Operators Whose Conjugation Orbits Satisfy Polynomial Growth Conditions

TL;DR

This paper characterizes operators whose conjugation orbits grow polynomially, especially when the spectrum of the base operator is a single point, linking to decomposability and providing commutator estimates.

Contribution

It provides a complete description of the class of operators with polynomial growth conjugation orbits when the spectrum is a single point, connecting to operator decomposability.

Findings

Complete characterization of the class _A^(\u211d) for single-point spectrum

Link between polynomial growth and operator decomposability

Estimates for the norm of the commutator in certain cases

Abstract

Let $A$ be a bounded linear operator on a complex Banach space $X.$ For a given $\alpha \geq 0,$ we consider the class $\mathcal{D}_{A}^{\alpha }\left( \mathbb{R} \right) $ of all bounded linear operators $T$ on $X$ for which there exists a constant $C_{T}>0$, such that \begin{equation*} \left\Vert e^{tA}Te^{-tA}\right\Vert \leq C_{T}\left( 1+\left\vert t\right\vert \right) ^{\alpha }, \text {} \forall t\in \mathbb{R} \end{equation*} We present complete description of the class $\mathcal{D}_{A}^{\alpha }\left( \mathbb{R} \right) $ in the case when the spectrum of $A$ consists of one point. These results are linked to the decomposability of $A.$ Some estimates for the norm of the commutator $AT-TA$ are obtained in the case $0\leq \alpha <1.$

Equations

$$\Vert e^{tA}T{e}^{-tA}\Vert \le C_T{\left(1+\left|t\right|\right)}^{\alpha },\forall t\in \mathbb{R}.$$

$$\text{Alg}\left\{E\right\}=\left\{T\in B\left(H\right):TE\subseteq E\text{, }\forall E\in \left\{E\right\}\right\}.$$

$${\mathcal{D}}_A:=\left\{T\in B\left(H\right):\underset{n\ge 0}{\sup }\Vert A^{n}T{A}^{-n}\Vert <\infty \right\}.$$

$$t\to \Vert e^{tA}T{e}^{-tA}\Vert$$

$$\Vert e^{tA}T{e}^{-tA}\Vert \le C_T{\left(1+\left|t\right|\right)}^{\alpha },\forall t\in \mathbb{R}.$$

$${\Delta }_A:T\mapsto AT-TA,T\in B\left(X\right).$$

$${\Delta }_A^n\left(T\right)=\sum _{k=0}^n{\left(-1\right)}^k\left(\genfrac{}{}{0ex}{}{n}{k}\right)A^{n-k}T{A}^k\left(n\in \mathbb{N}\right).$$

$${\mathcal{D}}_A^{\alpha }\left(\mathbb{R}\right)=\ker {\Delta }_A^{\left[\alpha \right]+1}.$$

$$r_T\left(x\right)=\sup \left\{\left|\lambda \right|:\lambda \in {\sigma }_T\left(x\right)\right\}.$$

$$r_T\left(x\right)=\underset{n\to \infty }{\overline{\lim }}{\Vert T^{n}x\Vert }^{\frac{1}{n}}$$

$${\Vert f\Vert }_{1,\omega }={\int }_{\mathbb{R}}\left|f\left(t\right)\right|\omega \left(t\right)dt<\infty .$$

$${\Vert g\Vert }_{\omega ,\infty }:=\text{ess}\underset{t\in \mathbb{R}}{\sup }\frac{\left|g\left(t\right)\right|}{\omega \left(t\right)}<\infty .$$

$$⟨g,f⟩={\int }_{\mathbb{R}}g\left(-t\right)f\left(t\right)dt\text{, }\forall f\in L_{\omega }^1\left(\mathbb{R}\right),\forall g\in L_{\omega }^{\infty }\left(\mathbb{R}\right).$$

$${\int }_{\mathbb{R}}\frac{\log \omega \left(t\right)}{1+t^2}dt<\infty .$$

$$\underset{t\to +\infty }{\lim }\frac{\log \omega \left(t\right)}{t}=\underset{t\to +\infty }{\lim }\frac{\log \omega \left(-t\right)}{t}=0.$$

$${\Vert \mu \Vert }_{1,\omega }:={\int }_{\mathbb{R}}\omega \left(t\right)d\left|\mu \right|\left(t\right)<\infty .$$

$$I_{\omega }\left(S\right):=\left\{f\in L_{\omega }^1\left(\mathbb{R}\right):\hat{f}\left(S\right)=\left\{0\right\}\right\}$$

$$J_{\omega }\left(S\right):=\overline{\left\{f\in L_{\omega }^1\left(\mathbb{R}\right):\text{supp}\hat{f}\text{ is compact and supp}\hat{f}\cap S=\varnothing \right\}}.$$

$$J_{\omega }\left(\left\{\infty \right\}\right)=\overline{\left\{f\in L_{\omega }^1\left(\mathbb{R}\right):\text{supp}\hat{f}\text{ is compact}\right\}}.$$

$${\text{sp}}_B\left\{M\right\}=\text{hull}\left({\mathcal{I}}_{\left\{M\right\}}\right)\text{,}$$

$${\mathcal{I}}_{\left\{M\right\}}=\left\{f\in L_{\omega }^1\left(\mathbb{R}\right):f\ast g=0\text{, }\forall g\in M\right\}$$

$${\text{sp}}_B\left\{M\right\}=\overline{\bigcup _{g\in M}{\text{sp}}_B\left\{g\right\}}\text{.}$$

$${\text{sp}}_B\left\{g^{\vee }\right\}=\left\{-\lambda :\lambda \in {\text{sp}}_B\left\{g\right\}\right\}.$$

$$G\left(z\right)=\left\{\begin{tabular}[]{l}$\int\limits_{0}^{\infty}e^{-zt}{g\left(t\right)dt\text{, }\ }${Re}${z>0\text{;}}$\\ ${-}\int\limits_{-\infty}^{0}e^{-zt}{g\left(t\right)dt\text{, \ Re}z<0\text{.}}$\end{tabular}\right.$$

