Some Convergence Theorems for Operator Sequences

TL;DR

This paper investigates conditions under which certain sequences of bounded linear operators on Banach spaces converge in norm, with applications to Toeplitz, composition, and model operators.

Contribution

It provides necessary and sufficient conditions for the norm convergence of operator sequences and applies these results to specific classes of operators.

Findings

Established criteria for convergence of operator sequences

Applied results to Toeplitz, composition, and model operators

Discussed related convergence problems

Abstract

Let $A,$ $T$ and $B$ be bounded linear operators on a Banach space. This paper is concerned mainly with finding some necessary and sufficient conditions for convergence in operator norm of the sequences $\left\{ A^{n}TB^{n}\right\} $ and $\left\{ \frac{1}{n}\sum_{i=0}^{n-1}A^{i}TB^{i}% \right\} $. These results are applied to the Toeplitz, composition and model operators. Some related problems are also discussed.

Equations

$$T_{\phi }f=P_{+}\left(\phi f\right),f\in H^2,$$

$$Sf\left(z\right)=zf\left(z\right)$$

$$S^{\ast }TS=T.$$

$$T=T_0+K,$$

$$⟨\left\{x_n\right\},\left\{y_n\right\}⟩={\text{l.i.m.}}_n⟨x_n,y_n⟩,$$

$$E:=\left\{\left\{x_n\right\}\in H_0:{\text{ l.i.m.}}_n{\Vert x_n\Vert }^2=0\right\},$$

$$⟨\left\{x_n\right\}+E,\left\{y_n\right\}+E⟩={\text{l.i.m.}}_n⟨\left\{x_n\right\},\left\{y_n\right\}⟩.$$

$$\Vert \left\{x_n\right\}+E\Vert ={\left({\text{l.i.m.}}_n{\Vert x_n\Vert }^2\right)}^{\frac{1}{2}}.$$

$$\hat{T}:\left\{x_n\right\}+E\mapsto \left\{T{x}_n\right\}+E.$$

$$\Vert \hat{T}\left(\left\{x_n\right\}+E\right)\Vert$$

$$\Vert \hat{T}\Vert =\Vert \hat{T}+\hat{K}\Vert \le \Vert T+K\Vert \text{, }\forall K\in K\left(H\right).$$

$$\Vert T+K\left(H\right)\Vert =\sup \left\{\overline{\underset{n\to \infty }{\lim }}\Vert T{x}_n\Vert :\Vert x_n\Vert =1,\forall n\in \mathbb{N}\text{ and }{x}_n\to 0\text{ weakly}\right\}.$$

$$\overline{\underset{n\to \infty }{\lim }}\Vert T{x}_n\Vert \ge \Vert T+K\left(H\right)\Vert -\epsilon .$$

$$\underset{k\to \infty }{\lim }\Vert T{x}_{n_k}\Vert \ge \Vert T+K\left(H\right)\Vert -\epsilon .$$

$$\Vert \hat{T}\Vert =\sup \left\{{\left({\text{l.i.m.}}_n{\Vert T{x}_n\Vert }^2\right)}^{\frac{1}{2}}:{\text{l.i.m.}}_n{\Vert x_n\Vert }^2=1\text{ and }{x}_n\to 0\text{ weakly}\right\}.$$

$$\Vert \hat{T}\Vert \ge \underset{k\to \infty }{\lim }\Vert T{x}_{n_k}\Vert \ge \Vert T+K\left(H\right)\Vert -\epsilon .$$

$$\underset{n\to \infty }{\lim }\Vert A^{n}K{B}^n\Vert =0.$$

$$\underset{n\to \infty }{\lim }\Vert A^{n}T{B}^n\Vert =0,$$

$$x\otimes y:z\mapsto ⟨z,y⟩x,z\in H.$$

$$K=\sum _{i=1}^N{x}_i\otimes y_i,$$

$$\Vert A^{n}K{B}^n\Vert =\Vert \sum _{i=1}^N{A}^n{x}_i\otimes B^{\ast n}{y}_i\Vert \le \sum _{i=1}^N\Vert A^n{x}_i\Vert \Vert B^{\ast n}{y}_i\Vert \to 0\left(n\to \infty \right).$$

$$\Vert \hat{T}\Vert =\Vert {\hat{A}}^n\hat{T}{\hat{B}}^n\Vert \le \Vert A^{n}T{B}^n\Vert \to 0.$$

$$A^{n}T{B}^n=T_0+A^{n}K{B}^n\text{, }\forall n\in \mathbb{N}.$$

$$\Vert A^{n+1}T{B}^{n+1}-A{T}_{0}B\Vert \to 0,$$

$$\Vert A^n(T-T_0)B^n\Vert \to 0,$$

$$T_{\psi }=\sum _{k=1}^N{c}_{-k}{S}^{\ast k}+\sum _{k=0}^N{c}_k{S}^k,$$

$$S^{\ast n}{T}_{\phi }{T}_{\psi }{S}^n=\sum _{k=1}^N{c}_{-k}{S}^{\ast k}{T}_{\phi }+\sum _{k=0}^N{c}_k{T}_{\phi }{S}^k,\forall n\ge N.$$

$$S^{\ast n}{T}_{\phi }{T}_{\psi }{S}^n=\sum _{k=1}^N{c}_{-k}{S}^{\ast k}{T}_{\psi }+\sum _{k=0}^N{c}_k{T}_{\psi }{S}^k,\forall n\ge N.$$

$$S^{\ast n}{T}_{\phi }{T}_{\psi }{S}^n$$

Full text

1. Introduction

Throughout this paper, $H$ will denote a complex separable infinite dimensional Hilbert space and $B\left(H\right)$, the algebra of all bounded linear operators on $H.$ The ideal of all compact operators on $H$ will be denoted by $K\left(H\right)$. The quotient algebra $B\left(H\right)\diagup K\left(H\right)$ is a $C^{\ast}-$algebra and called the Calkin algebra. As usual, $H^{2}$ will denote the classical Hardy space on the open unit disk $\mathbb{D}:=\left\{z\in\mathbb{C}:\left|z\right|<1\right\}.$ By $H^{\infty}$ we will denote the space of all bounded analytic functions on $\mathbb{D}.$

Let $\mathbb{T}$:$=\partial\mathbb{D}$ be the unit circle and let $m$ be the normalized Lebesgue measure on $\mathbb{T}$. Recall that for a given symbol $\varphi\in L^{\infty}:=L^{\infty}\left(\mathbb{T},m\right)$, the Toeplitz operator $T_{\varphi}$ on $H^{2}$ is defined by

[TABLE]

where $P_{+}$ is the orthogonal projection from $L^{2}\left(\mathbb{T},m\right)$ onto $H^{2}$. Let

[TABLE]

be the unilateral shift operator on $H^{2}.$ According to a theorem of Brown and Halmos [3], $T\in B\left(H^{2}\right)$ is a Toeplitz operator if and only if

[TABLE]

Barria and Halmos [1] examined the so-called strongly asymptotically Toeplitz operators $T$ on $H^{2}$ for which the sequence $\left\{S^{\ast n}TS^{n}\right\}$ converges strongly. This class includes the Hankel algebra, the operator norm-closed algebra generated by all Toeplitz and Hankel operators together [1].

