The Backward Shift and Two Infinite-Dimension Phenomena in Banach Spaces

Abstract

We consider the backward shift operator on a sequence Banach space in the context of two infinite-dimensional phenomena: the existence of topologically transitive operators, and the existence of entire analytic functions of the unbounded type. It is well known that the weighted backward shift (for an appropriated weight) is topologically transitive on $1\le p<\infty$ and on $c_0.$ We construct some generalizations of the weighted backward shift for non-separable Banach spaces, which remains topologically transitive. Also, we show that the backward shift, in some sense, generates analytic functions of the unbounded type. We introduce the notion of a generator of analytic functions of the unbounded type on a Banach space and investigate its properties. In addition, we show that, using this operator, one can obtain a quasi-extension operator of analytic functions in a germ of zero for entire analytic functions. The results are supported by examples.

1. Introduction

Let X be a real or complex Banach space with a Schauder (topological) basis $\left(e_n\right),$ and $\Lambda =\left({\lambda }_n\right)$ be a bounded sequence of positive numbers. An operator $T_{\Lambda }:X\to X$ of the form

$$T_{\Lambda }\left(x\right)=({\lambda }_1{x}_2,\dots ,{\lambda }_n{x}_{n+1},\dots ),x=(x_1,\dots ,x_n,\dots )=\sum _{n=1}^{\infty }{x}_n{e}_n$$

is called the weighted backward shift (with respect to the basis $\left(e_n\right)$) with the weight $\Lambda .$ It is easy to check that if $\left(e_n\right)$ is an unconditional basis, then $T_{\lambda }$ is a continuous operator and $\parallel T_{\Lambda }\parallel \le 2K,$ where K is the unconditional constant of the basis $\left(e_n\right)$ ([1] (p. 19)). Shift-type operators in Banach spaces are purely infinite-dimensional phenomena reflecting some self-similarity of infinity structures. Dynamics of weighted shifts on sequence spaces have been studied by many authors. In 1969, Rolewicz proved that the weighted backward shift on ${\mathsf{\ell }}_p,$ $1\le p<\infty$ or $c_0$ is hypercyclic for ${\lambda }_n=\lambda >1$ [2]. It means that there is $x\in X$ such that the set $\{T^n\left(x\right):n\in \mathbb{N}\}$ is dense in $X.$ Note that the first hypercyclic operator on the metric linear space $H\left(\mathbb{C}\right)$ of all analytic functions on the complex plane $\mathbb{C}$ was constructed by Birkhoff in 1929. Namely, in [3], Birkhoff shows that the operator $f\left(x\right)\mapsto f(x+a)$ is hypercyclic for every $a\ne 0.$ The hypercyclicity of weighted shift operators and their generalizations in various spaces was investigated in [4,5,6,7,8]. A relationship between the Birkhoff operator and the Rolewicz operator was established in [5,6]. In particular, a derivation D on a space of analytic functions can be represented as a weighted backward shift, while the Birkhoff operator is $e^D.$ Such an approach can also be used for spaces of analytic functions of infinitely many variables [9].

Clearly, to admit a hypercyclic operator, a Banach space must be infinite-dimensional and separable. Conversely, every separable infinite-dimensional Banach space (and even, more general, Fréchet space) admits a hypercyclic operator [10,11]. In [6] G. Godefroy and J. Shapiro suggested how to extend the notion of hypercyclicity to non-separable spaces using the concept of topological transitivity (see definition below). The property of the topological transitivity of some analogues of the weighted backward shift on non-separable Hilbert spaces were studied in [12,13,14]. Note that not every non-separable Banach space admits a topological transitive operator [15]. For example, ${\mathsf{\ell }}_{\infty }$ does not admit a topological transitive operator but admits weak-star hypercyclic operators [16].

Another infinite-dimensional phenomenon is the existence of analytic functions of the unbounded type on infinite-dimensional Banach spaces. Since any infinite-dimensional Banach space X is not locally compact, the continuity of a function does not guarantee its boundedness on a bounded subset of $X.$ Moreover, every infinite-dimensional Banach space X admits analytic functions on X that are unbounded on some bounded subsets (see [17], p. 157), so-called functions of the unbounded type. In [18,19], it was shown that for every finite collection of disjoint balls in a Banach space, X is an analytic function on X which is bounded on some prescribed subset of these balls and unbounded on the subset of other balls. Structures of the set of analytic functions of the unbounded type and linear subspaces in this set were considered in [20,21,22]. In particular, [22] considered conditions for subalgebras of analytic functions, providing the existence of analytic functions of the unbounded type. In [21,22], the authors constructed some linear spaces and algebras consisting (excepting zero) of analytic functions of the unbounded type on a Banach space.

In this paper, we consider some generalizations of the backward shift operator for (generally speaking, non-separable) Banach spaces admitting unconditional Schauder decompositions of a particular form (Section 2). In Section 3, using Theorem 1 in [20], we show that the backward shift operator generates, in some sense, analytic functions of the unbounded type. Thus, we have a link between the topological transitivity and analytic functions of the unbounded type. As an application of the obtained results, in Section 4, we show that the backward shift has some extension properties for analytic functions defined in a neighborhood of zero. We introduce a notion of a quasi-extension operator on a germ of analytic functions and construct some examples of such operators using the backward shift.

For general information on hypercyclic and topological transitive operators, we refer the readers to [23,24]. A background on analytic functions on infinite-dimensional spaces can be found in [17,25].

