Evolution Maps and Admissibility

Abstract

We show that the admissibility properties for a sequence of linear operators and the corresponding evolution maps are equivalent on various Banach spaces. We then use this information to obtain new descriptions for the hyperbolicity of a sequence of linear operators and the corresponding evolution maps.

1 Introduction

Our main objective is to establish the equivalence of the admissibility properties for a sequence of linear operators on a Banach space and the corresponding evolution maps on various Banach spaces. This includes spaces of bounded sequences, of sequence vanishing at infinity, and $\ell ^p$ spaces. We also consider the relation to hyperbolicity and, as an application, we use the correspondence of the admissibility properties to give new descriptions for the hyperbolicity of a sequence of linear operators.

The notion of admissibility goes back to Perron in [9]. A simple modification of his work for continuous time gives the following statement. Let $(A_m)_{m\in {\mathbb {Z}}}$ be a (two-sided) sequence of $n\times n$ matrices. If for each bounded sequence $(y_m)_{m\in {\mathbb {Z}}}$ in ${\mathbb {R}}^n$ there exists $x_0 \in {\mathbb {R}}^n$ such that the sequence

$$\begin{aligned} x_{m+1} =A_{m} x_{m} + y_{m+1} \end{aligned}$$

(1)

is bounded for $m \in {\mathbb {N}}$, then any bounded sequence $A_m\cdots A_1 x$ tends to zero when $m \rightarrow \infty$. Related results for discrete time were first obtained by Li in [6]. For some early contributions we refer the reader to the books [5, 7].

A general notion of admissibility can be introduced as follows. We say that a pair of Banach spaces (C, D) is admissible if for every sequence $(y_m)_{m\in {\mathbb {Z}}}$ in C there exists a unique sequence $(x_m)_{m\in {\mathbb {Z}}}$ in D satisfying (1). We consider this notion for various Banach spaces of sequences with values in a given Banach space X as well as their corresponding evolution maps. Namely, given a Banach space $Y\subset X^{\mathbb {Z}}$, we define a map $S :X^{\mathbb {Z}}\rightarrow X^{\mathbb {Z}}$ by

$$\begin{aligned} (S u)_{n}=A_{n-1} u_{n-1} \quad \text {for }n \in {\mathbb {Z}}\text { and }u=(u_m)_{m \in {\mathbb {Z}}} \in X^{\mathbb {Z}}. \end{aligned}$$

Provided that $S (Y)\subset Y$, the map S is called the evolution map of the sequence $(A_m)_{m\in {\mathbb {Z}}}$ on the Banach space Y.

As noted above, our main objective is to establish a faithful correspondence between the admissibility properties for some pairs of Banach spaces at the levels of sequences of linear operators and evolution maps, also on various Banach spaces. Evolution maps transfer the dynamics at the level of a sequence of linear maps to a dynamics on much larger space, although this new dynamics is autonomous, which often makes the approach much simpler. Furthermore, the properties of the dynamics are also transferred to those of the evolution map, and this often leads to much simpler proofs. In addition, the transference of properties is quite helpful in finding appropriate nonautonomous notions when they are not yet available in the nonautonomous case. An important example of such a correspondence is the study of hyperbolicity and its various variations that goes back to Mather in [8]. The theory of semigroups is nowadays an important tool in the theory of differential equations (see for example [10]).

To illustrate our results, we formulate a particular case of Theorem 3. Let $\ell ^\infty _0(X)$ be the set of all sequences in X vanishing at infinity and define $D_0(X)=\ell ^\infty _0(\ell ^\infty _0(X))$.

Theorem 1

Let $A=(A_m)_{m \in {\mathbb {Z}}}$ be a bounded sequence of linear maps. Then the following properties are equivalent:

• 1. for each $(y_m)_{m\in {\mathbb {Z}}} \in \ell ^\infty _0(X)$ there exists a unique $(x_m)_{m\in {\mathbb {Z}}} \in \ell ^\infty (X)$ satisfying

  $$\begin{aligned} x_{m+1} =A_{m} x_{m} + y_{m+1} \quad \text {for }m \in {\mathbb {Z}}; \end{aligned}$$

• 2. for each $(v_m)_{m\in {\mathbb {Z}}} \in D_0(X)$ there exists a unique $(u_m)_{m\in {\mathbb {Z}}} \in D_0(X)$ satisfying

  $$\begin{aligned} u_{m+1}=S u_m +v_{m+1} \quad \text {for }m \in {\mathbb {Z}}. \end{aligned}$$

Using the notion of admissibility, Theorem 1 can be reformulated as follows: for a bounded sequence A the following properties are equivalent:

• 1. the pair formed by the spaces $\ell ^\infty _0(X)$ and $\ell ^\infty (X)$ is admissible;

• 2. the pair formed by the spaces $D_0(X)$ and $D_0(X)$ is admissible.

Further pairs of spaces are considered in the paper. These include in particular spaces of sequences with bounded exponential growth and $\ell ^p$ spaces.

More generally, we consider families of norms $\Vert \cdot \Vert _m$, for $m\in {\mathbb {Z}}$. These norms play an essential role for example in smooth ergodic theory in the presence of nonuniform exponential behavior. Moreover, we use our results on the equivalence of admissibility properties for sequence of linear operators and evolution maps to give new descriptions for the hyperbolicity of a sequence of linear operators (see Sect. 6).

2 Evolution Maps

Let $X=(X,\Vert \cdot \Vert )$ be a Banach space. Given a sequence $A=(A_m)_{m \in {\mathbb {Z}}}$ of continuous maps on X, we define

$$\begin{aligned} U(m,n)={\left\{ \begin{array}{ll} A_{m-1} \cdots A_n &{} \text {if }m > n, \\ \text {Id}&{} \text {if }m=n \end{array}\right. } \end{aligned}$$

for each $m,n \in {\mathbb {Z}}$ with $m \ge n$. We shall only consider sequences A that are exponentially bounded with respect to a sequence of norms. Namely, let $\Vert \cdot \Vert _m$, for $m \in {\mathbb {Z}}$, be a sequence of norms on X such that

$$\begin{aligned} \Vert x \Vert \le \Vert x \Vert _m \le R_m \Vert x \Vert \quad \text {for }m \in {\mathbb {Z}}\text { and }x \in X \end{aligned}$$

and some sequence $(R_m)_{m\in {\mathbb {Z}}}$ in ${\mathbb {R}}^+$. We say that the sequence A is exponentially bounded with respect to the norms $\Vert \cdot \Vert _m$ if there exist $\alpha , \kappa >0$ such that

$$\begin{aligned} \Vert U(m,n) x\Vert _m \le \kappa e^{\alpha (m-n)} \Vert x_n \Vert _n \quad \text {for }m\ge n\text { and }x \in X. \end{aligned}$$

To each sequence $A=(A_m)_{m \in {\mathbb {Z}}}$ of continuous maps on X, we associate maps $S=S|_Y$ on certain Banach spaces $Y \subset X^{\mathbb {Z}}$. Namely, we define the evolution map of A on a space Y by

$$\begin{aligned} (S u)_n=A_{n-1} u_{n-1} \quad \text {for }n\in {\mathbb {Z}}\text { and }u=(u_m)_{m \in {\mathbb {Z}}} \in Y, \end{aligned}$$

(2)

whenever $S (Y)\subset Y$.