$$E_T^{\omega }:=\left\{x\in X:\exists C>0,\Vert e^{tT}x\Vert \le C\omega \left(t\right)\text{, }\forall t\in \mathbb{R}\right\}.$$

$$x_{\mu }={\int }_{\mathbb{R}}{e}^{tT}xd\mu \left(t\right)\text{.}$$

$$\Vert x_{\mu }\Vert \le C{\Vert \mu \Vert }_{1,\omega }\text{, }\forall \mu \in M_{\omega }\left(\mathbb{R}\right).$$

$$e^{tT}{x}_{\mu }={\int }_{\mathbb{R}}{e}^{\left(t+s\right)T}xd\mu \left(s\right),$$

$$\Vert e^{tT}{x}_{\mu }\Vert$$

Full text

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We present complete description of the class $\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right)$ in the case when the spectrum of $A$ consists of one point. These results are linked to the decomposability of $A.$ Some estimates for the norm of the commutator $AT-TA$ are obtained in the case $0\leq\alpha<1.$

1. Introduction

Let $H$ be an infinite dimensional separable Hilbert space and let $B\left(H\right)$ be the algebra of all bounded linear operators on $H.$ A family $\left\{E\right\}$ of subspaces of $H$ is called a nest if it is totally ordered by inclusion. Given a nest $\left\{E\right\},$ Ringrose [16] introduced the concept of the associated* nest algebra* Alg$\left\{E\right\}$ defined by

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In [10], Leobl and Muhly show that every nest algebra is the algebra of all analytic operators with respect to the one-parameter representation $T\rightarrow e^{-itA}Te^{itA}$ of inner $\ast-$automorphisms of $B\left(H\right),$ where $A$ is a self-adjoint operator on $H.$

For an invertible operator $A$ on $H$, Deddens [4] introduced the set

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Notice that $\mathcal{D}_{A}$ is a normed (not necessarily closed) algebra with identity and contains the commutant $\left\{A\right\}^{\prime}$ of $A.$ In the same paper, Deddens shows that if $A\geq 0,$ then $\mathcal{D}_{A}$ coincides with the nest algebra associated with the nest of spectral subspaces of $A.$ This gives a new and convenient characterization of nest algebras. In [4], Deddens conjectured that the equality $\mathcal{D}_{A}=\left\{A\right\}^{\prime}$ holds if the spectrum of $A$ is reduced to $\left\{1\right\}.$ In [17], Roth gave a negative answer to Deddens conjecture. He shows existence of a quasinilpotent operator $A$ for which $\mathcal{D}_{I+A}\neq\left\{I+A\right\}^{\prime}.$

Let $X$ be a complex Banach space and let $B\left(X\right)$ be the algebra of all bounded linear operators on $X.$ In [18], Williams proved that if the spectrum of an invertible operator $A\in B\left(X\right)$ is reduced to $\left\{1\right\}$ and $\sup_{n\in\mathbb{Z}}\left\|A^{n}TA^{-n}\right\|<\infty$, then $AT=TA$.

For a fixed $A\in B\left(X\right)$ and $\alpha\geq 0,$ we define the class $\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right)$ of all operators $T\in B\left(X\right)$ for which the growth of the function

[TABLE]

is at most polynomial in $t\in\mathbb{R}$, explicitly, there exists a constant $C_{T}>0$ such that

[TABLE]

Notice that $\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right)$ is a linear (not necessarily closed) subspace of $B\left(X\right)$ and contains the commutant of $A.$ In the case $\alpha=0,$ instead of $\mathcal{D}_{A}^{0}\left(\mathbb{R}\right)$ we will use the notation $\mathcal{D}_{A}\left(\mathbb{R}\right).$ Notice also that $\mathcal{D}_{A}\left(\mathbb{R}\right)$ is a normed (not necessarily closed) algebra with identity.

In Section 2, we give complete characterization of the class $\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right)$ in the case when the spectrum of $A$ consists of one point. Section 3 contains results related to the decomposability of $A$. In the case $0\leq\alpha<1,$ some estimates for the norm of the commutator $AT-TA$ are obtained in Section 4, where $T\in\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right).$

2. The class $\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right)$

In this section, we give complete characterization of the class $\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right)$ in the case when the spectrum of $A$ consists of one point. As usual, $\sigma\left(T\right)$ will denote the spectrum of the operator $T\in B\left(X\right).$ Throughout, $\left[\alpha\right]$ denotes the integer part of $\alpha\in\mathbb{R}.$

Let $A\in B\left(X\right)$ and $\Delta_{A}$ be the inner derivation on $B\left(X\right);$

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Then, we can write

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We have the following:

Theorem 2.1**.**

If the spectrum of the operator $A\in B\left(X\right)$ consists of one point, then

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In particular, if $0\leq\alpha<1,$ then $\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right)=\left\{A\right\}^{\prime}.$

Before to prove this theorem, we first prove the following:

Theorem 2.2**.**

If the spectrum of the operator $A\in B\left(X\right)$ lies on the real line, then

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In particular, if $0\leq\alpha<1,$ then $\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right)=\left\{A\right\}^{\prime}.$

For related results see also, [1, 13]. For the proof of Theorem 2.2, we need some preliminary results.

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 $\left(zI-T\right)u\left(z\right)=x$ for all $z\in U_{\lambda}$. 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 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 $x\in X\diagdown\left\{0\right\}$ [9, Proposition 1.2.16]. It can be seen that an operator $T\in B\left(X\right)$ having spectrum without interior points has the SVEP. Ample information about local spectra can be found in [2, 3, 5, 9].

The *local spectral radius *of $T\in B\left(X\right)$ at $x\in X$ is defined by

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If $T$ has SVEP, then

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[9, Proposition 3.3.13].