2. Backward Shifts for Non-Separable Spaces

Let X be a metric space and T be a continuous mapping $T:X\to X.$ T is called topologically transitive if, for any pair $U,$V of nonempty open subsets of X, there exists some integer $k\ge 0$ such that $T^k\left(U\right)\cap V\ne \varnothing .$

Let us recall that a sequence of closed subspaces $\left(X_n\right)$ of a Banach space X is a Schauder decomposition of X if every $x\in X$ has a unique representation as

$$x=\sum _{k=0}^{\infty }{x}_k,x_k\in X_k,$$

(1)

and the series (1) converges in $X.$ A Schauder decomposition $\left(X_n\right)$ is unconditional if (1) converges unconditionally.

In [15], it was observed that, similar to the case of hypercyclic operators, we have the following topologically transitive criterion.

Theorem 1.

(see for the proof e.g., [13]). Let T be a bounded linear operator on a Banach space X (not necessarily separable). Suppose that there exists a strictly increasing sequence $\left(n_k\right)$ of positive integers for which there are the following:

(i) 

A dense subset $Z_0\subset X$ such that $T^{n_k}\left(x\right)\to 0$ for every $x\in Z_0$ as $k\to \infty .$

(ii) 

A dense subset $Y_0\subset X$ and a sequence of mappings (not necessary linear and not necessary continuous) $S_k:Y_0\to X$ such that $S_k\left(y\right)\to 0$ for every $y\in Y_0$ and $T^{n_k}\circ S_k\left(y\right)\to y$ for every $y\in Y_0$ as $k\to \infty .$

Then, T is topologically transitive.

Let X be an infinite-dimensional (not necessary separable) Banach space which admits an unconditional Schauder decomposition to Banach spaces $X_k,$ $k=0,1,\dots .$ Let ${\left(F_k\right)}_{k=1}^{\infty }$ be a sequence of injective maps $F_k:X_{k+1}\to X_k$ with dense ranges and $\parallel F_k\parallel =1.$ We have the following shifts of spaces $X_k$ under maps $F_k$:

$$0⟵X_0\stackrel{F_1}{⟵}{X}_1\stackrel{F_2}{⟵}\cdots \stackrel{F_n}{⟵}{X}_n\cdots .$$

Let us define a weighted backward shift operator (associated with a Schauder decomposition $\left(X_n\right)$ of X) by

$$T_{\Lambda }\left(x\right)=\sum _{k=1}^{\infty }{\lambda }_k{F}_k\left(x_k\right),$$

(2)

$$T_{\Lambda }:(x_0,x_1,\dots ,x_n,\dots )\mapsto ({\lambda }_1{F}_1\left(x_1\right),{\lambda }_2{F}_2\left(x_2\right),\dots ,{\lambda }_n{F}_n\left(x_n\right),\dots ),$$

where $\Lambda =\left({\lambda }_k\right)$ is a sequence of positive numbers with ${sup}_k{\lambda }_k<\infty .$

Theorem 2.

Let X be a Banach space which admits an unconditional Schauder decomposition to Banach spaces $X_k,$ $k=0,1,\dots$ and $T_{\Lambda }$ a weighted backward shift, defined as in (2). Suppose that the following assumptions hold:

(i) 

The weight constants ${\lambda }_k$ are such that

$$\underset{n\to \infty }{lim\; sup}\prod _{k=1}^n{\lambda }_k=\infty .$$

(ii) 

There is a dense subspace $E_0\subset \mathrm{range}\left(F_1\right)\subset X_0$ such that for every $x\in E_0$ the set

$$\{F_n^{-1}\circ \cdots \circ F_1^{-1}\left(x\right),n\in \mathbb{N}\}$$

is bounded in $X.$

Then the operator T defined by (2) is topologically transitive.

Proof. 

Let $Z_0$ be subsets in X defined as

$$Z_0=\{(x_1,\dots ,x_m,0,\dots ):x_k\in \mathrm{range}\left(F_k\right),k=1,\dots ,m;m\in \mathbb{N}\}.$$

Since the range of $F_{k+1}$ is dense in $X_k,$ $Z_0$ is dense.

Since each map $F_k$ from $X_k$ to $X_{k-1}$ is injective and continuous, spaces $E_k=F_k^{-1}\left(E_{k-1}\right)$ are dense in $X_k,$ $k\in \mathbb{N}.$ We define a dense subset $Y_0\subset X$ as

$$Y_0=\{(y_0,y_1,\dots ,0,0,\dots ):y_k\in E_k,\mathrm{and}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{nonzero}{y}_k\mathrm{is}\mathrm{finite}\}.$$

For every $y\in Y_0$, we define

$$S\left(y\right)=\left({\lambda }_1^{-1}{F}_1^{-1}\left(y_0\right),\dots ,{\lambda }_n^{-1}{F}_n^{-1}\left(y_{(n-1)}\right),\dots \right).$$

Condition (i) of the theorem implies that for every $y\in Y_0,$ $S^{n_k}\left(y\right)\to 0$ as $k\to \infty$ for some subsequence $n_k\in \mathbb{N}.$ Indeed, from (i) it follows that there is a subsequence $n_k\in \mathbb{N}$ such that

$$\underset{k\to \infty }{lim}\prod _{j=1}^{n_k}{\lambda }_j=\infty .$$

Thus,

$$\parallel S^{n_k}\left(y\right)\parallel \le \frac{\parallel F_{n_k}^{-1}\circ \cdots \circ F_1^{-1}\left(y\right)\parallel }{{\prod }_{j=1}^{n_k}{\lambda }_j}\to 0ask\to \infty .$$

We set $S_k\left(y\right)=S^{n_k}\left(y\right).$ Also, $T^{n_k}\left(y\right)\to 0$ as $k\to \infty$ and $T^{n_k}\circ S_k\left(y\right)=y$ for every $y\in Y_0.$ Therefore, T satisfies the topologically transitive criterion (Theorem 1), and so is topologically transitive. □

The following example gives us a topologically transitive weighted backward shift on the space c of convergence sequences.