We also introduce a family of maps that will be used in the study of admissibility. Let $A=(A_m)_{m \in {\mathbb {Z}}}$ be a sequence of linear maps on X. Given a sequence $y=(y_n)_{n \in {\mathbb {Z}}}$ in X, we define a map $T_y :X^{\mathbb {Z}}\rightarrow X^{\mathbb {Z}}$ by

$$\begin{aligned} (T_y u)_n=A_{n-1} u_{n-1} + y_n \quad \text {for }n \in {\mathbb {Z}}\text { and }u=(u_m)_{m \in {\mathbb {Z}}} \in X^{\mathbb {Z}}. \end{aligned}$$

3 Admissibility for Bounded Sequences

In this section we consider a certain admissibility property on a space of bounded sequences for evolution maps.

3.1 Evolution Maps

Let $\ell ^\infty (X)$ be the set of all sequences $x=(x_m)_{m \in {\mathbb {Z}}}$ with values in X such that

$$\begin{aligned} \Vert x \Vert _\infty :=\sup _{m \in {\mathbb {Z}}} \Vert x_m \Vert _m < \infty . \end{aligned}$$

We note that $\ell ^\infty (X)$ is a Banach space when equipped with the norm $\Vert \cdot \Vert _\infty$. We also consider the closed subspace $\ell ^\infty _0(X) \subset \ell ^\infty (X)$ of those $x \in \ell ^\infty (X)$ such that

$$\begin{aligned} \lim _{|m |\rightarrow \infty } \Vert x_m \Vert _m =0. \end{aligned}$$

Proposition 1

Let $A=(A_m)_{m \in {\mathbb {Z}}}$ be a sequence of linear maps on X that is exponentially bounded with respect to the norms $\Vert \cdot \Vert _m$. Then the following properties hold:

• 1. for each $y \in \ell ^\infty (X)$ we have

  $$\begin{aligned} T_y(\ell ^\infty (X))\subset \ell ^\infty (X) ; \end{aligned}$$

• 2. for each $y \in \ell ^\infty _0(X)$ we have

  $$\begin{aligned} T_y (\ell ^\infty _0(X)) \subset \ell ^\infty _0(X) . \end{aligned}$$

Proof

Take $y,u \in \ell ^\infty (X)$. Then

$$\begin{aligned} \begin{aligned} \Vert T_y u\Vert _\infty&= \sup _{n \in {\mathbb {Z}}} \left\Vert A_{n-1} u_{n-1} +y_n \right\Vert _n \\&\le \kappa e^{\alpha } \sup _{n \in {\mathbb {Z}}} \Vert u_{n-1} \Vert _{n-1} + \sup _{n \in {\mathbb {Z}}} \Vert y_{n} \Vert _n \\&\le \kappa e^{\alpha } \Vert u\Vert _\infty + \Vert y\Vert _\infty <\infty \end{aligned} \end{aligned}$$

and so $T_y u \in \ell ^\infty (X)$. On the other hand, for $y,u \in \ell ^\infty _0(X)$ we have

$$\begin{aligned} \begin{aligned} \Vert (T_y u)_n \Vert _n&= \left\Vert A_{n-1} u_{n-1} +y_n \right\Vert _n \\&\le \kappa e^{\alpha } \Vert u_{n-1} \Vert _{n-1} +\Vert y_n \Vert _n \rightarrow 0 \end{aligned} \end{aligned}$$

when $|n |\rightarrow \infty$ and so $T_y u \in \ell ^\infty _0(X)$. $\square$

3.2 Admissibility Properties

We continue to consider a sequence of maps $A=(A_m)_{m \in {\mathbb {Z}}}$ that is exponentially bounded with respect to some norms $\Vert \cdot \Vert _m$. Taking $y=0$ in Proposition 1 we find that A generates the evolution map $S=T_0$ on $\ell ^\infty (X)$ given by

$$\begin{aligned} (S u)_n=A_{n-1} u_{n-1} \quad \text {for }n \in {\mathbb {Z}}\text { and }u=(u_m)_{m \in {\mathbb {Z}}} \in \ell ^\infty (X). \end{aligned}$$

Moreover, let D(X) be the set of all sequences $v=(v_m)_{m \in {\mathbb {Z}}}$ with values in $\ell ^\infty (X)$ such that

$$\begin{aligned} \Vert v \Vert _D:=\sup _{m \in {\mathbb {Z}}} \Vert v_m \Vert _\infty <\infty . \end{aligned}$$

We note that D(X) is a Banach space when equipped with the norm $\Vert \cdot \Vert _D$. We also consider the closed subspace $D_0(X) \subset D(X)$ of all sequences v with values in $\ell ^\infty _0(X)$ such that

$$\begin{aligned} \lim _{|m |\rightarrow \infty } \Vert v_m \Vert _\infty =0. \end{aligned}$$

The following theorem is our main result.

Theorem 2

Let $A=(A_m)_{m \in {\mathbb {Z}}}$ be a sequence of linear maps on X that is exponentially bounded with respect to the norms $\Vert \cdot \Vert _m$. Then the following properties are equivalent:

• 1. for each $y \in \ell ^\infty _0(X)$ there exists a unique $x \in \ell ^\infty (X)$ such that

  $$\begin{aligned} x_{m+1} =A_m x_m + y_{m+1} \quad \text {for }m \in {\mathbb {Z}}; \end{aligned}$$

  (3)

• 2. for each $v \in D_0(X)$ there exists a unique $u \in D(X)$ such that

  $$\begin{aligned} u_{m+1}=S u_m +v_{m+1} \quad \text {for }m \in {\mathbb {Z}}. \end{aligned}$$

  (4)

Proof

We first prove an auxiliary result. For each $u =(u_m)_{m \in {\mathbb {Z}}}\in D(X)$, we shall denote by $u_{m,k}$ the kth term of the sequence $u_m\in \ell ^\infty (X)$.

Lemma 1

Given $u,v \in D(X)$, Property (4) holds if and only if

$$\begin{aligned} u_{m+1-k,m+1}=A_m u_{m-k,m} +y_{k,m+1} \quad \text {for} \ m,k \in {\mathbb {Z}}, \end{aligned}$$

(5)

where $y_{k,n}=v_{n-k,n}$.