Recall that a weight function (shortly a weight) $\omega$ is a continuous function on $\mathbb{R}$ such that $\omega\left(t\right)\geq 1$ and $\omega\left(t+s\right)\leq\omega\left(t\right)\omega\left(s\right)$ for all $t,s\in\mathbb{R}.$ For a weight function $\omega$, by $L_{\omega}^{1}\left(\mathbb{R}\right)$ we will denote the Banach space of the functions $f\in L^{1}\left(\mathbb{R}\right)$ with the norm

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The space $L_{\omega}^{1}\left(\mathbb{R}\right)$ with convolution product and the norm $\left\|\cdot\right\|_{1,\omega}$ is a commutative Banach algebra and is called Beurling algebra. The dual space of $L_{\omega}^{1}\left(\mathbb{R}\right),$ denoted by $L_{\omega}^{\infty}\left(\mathbb{R}\right)$, is the space of all measurable functions $g$ on $\mathbb{R}$ such that

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The duality being implemented by the formula

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We say that the weight $\omega$ is regular if

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For example, $\omega\left(t\right)=\left(1+\left|t\right|\right)^{\alpha}$ $\left(\alpha\geq 0\right)$ is a regular weight and is called polynomial weight. If $\omega$ is a regular weight, then

[TABLE]

Consequently, the maximal ideal space of the algebra $L_{\omega}^{1}\left(\mathbb{R}\right)$ can be identified with $\mathbb{R}$ (for instance, see [6, 11, 14]). The Gelfand transform of $f\in L_{\omega}^{1}\left(\mathbb{R}\right)$ is just the Fourier transform of $f.$ Moreover, the algebra $L_{\omega}^{1}\left(\mathbb{R}\right)$ is regular in the Shilov sense [6, Ch.VI]. Notice also that $L_{\omega}^{1}\left(\mathbb{R}\right)$ is Tauberian, that is, the set $\left\{f\in L_{\omega}^{1}\left(\mathbb{R}\right):\text{supp}\widehat{f}\text{ is compact}\right\}$ is dense in $L_{\omega}^{1}\left(\mathbb{R}\right)$ [11, Ch.5]. Below, we will assume that $\omega$ is a regular weight.

Denote by $M_{\omega}\left(\mathbb{R}\right)$ the Banach algebra (with respect to convolution product) of all complex regular Borel measures on $\mathbb{R}$ such that

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The algebra $L_{\omega}^{1}\left(\mathbb{R}\right)$ is naturally identifiable with a closed ideal of $M_{\omega}\left(\mathbb{R}\right).$ By $\widehat{f}$ and $\widehat{\mu}$, we will denote the Fourier and the Fourier-Stieltjes transform of $f\in L_{\omega}^{1}\left(\mathbb{R}\right)$ and $\mu\in M_{\omega}\left(\mathbb{R}\right)$, respectively.

As usual, to any closed subset $S$ of $\mathbb{R},$ the following two closed ideals of $L_{\omega}^{1}\left(\mathbb{R}\right)$ associated:

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and

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The ideals $J_{\omega}\left(S\right)$ and $I_{\omega}\left(S\right)$ are respectively, the smallest and the largest closed ideals in $L_{\omega}^{1}\left(\mathbb{R}\right)$ with hull $S$. When these two ideals coincide, the set $S$ is said to be a set of synthesis for $L_{\omega}^{1}\left(\mathbb{R}\right)$ (for instance, see [8, Sect. 8.3]).

Notice that $I_{\omega}\left(\left\{\infty\right\}\right)=L_{\omega}^{1}\left(\mathbb{R}\right)$ and

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Since the algebra $L_{\omega}^{1}\left(\mathbb{R}\right)$ is Tauberian, we have $I_{\omega}\left(\left\{\infty\right\}\right)=J_{\omega}\left(\left\{\infty\right\}\right)$. Hence, $\left\{\infty\right\}$ is a set of synthesis for $L_{\omega}^{1}\left(\mathbb{R}\right).$ Notice also that if $\omega\left(t\right)=\left(1+\left|t\right|\right)^{\alpha}$ ($0\leq\alpha<1),$ then each point of $\mathbb{R}$ is a set of synthesis for $L_{\omega}^{1}\left(\mathbb{R}\right)$ [15, Ch.6].

Let $M$ be a non-void subset of $L_{\omega}^{\infty}\left(\mathbb{R}\right)$. A point $\lambda\in\mathbb{R}$ is said to be Beurling spectrum of $M$ if the character $e^{-i\lambda t}$ belongs to the weak∗-closed translation invariant subspace of $L_{\omega}^{\infty}\left(\mathbb{R}\right)$ generated by $M$. By sp${}_{B}\left\{M\right\},$ we will denote the set of all Beurling spectrum of $M$. It is easy to verify that

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where

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is a closed ideal of $L_{\omega}^{1}\left(\mathbb{R}\right)$. Notice also that

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For $g\in L_{\omega}^{\infty}\left(\mathbb{R}\right),$ we put $g^{\vee}\left(t\right):=g\left(-t\right).$ Clearly,

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Recall that the Carleman transform of $g\in L_{\omega}^{\infty}\left(\mathbb{R}\right)$ is defined as the analytic function $G\left(z\right)$ on $\mathbb{C}\diagdown i\mathbb{R},$ given by

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It is known [7] that $\lambda\in$sp${}_{B}\left\{g\right\}$ if and only if the Carleman transform $G\left(z\right)$ of $g$ has no analytic extension to a neighborhood of $i\lambda.$

Let $\omega$ be a weight function, $T\in B\left(X\right)$ and let

[TABLE]

Then, $E_{T}^{\omega}$ is a linear (non-closed, in general) subspace of $X$. If $x\in E_{T}^{\omega}$, then for an arbitrary $\mu\in M_{\omega}\left(\mathbb{R}\right),$ we can define $x_{\mu}\in X$ by

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Clearly, $\mu\mapsto x_{\mu}$ is a bounded linear map from $M_{\omega}\left(\mathbb{R}\right)$ into $X$;

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Further, from the identity

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we can write

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This shows that $x_{\mu}\in E_{T}^{\omega}$ for every $\mu\in M_{\omega}\left(\mathbb{R}\right).$ It is easy to check that