Example 1.

Note first that the vectors $e_n=(\underset{n}{\underbrace{0,\dots ,0,1}},0,\dots )$ do not form a basis in c. However, it is well known that c contains the following unconditional basis $\left(g_n\right)$ defined by

$$g_0=(1,1,\dots ),g_0^{*}\left(x\right)=\underset{n\to \infty }{lim}{x}_n,$$

and for $k>0,$

$$g_k=e_k,g_k^{*}\left(x\right)=x_k-g_0^{*}\left(x\right)$$

where $x=(x_1,\dots ,x_k,\dots )\in c,$ and $g_k^{*}$ are the coordinate functionals, that is,

$$x=\sum _{n=0}^{\infty }{g}_n^{*}\left(x\right)g_n,x\in c.$$

By Theorem 2, the operator

$$T_{\Lambda }:\sum _{n=0}^{\infty }{g}_n^{*}\left(x\right)g_n\mapsto \sum _{n=0}^{\infty }{\lambda }_n{g}_n^{*}\left(x\right)g_{n-1},{\lambda }_0=0$$

is topologically transitive and so hypercyclic (because c is separable) if Λ satisfies condition (i) in Theorem 2. The operator $T_{\Lambda }$ can be written as

$$T_{\Lambda }:(x_1,\dots ,x_k,\dots )\mapsto {\lambda }_1(x_1,x_1,x_1,\dots )+\left({\lambda }_2(x_2-g_0^{*}\left(x\right)),\dots ,{\lambda }_k(x_k-g_0^{*}\left(x\right)),\dots \right).$$

Note that condition (ii) in Theorem 2 is evidently true if $F_k$ are isomorphisms. Such topologically transitive operators for the case of Hilbert spaces were considered in [14]. The following example shows that mappings $F_k$ are not necessary isomorphisms.

Example 2.

Let X be the continual direct ${\mathsf{\ell }}_1$-sum of spaces $L_p[0,1].$ That is, every element in X can be represented as

$$x=\sum _{p\in [1,\infty )}{x}_p,\parallel x\parallel =\sum _{p\in [1,\infty )}{\parallel x_p\parallel }_{L_p}<\infty ,x_p\in L_p[0,1]$$

and only a countable number of terms are nonzero. We will use notation

$$X=\underset{p\in [1,\infty )}{⨁}{L}_p[0,1].$$

The space X can be represented as a Schauder decomposition $X={⨁}_{n=0}^{\infty }{X}_n,$ where

$$X_n=\underset{p\in [n+1,n+2)}{⨁}{L}_p[0,1].$$

For $k\ge 1$, we define $F_k:X_k\to X_{k-1}$ so that $F_k$ is the identical embedding of each $L_p[0,1]$ to $L_{p-1}[0,1]$ for $p\in [k+1,k+2).$ Since for every $r<p$ space, $L_p[0,1]$ is a dense subspace of $L_r[0,1],$ operators $F_k$ are well defined and have dense ranges. Let us define the space $E_0\subset X_0$ as the set of elements

$$\sum _{p\in [1,2)}{x}_{p}suchthat{x}_p\in L_{\infty }[0,1]and\sum _{p\in [1,2)}{\parallel x_p\parallel }_{L_{\infty }}<\infty .$$

Then

$$F_n^{-1}\circ \cdots \circ F_1^{-1}(\sum _{p\in [1,2)}{x}_p)=\parallel \sum _{p\in [1,2)}{x}_p{\parallel }_{X_n}\le \sum _{p\in [1,2)}{\parallel x_p\parallel }_{L_{\infty }}.$$

Thus, by Theorem 2, for every constant ${\lambda }_k$ such that $\underset{n\to \infty }{lim\; sup}{\prod }_{k=1}^n{\lambda }_k=\infty ,$ the operator

$$T_{\Lambda }\left(x\right)=\sum _{k=1}^{\infty }{\lambda }_k{F}_k\left(x_k\right)$$

is topologically transitive on $X.$

Let us recall that the von Neumann algebra can be defined as a $C^{*}$-algebra which is dual to a Banach space [26]. According to [15], if X is a non-reflexive quotient of a von Neumann algebra, then X does not admit a topologically transitive operator. For example, it is so if $X={\mathsf{\ell }}_{\infty }$ or X is the space of all bounded linear operators on a Hilbert space.

Corollary 1.

If a Banach space does not admit a topologically transitive operator, then it cannot be represented as a countable Schauder decomposition to a sequence of isomorphic Banach spaces.

It is easy to see that the ${\mathsf{\ell }}_p$-sum, ${\oplus }_{{\mathsf{\ell }}_p}X$ (or $c_0$-sum, ${\oplus }_{c_0}X$) of infinitely many copies of a Banach space X satisfies the conditions of Theorem 2 for a naturally defined backward shift if $1\le p<\infty$ because ${\oplus }_{{\mathsf{\ell }}_p}X$ (and ${\oplus }_{c_0}X$) is represented as a countable Schauder decomposition to isomorphic Banach spaces (c.f. [7]). Using the observation as in Example 1, we can construct also a backward shift operator for the c-sum of copies of a Banach space X.

Definition 1.