An operator $T\in B\left(H^{2}\right)$ is said to be* uniformly asymptotically Toeplitz *if the sequence $\left\{S^{\ast n}TS^{n}\right\}$ converges in the uniform operator topology. This class of operators is closed in operator norm and under adjoints. It contains both Toeplitz operators and the compact ones. Feintuch [8] proved that an operator $T\in B\left(H^{2}\right)$ is uniformly asymptotically Toeplitz if and only if it has the decomposition

[TABLE]

where $T_{0}$ is a Toeplitz operator, that is, $S^{\ast}T_{0}S=T_{0}$ and $K$ is a compact operator.

Recall that each holomorphic function $\phi:\mathbb{D\rightarrow D}$ induces a bounded linear composition operator $C_{\phi}$ on $H^{2}$ by $C_{\phi}f=f\circ\phi$ (for instance, see [16, Ch.5]). The only composition operator, which is also Toeplitz, is the identity operator [20]. Using Feintuch’s result, Nazarov and Shapiro [20, Theorem 1.1] proved that a composition operator on $H^{2}$ is uniformly asymptotically Toeplitz if and only if it is either compact or the identity operator.

Let $B\left(X\right)$ be the algebra of all bounded linear operators on a complex Banach space $X$ and let $A,$ $T$ and $B$ be in $B\left(X\right).$ The main purpose of this paper is to find necessary and sufficient conditions for convergence in operator norm of the sequences $\left\{A^{n}TB^{n}\right\}$ and $\left\{\frac{1}{n}\sum_{i=0}^{n-1}A^{i}TB^{i}\right\}$.

2. The sequence $\left\{A^{n}TB^{n}\right\}$

In this section, we give some results concerning convergence in operator norm of the sequence $\left\{A^{n}TB^{n}\right\}$ for Hilbert space operators.

Recall that an operator $T\in B\left(H\right)$ is said to be essentially isometric (resp. essentially unitary) if $I-T^{\ast}T\in K\left(H\right)$ (resp. $I-T^{\ast}T\in K\left(H\right)$ and $I-TT^{\ast}\in K\left(H\right)$).

We have the following:

Theorem 2.1**.**

Let $A$ and $B^{\ast}$ be two essentially isometric operators on $H$ such that $\left\|A^{n}x\right\|\rightarrow 0$ and $\left\|B^{\ast n}x\right\|\rightarrow 0,$ for all $x\in H$. If $T\in B\left(H\right),$ then the sequence $\left\{A^{n}TB^{n}\right\}$ converges in operator norm if and only if we have the decomposition

[TABLE]

where $AT_{0}B=T_{0}$ and $K\in K\left(H\right).$

For the proof, we need some preliminary results.

Let $H_{0}$ be the linear space of all weakly null sequences $\left\{x_{n}\right\}$ in $H.$ Let us define a semi-inner product in $H_{0}$ by

[TABLE]

where l.i.m. is a fixed Banach limit. If

[TABLE]

then $H_{0}\diagup E$ becomes a pre-Hilbert  space with respect to the inner product defined by

[TABLE]

Let $\widehat{H}$ be the Hilbert space defined by the completion of $H_{0}\diagup E$ with respect to the induced norm

[TABLE]

Now, for a given $T\in B\left(H\right)$ we define an operator $\widehat{T}$ on $H_{0}\diagup E$ by

[TABLE]

Consequently, we can write

[TABLE]

Since $H_{0}\diagup E$ is dense in $\widehat{H},$ the operator $\widehat{T}$ can be extended to the whole $\widehat{H}$ which we also denote by $\widehat{T}$. Clearly, $\left\|\widehat{T}\right\|\leq\left\|T\right\|.$ The operator $\widehat{T}$ will be called limit operator associated with $T.$

Proposition 2.2**.**

If $\widehat{T}$ is the limit operator associated with $T\in B\left(H\right)$, then:

$\left(a\right)$* The map $T\mapsto\widehat{T}$ is a linear contractive $\ast-$homomorphism.*

$\left(b\right)$* $T$ is a compact operator if and only if $\widehat{T}=0.$*

$\left(c\right)$* $T$ is an essentially isometry $($resp. essentially unitary$)$ if and only if $\widehat{T}$ is an isometry $($resp. unitary$).$*

$\left(d\right)$* For an arbitrary $T\in B\left(H\right),$ we have $\left\|\widehat{T}\right\|=\left\|T+K\left(H\right)\right\|.$*

Proof.

Proofs of the assertions (a), (b) and (c) are omitted, since they are clear. Let us prove (d). Let $\widehat{K}$ be the limit operator associated with $K\in K\left(H\right).$ Since $\widehat{K}=0,$ we get

[TABLE]

This implies $\left\|\widehat{T}\right\|\leq\left\|T+K\left(H\right)\right\|.$ For the reverse inequality, recall [2, p.94] that

[TABLE]

Therefore, for a given $\varepsilon>0$ there exists a sequence $\left\{x_{n}\right\}$ in $H$ such that $\left\|x_{n}\right\|=1$ $\left(\forall n\in\mathbb{N}\right),$ $x_{n}\rightarrow 0$ weakly and

[TABLE]

Consequently, there exists a subsequence $\left\{x_{n_{k}}\right\}$ of $\left\{x_{n}\right\}$ such that

[TABLE]

On the other hand,

[TABLE]

As l.i.m.${}_{k}\left\|x_{n_{k}}\right\|^{2}=1$ and $x_{n_{k}}\rightarrow 0$ $\left(k\rightarrow\infty\right)$ weakly, by the preceding identity we get

[TABLE]

Since $\varepsilon$ is arbitrary, we have $\left\|\widehat{T}\right\|\geq\left\|T+K\left(H\right)\right\|,$ as required.

Lemma 2.3**.**

$\left(a\right)$* Let $A,B\in B\left(H\right)$ and assume that $\left\|A^{n}x\right\|\rightarrow 0$ and $\left\|B^{\ast n}x\right\|\rightarrow 0,$ for all $x\in H.$ Then, for an arbitrary $K\in K\left(H\right),$ we have*

[TABLE]

$\left(b\right)$* If $A$ and $B^{\ast}$ are essentially isometric operators and*

[TABLE]

then $T$ is a compact operator.

Proof.

(a) For an arbitrary $x,y\in H,$ let $x\otimes y$ be the rank one operator on $H$;

[TABLE]

Since finite rank operators are dense (in operator norm) in $K\left(H\right)$, we may assume that $K$ is a finite rank operator, say,

[TABLE]

where $x_{i},y_{i}\in H$ $\left(i=1,...,N\right).$ Consequently, we can write

[TABLE]

(b) Let $\widehat{A}$, $\widehat{T}$ and $\widehat{B}$ be the limit operators associated with $A$, $T$ and $B,$ respectively. By Proposition 2.2, $\widehat{A}$ and $\widehat{B}^{\ast}$ are isometries. Since the map $T\mapsto\widehat{T}$ is a contractive homomorphism, for an arbitrary $n\in\mathbb{N}$ we get

[TABLE]

Hence $\widehat{T}=0.$ By Proposition 2.2, $T$ is a compact operator.

We are now in a position to prove Theorem 2.1.

Proof of Theorem 2.1.