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It follows that if $x\in E_{T}^{\omega},$ then

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is a closed ideal of $L_{\omega}^{1}\left(\mathbb{R}\right),$ where

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For a given $x\in E_{T}^{\omega},$ consider the function

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It follows from (2.1) that $u\left(z\right)$ is a function analytic on $\mathbb{C}\diagdown$i$\mathbb{R}$. Let $a:=\mathop{\mathrm{R}e}z>0.$ Then, for an arbitrary $s>0,$ we can write

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Since

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and

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we have

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Similarly,

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Hence

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This clearly implies that $\sigma_{T}\left(x\right)\subset i\mathbb{R}.$

Thus we have the following:

Proposition 2.3**.**

Let $\omega$ be a regular weight. Assume that $T\in B\left(X\right)$ and $x\in X$ satisfy the condition $\left\|e^{tT}x\right\|\leq C\omega\left(t\right)$ for all $t\in\mathbb{R}$ and for some $C>0.$ Then, $\sigma_{T}\left(x\right)\subset i\mathbb{R}.$

Now, assume that $T$ has SVEP. We claim that $\sigma_{T}\left(x\right)$ consists of all $\lambda\in i\mathbb{R}$ for which the function $u\left(z\right)$ has no analytic extension to a neighborhood of $\lambda$. To see this, let $v\left(z\right)$ be the analytic extension of $u\left(z\right)$ to a neighborhood $U_{\lambda}$ of $\lambda\in i\mathbb{R}.$ It follows from the identity (2.3) that the function

[TABLE]

vanishes on $U_{\lambda}^{+}:=\left\{z\in U_{\lambda}:\mathop{\mathrm{R}e}z>0\right\}$ and on $U_{\lambda}^{-}:=\left\{z\in U_{\lambda}:\mathop{\mathrm{R}e}z<0\right\}.$ By uniqueness theorem, $w\left(z\right)=0$ for all $z\in U_{\lambda}.$ So we have

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This shows that $\lambda\in\rho_{T}\left(x\right).$ If $\lambda\in\rho_{T}\left(x\right)\cap i\mathbb{R},$ then there exists a neighborhood $U_{\lambda}$ of $\lambda$ with $v\left(z\right)$ analytic on $U_{\lambda}$ having values in $X$ such that

[TABLE]

By (2.3),

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Since $T$ has SVEP, we have

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This shows that $u\left(z\right)$ can be analytically extended to a neighborhood of $\lambda$.

Let $x\in E_{T}^{\omega}.$ For a given $\varphi\in X^{\ast},$ define a function $\varphi_{x}$ on $\mathbb{R}$ by

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Then, $\varphi_{x}$ is continuous and

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Consequently, $\varphi_{x}\in L_{\omega}^{\infty}\left(\mathbb{R}\right)$. Taking into account the identity (2.2), we have

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This shows that the function $z\rightarrow\langle\varphi,u\left(z\right)\rangle$ is the Carleman transform of ${\varphi_{x}.}$ It follows that

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and so

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To show the reverse inclusion, assume that $\lambda_{0}\in\mathbb{R}$ and

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Then, there exist a neighborhood $U$ of $\overline{\bigcup_{\varphi\in X^{\ast}}\text{sp}_{B}\left\{\varphi_{x}\right\}}$ and $\varepsilon>0$ such that $\left(\lambda_{0}-\varepsilon,\lambda_{0}+\varepsilon\right)\cap\overline{U}=\emptyset.$ Since the algebra $L_{\omega}^{1}\left(\mathbb{R}\right)$ is regular, there exists a function $f\in L_{\omega}^{1}\left(\mathbb{R}\right)$ such that $\widehat{f}=1$ on $\left[\lambda_{0}-\varepsilon/2,\lambda_{0}+\varepsilon/2\right]$ and $\widehat{f}=0$ on $\overline{U}.$ Notice that $\widehat{f}$ vanishes in a neighborhood of sp${}_{B}\left\{\varphi_{x}\right\}$ and supp$\widehat{f}\subseteq\left[\lambda_{0}-\varepsilon,\lambda_{0}+\varepsilon\right].$ Consequently, $f$ belongs to the smallest ideal of $L_{\omega}^{1}\left(\mathbb{R}\right)$ whose hull is sp${}_{B}\left\{\varphi_{x}\right\}.$ Since

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we have $\left(\lambda_{0}-\varepsilon/2,\lambda_{0}+\varepsilon/2\right)\subset\mathbb{R}\setminus$sp${}_{B}\left\{\varphi_{x}\right\}$ for every $\varphi\in X^{\ast}.$ It follows that the function $z\rightarrow\langle\varphi,u\left(z\right)\rangle$ can be analytically extended to $i\left(\lambda_{0}-\varepsilon/2,\lambda_{0}+\varepsilon/2\right)$ for every $\varphi\in X^{\ast}.$ Hence, $u\left(z\right)$ can be analytically extended to $i\left(\lambda_{0}-\varepsilon/2,\lambda_{0}+\varepsilon/2\right)$ and therefore, $i\lambda_{0}\notin\sigma_{T}\left(x\right)$ or $\lambda_{0}\notin-i\sigma_{T}\left(x\right).$ Thus we have

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Further, it is easy to check that

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where

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Taking into account that

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we can write

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Thus we have the following:

Proposition 2.4**.**

Let $\omega$ be a regular weight. Assume that $T\in B\left(X\right)$ has SVEP and $x\in X$ satisfies the condition $\left\|e^{tT}x\right\|\leq C\omega\left(t\right)$ for all $t\in\mathbb{R}$ and for some $C>0.$ Then,

[TABLE]