Let X be a Banach space. We denote by $W={\oplus }_{c}X$ the c-sum of copies of X that is the Banach space containing all elements of the form

$$w=(x_1,x_2,\dots ,x_k,\dots ),x_k\in X$$

such that there exists a limit

$$\underset{n\to \infty }{lim}{x}_n\in X,$$

and endowed with the norm

$$\parallel w\parallel =\underset{n}{sup}{\parallel x_n\parallel }_X.$$

Note that the representation $W={\oplus }_{c}X$ is not a Schauder decomposition because the formal sum ${\sum }_{n=1}^{\infty }{x}_n$ does not converge (in general) to w. But, like in Example 1, we can construct a countable Schauder decomposition of $W.$

Example 3.

Let

$$G_0\left(w\right)=\underset{n\to \infty }{lim}{x}_n.$$

Clearly, $G_0$ is a surjection onto $X.$ Any element $w=(x_1,x_2,\dots ,x_k,\dots )\in W$ can be represented as

$$w=(G_0\left(w\right),G_0\left(w\right),\dots )+(x_1-G_0\left(w\right),x_2-G_0\left(w\right),\dots ,x_k-G_0\left(w\right),\dots ).$$

Since $(x_1-G_0\left(w\right),x_2-G_0\left(w\right),\dots ,x_k-G_0\left(w\right),\dots )\in {\oplus }_{c_0}X,$ it follows that

$$W=X\oplus \left({\oplus }_{c_0}X\right)={\oplus }_{c_0}X$$

is a countable Schauder decomposition of $W.$ If $\Lambda =\left({\lambda }_n\right)$ satisfies condition (i) of Theorem 2, then

$$T_{\Lambda }:(x_1,\dots ,x_k,\dots )\mapsto {\lambda }_1(x_1,x_1,x_1,\dots )+\left({\lambda }_2(x_2-G_0\left(x\right)),\dots ,{\lambda }_k(x_k-G_0\left(x\right)),\dots \right)$$

is topologically transitive.

3. The Backward Shift and Analytic Functions of Unbounded Type

In this section, we consider another generalization of the backward shift operator and its application to the analytic function of the unbounded type.

Let us recall that $P_n$ is an n-homogeneous (scalar-valued) polynomial on a Banach space X if there exists a (necessary unique) n-linear symmetric form $B_{P_n}$ on the n-th Cartesian product of X such that

$$P_n\left(x\right)=B_{P_n}(x,\dots ,x).$$

It is well known that $P_n$ is continuous if and only if $B_{P_n}$ is continuous and it is equivalent that $P_n$ is bounded on every bounded subset of $X.$ A finite sum of homogeneous polynomials is just a polynomial. The Banach space of continuous n-homogeneous polynomials on $X,$ endowed with the norm

$$\parallel P_n\parallel =\underset{\parallel x\parallel \le 1}{sup}\left|P_n\left(x\right)\right|$$

is denoted by $\mathcal{P}{(}^{n}X).$ Zero-homogeneous polynomials coincide with the constant functions.

Functions on X of the form

$$f\left(x\right)=\sum _{n=0}^{\infty }{f}_n\left(x\right),f_n\in \mathcal{P}{(}^{n}X)$$

are called (entire) analytic functions on $X.$ An analytic function f on a Banach space X is said to be a function of the bounded type if it is bounded on all bounded subsets of $X.$ We denote by $H\left(X\right)$ the space of all analytic functions on X and by $H_b\left(X\right)$ the subspace of analytic functions of the bounded type. It is well known that if X is infinite-dimensional, then $H_b\left(X\right)$ is a proper subset of $H\left(X\right).$ Elements of $H\left(X\right)\setminus H_b\left(X\right)$ are called analytic functions of the unbounded type.

In [20], the following theorem was proved.

Theorem 3.

Let us suppose that there is a dense subset $\Omega \subset X$ and a sequence of polynomials $P_n\in \mathcal{P}{(}^{n}X),$ ${lim\; sup}_{n\to \infty }{\parallel P_n\parallel }^{1/n}=c,$ $0<c<\infty$ such that for every $z\in \Omega$ there exists $m\in \mathbb{N}$ with the property that for every $y\in X,$

$$B_{P_n}(\underset{k}{\underbrace{z,\dots ,z}},\underset{n-k}{\underbrace{y,\dots ,y}})=0$$

for all $k>m$ and $n>k,$ where $B_{P_n}$ is the symmetric n-linear mapping associated with $P_n.$ Then

$$g\left(x\right)=\sum _{n=1}^{\infty }{P}_n\left(x\right)\in H\left(X\right)\setminus H_b\left(X\right).$$

Theorem 4.

Let $P_n$ be a sequence of n-homogeneous polynomials on a Banach space X with $\parallel P_n\parallel =1$ and $T:X\to X$, a bounded linear operator satisfying

$$0<\underset{n\to \infty }{lim\; sup}{\parallel P_n\circ T^n\parallel }^{1/n}<\infty .$$

Suppose that there exists a dense subspace $Z_0\subset X$ such that for every $z\in Z_0$, there is a number N such that $T^N\left(z\right)=0.$ Then

$$f\left(x\right)=\sum _{n=1}^{\infty }{P}_n\circ T^n\left(x\right)$$

is an analytic function of unbounded type on $X.$

Proof. 

Let $B_{P_n}$ be the symmetric n-linear form associated with $P_n.$ Then for $f_n=P_n\circ T^n\left(x\right)$ we have

$$B_{f_n}(x_1,\dots ,x_n)=B_{P_n}(T^n\left(x_1\right),\dots ,T^n\left(x_n\right)).$$

Hence, for every $z\in Z_0$ there exists a number N such that for every $x\in X$ and $n>N,$

$$B_{f_n}(\underset{m}{\underbrace{T^n\left(z\right),\dots ,T^n\left(z\right)}},\underset{n-m}{\underbrace{x,\dots ,x}})=0,0<m\le n.$$

By Theorem 3, f is of the unbounded type. □

Definition 2.