If $T=T_{0}+K,$ where $AT_{0}B=T_{0}$ and $K\in K\left(H\right),$ then

[TABLE]

By Lemma 2.3, $\left\|A^{n}KB^{n}\right\|\rightarrow 0$ and therefore $\left\|A^{n}TB^{n}-T_{0}\right\|\rightarrow 0.$ Now, assume that there exists $T_{0}\in B\left(H\right)$ such that $\left\|A^{n}TB^{n}-T_{0}\right\|\rightarrow 0.$ Since

[TABLE]

we have $AT_{0}B=T_{0}$ which implies $A^{n}T_{0}B^{n}=T_{0}$ for all $n\in\mathbb{N}.$ Also, since

[TABLE]

by Lemma 2.3, $T-T_{0}$ is a compact operator. So we have $T=T_{0}+K,$ where $K\in K\left(H\right).$

As a consequence of Theorem 2.1 we have the following:

Corollary 2.4**.**

Let $A\in B\left(H\right)$ and assume that $I-AA^{\ast}\in K\left(H\right)$ and $\left\|A^{\ast n}x\right\|\rightarrow 0$ for all $x\in H.$ If $T\in B\left(H\right),$ then the sequence $\left\{A^{\ast n}TA^{n}\right\}$ converges in operator norm if and only if we have the decomposition $T=T_{0}+K,$ where $A^{\ast}T_{0}A=T_{0}$ and $K\in K\left(H\right).$

If $S$ is the unilateral shift on $H^{2}$, then the operator $I-SS^{\ast}$ is one dimensional and $\left\|S^{\ast n}f\right\|\rightarrow 0$ for all $f\in H^{2}.$ By taking $A=S$ in Corollary 2.4, we obtain Feintuch’s result mentioned above.

Let an arbitrary $\varphi,\psi\in L^{\infty}$ be given. As we have noted in the Introduction, $T_{\varphi}T_{\psi}$ is a strongly asymptotically Toeplitz operator, that is, $S^{\ast n}T_{\varphi}T_{\psi}S^{n}\rightarrow T_{\varphi\psi}$ strongly [1, Theorem 4]. From this and from Corollary 2.4 it follows that $T_{\varphi}T_{\psi}$ is a uniformly asymptotically Toeplitz operator if and only if $T_{\varphi}T_{\psi}$ is a compact perturbation of the Toeplitz operator $T_{\varphi\psi}.$ Now, assume that one of the functions $\varphi,\psi$ is a trigonometric polynomial, say, $\psi=\sum_{-N}^{N}c_{k}e^{ik\theta}.$ Then as

[TABLE]

$S^{\ast n}T_{\varphi}S^{\ast k}S^{n}=S^{\ast k}T_{\varphi}$ $\left(\forall n\geq k\right)$ and $S^{\ast n}T_{\varphi}S^{k}S^{n}=T_{\varphi}S^{k}$ $\left(\forall k\geq 0\right),$ we have

[TABLE]

If $\varphi=\sum_{-N}^{N}c_{k}e^{ik\theta},$ then as $S^{\ast n}S^{\ast k}T_{\psi}S^{n}=S^{\ast k}T_{\psi}$ $\left(\forall k\geq 0\right)$ and $S^{\ast n}S^{k}T_{\psi}S^{n}=T_{\psi}S^{k}$ $\left(\forall n\geq k\right),$ we have

[TABLE]

Therefore, if one of the functions $\varphi,\psi$ is continuous, then $T_{\varphi}T_{\psi}$ is a uniformly asymptotically Toeplitz operator. Further, if $\psi$ has the form $\psi=h+f$, where $h\in H^{\infty}$ and $f\in C\left(\mathbb{T}\right),$ then as $T_{\varphi}T_{h}=T_{\varphi h}$ we get

[TABLE]

It follows that $T_{\varphi}T_{\psi}$ is a uniformly asymptotically Toeplitz operator for all $\varphi\in L^{\infty}$ and $\psi\in H^{\infty}+C\left(\mathbb{T}\right)$ (recall that the algebraic sum $H^{\infty}+C\left(\mathbb{T}\right)$ is a uniformly closed subalgebra of $L^{\infty}$ and sometimes called a Douglas algebra). Consequently, $T_{\varphi}T_{\psi}$ is a compact perturbation of the Toeplitz operator $T_{\varphi\psi}$ for all $\varphi\in L^{\infty}$ and $\psi\in H^{\infty}+C\left(\mathbb{T}\right).$ Similarly, we can see that if $\varphi$ has the form $\varphi=\overline{h}+f$, where $h\in H^{\infty}$ and $f\in C\left(\mathbb{T}\right),$ then $T_{\varphi}T_{\psi}$ is a uniformly asymptotically Toeplitz operator.

Note that in Corollary 2.4, compactness condition of the operator $I-AA^{\ast}$ is essential. To see this, let $A=V$ be the Volterra integral operator on $H=L^{2}\left[0,1\right].$ Then, $I-VV^{\ast}\notin K\left(H\right)$ and as $\left\|V^{n}\right\|\rightarrow 0,$ we have $\left\|V^{\ast n}x\right\|\rightarrow 0$ for all $x\in H.$ Since $\left\|V^{\ast n}TV^{n}\right\|\rightarrow 0$ for all $T\in B\left(H\right),$ the equation $V^{\ast}T_{0}V=T_{0}$ has only zero solution. If the conclusion of Corollary 2.4 were true, we would get $B\left(H\right)\subseteq K\left(H\right),$ which is a contradiction.

Let $H^{2}\left(E\right)$ be the Hardy space of all analytic functions on $\mathbb{D}$ with values in a Hilbert space $E.$ Let $A\in B\left(H\right)$ be a contraction, $E:=\overline{\left(I-AA^{\ast}\right)H}$ and assume that $\left\|A^{\ast n}x\right\|\rightarrow 0$ for all $x\in H.$ By the Model Theorem of Nagy-Foiaş (see, [19, Ch.VI, Theorem 2.3] and [21]), $A$ is unitary equivalent to its model operator

[TABLE]

where $\mathcal{K}=H^{2}\left(E\right)\ominus\Theta H^{2}\left(F\right),$ $F$ is a subspace of $E,$ $\Theta$ is a bounded analytic function on $\mathbb{D}$ with values in $B\left(F,E\right),$ the space of all bounded linear operators from $F$ into $E$ ($\Theta\left(\xi\right)$ is an isometry for almost all $\xi\in\mathbb{T}$), $P_{\mathcal{K}}$ is the orthogonal projection from $H^{2}\left(E\right)$ onto $\mathcal{K}$ and $S_{E}$ is the unilateral shift operator on $H^{2}\left(E\right).$ Notice also that $A_{\Theta}^{\ast}=S_{E}^{\ast}\mid_{\mathcal{K}}$. Consequently, Corollary 2.4 can be applied to the model operator $A_{\Theta}$ in the case when the operator $A$ satisfies the following conditions: 1) $A$ is a contraction; 2) $\left\|A^{\ast n}x\right\|\rightarrow 0$ for all $x\in H;$ 3) The defect operator $\mathcal{D}_{A^{\ast}}:=(I-AA^{\ast})^{\frac{1}{2}}$ is compact.

In addition, assume that $\left\|A^{n}x\right\|\rightarrow 0$ for all $x\in H.$ In this case, the subspace $E$ can be identified with $F$ and $\Theta\left(\xi\right)$ becomes unitary for almost all $\xi\in\mathbb{T}$. Consequently, Proposition 2.5 (shown below) is applicable to the model operator $A_{\Theta}$ in the case when the operator $A$ satisfies the following conditions: 1) $A$ is a contraction; 2) $\left\|A^{n}x\right\|\rightarrow 0$ and $\left\|A^{\ast n}x\right\|\rightarrow 0$ for all $x\in H;$ 3) the defect operator $\mathcal{D}_{A^{\ast}}$ is compact.