For $f\in L_{\omega}^{1}\left(\mathbb{R}\right)$ and $s\in\mathbb{R},$ let $f_{s}\left(t\right):=f\left(t-s\right).$ Let $e_{n}:=2n\chi_{\left[-1/n,1/n\right]}$ $\left(n\in\mathbb{N}\right)$, where $\chi_{\left[-1/n,1/n\right]}$ is the characteristic function of the interval $\left[-1/n,1/n\right].$ If $K:=\sup_{t\in\left[-1,1\right]}\omega\left(t\right),$ then $\left\|e_{n}\right\|_{\omega}\leq K$ for all $n\in\mathbb{N}.$ On the other hand, by continuity of the mapping $s\mapsto f_{s}$, we have

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Consequently, $\left\{e_{n}\right\}$ is a bounded approximate identity (b.a.i.) for $L_{\omega}^{1}\left(\mathbb{R}\right).$ If $x\in E_{T}^{\omega},$ then from the identity

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it follows that $x_{e_{n}}\rightarrow x.$ Similarly, $x_{f\ast e_{n}}\rightarrow x_{f}$ for all $f\in L_{\omega}^{1}\left(\mathbb{R}\right).$

Proposition 2.5**.**

Let $\omega$ be a regular weight and $x\in X.$ Assume that $T\in B\left(X\right)$ has SVEP and $\left\|e^{tT}x\right\|\leq C\omega\left(t\right)$ for all $t\in\mathbb{R}$ and for some $C>0.$ For an arbitrary $f\in L_{\omega}^{1}\left(\mathbb{R}\right)$, the following assertions hold:

$a)$* If $x_{f}=0,$ then $\widehat{f}$ vanishes on $i\sigma_{T}\left(x\right).$*

$b)$* If $\widehat{f}$ vanishes in a neighborhood of $i\sigma_{T}\left(x\right),$ then $x_{f}=0$.*

$c)$* If $\widehat{f}=1$ in a neighborhood of $i\sigma_{T}\left(x\right),$ then $x_{f}=x$.*

Proof.

a) By Proposition 2.4, $i\sigma_{T}\left(x\right)=$hull$\left(I_{x}\right)$ and therefore $I_{x}\subseteq I_{\omega}\left(i\sigma_{T}\left(x\right)\right).$ This clearly implies a).

b) Let $g\in L_{\omega}^{1}\left(\mathbb{R}\right)$ be such that supp$\widehat{g}$ is compact. Then, $f\ast g\in J_{\omega}\left(i\sigma_{T}\left(x\right)\right)$ and therefore, $f\ast g\in I_{x}.$ So we have $x_{f\ast g}=0.$ Since the algebra $L_{\omega}^{1}\left(\mathbb{R}\right)$ is Tauberian, $x_{f\ast g}=0$ for all $g\in L_{\omega}^{1}\left(\mathbb{R}\right).$ It follows that $x_{f\ast e_{n}}=0$ for all $n,$ where $\left\{e_{n}\right\}$ be a b.a.i. for $L_{\omega}^{1}\left(\mathbb{R}\right).$ As $n\rightarrow\infty,$ we have $x_{f}=0.$

c) Since the Fourier transform of $f\ast e_{n}-e_{n}$ vanishes in a neighborhood of $i\sigma_{T}\left(x\right),$ by b), $x_{f\ast e_{n}}=x_{e_{n}}$. As $n\rightarrow\infty,$ we have $x_{f}=x.$

By $S\left(\mathbb{R}\right),$ we denote the set of all rapidly decreasing functions on $\mathbb{R},$ i.e. the set of all infinitely differentiable functions $\phi$ on $\mathbb{R}$ such that

[TABLE]

(in this definition, $n$ can be replaced by any $\alpha\geq 0$). It can be seen that if $\omega$ is a polynomial weight, then $S\left(\mathbb{R}\right)\subset L_{\omega}^{1}\left(\mathbb{R}\right).$

Lemma 2.6**.**

Assume that $T\in B\left(X\right)$ and $x\in X$ satisfy the condition $\left\|e^{tT}x\right\|\leq C\left(1+\left|t\right|\right)^{\alpha}$ $\left(\alpha\geq 0\right)$ for all $t\in\mathbb{R}$ and for some $C>0.$ Then, for an arbitrary $\phi\in S\left(\mathbb{R}\right),$ we have

[TABLE]

Proof.

For an arbitrary $a,b\in\mathbb{R}$ $\left(a<b\right),$ we can write

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On the other hand,

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Since $\phi\in S\left(\mathbb{R}\right),$ it follows that

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Hence $x_{\phi^{\prime}}=-Tx_{\phi}.$ By induction we obtain our result.

Next, we have the following:

Proposition 2.7**.**

Let $\omega\left(t\right)=\left(1+\left|t\right|\right)^{\alpha}$ $\left(\alpha\geq 0\right).$ Assume that $T\in B\left(X\right)$ has SVEP and $x\in X$ satisfies the condition $\left\|e^{tT}x\right\|\leq C\omega\left(t\right)$ for all $t\in\mathbb{R}$ and for some $C>0.$ If $\sigma_{T}\left(x\right)=\left\{0\right\}$, then for an arbitrary $f\in L_{\omega}^{1}\left(\mathbb{R}\right),$ we have

[TABLE]

where $k=[\alpha].$ In particular, we have $T^{k+1}x=0.$

Proof.

We know (for instance, see [6, Ch.VI, §41] and [19, Theorem 3.2]) that if $f\in L_{\omega}^{1}\left(\mathbb{R}\right),$ then the first $k$ derivatives of the Fourier transform of $f$ exist and

[TABLE]

where $k=\left[\alpha\right]$. Recall that $J_{\omega}\left(\left\{0\right\}\right)$ is the smallest closed ideal of $L_{\omega}^{1}\left(\mathbb{R}\right)$ whose hull is $\left\{0\right\}.$ On the other hand, by Proposition 2.4, hull$\left(I_{x}\right)=\left\{0\right\}.$ Hence we have $J_{\omega}\left(\left\{0\right\}\right)\subseteq I_{x}$.