We say that a bounded operator $A:X\to X$ is a generator of analytic functions of the unbounded type if for every sequence of n-homogeneous polynomials $P_n$ on $X,$ $\parallel P_n\parallel =1,$

$$f\left(x\right)=\sum _{n=1}^{\infty }{P}_n\circ A^n\left(x\right)$$

is an analytic function of the unbounded type on $X.$

It is easy to check that if A is a generator of analytic functions of the unbounded type on $X,$ then for every sequence of $n_k$-homogeneous polynomials $P_{n_k}$ on $X,$ $\parallel P_{n_k}\parallel =1,$ the function

$$f\left(x\right)=\sum _{k=1}^{\infty }{P}_{n_k}\circ A^{n_k}\left(x\right)$$

is in $H\left(X\right)\setminus H_b\left(X\right),$ where $\left(n_k\right)$ is a strictly increasing sequence of positive integers.

Proposition 1.

Let $X={\mathsf{\ell }}_p,$ $1\le p<\infty$ or $X=c_0,$ and $T=T_1$ be the backward shift. Then T is a generator of analytic functions of the unbounded type.

Proof. 

According to Theorem 4, it is enough to show that for every sequence of n-homogeneous polynomials $\left(P_n\right)$ on X with $\parallel P_n\parallel =1$, we have

$$0<\underset{n\to \infty }{lim\; sup}{\parallel P_n\circ T^n\parallel }^{1/n}<\infty .$$

But if $\parallel x\parallel =1,$ $x=(x_1,x_2,\dots )\in X,$ then $\parallel (0,\dots ,0,x_1,x_2,\dots )\parallel =\parallel x\parallel =1,$ and $x=T^n(0,\dots ,0,x_1,x_2,\dots ).$ Thus, $\parallel P_n\circ T^n\parallel =\parallel P_n\parallel =1$ and so

$$\underset{n\to \infty }{lim\; sup}{\parallel P_n\circ T^n\parallel }^{1/n}=1.$$

□

Note that, in a similar way, the backward shift associated with a Schauder decomposition of a Banach space satisfies the conditions of Theorem 4.

Example 4.

Let X be $c_0$ or ${\mathsf{\ell }}_p$ for $1\le p<\infty ,$ $T=T_1$ be the backward shift, and $P_n={\left(e_1^{*}\right)}^n,$ where $e_1^{*}$ is the first coordinate functional. Then $f_n\left(x\right)=P_n\circ T^n\left(x\right)=x_n^n$, and so

$$f\left(x\right)=\sum _{n=1}^{\infty }{x}_n^n$$

is a well-known example of a function of the unbounded type.

Example 5.

Let $X={\mathsf{\ell }}_1,$ $T=T_1$ and

$$P_n\left(x\right)=\sum _{k=1}^{\infty }{x}_k^n.$$

Then

$$f_n\left(x\right)=P_n\circ T^n\left(x\right)=\sum _{k=n+1}^{\infty }{x}_k^{n}andf\left(x\right)=\sum _{n=1}^{\infty }{f}_n\left(x\right)\in H\left(X\right)\setminus H_b\left(X\right).$$

Proposition 2.

Let A be a generator of analytic functions of the unbounded type on $X.$ Then the point spectrum ${\sigma }_p\left(A^{*}\right)$ of the adjoint operator $A^{*}$ is empty.

Proof. 

Suppose that there is a functional $\phi \in X^{*},$ $\parallel \phi \parallel =1$ such that $A^{*}\left(\phi \right)=\lambda \phi$ for some $\lambda \in \mathbb{C}.$ Set $P_n={\phi }^n.$ Then

$$\sum _{n=1}^{\infty }{\phi }^n\circ A^n\left(x\right)=\sum _{n=1}^{\infty }{\left(\phi \circ A^n\left(x\right)\right)}^n=\sum _{n=1}^{\infty }{\left({\lambda }^n\right)}^n{\left(\phi \left(x\right)\right)}^n$$

is an analytic function of the bounded type if $\left|\lambda \right|<1$ and the series diverges if $\left|\lambda \right|\ge 1$ at any x such that $\left|\phi \right(x\left)\right|>1$: a contradiction. □

Corollary 2.

Let X be a non-reflexive quotient of a von Neumann algebra. Then X does not support a generator of analytic functions of the unbounded type.

Proof. 

By Proposition 3, the point spectrum of a such operator should be empty. But, according to ([15] (Th. 2.4)), the point spectrum of any operator on a non-reflexive quotient of a von Neumann algebra is nonempty. □

Let us consider some properties of generators of analytic functions of the unbounded type.

Proposition 3.

Let A be a generator of analytic functions of the unbounded type and λ be a number such that $\left|\lambda \right|\ne 1.$ Then $\lambda A$ is not a generator of analytic functions of the unbounded type.

Proof. 