Proposition 2.5**.**

Let $A\in B\left(H\right)$ and assume that $I-AA^{\ast}\in K\left(H\right),$ $\left\|A^{n}x\right\|\rightarrow 0$ and $\left\|A^{\ast n}x\right\|\rightarrow 0$ for all $x\in H.$ For an arbitrary $T\in B\left(H\right),$ the following assertions are equivalent:

$\left(a\right)$* The sequence $\left\{A^{\ast n}TA^{n}\right\}$ converges in operator norm.*

$\left(b\right)$* $A^{\ast n}TA^{n}\rightarrow 0$ in operator norm.*

$\left(c\right)$* $T$ is a compact operator.*

Proof.

(a)$\Rightarrow$(b) By Corollary 2.4, $T=T_{0}+K$, where $A^{\ast}T_{0}A=T_{0}$ and $K\in K\left(H\right).$ On the other hand, by Lemma 2.3, $\left\|A^{\ast n}KA^{n}\right\|\rightarrow 0$. It remains to show that $T_{0}=0.$ Indeed, for an arbitrary $x,y\in H,$ from the identity $A^{\ast n}T_{0}A^{n}=T_{0}$ $\left(\forall n\in\mathbb{N}\right)$, we can write

[TABLE]

Hence $T_{0}=0.$

(b)$\Rightarrow$(c)$\Rightarrow$(a) are obtained from Lemma 2.3.

Recall that an operator $T\in B\left(X\right)$ is said to be almost periodic if for every $x\in X$, the orbit $\left\{T^{n}x:n\in\mathbb{N}\right\}$ is relatively compact. Clearly, an almost periodic operator is power bounded, that is,

[TABLE]

If $T\in B\left(X\right)$ is an almost periodic operator, then by the Jacobs-Glicksberg-de Leeuw decomposition theorem [7, Ch.I, Theorem 1.15], every $x\in X$ can be written as $x=x_{0}+x_{1},$ where $\left\|T^{n}x_{0}\right\|\rightarrow 0$ and $x_{1}\in\overline{\text{span}}\left\{y\in X:\exists\xi\in\mathbb{T},\text{ }Ty=\xi y\right\}.$

From now on, for a given $T\in B\left(X\right)$ the left and right multiplication operators on $B\left(X\right)$ will be denoted by $L_{T}$ and $R_{T},$ respectively.

The following result is an improvement of Proposition 2.5.

Proposition 2.6**.**

Let $A\in B\left(H\right)$ and assume that $I-AA^{\ast}\in K\left(H\right),$ $\left\|A^{n}x\right\|\rightarrow 0$ and $\left\|A^{\ast n}x\right\|\rightarrow 0$ for all $x\in H.$ For an arbitrary $T\in B\left(H\right)$ the following assertions are equivalent:

$\left(a\right)$* $\left\{A^{\ast n}TA^{n}:n\in\mathbb{N}\right\}$ is relatively compact in the operator norm topology.*

$\left(b\right)$* $A^{\ast n}TA^{n}\rightarrow 0$ in operator norm.*

$\left(c\right)$* $T$ is a compact operator.*

Proof.

(a)$\Rightarrow$(b) Let $E$ be the set of all $Q\in B\left(H\right)$ such that

[TABLE]

is relatively compact in the operator norm topology. By the uniform boundedness principle, the operator $L_{A^{\ast}}R_{A}$ is power bounded and therefore $E$ is a closed (in operator norm) $L_{A^{\ast}}R_{A}-$invariant subspace. Consequently, $L_{A^{\ast}}R_{A}\mid_{E},$ the restriction of $L_{A^{\ast}}R_{A}$ to $E$ is an almost periodic operator. Since $T\in E,$ by the Jacobs-Glicksberg-de Leeuw decomposition theorem, $T=T_{0}+T_{1},$ where

[TABLE]

and

[TABLE]

We must show that $T_{1}=0.$ For this, it suffices to show that the identity $A^{\ast}QA=\xi Q$ $\left(\xi\in\mathbb{T}\right)$ implies $Q=0.$ Indeed, since

[TABLE]

we get

[TABLE]

Hence $Q=0$.

(b)$\Rightarrow$(c)$\Rightarrow$(a) are obtained from Lemma 2.3.

Next, we have the following:

Theorem 2.7**.**

Let $A$ and $B^{\ast}$ be two essentially isometric contractions on $H$ and assume that $\left\|A^{n}x\right\|\rightarrow 0$ and $\left\|B^{\ast n}x\right\|\rightarrow 0$ for all $x\in H.$ Then, for an arbitrary $T\in B\left(H\right)$ we have

[TABLE]

Proof.

If $K\in K\left(H\right),$ then by Lemma 2.3, $\left\|A^{n}KB^{n}\right\|\rightarrow 0.$ Since

[TABLE]

we have

[TABLE]

For the reverse inequality, let $\widehat{A}$, $\widehat{T}$ and $\widehat{B}$ be the limit operators associated with $A$, $T$ and $B,$ respectively. By Proposition 2.2, $\widehat{A}$ and $\widehat{B}^{\ast}$ are isometries. By using the same proposition again, we can write

[TABLE]

Thus we have

[TABLE]

We know [5, Corollary 7.13] that every Toeplitz operator $T_{\varphi}$ with symbol $\varphi\in L^{\infty}$ satisfies

[TABLE]

As a consequence of Theorem 2.7, we have the following generalization of the preceding formula.

Corollary 2.8**.**

Let $A\in B\left(H\right)$ be a contraction and assume that $I-AA^{\ast}\in K\left(H\right)$ and $\left\|A^{\ast n}x\right\|\rightarrow 0$ for all $x\in H.$ Then, for an arbitrary $T\in B\left(H\right)$ we have

[TABLE]

For an arbitrary $A,B\in B\left(H\right),$ we put

[TABLE]

Proposition 2.9**.**

Assume that the operators $A,B$ satisfy the hypotheses of Theorem 2.7. Then, for an arbitrary $K\in K\left(H\right)$ we have

[TABLE]

In the case $AB=I,$ this estimate is the best possible.

Proof.

Assume that there exists $K\in K\left(H\right)$ such that

[TABLE]

Then there exists $T\in\mathcal{I}_{A,B}$ such that

[TABLE]

By Theorem 2.7, $\left\|T\right\|=\left\|T+K\left(H\right)\right\|$ which implies $\left\|T\right\|\leq\left\|K+T\right\|$. Consequently, we can write

[TABLE]

which is a contradiction.

In the case $AB=I,$ we have $I\in\mathcal{I}_{A,B}.$ If $K=x\otimes x,$ where $\left\|x\right\|=1$ and $T=-\frac{1}{2}I$, then $\left\|K+T\right\|=\frac{1}{2}.$

Let $\mathcal{T}$ be the space of all Toeplitz operators. By taking $A=S^{\ast}$ and $B=S$ in Proposition 2.9, we have

[TABLE]

where this estimate is the best possible.

3. One dimensional model and the Hartman-Sarason theorem

Using the results of the preceding section, here we give a quantitative generalization of the Hartman-Sarason theorem.