Let $\phi\in S\left(\mathbb{R}\right)$ be such that $\widehat{\phi}\left(\lambda\right)=1$ in a neighborhood of $0.$ For a given $f\in L_{\omega}^{1}\left(\mathbb{R}\right),$ consider the function

[TABLE]

As

[TABLE]

we have

[TABLE]

It can be seen that the first $k$ derivatives of $\widehat{g}$ at [math] are zero and therefore $g\in J_{\omega}\left(\left\{0\right\}\right).$ Consequently, $g\in I_{x}$ and so $x_{g}=0.$ On the other hand, by Lemma 2.6 and Proposition 2.5,

[TABLE]

which implies

[TABLE]

Hence,

[TABLE]

If $f:=\phi^{\left(k+1\right)},$ then as

[TABLE]

we get

[TABLE]

Now, we are in a position to prove Theorem 2.2.

Proof of Theorem 2.2.

If $T\in\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right),$ then as

[TABLE]

we have

[TABLE]

By Proposition 2.3, $\sigma_{\Delta_{A}}\left(T\right)\subset i\mathbb{R}.$ Further, since

[TABLE]

[9, Theorem 3.5.1] and $\sigma\left(A\right)\subset\mathbb{R},$ we have $\sigma\left(\Delta_{A}\right)\subset\mathbb{R}$. Therefore, $\Delta_{A}$ has SVEP. On the other hand, as

[TABLE]

we obtain that

[TABLE]

Since $\Delta_{A}$ has SVEP, $\sigma_{\Delta_{A}}\left(T\right)\neq\emptyset$, so that $\sigma_{\Delta_{A}}\left(T\right)=\left\{0\right\}.$ Applying now Proposition 2.7 to the operator $\Delta_{A}$ on the space $B\left(X\right),$ we get

[TABLE]

For the reverse inclusion, assume that $T\in B\left(X\right)$ satisfies the equation $\Delta_{A}^{n}\left(T\right)=0$ for some $n\in\mathbb{N}.$ Then, we can write

[TABLE]

This shows that $T\in\mathcal{D}_{A}^{n-1}\left(\mathbb{R}\right).$ The proof is complete.

Next, we will prove Theorem 2.1.

Proof of Theorem 2.1.

Assume that $\sigma\left(A\right)=\left\{\lambda\right\}.$ If $T\in\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right),$ then $T\in\mathcal{D}_{B}^{\alpha}\left(\mathbb{R}\right),$ where $B=A-\lambda I.$ Since $\sigma\left(B\right)=\left\{0\right\}$ and $\Delta_{B}^{n}\left(T\right)=\Delta_{A}^{n}\left(T\right)$ $\left(\forall n\in\mathbb{N}\right)$, by Theorem 2.2 we obtain our result.

Note that if $\alpha\geq 1,$ then $\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right)\neq\left\{A\right\}^{\prime},$ in general. To see this, let A=\left(\begin{array}[]{cc}0&0\\ 1&0\end{array}\right) and T=\left(\begin{array}[]{cc}1&0\\ 0&0\end{array}\right) be two $2\times 2$ matrices on $2-$dimensional Hilbert space. As $A^{2}=0,$ we have $\sigma\left(A\right)=\left\{0\right\}$ and

[TABLE]

Since

[TABLE]

we have $T\in\mathcal{D}_{A}^{1}\left(\mathbb{R}\right),$ but $AT\neq TA.$

For a given $\alpha\geq 0,$ we define the class $\mathcal{D}_{A}^{\alpha}\left(\mathbb{Z}\right)$ of all operators $T\in B\left(X\right)$ for which there exists a constant $C_{T}>0$ such that

[TABLE]

Clearly, $\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right)\subseteq\mathcal{D}_{A}^{\alpha}\left(\mathbb{Z}\right).$ We claim that $\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right)=\mathcal{D}_{A}^{\alpha}\left(\mathbb{Z}\right).$ Indeed, if $T\in\mathcal{D}_{A}^{\alpha}\left(\mathbb{Z}\right)$ and $t\in\mathbb{R},$ then $t=n+r,$ where $n\in\mathbb{Z}$, $\left|r\right|<1$ and $\left|n\right|\leq\left|t\right|.$ Consequently, we can write

[TABLE]

Therefore, in Theorems 2.1 and 2.2, the class $\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right)$ can be replaced by $\mathcal{D}_{A}^{\alpha}\left(\mathbb{Z}\right).$

As a consequence of Proposition 2.7, we will need the following:

Corollary 2.8**.**

Assume that $T\in B\left(X\right)$ has SVEP and $x\in X$ satisfies the condition

[TABLE]

for all $n\in\mathbb{Z}$ and for some $C>0.$ If $\sigma_{T}\left(x\right)=\left\{0\right\}$, then $T^{\left[\alpha\right]+1}x=0.$

By $K\left(X\right)$ we will denote the space of compact operators on a Banach space $X.$

Next, we have the following:

Proposition 2.9**.**

Assume that the spectrum of the operator $A\in B\left(X\right)$ consists of one point. If $K\left(X\right)\subset\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right),$ then

[TABLE]

Proof.

We have

[TABLE]

Applying uniform boundedness principle to the sequence of operators

[TABLE]

we obtain existence of a constant $C>0$ such that

[TABLE]

Consequently, we have

[TABLE]

For a given $x\in X$ and $\varphi\in X^{\ast}$, let $x\otimes\varphi$ be the one dimensional operator on $X;$

[TABLE]

By taking $T=x\otimes\varphi$ in the preceding inequality, we can write

[TABLE]

which implies

[TABLE]

Now, assume that $\sigma\left(A\right)=\left\{\lambda\right\}.$ Then as $\sigma\left(e^{n\left(A-\lambda I\right)}\right)=\left\{1\right\},$ we have

[TABLE]

so that

[TABLE]

Thus, we obtain that

[TABLE]

By Corollary 2.8, $\left(A-\lambda I\right)^{k+1}=0,$ where $k=\left[\alpha\right].$ If $N:=A-\lambda I,$ then $A=\lambda I+N$, where $N$ is a nilpotent of degree $\leq k+1.$ Further, for an arbitrary $T\in B\left(X\right)$ and $t\in\mathbb{R},$ from the identity

[TABLE]

we can write

[TABLE]

where the functions $f_{1},...,f_{2k}$ do not depend from $t.$ It follows that

[TABLE]

and so $T\in\mathcal{D}_{A}^{2k}\left(\mathbb{R}\right).$ Thus we have $\mathcal{D}_{A}^{2k}\left(\mathbb{R}\right)=B\left(X\right).$

As a consequence of Proposition 2.9, we have the following.