Let $P_n$ be a sequence of n-homogeneous polynomials on X and $\parallel P_n\parallel =1.$ Then

$$\left(P_n\circ {\left(\lambda A\right)}^n\right)\left(x\right)={\left({\lambda }^n\right)}^n{P}_n\circ A^n\left(x\right).$$

Since A is a generator of analytic functions of the unbounded type,

$$f\left(x\right)=\sum _{n=1}^{\infty }{P}_n\circ A^n\left(x\right)$$

is an analytic function of the unbounded type and so

$$0<\underset{n}{sup}{\parallel P_n\circ A^n\parallel }^{1/n}<\infty .$$

Thus,

$$\underset{n}{sup}\parallel P_n\circ {\left(\lambda A\right)}^n{\parallel }^{1/n}=\underset{n}{sup}{\lambda }^n{\parallel P_n\circ A^n\parallel }^{1/n}=\left\{\begin{array}{cc}0& \mathrm{if}\left|\lambda \right|<1,\\ \infty & \mathrm{if}\left|\lambda \right|>1.\end{array}\right.$$

□

Note that Proposition 3 does not imply that the norm of a generator of analytic functions of the unbounded type must be equal to one.

Example 6.

Let $\Lambda =({\lambda }_1,1,1,\dots ),$ and ${\lambda }_1>1.$ Then the weighted backward shift $T_{\Lambda }$ on ${\mathsf{\ell }}_p,$ $1\le p<\infty$ or $c_0$ satisfies the conditions of Theorem 4 but $\parallel T_{\Lambda }\parallel ={\lambda }_1>1.$

Proposition 4.

Let A be a generator of analytic functions of the unbounded type on a complex Banach space $X.$ Then $A^n\left(x_0\right)\to 0$ as $n\to \infty$ for every $x_0\in X.$

Proof. 

Let $x_0\ne 0$ and ${\phi }_n$ be a norm one linear functional such that ${\phi }_n\left(x_0\right)=\parallel A^n\left(x_0\right)\parallel .$ Such a functional must exist for every $n\in \mathbb{N}$ by the Hahn–Banach theorem. Put $P_n={\phi }_n^n.$ Then $\parallel P_n\parallel =1$ and

$$\sum _{n=1}^{\infty }{P}_n\circ A^n\left(x\right)=\sum _{n=1}^{\infty }{\left({\phi }_n\circ A^n\left(x\right)\right)}^n$$

is a function of the unbounded type and, in particular, the series converges for every $x\in X.$ Setting $x=t{x}_0,$ we have that

$$\sum _{n=1}^{\infty }{t}^n{\left({\phi }_n\circ A^n\left(x_0\right)\right)}^n=\sum _{n=1}^{\infty }{t}^n{\parallel A^n\left(x_0\right)\parallel }^n$$

converges at any $t\in \mathbb{C}.$ From the Cauchy–Hadamard theorem, $\parallel A^n\left(x_0\right)\parallel \to 0$ as $n\to \infty$. □

The following theorem shows that under some conditions, $\lambda A$ will be topologically transitive for a generator of analytic functions of the unbounded type $A.$

Theorem 5.

Let X be a Banach space and A a generator of analytic functions of the unbounded type on $X.$ Suppose that A is surjective, $\parallel A\parallel \ge 1$ and there is a dense subspace $Z_0\subset X$ such that

$$C^n{A}^n\left(x\right)\to 0asn\to \infty$$

for every $C>0$ and $x\in Z_0.$ Then there is a constant c such that for every $\lambda >c,$ the operator $\lambda A$ is topologically transitive.

Proof. 

Since A is surjective, by the open mapping theorem, there is a right inverse continuous operator $D,$ that is, $AD\left(x\right)=x,$ $x\in X.$

Let $T=\lambda A,$ where $\lambda >c:=\parallel D\parallel ,$ and $S=D/\lambda .$ Then T obviously satisfies the conditions of Theorem 1 for $n_k=k,$ $S_k=S^k,$ $Y_0=X.$ Thus, $T=\lambda A$ is topologically transitive. □

4. Quasi-Extension of Analytic Functions

Let us denote by $Hol\left(0\right)$ the linear space of germs of analytic functions at zero in a complex Banach space $X.$ In other words, $Hol\left(0\right)$ consists of pairs $(f,\mathrm{dom}f),$ where $\mathrm{dom}f$ is an open set containing zero and f is analytic on the domain $\mathrm{dom}f.$ It is well known that any analytic function on an open set is locally bounded. In particular, for every $f\in Hol\left(0\right)$, there exists $r>0$ such that f is bounded on the open ball $B_{r,0}$ of radius r centered at the origin. The supremum over all r such that f is bounded on $B_{r,0}$ is called the radius of boundedness of f and denoted by ${\varrho }_0\left(f\right).$ The function f has the Taylor series representation in the open ball of radius ${\varrho }_0\left(f\right),$

$$f\left(x\right)=\sum _{n=0}^{\infty }{f}_n\left(x\right),f_n\in \mathcal{P}{(}^{n}X),$$

and the series converges absolutely and uniformly on the closed ball ${\overline{B}}_{r,0}$ for every $0<r<{\varrho }_0\left(f\right)$. The radius of the boundedness of f can be computed by the following infinite-dimensional analogue of the Cauchy–Hadamard formula

$${\varrho }_0\left(f\right)={\left(\underset{n\to \infty }{lim\; sup}{\parallel f_n\parallel }^{1/n}\right)}^{-1}.$$

Note that in contrast with the finite-dimensional case, the Taylor series of f may converge pointwise to f outside of the ball ${\overline{B}}_{{\varrho }_0\left(f\right),0}.$ For example, if f is an entire function of the unbounded type, then ${\varrho }_0\left(f\right)<\infty$, while f is well defined at each point of X by its Taylor series expansion. Germs of analytic functions on compact subsets of a Banach space were studied in [27].