Proof of the lemma

First assume that property (4) holds. By the definition of S we have

$$\begin{aligned} (S u_p)_{m+1}=A_{m} u_{p,m} \end{aligned}$$

(where $u_{p,m}$ is the mth term of the sequence $u_p$) and so

$$\begin{aligned} u_{p+1,m+1} =(S u_p)_{m+1} +v_{p+1,m+1} = A_m u_{p,m} + v_{p+1,m+1} \end{aligned}$$

for each $m,p \in {\mathbb {Z}}$. Taking $p=m-k$, we obtain

$$\begin{aligned} u_{m+1-k,m+1} = A_m u_{m-k,m}+ v_{m+1-k,m+1} = A_m u_{m-k,m} + y_{k,m+1}. \end{aligned}$$

(6)

Now assume that property (5) holds. Proceeding as in (6) and again by the definition of S, we have

$$\begin{aligned} \begin{aligned} u_{m+1-k,m+1}&= A_m u_{m-k,m}+y_{k,m+1} \\&= A_m u_{m-k,m} + v_{m+1-k,m+1}\\&= (S u_{m-k})_{m+1} + (v_{m+1-k})_{m+1}. \end{aligned} \end{aligned}$$

Since m and k are arbitrary, this yields property (4). Indeed, replacing $m-k$ by p gives

$$\begin{aligned} \begin{aligned} (u_{p+1})_{m+1}&=u_{p+1,m+1}\\&= (S u_p)_{m+1} + (v_{p+1})_{m+1} \\&= (S u_p+v_{p+1})_{m+1}. \end{aligned} \end{aligned}$$

Finally, since m is arbitrary, we obtain

$$\begin{aligned} u_{p+1}= S u_p+v_{p+1} \end{aligned}$$

and property (4) follows from the arbitrariness of p (since k is arbitrary, for a given m one can choose k such that $p=m-k$ takes any desired value). $\square$

We proceed with the proof of the theorem.

$(1 \Rightarrow 2)$. Take $v \in D_0(X)$. For each $k \in {\mathbb {Z}}$, we define $y^{(k)} \in X^{\mathbb {Z}}$ by

$$\begin{aligned} y^{(k)}_m=v_{m-k,m} \quad \text {for} \ m \in {\mathbb {Z}}\end{aligned}$$

(7)

(in a similar manner to that in Lemma 1). Since $v \in D_0(X)$, we have

$$\begin{aligned} \lim _{|m |\rightarrow \infty } \Vert y^{(k)}_m \Vert _m \le \lim _{|m |\rightarrow \infty } \Vert v_{m-k}\Vert _\infty =0 \end{aligned}$$

and so $y^{(k)} \in \ell ^\infty _0(X)$. By property 1, there exists a unique solution $x^{(k)} \in \ell ^\infty (X)$ of Eq.  (3) with $y=y^{(k)}$, for each $k \in {\mathbb {Z}}$.

By Lemma 1, if $u\in D(X)$ is a solution of Eq. (4), then $(u_{m-k,m})_{m \in {\mathbb {Z}}}$ is a solution of Eq. (3) with $y=y^{(k)}$, for each $k \in {\mathbb {Z}}$, that is,

$$\begin{aligned} u_{m+1-k,m+1} =A_m u_{m-k,m} + v_{m+1-k,m+1} \quad \text {for }m \in {\mathbb {Z}}. \end{aligned}$$

Therefore, necessarily $u_{m-k,m}=x^{(k)}_m$ for $m,k \in {\mathbb {Z}}$, which is equivalent to

$$\begin{aligned} u_{m,n}=x^{(n-m)}_n \quad \text {for }m,n \in {\mathbb {Z}}. \end{aligned}$$

(8)

This shows that any solution of Eq. (4) is given by (8) and so, in particular, it is unique. We show below that the sequence u defined by (8) belongs to D(X).

Let E be the Banach space of all sequences $x \in \ell ^\infty (X)$ for which there exists $y \in \ell ^\infty _0(X)$ satisfying (3) and define a linear operator $R:E\rightarrow \ell ^\infty _0 (X)$ by

$$\begin{aligned} (Rx)_{m+1} =x_{m+1} -A_m x_m \quad \text {for }m \in {\mathbb {Z}}. \end{aligned}$$

(9)

We show that R is closed. Let $({\bar{x}}^{(i)})_{i \in {\mathbb {N}}}$ be a sequence in E converging to $x \in \ell ^\infty (X)$ such that ${\bar{y}}^{(i)}=R{\bar{x}}^{(i)}$ converges to $y \in \ell ^\infty _0(X)$. Then

$$\begin{aligned} \begin{aligned} x_{m+1}-A_m x_m&= \lim _{i \rightarrow \infty } \bigl ( {\bar{x}}^{(i)}_{m+1} -A_m {\bar{x}}^{(i)}_{m} \bigr ) \\&= \lim _{i \rightarrow \infty } (R{\bar{x}}^{(i)})_{m+1} = y_{m+1} \end{aligned} \end{aligned}$$

for $m \in {\mathbb {Z}}$. This shows that $Rx =y$ and so $x\in E$. Hence, the operator R is closed. By the closed graph theorem, R is bounded. Moreover, by property 1 the operator R is onto and invertible. It follows from the open mapping theorem that it has a bounded inverse.

Now we show that $u \in D(X)$. First observe that for a fixed m, replacing n by $m+k$ we have

$$\begin{aligned} \sup _{n \in {\mathbb {Z}}} \Vert x^{(n-m)}_n \Vert _n = \sup _{k \in {\mathbb {Z}}} \Vert x^{(k)}_{m+k} \Vert _{m+k} \le \sup _{k \in {\mathbb {Z}}} \Vert x^{(k)} \Vert _\infty . \end{aligned}$$

Since $R x^{(k)}=y^{(k)}$, we obtain

$$\begin{aligned} \Vert u_m\Vert _\infty = \sup _{n \in {\mathbb {Z}}} \Vert x^{(n-m)}_n \Vert _n \le \sup _{k \in {\mathbb {Z}}} \Vert x^{(k)} \Vert _\infty \le \Vert R^{-1} \Vert \sup _{k \in {\mathbb {Z}}} \Vert y^{(k)} \Vert _\infty . \end{aligned}$$

Moreover,

$$\begin{aligned} \sup _{m \in {\mathbb {Z}}} \Vert v_{m-k,m}\Vert _m \le \sup _{m,n \in {\mathbb {Z}}} \Vert v_{m,n} \Vert _n \end{aligned}$$

(10)

since the pairs $(m-k,m)$ with $m\in {\mathbb {Z}}$ form a subset of the pairs (m, n) with $m,n\in {\mathbb {Z}}$, and so

$$\begin{aligned} \begin{aligned} \Vert y^{(k)} \Vert _\infty&= \sup _{m \in {\mathbb {Z}}} \Vert v_{m-k,m}\Vert _m \le \sup _{m,n \in {\mathbb {Z}}} \Vert v_{m,n} \Vert _n \\&= \sup _{m \in {\mathbb {Z}}} \Vert v_m \Vert _\infty = \Vert v \Vert _D < \infty . \end{aligned} \end{aligned}$$

This shows that

$$\begin{aligned} \Vert u_m\Vert _\infty \le \Vert R^{-1} \Vert \cdot \Vert v \Vert _D < \infty \end{aligned}$$

and so $u_m \in \ell ^\infty (X)$. Finally, we also have

$$\begin{aligned} \sup _{m \in {\mathbb {Z}}} \Vert u_m\Vert _\infty \le \Vert R^{-1} \Vert \cdot \Vert v \Vert _D < +\infty \end{aligned}$$

and so $u \in D(X)$.