Corollary 2.10**.**

Assume that $\Delta_{A}$ is a quasinilpotent for some $A\in B\left(X\right).$ If $K\left(X\right)\subset\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right),$ then $\Delta_{A}$ is a nilpotent of degree $\leq 2\left[\alpha\right]+1.$

Proof.

It follows from the identity

[TABLE]

that $\sigma\left(A\right)$ consists of one point. By Proposition 2.9, $\mathcal{D}_{A}^{2\left[\alpha\right]}\left(\mathbb{R}\right)=B\left(X\right).$ On the other hand, by Theorem 2.1,

[TABLE]

We conclude this section with the following result:

Proposition 2.11**.**

The following assertions hold:

$a)$* For an arbitrary $T\in D_{A}\left(\mathbb{R}\right),$*

[TABLE]

$b)$* If $D_{A}\left(\mathbb{R}\right)$ is closed, then there exists a constant $C>0$ such that*

[TABLE]

Proof.

a) Let $T\in D_{A}\left(\mathbb{R}\right)$ and $\delta:=\sup_{t\in\mathbb{R}}\left\|T-e^{tA}Te^{-tA}\right\|.$ Define a mapping $\pi:D_{A}\left(\mathbb{R}\right)\rightarrow B\left(X\right)$ by

[TABLE]

where $\Phi$ is a fixed invariant mean on $\mathbb{Z}.$ We claim that $\pi\left(T\right)\in\left\{A\right\}^{\prime}.$ Indeed, for an arbitrary $s\in\mathbb{R},$ from the identities

[TABLE]

we have $e^{sA}\pi\left(T\right)=\pi\left(T\right)e^{sA}.$ This clearly implies $A\pi\left(T\right)=\pi\left(T\right)A.$ Notice that $\langle\pi\left(T\right)x,\varphi\rangle$ belongs to the closure of convex combination of the set

[TABLE]

Now, from the inequality

[TABLE]

we have

[TABLE]

Hence $\left\|T-\pi\left(T\right)\right\|\leq\delta.$ Since $\pi\left(T\right)\in\left\{A\right\}^{\prime}$, the result follows.

b) Applying uniform boundedness principle to the family of the operators

[TABLE]

we obtain existence of a constant $K>0$ such that

[TABLE]

Now if $T\in D_{A}\left(\mathbb{R}\right)$ and $S\in\left\{A\right\}^{\prime}$, then as $T-S\in D_{A}\left(\mathbb{R}\right),$ from the relations

[TABLE]

we have

[TABLE]

3. Decomposability

Recall that an operator $A\in B\left(X\right)$ is called decomposable if, for every open covering $\left\{U,W\right\}$ of the complex plane, there exist a pair of $T-$invariant closed linear subspaces $E$ and $F$ of $X$ such that $E+F=X$, $\sigma\left(A\mid_{E}\right)\subset U$ and $\sigma\left(A\mid_{F}\right)\subset W$ (for instance, see [3] and [9]). For a closed subset $F$ of $\mathbb{C},$ the local spectral subspace $X_{A}\left(F\right)$ of $A\in B\left(X\right)$ is defined by

[TABLE]

If $A$ is decomposable, then $X_{A}\left(F\right)$ is closed for every closed set $F\subset\mathbb{C}$ [9, Sect. 1.2]. Moreover,

[TABLE]

for every closed set $F\subset\mathbb{C}$, where $f\left(A\right)$ is defined by the Riesz-Dunford functional calculus [9, Theorem 3.3.6].

Proposition 3.1**.**

Assume that $A\in B\left(X\right)$ is decomposable and $T\in B\left(X\right)$ satisfies the condition

[TABLE]

for all $n\in\mathbb{Z}$ and for some $C_{T}>0$. Then the following conditions are equivalent:

$\left(a\right)$* $TX_{A}\left(F\right)\subseteq X_{A}\left(F\right)$ for every closed set $F\subset\mathbb{C}.$*

$\left(b\right)\mathcal{\ }\Delta_{A}^{\left[\alpha\right]+1}\left(T\right)=0.$**

In particular, if $0\leq\alpha<1,$ then $AT=TA$ if and only if $TX_{A}\left(F\right)\subseteq X_{A}\left(F\right),$ for every closed set $F\subset\mathbb{C}.$

Proof.

(a)$\Rightarrow$(b) We have

[TABLE]

Since $A$ is decomposable, $\Delta_{A}$ has SVEP [9, Proposition 3.4.6] and therefore,

[TABLE]

On the other hand, $TX_{A}\left(F\right)\subseteq X_{A}\left(F\right)$ for every closed set $F\subset\mathbb{C}$ if and only if

[TABLE]

[9, Corollary 3.4.5]. Now, since $\sigma_{\Delta_{A}}\left(T\right)=\left\{0\right\}$, by Corollary 2.8, $\Delta_{A}^{\left[\alpha\right]+1}\left(T\right)=0.$

In fact, (b)$\Rightarrow$(a) follows from [9, Proposition 3.4.2]. Here, we present more simple proof. Now, it suffices to show that $\sigma_{A}\left(Tx\right)\subseteq\sigma_{A}\left(x\right)$ for every $x\in X.$ If $x\in X$ and $\lambda\in\rho_{A}\left(x\right),$ then there is a neighborhood $U_{\lambda}$ of $\lambda$ with $u\left(z\right)$ analytic on $U_{\lambda}$ having values in $X$, such that

[TABLE]

Using this identity, it is easy to check that the function

[TABLE]

satisfies the equation

[TABLE]

This shows that $\lambda\in\rho_{A}\left(Tx\right).$

Next, we have the following:

Proposition 3.2**.**

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

$\left(i\right)$* $A$ is decomposable and $\sigma\left(A\right)\subset\left\{z\in\mathbb{C}:\mathop{\mathrm{R}e}z>0\right\};$*

$\left(ii\right)$* $\left\|A^{n}TA^{-n}\right\|\leq C_{T}\left(1+\left|n\right|\right)^{\alpha}$ $\left(0\leq\alpha<1\right)$ for all $n\in\mathbb{Z}$ and for some $C_{T}>0$.*

Then, $AT=TA$ if and only if $TX_{A}\left(F\right)\subseteq X_{A}\left(F\right)$ for every closed set $F\subset\mathbb{C}$.