Let A be a generator of analytic functions of the unbounded type on X and $f={\sum }_{n=0}^{\infty }{f}_n$ be an analytic function, defined on a neighborhood of the origin. Suppose that the radius of boundedness of f at zero is equal to $r.$ Set

$$\mathcal{A}\left(f\right)=\sum _{n=0}^{\infty }{f}_n\circ A^n.$$

Then, by the definition of the generator of analytic functions of unbounded type, $\mathcal{A}\left(f\right)\in H\left(X\right)$ and $\mathcal{A}\left(f\right)$ is of the unbounded type if $r<\infty .$ More specifically, we have the following theorem.

Theorem 6.

(i) 

If $0<{\varrho }_0\left(f\right)=r<\infty ,$ then ${\varrho }_0\left(\mathcal{A}\left(f\right)\right)<\infty$ and so $\mathcal{A}\left(f\right)$ is an analytic function of the unbounded type.

(ii) 

$\mathcal{A}$is a linear operator from$Hol\left(0\right)$ to $H\left(X\right).$

Proof. 

(i). Let $\parallel f_n\parallel =c_n.$ Set $P_n=f_n/c_n$ if $c_n\ne 0$ and $P_n=0$ if $c_n=0.$ Since ${\varrho }_0\left(f\right)=r,$

$$\underset{n\to \infty }{lim\; sup}{\parallel f_n\parallel }^{1/n}=\underset{n\to \infty }{lim\; sup}{c}_n^{1/n}=\frac{1}{r},n\in \mathbb{N}.$$

Let $n_k$ be a subsequence of natural numbers such that

$$\underset{k\to \infty }{lim}{c}_{n_k}^{1/n_k}=\frac{1}{r}.$$

On the other hand, by the definition of a generator of analytic functions of the unbounded type, the function

$$g\left(x\right)=\sum _{n=1}^{\infty }{P}_{n_k}\circ A^{n_k}\left(x\right)$$

is in $H\left(X\right)\setminus H_b\left(X\right).$ Thus, g is well defined and analytic on X and the radius of boundedness ${\varrho }_0\left(g\right)$ of g is finite and positive. So,

$${\left({\varrho }_0\left(\mathcal{A}\left(f\right)\right)\right)}^{-1}=\underset{k\to \infty }{lim\; sup}\parallel c_{n_k}{P}_{n_k}\circ A^{n_k}{\parallel }^{1/n_k}=\frac{1}{r{\varrho }_0\left(g\right)},$$

that is, $0<{\varrho }_0\left(\mathcal{A}\left(f\right)\right)=r{\varrho }_0\left(g\right)<\infty .$

Let us show that $\mathcal{A}\left(f\right)$ is well defined at any point $x\in X.$ Since ${\sum }_{n=1}^{\infty }{P}_n\circ A^n\in H\left(X\right),$ the function of a complex variable $t,$

$$g_x\left(t\right):=\sum _{n=1}^{\infty }{P}_n\circ A^n\left(tx\right)=\sum _{n=1}^{\infty }{t}^n{P}_n\circ A^n\left(x\right),$$

is an analytic function on $\mathbb{C}$ for the fixed $x\in X.$ Thus, by the Cauchy–Hadamard theorem,

$$\underset{n\to \infty }{lim\; sup}|P_n\circ A^n\left(x\right){|}^{1/n}=0.$$

Hence,

$$\underset{n\to \infty }{lim\; sup}{c}_n^{1/n}|P_n\circ A^n\left(x\right){|}^{1/n}=0/r=0,$$

and so

$$\mathcal{A}\left(f\right)\left(tx\right)=\sum _{n=0}^{\infty }{t}^n{f}_n\circ A^n\left(x\right)$$

is well defined for every $t\in \mathbb{C}$ (in particular, for $t=1$) and for every $x\in X.$

(ii). Every $f\in Hol\left(0\right)$ is an analytic function in a neighborhood of zero. By item (i) above, $\mathcal{A}$ maps f to $H\left(X\right).$ Thus, $\mathcal{A}$ is an operator from $Hol\left(0\right)$ to $H\left(X\right).$ It is easy to check that $\mathcal{A}$ is linear. □

Therefore, the function $\mathcal{A}\left(f\right)$ is well defined on the whole space X, while f is not, in the general case. In this way, we can consider the operator $\mathcal{A}$ as some kind of universal “extension” operator. It is not a real extension because the restriction of $\mathcal{A}\left(f\right)$ to the domain of f does not coincide with $f.$ Moreover, $\mathcal{A}\left(f\right)$ is not multiplicative. Note that spaces $Hol\left(0\right)$ and $H\left(X\right)$ can be naturally topologized [27], and we can ask about the continuity of $\mathcal{A}\left(f\right).$ But this question is outside of the scope of our paper.

Definition 3.

Let $H_0$ be a linear subspace of $Hol\left(0\right).$ We say that an operator $\mathcal{B}:H_0\to H\left(X\right)$ is a quasi-extension operator if for every $x\in X$ and $r>\parallel x\parallel$ there is a sequence $x_k$ in the open ball $B_{r,0}$ of radius r centered at zero such that for every $f\in H_0$ with ${\varrho }_0\left(f\right)>r$, we have

$$f\left(x\right)=\underset{k\to \infty }{lim}\mathcal{B}\left(f\right)\left(x_k\right).$$

The following theorem shows that under some condition, a subspace $H_0\subset Hol\left(0\right)$ admits a quasi-extension operator.

Theorem 7.

Let A, $\parallel A\parallel =1$, be a generator of analytic functions of the unbounded type on a Banach space $X.$ Suppose that there exists a right inverse operator R to A with $\parallel R\parallel =1.$ Let $H_0$ be a subspace of $Hol\left(0\right)$ consisting of the R-invariant function. That is, if $f\in H_0,$ then $R\left(x\right)\in \mathrm{dom}f$ for every $x\in \mathrm{dom}f$ and $f\circ R\left(x\right)=f\left(x\right).$ Then the restriction of $\mathcal{A}$ to $H_0$ is a quasi-extension operator on $H_0.$

Proof. 