$(2 \Rightarrow 1)$. Take $y \in \ell ^\infty _0(X)$ and define a sequence $(v_m)_{m \in {\mathbb {Z}}}$ with values in $\ell ^\infty _0(X)$ by

$$\begin{aligned} v_{m,n}=\frac{y_n}{1+(m-n)^2} \quad \text {for }m,n \in {\mathbb {Z}}. \end{aligned}$$

(11)

Note that

$$\begin{aligned} \Vert v_m \Vert _\infty =\sup _{n \in {\mathbb {Z}}} \frac{\Vert y_n \Vert _n}{1+(m-n)^2} . \end{aligned}$$

Given $\varepsilon >0$, take $\rho >0$ such that $\Vert y_n \Vert _n<\varepsilon$ whenever $|n|\ge \rho$. Then

$$\begin{aligned} \Vert v_m \Vert _\infty \le \sup _{n \in [-\rho ,\rho ]} \frac{\Vert y\Vert _\infty }{1+(m-n)^2} +\varepsilon \rightarrow \varepsilon \end{aligned}$$

when $|m|\rightarrow \infty$. It follows from the arbitrariness of $\varepsilon$ that $v\in D_0(X)$.

By property 2, there exists a unique $u \in D(X)$ satisfying (4). In view of Lemma 1, for each $k\in {\mathbb {R}}$ the sequence $x^{(k)}=(u_{m-k,m})_{m\in {\mathbb {Z}}}$ satisfies the equation

$$\begin{aligned} x^{(k)}_{m+1} =A_m x^{(k)}_{m} + y^{(k)}_{m+1} \quad \text {for }m \in {\mathbb {Z}}, \end{aligned}$$

where

$$\begin{aligned} y^{(k)}_{n}=v_{n-k,n}=\frac{y_n}{1+k^2} \quad \text {for} \ n\in {\mathbb {Z}}. \end{aligned}$$

Therefore, ${\bar{x}}^{(k)}=(1+k^2)x^{(k)}$ satisfies Eq. (3) for each $k\in {\mathbb {Z}}$. Moreover, proceeding as in (10) we obtain

$$\begin{aligned} \begin{aligned} \Vert x^{(k)}\Vert _\infty&= \sup _{m\in {\mathbb {Z}}}\Vert u_{m-k,m}\Vert _m \le \sup _{m,n \in {\mathbb {Z}}} \Vert u_{m,n} \Vert _n \\&= \sup _{m \in {\mathbb {Z}}} \Vert u_m\Vert _\infty = \Vert u \Vert _D < \infty \end{aligned} \end{aligned}$$

and so ${\bar{x}}^{(k)} \in \ell ^\infty (X)$. We also show that ${\bar{x}}^{(k)}$ is independent of k. Given $p \in {\mathbb {Z}}$, we define a sequence ${\bar{u}}=({\bar{u}}_m)_{m \in {\mathbb {Z}}}$ in D(X) by

$$\begin{aligned} {\bar{u}}_{m,n}=\frac{{\bar{x}}^{(p)}_{n}}{1+(n-m)^2}\quad \text {for} \ m,n\in {\mathbb {Z}}. \end{aligned}$$

(12)

Then ${\bar{u}}_{m-k,m}={\bar{x}}^{(p)}_{m}/(1+k^2)$ satisfies equation (5) for all k and so by Lemma 1, ${\bar{u}}$ is a solution of Eq. (4). But by property 2, we must have ${\bar{u}}=u$. Therefore, for each $q \in {\mathbb {Z}}$ we have

$$\begin{aligned} x^{(q)}_{m}=u_{m-q,m}={\bar{u}}_{m-q,m}=\frac{{\bar{x}}^{(p)}_{m}}{1+q^2} \end{aligned}$$

for all $m \in {\mathbb {Z}}$ and so ${\bar{x}}^{(q)} ={\bar{x}}^{(p)}$. This shows that ${\bar{x}}:={\bar{x}}^{(k)} \in \ell ^\infty (X)$, which is a solution of Eq. (3), is independent of k.

To establish property 1, it remains to show that Eq. (3) has a unique solution. Assume that $z \in \ell ^\infty (X)$ was a solution different from ${\bar{x}}$. We define a sequence $w\in D(X)$ by

$$\begin{aligned} w_{m,n}=\frac{z_n}{1+(n-m)^2}\quad \text {for} \ m,n\in {\mathbb {Z}}. \end{aligned}$$

Then $w_{m-k,m}=z_m/(1+k^2)$ satisfies Eq. (5) for all k, that is,

$$\begin{aligned} w_{m+1-k,m+1}=A_m w_{m-k,m} +y^{(k)}_{m+1} \quad \text {for} \ m,k \in {\mathbb {Z}}. \end{aligned}$$

It follows from Lemma 1 that w is a solution of Eq. (4). But then both $u,w\in D(X)$ are solutions of Eq. (4), which by hypothesis has a single solution. Therefore, since ${\bar{u}}=u$ and ${\bar{x}}^{(p)}={\bar{x}}$, it follows from (12) that

$$\begin{aligned} \frac{{\bar{x}}_n}{1+(n-m)^2}=u_{m,n}=w_{m,n}=\frac{z_n}{1+(n-m)^2} \end{aligned}$$

for $m,n\in {\mathbb {Z}}$, which readily implies that ${\bar{x}}=z$. This contradiction shows that Eq. (3) has a unique solution. $\square$

4 Admissibility with Exponential Growth

In this section we consider the same admissibility property as before but for spaces of sequences with bounded exponential growth.

4.1 Evolution Maps

Given $c\ge 0$, let $E^c(X)$ be the set of all sequences $(x_m)_{m \in {\mathbb {Z}}}$ with values in X such that the sequence $x^c=(x^c_m)_{m\in {\mathbb {Z}}}$ defined by $x^c_m=e^{-c|m|}x_m$ for $m\in {\mathbb {Z}}$ is in $\ell ^\infty _0(X)$. We note that $E^c(X)$ is a Banach space when equipped with the norm

$$\begin{aligned} \Vert x \Vert _{E^c}:=\Vert x^c \Vert _\infty . \end{aligned}$$

Proposition 2

Let $A=(A_m)_{m \in {\mathbb {Z}}}$ be a sequence of linear maps on X that is exponentially bounded with respect to the norms $\Vert \cdot \Vert _m$. Then for each $c\ge 0$ and $y \in E^c(X)$ we have

$$\begin{aligned} T_y(E^c(X)) \subset E^c(X). \end{aligned}$$

Proof

Take $y,u \in E^c(X)$. Then

$$\begin{aligned} \begin{aligned} e^{-c|n|}\Vert (T_y u)_n \Vert _n&= e^{-c|n|}\left\Vert A_{n-1} u_{n-1} + y_n\right\Vert _n \\&\le \kappa e^{\alpha +c}e^{-c|n-1|} \Vert u_{n-1} \Vert _{n-1} + e^{-c |n|} \Vert y_n \Vert _n \rightarrow 0 \end{aligned} \end{aligned}$$

when $|n |\rightarrow \infty$ and so $T_y u \in E^c(X)$. $\square$

4.2 Admissibility Properties

As a preparation for the result relating admissibility properties using the spaces $E^c(X)$, we first establish a version of Theorem 2 in which we consider the same spaces for the perturbations and for the solutions.