Recall that a contraction $T$ on $H$ is said to be completely non-unitary if it has no proper reducing subspace on which it acts as a unitary operator. If $T$ is a completely non-unitary contraction, then $f\left(T\right)$ $\left(f\in H^{\infty}\right)$ can be defined by the Nagy-Foiaş functional calculus [19, Ch.III].

Let $T$ be a contraction on $H$ and assume that

[TABLE]

In addition, if

[TABLE]

then by the Model Theorem of Nagy-Foiaş [19, Ch.VI, Theorem 2.3] (see also, [21]) $T$ is unitary equivalent to its model operator

[TABLE]

acting on the model space

[TABLE]

where $\theta$ is an inner function (a function $\theta$ in $H^{\infty}$ is an inner function if $\left|\theta\right|=1$ a.e. on $\mathbb{T}$) and $P_{\theta}$ is the orthogonal projection from $H^{2}$ onto $H_{\theta}^{2}$. Beurling’s theorem [5, Corollary 6.11] says that these spaces are generic invariant subspaces for the backward shift operator

[TABLE]

Notice that

[TABLE]

Let $\theta$ be an inner function and let $S_{\theta}$ be the model operator on the model space $H_{\theta}^{2}.$ For an arbitrary $f\in H^{\infty}$, we can define the operator

[TABLE]

which is unitary equivalent to $f\left(T\right).$ The map $f\mapsto f\left(S_{\theta}\right)$ is linear, multiplicative and by the Nehari formula [21, p.235],

[TABLE]

Let us mention Sarason’s theorem [21, p.230] which asserts that an operator $Q\in B\left(H_{\theta}^{2}\right)$ is a commutant of $S_{\theta}$ if and only if $Q=f\left(S_{\theta}\right)$ for some $f\in H^{\infty}$.

Let us also mention that the classical theorem of Hartman and Sarason [21, p.235] classifies compactness of the operators $f\left(S_{\theta}\right).$ The operator $f\left(S_{\theta}\right)$ $\left(f\in H^{\infty}\right)$ is compact if and only if $\overline{\theta}f\in H^{\infty}+C\left(\mathbb{T}\right).$

We have the following quantitative generalization of the Hartman-Sarason theorem.

Theorem 3.1**.**

Let $\theta$ be an inner function and let $S_{\theta}$ be the model operator on the model space $H_{\theta}^{2}.$ Then, for an arbitrary $f\in H^{\infty}$ we have

[TABLE]

For the proof, we need several lemmas.

Lemma 3.2**.**

Let $\left\{E_{n}\right\}$ be an increasing sequence of closed subspaces of a Banach space $X.$ Then, for an arbitrary $x\in X$ we have

[TABLE]

Proof.

If $x\in X,$ then the sequence $\left\{\text{dist}\left(x,E_{n}\right)\right\}$ is decreasing. Let

[TABLE]

Since

[TABLE]

we have

[TABLE]

which implies

[TABLE]

If

[TABLE]

then $\left\|x-x_{0}\right\|<\alpha$ for some $x_{0}\in\bigcup_{n=1}^{\infty}E_{n}.$ Consequently, $x_{0}\in E_{n_{0}}$ for some $n_{0}.$ Hence dist$\left(x,E_{n_{0}}\right)<\alpha.$ This contradicts dist$\left(x,E_{n_{0}}\right)\geq\alpha.$

Lemma 3.3**.**

For an arbitrary $\varphi\in L^{\infty}$ we have

[TABLE]

Proof.

We know [5, Proposition 6.36] that $H^{\infty}+C\left(\mathbb{T}\right)$ is a uniformly closed subalgebra* *of $L^{\infty}$ generated by $\overline{z}$ and $H^{\infty}.$ If $E_{n}:=\overline{z}^{n}H^{\infty},$ then $\left\{E_{n}\right\}$ is an increasing sequence of closed subspaces of $L^{\infty}$. Since

[TABLE]

and

[TABLE]

we have

[TABLE]

Applying Lemma 3.2 to the subspaces $\left\{E_{n}\right\},$ we obtain our result.

Now, we can prove Theorem 3.1.

Proof of Theorem 3.1.

As we have noted above, the model operator $S_{\theta}$ is an essentially unitary contraction. Moreover, $\left\|S_{\theta}^{n}h\right\|\rightarrow 0$ and $\left\|S_{\theta}^{\ast n}h\right\|\rightarrow 0$ for all $h\in H_{\theta}^{2}.$ If $T\in B\left(H_{\theta}^{2}\right),$ then by Theorem 2.7,

[TABLE]

which implies

[TABLE]

If $K\in K\left(H_{\theta}^{2}\right),$ then as $\left\|KS_{\theta}^{n}\right\|\rightarrow 0$ (see, the proof of Lemma 2.3) we get

[TABLE]

It follows that

[TABLE]

Thus we have

[TABLE]

In particular, taking $T=f\left(S_{\theta}\right)$ we obtain

[TABLE]

Further, by the Nehari formula we can write

[TABLE]

On the other hand, by Lemma 3.3,

[TABLE]

Now, taking into account (3.1), finally we obtain

[TABLE]

The proof is complete.

Below, we present some applications of Theorem 3.1.

Let $X$ be a Banach space. As usual, $\sigma\left(T\right)$ will denote the spectrum of the operator $T\in B\left(X\right).$ Given $T\in B\left(X\right),$ we let $A_{T}$ denote the closure in the uniform operator topology of all polynomials in $T.$ Then, $A_{T}$ is a commutative unital Banach algebra. The Gelfand space of $A_{T}$ can be identified with $\sigma_{A_{T}}\left(T\right)$, the spectrum of $T$ with respect to the algebra $A_{T}$. Since $\sigma\left(T\right)$ is a (closed) subset of $\sigma_{A_{T}}\left(T\right),$ for every $\lambda\in\sigma\left(T\right),$ there is a multiplicative functional $\phi_{\lambda}$ on $A_{T}$ such that $\phi_{\lambda}\left(T\right)=\lambda$. By $\widehat{Q}$ we will denote the Gelfand transform of $Q\in A_{T}$. Instead of $\widehat{Q}\left(\phi_{\lambda}\right)\left(=\phi_{\lambda}\left(Q\right)\right),$ where $\lambda\in\sigma\left(T\right),$ we will use the notation $\widehat{Q}\left(\lambda\right).$ It follows from the Shilov Theorem [5, Theorem 2.54] that if $T$ is a contraction, then

[TABLE]

The following result was obtained in [17].

