq+r-k)(k-jq)[q-(k-jq)]T(k-jq)x \\&= \sum _{j=0}^{N-1}T(Nq+r-jq) \sum _{k=1+jq}^{(q-1+jq)}e^{i\mu k}(k-jq)[q-(k-jq)]x \\&= \sum _{j=0}^{N-1}T(Nq+r-jq)e^{i\mu jq} \sum _{\nu =1}^{q-1}e^{i\mu \nu }\nu (q-\nu )x \\&= \sum _{j=0}^{N-1}e^{-i\mu (Nq+r-jq)}T(Nq+r-jq) e^{i\mu (Nq+r)}\sum _{\nu =1}^{q-1}e^{i\mu \nu }\nu (q-\nu )x \\&= \sum _{\omega =r+q}^{r+Nq}e^{-i\mu \omega }T^{\omega }(1) e^{i\mu n}\sum _{\nu =1}^{q-1}e^{i\mu \nu }\nu (q-\nu )x\\&= \sum _{\omega =r+q}^{n}e^{-i\mu \omega }T^{\omega }(1)S(x) \end{aligned}$$

with \(S(x)=e^{i\mu n}\sum _{\nu =1}^{q-1}e^{i\mu \nu }\nu (q-\nu )x\) with the assumption that \(S(x)\ne 0\) for each non zero \(x\) in \(X\).

And

$$\begin{aligned} J_{2}&= \sum _{\rho =0}^{r-1}e^{i\mu (Nq+r-\rho )}T(\rho )z(Nq+r-\rho )x \\&= \sum _{\rho =0}^{r-1}e^{i\mu (Nq+r-\rho )}T(\rho )z(r-\rho )x. \end{aligned}$$

Taking norm of both sides

$$\begin{aligned} \Vert J_{2}\Vert&= \Vert \sum _{\rho =0}^{r-1}e^{i\mu (Nq+r-\rho )}T(\rho )z(r-\rho )x\Vert \\&\le \Vert \sum _{\rho =0}^{r-1}T(\rho )x\Vert \Vert z(r-\cdot )\Vert _{\infty }\\&\le M, \ \hbox {where M is some constant} \end{aligned}$$

i.e. \(J_{2}\) is bounded.

Hence,

$$\begin{aligned} \sum _{k=0}^{n}e^{i\mu k}T(n-k)z(k)=\sum _{\omega =r+q}^{n}e^{-i\mu \omega }T^{\omega }(1)S(x)+\sum _{\rho =0}^{r-1}e^{i\mu (n-\rho )}T(\rho )z(r-\rho )x. \end{aligned}$$

Now by our assumption \((y_n)\) is bounded i.e.

$$\begin{aligned} \sup _{n\ge 0}\Vert y_n\Vert =\sup _{n\ge 0}\Vert \sum _{k=0}^{n}e^{i\mu k}T(n-k)z(k)\Vert < \infty . \end{aligned}$$

Thus

$$\begin{aligned} \sup _{n\ge 0}\Vert \sum _{\omega =r+q}^{n}e^{-i\mu \omega }T^{\omega }(1)S(x)+\sum _{\rho =0}^{r-1}e^{i\mu (n-\rho )}T(\rho )z(r-\rho )x\Vert < \infty . \end{aligned}$$

Which implies that

$$\begin{aligned} \sup _{n\ge 0}\Vert \sum _{\omega =r+q}^{n}e^{-i\mu \omega }T^{\omega }(1)S(x)\Vert <\infty . \end{aligned}$$

i.e.

$$\begin{aligned} \sup _{n\ge 0}\Vert \sum _{\omega =r+q}^{n}e^{-i\mu \omega }T^{\omega }(1)\Vert <\infty . \end{aligned}$$

Thus by Lemma 3.4, \(\mathbb {T}\) is uniformly exponentially stable. \(\square \)

Corollary 3.3

The system \(x_{n+1}=T(1)x_n\) is uniformly exponentially stable if and only if for each real number \(\mu \) and each \(q\)-periodic bounded sequence \(z(n)\) with \(z(0)=0\) the unique solution of the Cauchy Problem \((T(1), \mu , 0)\) is bounded.