Theorem 3

Let $A=(A_m)_{m \in {\mathbb {Z}}}$ be a sequence of linear maps that is exponentially bounded with respect to the norms $\Vert \cdot \Vert _m$. Then the following properties are equivalent:

• 1. for each $y \in \ell ^\infty _0(X)$ there exists a unique $x \in \ell ^\infty _0(X)$ satisfying (3);

• 2. for each $v \in D_0(X)$ there exists a unique $u \in D_0(X)$ satisfying (4).

Proof

$(1 \Rightarrow 2)$. Take $v \in D_0(X)$ and consider the sequences $y^{(k)} \in \ell ^\infty _0(X)$ defined by (7) for each $k \in {\mathbb {Z}}$. By property 1, there exists a unique solution $x^{(k)} \in \ell ^\infty _0(X)$ of equation (3) with $y=y^{(k)}$. Again we define $u_{m,n}$ as in (8). We will show that $u \in D_0(X)$. As in the proof of Theorem 2, u is then the unique solution of Eq. (4) in $D_0(X)$.

We already know from the proof of Theorem 2 that $u \in D(X)$ and so it remains to verify that $u_m\in \ell ^\infty _0(X)$ for each $m\in {\mathbb {Z}}$ and that $\Vert u_m\Vert _\infty \rightarrow 0$ when $|m|\rightarrow \infty$. Since

$$\begin{aligned} \lim _{|m|\rightarrow \infty }\Vert v_m\Vert _\infty =0, \end{aligned}$$

for each $\varepsilon >0$ there exists $M \in {\mathbb {N}}$ such that

$$\begin{aligned} \Vert v_{m,n}\Vert _n<\varepsilon \quad \text {whenever} \ |m|> M \ \text {and} \ n\in {\mathbb {Z}}. \end{aligned}$$

On the other hand, for each $m\in [-M,M]\cap {\mathbb {Z}}$ there exists $n_m \in {\mathbb {N}}$ such that

$$\begin{aligned} \Vert v_{m,n}\Vert _n<\varepsilon \quad \text {whenever} \ |n|\ge n_m. \end{aligned}$$

Letting

$$\begin{aligned} N=\max \bigl \{n_{-M},\ldots ,n_M\bigr \}, \end{aligned}$$

we obtain

$$\begin{aligned} \Vert v_{m,n}\Vert _n<\varepsilon \quad \text {whenever} \ m\in [-M,M]\cap {\mathbb {Z}}\ \text {and} \ |n|\ge N. \end{aligned}$$

(13)

This readily implies that

$$\begin{aligned} \Vert y^{(k)}\Vert _\infty =\sup _{m \in {\mathbb {Z}}} \Vert v_{m-k,m} \Vert _m <\varepsilon \end{aligned}$$

(14)

for any sufficiently large |k| since then the line $\{(m-k,m):m\in {\mathbb {Z}}\}$ does not intersect the rectangle $[-M,M]\times [-N,N]$. Hence, it follows from the arbitrariness of $\varepsilon$ that

$$\begin{aligned} \lim _{|k|\rightarrow \infty }\Vert y^{(k)}\Vert _\infty =0. \end{aligned}$$

(15)

Let $E_0$ be the Banach space of all sequences $x \in \ell ^\infty _0(X)$ for which there exists $y \in \ell ^\infty _0(X)$ satisfying (3) and define a linear operator $R:E_0\rightarrow \ell ^\infty _0 (X)$ by (9). One can show as in the proof of Theorem 2 that R has a bounded inverse. By (15) we have

$$\begin{aligned} \begin{aligned} \lim _{|n|\rightarrow \infty }\Vert u_{m,n}\Vert _n&= \lim _{|n|\rightarrow \infty }\Vert x^{(n-m)}_n\Vert _n\le \lim _{|n|\rightarrow \infty }\Vert x^{(n-m)}\Vert _\infty \\&\le \Vert R^{-1}\Vert \lim _{|n|\rightarrow \infty }\Vert y^{(n-m)}\Vert _\infty =0 \end{aligned} \end{aligned}$$

and so $u_m\in \ell ^\infty _0(X)$ for each $m\in {\mathbb {Z}}$. Moreover, since

$$\begin{aligned} \Vert x^{(k)}\Vert _\infty \le \Vert R^{-1}\Vert \cdot \Vert y^{(k)}\Vert _\infty , \end{aligned}$$

it follows from (15) that for each $\varepsilon >0$ there exists $K \in {\mathbb {N}}$ such that

$$\begin{aligned} \Vert x^{(k)}_n\Vert _n<\varepsilon \quad \text {whenever} \ |k|> K \ \text {and} \ n\in {\mathbb {Z}}. \end{aligned}$$

Since $x^{(k)} \in \ell ^\infty _0(X)$, for each $k\in [-K,K]\cap {\mathbb {Z}}$ there exist $n_k \in {\mathbb {N}}$ such that

$$\begin{aligned} \Vert x^{(k)}_n\Vert _n<\varepsilon \quad \text {whenever} \ |n|\ge n_k. \end{aligned}$$

So, there exists $N \in {\mathbb {N}}$ such that

$$\begin{aligned} \Vert x^{(k)}_n\Vert _n<\varepsilon \quad \text {whenever} \ k\in [-K,K]\cap {\mathbb {Z}}\ \text {and} \ |n|\ge N. \end{aligned}$$

This implies that $\sup _{n\in {\mathbb {Z}}}\Vert x^{(n-m)}_n\Vert _n<\varepsilon$ for any sufficiently large |m| since then the line $\{(n-m,n):m\in {\mathbb {Z}}\}$ does not intersect $[-K,K]\times [-N,N]$. Hence, it follows from the arbitrariness of $\varepsilon$ that

$$\begin{aligned} \lim _{|m|\rightarrow \infty }\Vert u_m\Vert _\infty = \lim _{|m|\rightarrow \infty }\sup _{n\in {\mathbb {Z}}}\Vert x^{(n-m)}_n\Vert _n=0. \end{aligned}$$

and so $u \in D_0(X)$.

$(2 \Rightarrow 1)$. Take $y \in \ell ^\infty _0(X)$ and consider the sequence $(v_m)_{m \in {\mathbb {Z}}} \in D_0(X)$ defined by (11). By property 2, there exists a unique $u \in D_0(X)$ satisfying (4). We already know from the proof of Theorem 2 that the sequence $x=(x_m)_{m\in {\mathbb {Z}}}$ with

$$\begin{aligned} x_m=(1+k^2)u_{m-k,m}\quad \text {for} \ m\in {\mathbb {Z}}\end{aligned}$$

is independent of $k\in {\mathbb {Z}}$ and that it is the unique solution of equation (3) in $\ell ^\infty (X)$. It remains to verify that $x\in \ell ^\infty _0(X)$.