Theorem 3.4**.**

If $T$ is a contraction on a Hilbert space, then for an arbitrary $Q\in A_{T}$ we have

[TABLE]

For a non-empty closed subset $\Gamma$ of $\mathbb{T}$, by $H_{\Gamma}^{\infty}$ we will denote the set of all those functions $f$ in $H^{\infty}$ that have a continuous extension $\widetilde{f}$ to $\mathbb{D}\cup\Gamma$. Clearly, $H_{\Gamma}^{\infty}$ is a closed subspace of $H^{\infty}.$ It follows from the general theory of $H^{p}$ spaces that if $\Gamma$ has positive Lebesgue measure and $f\in H_{\Gamma}^{\infty}$ is not identically zero, then $\widetilde{f}$ cannot vanish identically on $\Gamma.$

If $T$ is a contraction on a Hilbert space $H,$ then there is a canonical decomposition of $H$ into two $T-$reducing subspaces $H=H_{0}\oplus H_{u}$ such that $T_{0}:=T\mid_{H_{0}}$ is completely non-unitary and $T_{u}:=T\mid_{H_{u}}$ is unitary [19, Ch.I, Theorem 3.2]. It can be seen that

[TABLE]

Let $f$ be in $H_{\sigma\left(T\right)\cap\mathbb{T}}^{\infty}$ with continuous extension $\widetilde{f}$ to $\mathbb{D}\cup(\sigma\left(T\right)\cap\mathbb{T)}$. As in [10], we can define $f\left(T\right)\in B\left(H\right)$ by

[TABLE]

where $f\left(T_{0}\right)$ is given by the Nagy-Foias functional calculus and

[TABLE]

It can be seen that

[TABLE]

Further, by the Gamelin-Garnett theorem [9], there exists a sequence $\left\{f_{n}\right\}$ in $H^{\infty}$ such that each $f_{n}$ has an analytic extension $g_{n}$ to a neighborhood $O_{n}$ of $\mathbb{D}\cup(\sigma\left(T\right)\cap\mathbb{T)}$ and

[TABLE]

Then, $g_{n}\left(T\right)$ can be defined by the Riesz-Dunford functional calculus. Since $f_{n}\left(T\right)=g_{n}\left(T\right)\in A_{T}$ and

[TABLE]

we have that $f\left(T\right)\in A_{T}$. Moreover,

[TABLE]

As a consequence of Theorem 3.4 we have the following:

Corollary 3.5**.**

Let $T$ be a contraction on a Hilbert space. If $f\in H_{\sigma\left(T\right)\cap\mathbb{T}}^{\infty}$ with continuous extension $\widetilde{f}$ to $\mathbb{D}\cup(\sigma\left(T\right)\cap\mathbb{T)},$ then

[TABLE]

Now, let $\theta$ be an inner function and let $S_{\theta}$ be the model operator on the model space $H_{\theta}^{2}.$ We put

[TABLE]

It follows from the Lipschitz-Moeller theorem [21, p.81] that

[TABLE]

If $f\in H_{\Sigma_{u}\left(\theta\right)}^{\infty}$ with continuous extension $\widetilde{f}$ to $\mathbb{D}\cup\Sigma_{u}\left(\theta\right),$ then by Corollary 3.5,

[TABLE]

On the other hand, by (3.1),

[TABLE]

Thus we have

[TABLE]

From Theorem 3.1 and from the preceding identity we have the following:

Corollary 3.6**.**

Let $\theta$ be an inner function and let $S_{\theta}$ be the model operator on the model space $H_{\theta}^{2}.$ For an arbitrary $f\in H_{\Sigma_{u}\left(\theta\right)}^{\infty}$ with continuous extension $\widetilde{f}$ to $\mathbb{D}\cup\Sigma_{u}\left(\theta\right),$ we have

[TABLE]

4. The sequence $\left\{\frac{1}{n}\sum_{i=0}^{n-1}A^{i}TB^{i}\right\}$

In this section, we give some results concerning convergence in operator norm of the sequence $\left\{\frac{1}{n}\sum_{i=0}^{n-1}A^{i}TB^{i}\right\}$ for Hilbert space operators.

Let $X$ be a Banach space. It is easy to check that if $T\in B\left(X\right)$ is power bounded, then

[TABLE]

The following result is well known (for instance, see [12, Ch.2, §2.1, Theorems 1.2 and 1.3]).

Proposition 4.1**.**

Let $T\in B\left(X\right)$ be power bounded and let $E$ be the set of all $x\in X$ such that the sequence $\left\{\frac{1}{n}\sum_{i=0}^{n-1}T^{i}x\right\}$ converges strongly. Then, we have the decomposition

[TABLE]

If $X$ is reflexive, then $E=X.$

Applying Proposition 4.1 to the operator $L_{A}R_{B}$ on the space $B\left(X\right)$, we have the following:

Corollary 4.2**.**

Let $A,B\in B\left(X\right)$ be two operators such that $\sup_{n\geq 0}\left(\left\|A^{n}\right\|\left\|B^{n}\right\|\right)<\infty$ and $T\in B\left(X\right).$ Then, the sequence $\left\{\frac{1}{n}\sum_{i=0}^{n-1}A^{i}TB^{i}\right\}$ converges in operator norm if and only if we have the decomposition $T=T_{0}+Q,$ where

[TABLE]

Lemma 4.3**.**

Let $T\in B\left(X\right)$ be power bounded, $x\in X$ and assume that

[TABLE]

$\left(a\right)$* If the sequence $\left\{\frac{1}{n}\sum_{i=0}^{n-1}T^{i}x\right\}$ converges strongly, then the sequence $\left\{T^{n}x\right\}$ converges strongly $($to same element$),$ too.*

$\left(b\right)$* If $X$ is reflexive, then the sequence $\left\{T^{n}x\right\}$ converges strongly.*

Proof.

(a) Notice that

[TABLE]

is a closed $T-$invariant subspace and $x\in F.$ Since $T$ is power bounded and

[TABLE]

we have $\left\|T^{n}y\right\|\rightarrow 0$ for all $y\in\overline{\left(T-I\right)F}.$ Now, let $E$ be the set of all $y\in F$ such that the sequence $\left\{\frac{1}{n}\sum_{i=0}^{n-1}T^{i}y\right\}$ converges strongly. Since $x\in E,$ by Proposition 4.1 we have the decomposition $x=x_{0}+y_{0},$ where $x_{0}\in\overline{\left(T-I\right)F}$ and $Ty_{0}=y_{0}.$ As $T^{n}x=T^{n}x_{0}+y_{0}$ and $\left\|T^{n}x_{0}\right\|\rightarrow 0,$ we have $\left\|T^{n}x-y_{0}\right\|\rightarrow 0.$ Clearly, $\frac{1}{n}\sum_{i=0}^{n-1}T^{i}x\rightarrow y_{0}$ strongly.

(b) If $X$ is reflexive, then by Proposition 4.1 the sequence $\left\{\frac{1}{n}\sum_{i=0}^{n-1}T^{i}x\right\}$ converges strongly for every $x\in X.$ By (a), the sequence $\left\{T^{n}x\right\}$ converges strongly.

Next, we have the following:

Theorem 4.4**.**

Let $A$ and $B^{\ast}$ be two essentially isometric operators on $H$ and $T\in B\left(H\right).$ Assume that:

$\left(i\right)\ \left\|A^{n}x\right\|\rightarrow 0$* and $\left\|B^{\ast n}x\right\|\rightarrow 0$ for all $x\in H;$*

$\left(ii\right)$* $ATB-T\in K\left(H\right).$*

Then, the sequence $\left\{\frac{1}{n}\sum_{i=0}^{n-1}A^{i}TB^{i}\right\}$ converges in operator norm if and only if we have the decomposition $T=T_{0}+K,$ where $AT_{0}B=T_{0}$ and $K\in K\left(H\right).$

Proof.