Proof.

Assume that $TX_{A}\left(F\right)\subseteq X_{A}\left(F\right)$ for every closed set $F\subset\mathbb{C}$. We can write $A=e^{B},$ where $B=\log A.$ Then, $B$ is decomposable [9, Theorem 3.3.6] and

[TABLE]

Moreover, for every closed set $F\subset\mathbb{C},$

[TABLE]

where $f\left(z\right)=\log z.$ By Proposition 3.1, $BT=TB$ which implies $e^{B}T=Te^{B}.$ Hence $AT=TA.$

Assume that $AT=TA.$ It suffices to show that $\sigma_{A}\left(Tx\right)\subseteq\sigma_{A}\left(x\right)$ for every $x\in X.$ If $x\in X$ and $\lambda\in\rho_{A}\left(x\right),$ then there is a neighborhood $U_{\lambda}$ of $\lambda$ with $u\left(z\right)$ analytic on $U_{\lambda}$ having values in $X$, such that

[TABLE]

It follows that

[TABLE]

This shows that $\lambda\in\rho_{A}\left(Tx\right).$

4. The norm of the commutator $AT-TA$

In this section, we give some estimates for the norm of the commutator $AT-TA,$ where $T\in\mathcal{D}_{A}^{\alpha}\left(\mathbb{R}\right)$ $\left(0\leq\alpha<1\right).$

Lemma 4.1**.**

Let $\mu\in M_{\omega}\left(\mathbb{R}\right),$ where $\omega\left(t\right)=\left(1+\left|t\right|\right)^{\alpha}$ $\left(\alpha\geq 0\right).$ Assume that $T\in B\left(X\right)$ and $x\in X$ satisfy the following conditions:

$\left(i\right)$* $\left\|e^{tT}x\right\|\leq C\omega\left(t\right)$ for all $t\in\mathbb{R}$ and for some $C>0;$*

$\left(ii\right)$* $T$ has SVEP.*

If $\widehat{\mu}\left(\lambda\right)=\lambda$ in a neighborhood of $i\sigma_{T}\left(x\right),$ then

[TABLE]

Proof.

Let $g\in S\left(\mathbb{R}\right)$ be such that $\widehat{g}\left(\lambda\right)=1$ in a neighborhood of $i\sigma_{T}\left(x\right).$ By Proposition 2.5, $x_{g}=x.$ On the other hand, by Lemma 2.6,

[TABLE]

Since

[TABLE]

the Fourier transform of the function $-ig^{\prime}-\mu\ast g$ vanishes in a neighborhood of $i\sigma_{T}\left(x\right).$ By Proposition 2.5,

[TABLE]

Hence $x_{\mu}=iTx.$

Note that in the preceding lemma, the weight function $\omega\left(t\right)=\left(1+\left|t\right|\right)^{\alpha}$ $\left(\alpha\geq 0\right)$ can be replaced by the weight $\omega\left(t\right)=1+\left|t\right|^{\alpha}$ .

Theorem 4.2**.**

Assume that $T\in B\left(X\right)$ has SVEP and $x\in X$ satisfies the condition

[TABLE]

for all $t\in\mathbb{R}$ and for some $C>0.$ Then we have

[TABLE]

where

[TABLE]

Proof.

We basically follow the proof of Lemma 3.4 in [12]. Let an arbitrary $a>r_{T}\left(x\right)$ be fixed. Consider the function $f,$ defined by $f\left(\lambda\right)=\lambda$ for $-a\leq\lambda\leq a$ and $f\left(\lambda\right)=2a-\lambda$ for $a\leq\lambda\leq 3a.$ We extend this function periodically to the real line by putting $f\left(\lambda+4a\right)=f\left(\lambda\right)$ $\left(\lambda\in\mathbb{R}\right).$ A few lines of computation show that the Fourier coefficients of $f$ are given by the equalities:

[TABLE]

Let $\mu$ be a discrete measure on $\mathbb{R}$ concentrated at the points

[TABLE]

with the corresponding weights

[TABLE]

Since

[TABLE]

it follows from the uniqueness theorem that

[TABLE]

Now, if $\omega\left(t\right):=1+\left|t\right|^{\alpha}$ $\left(0\leq\alpha<1\right),$ then as

[TABLE]

we can write

[TABLE]

where

[TABLE]

Since $\widehat{\mu}\left(\lambda\right)=\lambda$ in a neighborhood of $i\sigma_{T}\left(x\right),$ by Lemma 4.1, $x_{\mu}=iTx$. Therefore, we get

[TABLE]

Since $a>r_{T}\left(x\right)$ is arbitrary, we obtain our result.

As an application of Theorem 4.2, we have the following quantitative version of Theorem 2.1 in the case $0\leq\alpha<1.$

Corollary 4.3**.**

Let $A\in B\left(X\right)$ and assume that $\Delta_{A}$ has SVEP. If $T\in B\left(X\right)$ satisfies the condition

[TABLE]

for all $t\in\mathbb{R}$ and for some $C_{T}>0,$ then

[TABLE]

where $C\left(\alpha\right)$ is defined by $\left(4.1\right).$

Proof.

Noting that

[TABLE]

by Theorem 4.2,

[TABLE]