As in the proof of the implication $1 \Rightarrow 2$ (see (13)), since $u\in D_0(X)$, for each $\varepsilon >0$ there exist $M,N \in {\mathbb {N}}$ such that

$$\begin{aligned} \Vert u_{m,n}\Vert _n<\varepsilon \quad \text {whenever} \ m\in [-M,M]\cap {\mathbb {Z}}\ \text {and} \ |n|\ge N. \end{aligned}$$

Also as before (see (14)), this implies that $\sup _{m\in {\mathbb {Z}}}\Vert u_{m-k,m}\Vert _m<\varepsilon$ for any sufficiently large |k|. It thus follows from the arbitrariness of $\varepsilon$ that $x\in \ell ^\infty _0(X)$. This completes the proof of the theorem. $\square$

Using this result we are able to consider the space $E^c(X)$ for an arbitrary constant $c\ge 0$. Given $c\ge 0$ and taking $y=0$ in Proposition 2, we find that A generates the evolution map $S=T_0$ on $E^c(X)$ given by

$$\begin{aligned} (S u)_n=A_{n-1} u_{n-1} \quad \text {for }n \in {\mathbb {Z}}\text { and }u \in E^c(X). \end{aligned}$$

Moreover, let $F^c(X)$ be the set of all sequences $(v_m)_{m \in {\mathbb {Z}}}$ with values in $E^c(X)$ such that

$$\begin{aligned} \lim _{|m |\rightarrow \infty } \Vert v_m \Vert _{E^c} =0. \end{aligned}$$

We note that $E^c(X)$ is a Banach space when equipped with the norm

$$\begin{aligned} \Vert v \Vert _{F^c}:=\sup _{m \in {\mathbb {Z}}} \Vert v_m \Vert _{E^c}<\infty . \end{aligned}$$

Theorem 4

Let $A=(A_m)_{m \in {\mathbb {Z}}}$ be a sequence of linear maps on X that is exponentially bounded with respect to the norms $\Vert \cdot \Vert _m$. Then for each $c\ge 0$ the following properties are equivalent:

• 1. for each $y \in E^c(X)$ there exists a unique $x \in E^c(X)$ satisfying (3);

• 2. for each $v \in F^c(X)$ there exists a unique $u \in F^c(X)$ satisfying (4).

Proof

Take $y,x \in E^c(X)$. We consider the sequences $y^c,x^c\in \ell ^\infty _0(X)$ defined by

$$\begin{aligned} y^c_m=e^{-c|m|}y_m \quad \text {and}\quad x^c_m=e^{-c|m|}x_m \end{aligned}$$

for $m\in {\mathbb {Z}}$. Note that property (3) holds if and only if

$$\begin{aligned} x^c_{m+1}=A_{m}^c x_m^c +y_{m+1}^c \quad \text {for }m \in {\mathbb {Z}}, \end{aligned}$$

(16)

where

$$\begin{aligned} A_m^c = e^{-c|m+1| + c |m|} A_m. \end{aligned}$$

Therefore, property 1 holds if and only if for each $f \in E^c(X)$ there exists a unique $x \in E^c(X)$ satisfying (16) (using the definitions of $y^c$ and $x^c$).

Notice that the sequence $A^c=(A_m^c)_{m \in {\mathbb {Z}}}$ is also exponentially bounded with respect to the norms $\Vert \cdot \Vert _m$. Since the maps $y\mapsto y^c$ and $x\mapsto x^c$ are bijections from $E^c(X)$ onto $\ell ^\infty _0(X)$, it follows from Theorem 3 that property 1 holds if and only if for each $F \in D_0(X)$ there exists a unique $u \in D_0(X)$ satisfying

$$\begin{aligned} u_{m+1}=S^c u_m +v_{m+1} \quad \text {for }m \in {\mathbb {Z}}, \end{aligned}$$

(17)

where

$$\begin{aligned} (S^c v)_m =A_{m-1}^c v_{m-1} \quad \text {for }m\in {\mathbb {Z}}\text { and }v \in \ell ^\infty _0(X). \end{aligned}$$

We have

$$\begin{aligned} e^{c|m|} (S^c v)_m=A_{m-1} e^{c|m-1|}v_{m-1}, \end{aligned}$$

that is,

$$\begin{aligned} \gamma \circ S^c =S\circ \gamma ,\quad \text {with} \ \gamma (v)_n=e^{c|n|}v_n. \end{aligned}$$

Letting $u^c_m=\gamma (u_m )$, we obtain

$$\begin{aligned} \gamma (S^c u_m) =S (\gamma (u_m))=S u^c_m \end{aligned}$$

and so property (17) is equivalent to

$$\begin{aligned} u^c_{m+1}=S u^c_m +v^c_{m+1}, \end{aligned}$$

(18)

where

$$\begin{aligned} (v^c_{m})_n=\gamma (v_m)_n=e^{c|n|}v_{m,n}. \end{aligned}$$

Since the maps $v \mapsto v^c=(v^c_m)_{m \in {\mathbb {Z}}}$ and $u\mapsto u^c=(u^c_m)_{m \in {\mathbb {Z}}}$ are bijections from $D_0(X)$ onto $F^c(X)$, it follows from (18) that property 1 holds if and only if property 2 holds. $\square$

5 Admissibility on $\ell ^p$ Spaces

In this section we consider once more an admissibility property, now for evolution maps on $\ell ^p$ spaces.

5.1 Evolution Maps

For each $p\in [1,+\infty )$, let $\ell ^p(X)$ be the set of all sequences $x=(x_m)_{m \in {\mathbb {Z}}}$ with values in X such that

$$\begin{aligned} \Vert x \Vert _{\ell ^p}=\left( \sum _{m \in {\mathbb {Z}}} \Vert x_m \Vert _m^p \right) ^{1/p}< \infty . \end{aligned}$$

We note that ${\ell ^p}(X)$ is a Banach space when equipped with the norm $\Vert \cdot \Vert _{\ell ^p}$.