Assume that the sequence $\left\{\frac{1}{n}\sum_{i=0}^{n-1}A^{i}TB^{i}\right\}$ converges in operator norm. Since $ATB-T\in K\left(H\right),$ by Lemma 2.3,

[TABLE]

Notice also that the operator $L_{A}R_{B}$ is power bounded. Applying Lemma 4.3 to the operator $L_{A}R_{B}$ on the space $B\left(X\right)$, we obtain that the sequence $\left\{A^{n}TB^{n}\right\}$ converges in operator norm. By Theorem 2.1, $T=T_{0}+K,$ where $AT_{0}B=T_{0}$ and $K\in K\left(H\right).$

If $T=T_{0}+K,$ where $AT_{0}B=T_{0}$ and $K\in K\left(H\right),$ then we have

[TABLE]

By Lemma 2.3, $\left\|A^{n}KB^{n}\right\|\rightarrow 0$ and therefore $\left\|\frac{1}{n}\sum_{i=0}^{n-1}A^{i}KB^{i}\right\|\rightarrow 0.$ Thus

[TABLE]

Corollary 4.5**.**

Assume that the operators $A,T\in B\left(H\right)$ satisfy the following conditions:

$\left(i\right)$* $I-AA^{\ast}\in K\left(H\right);$*

$\left(ii\right)$* $\left\|A^{\ast n}x\right\|\rightarrow 0$ for all $x\in H;$*

$\left(iii\right)$* $A^{\ast}TA-T\in K\left(H\right).$*

Then, the sequence $\left\{\frac{1}{n}\sum_{i=0}^{n-1}A^{\ast i}TA^{i}\right\}$ converges in operator norm if and only if we have the decomposition $T=T_{0}+K,$ where $A^{\ast}T_{0}A=T_{0}$ and $K\in K\left(H\right).$

The proof of the following lemma is straightforward and will be omitted.

Lemma 4.6**.**

Let $U$ be an essentially unitary operator on $H.$ Then, $T\in B\left(H\right)$ is an essential commutant of $U$ if and only if $U^{\ast}TU-T\in K\left(H\right).$

Recall that $T\in B\left(H^{2}\right)$ is an essentially Toeplitz operator if

[TABLE]

By Lemma 4.6, $T\in B\left(H^{2}\right)$ is an essentially Toeplitz operator if and only if $T$ is an essential commutant of the unilateral shift operator $S.$ On the other hand, essential commutant of the unilateral shift is a $C^{\ast}-$algebra. Consequently, the set of all essentially Toeplitz operators is a $C^{\ast}-$algebra and therefore contains the $C^{\ast}-$algebra generated by all Toeplitz operators.

Corollary 4.7**.**

An essentially Toeplitz operator $T$ is a compact perturbation of a Toeplitz operator if and only if the sequence $\left\{\frac{1}{n}\sum_{i=0}^{n-1}S^{\ast i}TS^{i}\right\}$ converges in operator norm.

In [20], it was proved that if the composition operator $C_{\phi}$ on $H^{2}$ is neither compact nor the identity, then $C_{\phi}$ cannot be compact perturbation of a Toeplitz operator.

Corollary 4.8**.**

If $C_{\phi}$ is a composition operator on $H^{2},$ then the sequence

[TABLE]

converges in operator norm if and only if either $C_{\phi}$ is compact or the identity operator.

Recall that the class of compact composition operators are sufficiently large (for instance, see [16]).

Following [15], we could define an asymptotic Toeplitz operator in the Calkin algebra as an operator $T\in B\left(H^{2}\right)$ such that the sequence $\left\{S^{\ast n}TS^{n}\right\}$ converges in the Calkin algebra.

The following result, which seems to be unnoticed (see, [15, p.745]).

Proposition 4.9**.**

Every asymptotic Toeplitz operator in the Calkin algebra is an essentially Toeplitz operator.

Proof.

If $T\in B\left(H^{2}\right)$ is an asymptotic Toeplitz operator in the Calkin algebra,* *then there is an operator $Q\in B\left(H^{2}\right)$ such that

[TABLE]

Let $\widehat{S^{\ast}}$, $\widehat{T}$, $\widehat{S}$ and $\widehat{Q}$ be the limit operators associated with $S^{\ast}$, $T$, $S$ and $Q,$ respectively. By Proposition 2.2,

[TABLE]

Since

[TABLE]

we have $\widehat{S^{\ast}}\widehat{Q}\widehat{S}=\widehat{Q}.$ By using the same proposition again, we obtain that $S^{\ast}QS-Q\in K\left(H^{2}\right).$

5. Banach space operators

In this section, we study convergence in operator norm of the sequence $\left\{A^{n}TB^{n}\right\}$ for Banach space operators.

Let $X$ be a Banach space. For an arbitrary $T\in B\left(X\right)$ and $x\in X$, we define $\rho_{T}\left(x\right)$ to be the set of all $\lambda\in\mathbb{C}$ for which there exists a neighborhood $U_{\lambda}$ of $\lambda$ with $u\left(z\right)$ analytic on $U_{\lambda}$ having values in $X$ such that

[TABLE]

This set is open and contains the resolvent set $\rho\left(T\right)$ of $T$. By definition, the local spectrum of $T$ at $x\in X$, denoted by $\sigma_{T}\left(x\right),$ is the complement of $\rho_{T}\left(x\right)$, so it is a compact subset of $\sigma\left(T\right)$. This object is the most tractable if the operator $T$ has the *single-valued extension property *(SVEP), i.e., for every open set $U$ in $\mathbb{C},$ the only analytic function $u:U\rightarrow X$ for which the equation $\left(zI-T\right)u\left(z\right)=0$ holds is the constant function $u\equiv 0$. If $T$ has SVEP, then $\sigma_{T}\left(x\right)\neq\emptyset,$ whenever $x\in X\diagdown\left\{0\right\}$ [13, Proposition 1.2.16]. Note that the local spectrum of $T$ may be ”very small” with respect to its usual spectrum. To see this, let $\sigma$ be a ”small” clopen part of $\sigma\left(T\right)$. Let $P_{\sigma}$ be the spectral projection associated with $\sigma$ and $X_{\sigma}:=P_{\sigma}X$. Then, $X_{\sigma}$ is a closed $T-$invariant subspace of $X$ and $\sigma\left(T\mid_{X_{\sigma}}\right)=\sigma$. It is easy to see that $\sigma_{T}\left(x\right)\subseteq\sigma$ for every $x\in X_{\sigma}$.

If $T$ is power bounded, then clearly, $\sigma\left(T\right)\subset\overline{\mathbb{D}\text{ }}$and $\sigma_{T}\left(x\right)\cap\mathbb{T}$ consists of all $\xi\in\mathbb{T}$ such that the function $z\rightarrow\left(zI-T\right)^{-1}x$ $\left(\left|z\right|>1\right)$ has no analytic extension to a neighborhood of $\xi$.

Lemma 5.1**.**

Let $T\in B\left(X\right)$, $x\in X$ and assume that $\sup_{n\geq 0}\left\|T^{n}x\right\|<\infty.$ Then, $\sigma_{T}\left(x\right)\subseteq\overline{\mathbb{D}}.$

Proof.

Consider the function

[TABLE]

which is analytic on $\mathbb{C}\diagdown\overline{\mathbb{D}}$ and $\left(zI-T\right)u\left(z\right)=x$ for all $z\in\mathbb{C}\diagdown\overline{\mathbb{D}}.$ This shows that $\mathbb{C}\diagdown\overline{\mathbb{D}}\subseteq\rho_{T}\left(x\right)$ and therefore $\sigma_{T}\left(x\right)\subseteq\overline{\mathbb{D}}.$

We mention the following classical result of Katznelson and Tzafriri [11, Theorem 1]: If $T\in B\left(X\right)$ is power bounded, then $\lim_{n\rightarrow\infty}\left\|T^{n+1}-T^{n}\right\|=0$ if and only if $\sigma\left(T\right)\cap\mathbb{T}\subseteq\left\{1\right\}.$

We have the following local version of the Katznelson-Tzafriri theorem [18, Theorem 4.2].