Proposition 3

Let $A=(A_m)_{m \in {\mathbb {Z}}}$ be a sequence of linear maps on X that is exponentially bounded with respect to the norms $\Vert \cdot \Vert _m$. Then for each $y \in {\ell ^p}(X)$, we have

$$\begin{aligned} T_y({\ell ^p}(X))\subset {\ell ^p}(X). \end{aligned}$$

Proof

Take $y,u\in \ell ^p(X)$. By Minkowski’s inequality we have

$$\begin{aligned} \Vert T_y u \Vert _{\ell ^p}&=\left( \sum _{n \in {\mathbb {Z}}} \Vert A_{n-1} u_{n-1} + y_n\Vert _{n}^p\,ds\right) ^{1/p}\\&\le \left( \sum _{n \in {\mathbb {Z}}} \Vert A_{n-1} u_{m-1}\Vert _{n}^p\,ds\right) ^{1/p}+\left( \sum _{n \in {\mathbb {Z}}} \Vert y_n\Vert _p\right) ^{1/p} \\&\le \kappa e^{\alpha } \left( \sum _{n \in {\mathbb {Z}}} \Vert u_{n-1}\Vert _{n-1}^p\right) ^{1/p} +\Vert y\Vert _{\ell ^p} \\&= \kappa e^{\alpha }\Vert u\Vert _{\ell ^p} +\Vert y\Vert _{\ell ^p} < \infty , \end{aligned}$$

which shows that $T_y u \in \ell ^p(X)$. $\square$

5.2 Admissibility Properties

Taking $y=0$ in Proposition 3, we find that A generates the evolution map $S=T_0$ on $\ell ^p(X)$ given by

$$\begin{aligned} (S u)_n=A_{n-1} u_{n-1} \quad \text {for }n \in {\mathbb {Z}}\text { and }u \in \ell ^p(X). \end{aligned}$$

Moreover, for each $p\in [1,+\infty )$ let $M^p(X)=\ell ^p(\ell ^p(X))$ be the set of all sequences $v=(v_n)_{n \in {\mathbb {Z}}}$ with $v_n \in \ell ^p(X)$ such that

$$\begin{aligned} \Vert v \Vert _{M^p}:=\left( \sum _{m \in {\mathbb {Z}}} \Vert v_m \Vert _{\ell ^p}^p \right) ^{1/p}=\left( \sum _{m \in {\mathbb {Z}}} \sum _{n \in {\mathbb {Z}}} \Vert v_{m,n} \Vert _n^p \right) ^{1/p}< \infty . \end{aligned}$$

We note that $M^p(X)$ is a Banach space when equipped with the norm $\Vert \cdot \Vert _{M^p}$.

Theorem 5

Let $A=(A_m)_{m\in {\mathbb {Z}}}$ be a sequence of linear maps on X that is exponentially bounded with respect to the norms $\Vert \cdot \Vert _m$. Then the following properties are equivalent:

• 1. for each $y \in \ell ^p(X)$ there exists a unique $x \in \ell ^p(X)$ satisfying (3);

• 2. for each $v \in M^p(X)$ there exists a unique $u \in M^p(X)$ satisfying (4).

Proof

$(1 \Rightarrow 2)$. Take $v \in M^p(X)$. For each $k \in {\mathbb {R}}$ we consider the sequence $y^{(k)} \in X^{\mathbb {Z}}$ defined by (7). We have

$$\begin{aligned} \begin{aligned} \sum _{k \in {\mathbb {Z}}} \Vert y^{(k)}_n\Vert _{\ell ^p}^p&=\sum _{k \in {\mathbb {Z}}} \sum _{n \in {\mathbb {Z}}} \Vert v_{n-k,n} \Vert _n^p \\&= \sum _{m \in {\mathbb {Z}}} \sum _{j \in {\mathbb {Z}}} \Vert v_{m,j}\Vert _j^p =\Vert v\Vert _{M^p}^p < \infty \end{aligned} \end{aligned}$$

(19)

and so $y^{(k)} \in \ell ^p(X)$. By property 1, there exists a unique solution $x^{(k)}\in \ell ^p(X)$ of Eq. (3) with $y=y^{(k)}$. Again we define $u_{m,n}$ as in (8). By Lemma 1, u is a solution of Eq. (4) and as in the proof of Theorem 2 it is automatically unique. We will show that $u \in M^p(X)$.

Let F be the Banach space of all sequences $x \in \ell ^p (X)$ for which there exists $y \in \ell ^p (X)$ satisfying (3) and define a linear operator $R:F\rightarrow \ell ^p (X)$ by (9). One can show as in the proof of Theorem 2 that R has a bounded inverse.

Now we show that $u \in M^p(X)$. We have

$$\begin{aligned} \begin{aligned} \sum _{m \in {\mathbb {Z}}} \Vert u_m\Vert _{\ell ^p}^p&=\sum _{m \in {\mathbb {Z}}} \sum _{n \in {\mathbb {Z}}} \Vert x^{(n-m)}_n \Vert _n^p \\&= \sum _{k \in {\mathbb {Z}}} \sum _{j\in {\mathbb {Z}}} \Vert x^{(k)}_j\Vert _j^p =\sum _{k \in {\mathbb {Z}}} \Vert x^{(k)}\Vert _{\ell ^p}^p. \end{aligned} \end{aligned}$$

(20)

Since

$$\begin{aligned} \Vert x^{(k)}\Vert _{\ell ^p}\le \Vert R^{-1} \Vert \cdot \Vert y^{(k)} \Vert _{\ell ^p}, \end{aligned}$$

it follows from (19) that

$$\begin{aligned} \begin{aligned} \sum _{m \in {\mathbb {Z}}} \Vert u_m\Vert _{L^p}^p&=\sum _{k \in {\mathbb {Z}}} \Vert x^{(k)}\Vert _{L^p}^p\\&\le \Vert R^{-1} \Vert ^p \sum _{k \in {\mathbb {Z}}}\Vert y^{(k)}\Vert _{\ell ^p}^p\\&= \Vert R^{-1} \Vert ^p \Vert v\Vert _{M^p}^p<\infty \end{aligned} \end{aligned}$$

and so $u \in M^p(X)$.

$(2 \Rightarrow 1)$. Take $y \in \ell ^p(X)$ and define $v=(v_m)_{m\in {\mathbb {Z}}}$ by

$$\begin{aligned} v_{m,n}=\frac{y_n}{1+(m-n)^2} \quad \text {for }m,n \in {\mathbb {Z}}\text {.} \end{aligned}$$

Note that $v \in M^p(X)$. Indeed,

$$\begin{aligned} \Vert v \Vert _{M^p}^p&= {} \sum _{m \in {\mathbb {Z}}} \sum _{n \in {\mathbb {Z}}} \Vert v_{m,n} \Vert _n^p \\ {}&=\sum _{n \in {\mathbb {Z}}} \sum _{m \in {\mathbb {Z}}} \Vert v_{m,n} \Vert _n^p \\ {}&=\sum _{n \in {\mathbb {Z}}} \Vert y_n\Vert _n^p \sum _{m \in {\mathbb {Z}}} \frac{1}{(1+(m-n)^2)^p}\\ {}&=\sum _{n \in {\mathbb {Z}}} \Vert y_n\Vert _n^p \sum _{m \in {\mathbb {Z}}} \frac{1}{(1+m^2)^p} \\ {}&=c_p\Vert y\Vert _{\ell ^p}^p<\infty \end{aligned}$$

for some constant $c_p>0$ that depends only on p. By property 2, there exists a unique $u \in M^p(X)$ satisfying (4). By Lemma 1, for each $k\in {\mathbb {R}}$ the sequence $x^{(k)}$ defined by

$$\begin{aligned} x^{(k)}_m=u_{m-k,m} \quad \text {for }m \in {\mathbb {Z}}\end{aligned}$$

satisfies Eq. (3) with y replaced by $y^{(k)}=(y^{(k)}_m)_{m\in {\mathbb {Z}}}$ with

$$\begin{aligned} y^{(k)}_m=v_{m-k,m} =\frac{y_m}{1+k^2}\quad \text {for} \ m\in {\mathbb {Z}}. \end{aligned}$$