Theorem 5.2**.**

Let $T\in B\left(X\right)$, $x\in X$ and assume that $\sup_{n\geq 0}\left\|T^{n}x\right\|<\infty.$ If $\sigma_{T}\left(x\right)\cap\mathbb{T}\subseteq\left\{1\right\},$ then

[TABLE]

Note that in contrast with the Katznelson-Tzafriri theorem, the converse of Theorem 5.2 does not hold, in general. Indeed, if $S^{\ast}$ is the backward shift operator on $H^{2},$ then as $\left\|S^{\ast n}f\right\|\rightarrow 0,$ we have

[TABLE]

On the other hand, since

[TABLE]

$\sigma_{S^{\ast}}\left(f\right)\cap\mathbb{T}$ consists of all $\xi\in\mathbb{T}$ for which the function $f$ has no analytic extension to a neighborhood of $\xi$ (see, [6, p.24]).

Theorem 5.2 combined with Lemma 4.3 yields the next result.

Theorem 5.3**.**

Assume that $T\in B\left(X\right)$ and $x\in X$ satisfy the following conditions:

$\left(i\right)$* $\sup_{n\geq 0}\left\|T^{n}x\right\|<\infty$;*

$\left(ii\right)$* $\sigma_{T}\left(x\right)\cap\mathbb{T}\subseteq\left\{1\right\}.$*

If the sequence $\left\{\frac{1}{n}\sum_{i=0}^{n-1}T^{i}x\right\}$ converges strongly to $y\in X,$ then $T^{n}x\rightarrow y$ strongly.

Corollary 5.4**.**

Let $T\in B\left(X\right)$ and let $x\in X$ be such that $\sup_{n\geq 0}\left\|T^{n}x\right\|<\infty.$ Let

[TABLE]

and assume that the sequence $\left\{\frac{1}{n}\sum_{i=0}^{n-1}S^{i}x\right\}$ converges strongly to $y\in X.$ Then, $S^{n}x\rightarrow y$ strongly.

Proof.

It is easy to check that

[TABLE]

Notice also that if

[TABLE]

then $f\left(1\right)=1$ and $\left|f\left(z\right)\right|<1$ for all $z\in\overline{\mathbb{D}}\diagdown\left\{1\right\}.$ On the other hand, by [13, Theorem 3.3.8],

[TABLE]

Since $\sigma_{T}\left(x\right)\subseteq\overline{\mathbb{D}}$ (Lemma 5.1), we have $\sigma_{S}\left(x\right)\cap\mathbb{T}\subseteq\left\{1\right\}.$ By Theorem 5.3, $S^{n}x\rightarrow y$ strongly.

We put

[TABLE]

As another application of Theorem 5.3, we have the following:

Theorem 5.5**.**

Assume that the operators $A,T,B\in B\left(X\right)$ satisfy the following conditions:

$\left(i\right)\sup_{n\geq 0}\left\|A^{n}TB^{n}\right\|<\infty;$**

$\left(ii\right)$* either $\sigma\left(A\right)\subset D_{+}$ and $\sigma\left(B\right)\subset D_{-}$ or $\sigma\left(A\right)\subset D_{-}$ and $\sigma\left(B\right)\subset D_{+}.$*

If the sequence $\left\{\frac{1}{n}\sum_{i=0}^{n-1}A^{i}TB^{i}\right\}$ converges in operator norm to $Q\in B\left(X\right),$ then $A^{n}TB^{n}\rightarrow Q$ in operator norm.

Proof.

Since

[TABLE]

by Lemma 5.1,

[TABLE]

On the other hand, by the Lumer-Rosenblum theorem [14, Theorem 10],

[TABLE]

which implies

[TABLE]

Thus we have

[TABLE]

Applying Theorem 5.3 to the operator $L_{A}R_{B}$ on the space $B\left(X\right)$, we obtain that

[TABLE]

Next, we will show that the hypothesis $\sigma_{T}\left(x\right)\cap\mathbb{T}\subseteq\left\{1\right\}$ in Theorem 5.3 is the best possible, in general.

Let $N$ be a normal operator on a Hilbert space $H$ with the spectral measure $P$ and $x\in H.$ Define a measure $\mu_{x}$ on $\sigma\left(N\right)$ by

[TABLE]

It follows from the Spectral Theorem that $\sigma\left(N\right)=$supp$P$ and $\sigma_{N}\left(x\right)=$supp$\mu_{x}$. It is easy to check that if $N$ is a contraction (a normal operator is power bounded if and only if it is a contraction) then,

[TABLE]

Proposition 5.6**.**

Let $N$ be a normal contraction operator on $H$ with the spectral measure $P$ and $x\in H.$ The sequence $\left\{N^{n}x\right\}$ converges strongly if and only if

[TABLE]

In this case, $N^{n}x\rightarrow P\left(\left\{1\right\}\right)x$ strongly.

Proof.

Let $\mu_{x}$ be the measure on $\sigma\left(N\right)$ defined by (5.1). We can write

[TABLE]

It follows that $\left\|N^{n+1}x-N^{n}x\right\|\rightarrow 0$ if and only if

[TABLE]

By Lemma 4.3 the sequence $\left\{N^{n}x\right\}$ converges strongly if and only if

[TABLE]

By (5.2),

[TABLE]

Let $W^{\ast}\left(N\right)$ be the von Neumann algebra generated by $N.$ Recall that $x\in H$ is a separating vector for $N$ if the only operator $A$ in $W^{\ast}\left(N\right)$ such that $Ax=0$ is $A=0$. As is known [4, Ch.IX, Section 8.1], each normal operator has a separating vector. If $x\in H$ is a separating vector for $N,$ then the spectral measure of $N$ and the measure $\mu_{x}$ are mutually absolutely continuous [4, Ch.IX, Proposition 8.3], where $\mu_{x}$ is defined by (5.1).

Corollary 5.7**.**

If $x$ is a separating vector for $N,$ then the sequence $\left\{N^{n}x\right\}$ converges strongly if and only if

[TABLE]

Now, let $K$ be a compact subset of $\overline{\mathbb{D}}$ such that $1\in K$ and let $\nu$ be a regular positive Borel measure in $\mathbb{C}$ with support $K.$ Define the operator $N$ on $L^{2}\left(K,\nu\right)$ by $Nf=zf.$ Then, $N$ is a normal contraction on $L^{2}\left(K,\nu\right)$ and $\sigma\left(N\right)=K.$ Moreover,

[TABLE]

where $\chi_{\Delta}$ is the characteristic function of $\Delta.$ It can be seen that the identity one function $\mathbf{1}$ on $K$ is a separating vector for $N$ and $\sigma\left(N\right)=\sigma_{N}\left(\mathbf{1}\right).$ By (5.3), the sequence $\left\{N^{n}\mathbf{1}\right\}$ converges strongly if and only if $\chi_{\sigma_{N}\left(\mathbf{1}\right)\cap\mathbb{T}}=\chi_{\left\{1\right\}}$ or $\sigma_{N}\left(\mathbf{1}\right)\cap\mathbb{T}=\left\{1\right\}.$