Proceeding as in (20), we obtain

$$\begin{aligned} \sum _{k \in {\mathbb {Z}}} \Vert x^{(k)}\Vert _{\ell ^p}^p = \sum _{m \in {\mathbb {Z}}}\Vert u_m\Vert _{\ell ^p}^p=\Vert u\Vert _{M^p}^p<\infty \end{aligned}$$

and so $x^{(k)}\in \ell ^p(X)$. One can then show in a similar manner to that in the proof of Theorem 2 that $x=(x_m)_{m\in {\mathbb {Z}}}$ with

$$\begin{aligned} x_m=(1+k^2)x^{(k)}_m\quad \text {for} \ m\in {\mathbb {Z}}\end{aligned}$$

is independent of k and that it is the unique solution of Eq. (3) in $\ell ^p(X)$. This concludes the proof of the theorem. $\square$

6 Hyperbolicity

In this section we discuss the relation of hyperbolicity with the admissibility properties considered in the former sections.

Let $\Vert \cdot \Vert _m$, for $m \in {\mathbb {Z}}$, be a family of norms on a Banach space X. We say that a sequence $(A_m)_{m\in {\mathbb {Z}}}$ of linear maps on X is hyperbolic with respect to the norms $\Vert \cdot \Vert _m$ if:

• 1. there exist projections $P_n$ for $n\in {\mathbb {Z}}$ such that $P_{n+1} A_n =A_n P_n$ and the map

  $$\begin{aligned} A_n |_{{{\,\mathrm{Im}\,}}Q_n}:{{\,\mathrm{Im}\,}}Q_n\rightarrow {{\,\mathrm{Im}\,}}Q_{n+1}, \end{aligned}$$

  where $Q_n =\text {Id}-P_n$, is onto and invertible for each $n\in {\mathbb {Z}}$;

• 2. there exist constants $\lambda , N>0$ such that for each $x\in X$ we have

  $$\begin{aligned} \Vert U(m,n)P_n x\Vert _m \le N e^{-\lambda (m-n)} \Vert x \Vert _n \quad \text {for }m \ge n \end{aligned}$$

  and

  $$\begin{aligned} \Vert {\bar{U}}(m,n)Q_n x\Vert _m \le N e^{-\lambda (n-m)} \Vert x \Vert _n \quad \text {for }m \le n, \end{aligned}$$

  where ${\bar{U}}(n,m)=(U(m,n)|_{{{\,\mathrm{Im}\,}}Q_n})^{-1}$.

The following proposition is a particular case of more general results in [1] that relate hyperbolicity with admissibility.

Proposition 4

Let $A=(A_m)_{m\in {\mathbb {Z}}}$ be a sequence of linear maps on X that is exponentially bounded with respect to the norms $\Vert \cdot \Vert _m$. Then the following properties are equivalent:

• 1. the sequence $(A_m)_{m\in {\mathbb {Z}}}$ is hyperbolic with respect to the norms $\Vert \cdot \Vert _m$;

• 2. for each $y \in \ell ^\infty _0(X)$ there exists a unique $x \in \ell ^\infty _0(X)$ satisfying (3);

• 3. for each $y \in \ell ^p(X)$ there exists a unique $x \in \ell ^p(X)$ satisfying (3).

We refer the reader to [2] for a detailed list of references on further related results, including specifically for the family of norms $\Vert \cdot \Vert _m=\Vert \cdot \Vert$.

The following statement is a simple consequence of Theorems 3 and 5 together with Proposition 4.

Theorem 6

Let $A=(A_m)_{m\in {\mathbb {Z}}}$ be a sequence of linear maps on X that is exponentially bounded with respect to the norms $\Vert \cdot \Vert _m$. Then the following properties are equivalent:

• 1. the sequence $(A_m)_{m\in {\mathbb {Z}}}$ is hyperbolic with respect to the norms $\Vert \cdot \Vert _m$;

• 2. for each $v \in D_0(X)$ there exists a unique $u \in D_0(X)$ satisfying (4);

• 3. for each $v \in M^p(X)$ there exists a unique $u \in M^p(X)$ satisfying (4).

One can also consider the hyperbolicity of the evolution map. We recall that a map T on a Banach space Y is said to be hyperbolic if:

• 1. there exists a projection P satisfying $P T=T P$ and the map

  $$\begin{aligned} T |_{{{\,\mathrm{Im}\,}}Q} :{{\,\mathrm{Im}\,}}Q \rightarrow {{\,\mathrm{Im}\,}}Q, \end{aligned}$$

  where $Q=\text {Id}-P$, is onto and invertible;

• 2. there exist $\lambda ,N > 0$ such that

  $$\begin{aligned} \Vert T^m P\Vert \le N e^{-\lambda m} \quad \text {and}\quad \Vert S^m Q\Vert \le N e^{-\lambda m} \end{aligned}$$

  for $m\ge 0$, where $S =(T |_{{{\,\mathrm{Im}\,}}Q})^{-1}$.

In particular, the equivalence of the notions of hyperbolicity for a sequence $(A_m)_{m\in {\mathbb {Z}}}$ and its evolution map on $\ell ^\infty _0(X)$ and on $\ell ^p(X)$ lead to further equivalences to the former admissibility properties.

In particular, we have the following result.

Theorem 7

[3] Let $A=(A_m)_{m\in {\mathbb {Z}}}$ be a sequence of linear maps on X that is exponentially bounded with respect to the norms $\Vert \cdot \Vert _m$. Then $(A_m)_{m\in {\mathbb {Z}}}$ is hyperbolic with respect to the norms $\Vert \cdot \Vert _m$ if and only if the evolution map S on $Y=\ell ^\infty _0(X)$ given by (2) is hyperbolic.

In addition, one can replace the space Y in Theorem 7 by many other Banach spaces, including $\ell ^p(X)$ with $p\in [1,+\infty )$ (see [4] for details).

The following statement is a simple consequence of the former results.

Theorem 8

Let $A=(A_m)_{m\in {\mathbb {Z}}}$ be a sequence of linear maps on X that is exponentially bounded with respect to the norms $\Vert \cdot \Vert _m$. Then the following properties are equivalent:

• 1. the sequence $(A_m)_{m\in {\mathbb {Z}}}$ is hyperbolic with respect to the norms $\Vert \cdot \Vert _m$;

• 2. the pair formed by the spaces $\ell ^\infty _0(X)$ and $\ell ^\infty (X)$ is admissible;

• 3. the pair formed by the spaces $\ell ^p(X)$ and $\ell ^p(X)$ is admissible;

• 4. the evolution map S on $Y=\ell ^\infty _0(X)$ or on $Y=\ell ^p(X)$ is hyperbolic;

• 5. the pair formed by the spaces $D_0(X)$ and $D_0(X)$ is admissible;

• 6. the pair formed by the spaces $M^p(X)$ and $M^p(X)$ is admissible.