Let $x\in B_{r,0}.$ We define $x_k=R^k\left(x\right).$ Let $f\in H_0,$ $r<{\varrho }_0\left(f\right),$ and

$$f\left(x\right)=\sum _{n=0}^{\infty }{f}_n\left(x\right),f_n\in \mathcal{P}{(}^{n}X)$$

be the Taylor series expansion of f in $B_{{\varrho }_0\left(f\right),0}.$ Since $\parallel R\parallel =1,$ it follows that $x_k\in B_{r,0}.$ According to the assumption of the theorem, f is invariant with respect to the action of $R.$ But R is linear and so $deg{f}_n\circ R=deg{f}_n=n.$ Thus, each polynomial $f_n$ must be R-invariant. Hence,

$$\mathcal{A}\left(f\right)\left(x_k\right)=\sum _{n=0}^{\infty }{f}_n\circ A^n\circ R^k\left(x\right)=\sum _{n=0}^k{f}_n\left(x\right)+\sum _{n=k+1}^{\infty }{f}_n\circ A^{n-k}\left(x\right)\to f\left(x\right)$$

as $k\to \infty$ because from the assumption $\parallel A\parallel =1$, it follows that

$$|\sum _{n=k+1}^{\infty }{f}_n\circ A^{n-k}\left(x\right)|\le \sum _{n=k+1}^{\infty }\parallel f_n{\parallel \parallel x\parallel }^n\to 0$$

as $k\to \infty .$ □

Next we construct a subspace $H_0$ of germs of analytic functions on ${\mathsf{\ell }}_p$ which satisfies the conditions of Theorem 7 in the case when $A=T$ is the backward shift on ${\mathsf{\ell }}_p,$ $1\le p<\infty .$

Let us recall that a polynomial P on ${\mathsf{\ell }}_p,$ $1\le p<\infty$ is symmetric if it is invariant with respect to all permutations of basis vectors of the canonical basis $e_n=(\underset{n}{\underbrace{0,\dots ,0,1}},0,\dots )$ in ${\mathsf{\ell }}_p.$ It is well known (see [28,29]) that polynomials

$$P_k=\sum _{n=1}^{\infty }{x}_n^k,k\ge \lceil p\rceil$$

form an algebraic basis in the algebra of all symmetric polynomials on ${\mathsf{\ell }}_p,$ that is, every symmetric polynomial can be uniquely represented as a finite algebraic combination of $\left(P_k\right),$ $k\ge \lceil p\rceil ,$ where $\lceil p\rceil$ is the minimal integer, which is greater than or equal to $p.$ A polynomial P on ${\mathsf{\ell }}_p$ is subsymmetric if it is invariant with respect to the following operators ${\mathcal{C}}_j$ on ${\mathsf{\ell }}_p$:

$${\mathcal{C}}_j:(x_1,\dots ,x_n,\dots )\mapsto (x_1,\dots ,x_{j-1},0,x_j,x_{j+1}\dots ),j\in \mathbb{N}.$$

The existence of an algebraic basis in the algebra of all subsymmetric polynomials on ${\mathsf{\ell }}_p$ [30,31] is unknown, but polynomials

$$P_{k_1,\dots ,k_n}\left(x\right)=\sum _{i_1<\cdots <i_n}{x}_{i_1}^{k_1}\cdots x_{i_n}^{k_n},k_j\ge \lceil p\rceil ,k_1+\cdots +k_n=n$$

form a linear basis in the linear space of n-homogeneous subsymmetric polynomials on ${\mathsf{\ell }}_p$ [32]. More information about spaces and algebras generated by symmetric and subsymmetric polynomials can be found in [33,34,35,36,37] and the references therein. It is easy to see that if P is symmetric, then it is subsymmetric, and the inverse statement is not true. Clearly, every subsymmetric polynomial is invariant with respect to the forward shift $R={\mathcal{C}}_1.$ However, the following polynomial Q is R-invariant but not subsymmetric on ${\mathsf{\ell }}_1$:

$$Q\left(x\right)=\sum _{i=1}^{\infty }{x}_i{x}_{i+1}^2.$$

Example 7.

Let $X={\mathsf{\ell }}_p,$ $1\le p<\infty ,$ $A=T$ be the backward shift, and $H_0$ be the subspace of all R-invariant analytic germs in $Hol\left(0\right),$ where $R\left(x\right)=(0,x_1,x_2,\dots )$ is the forward shift in ${\mathsf{\ell }}_p.$ Then, R is the right inverse to T and $H_0$ is nontrivial because, as it is mentioned above, it contains subsymmetric polynomials. Thus, by Theorem 7, $\mathcal{A}$ is a quasi-extension operator on $H_0.$

The following example shows that $\mathcal{A}$ can be a quasi-extension operator on $H_0$ even if $H_0$ is not R invariant.

Example 8.

Let $X=c_0$ and $A=T$ be the backward shift. Let $Ho{l}_{\mathbb{C}}\left(0\right)$ be the space of all analytic germs of a complex variable. For every

$$\gamma \left(z\right)=\sum _{n=0}^{\infty }{\gamma }_n{z}^n\in Ho{l}_{\mathbb{C}}\left(0\right),z\in \mathbb{C}$$

we assign a function $f_{\gamma }\left(x\right)\in Hol\left(0\right)$ by

$$f_{\gamma }(x_1,x_2,\dots )=\gamma \left(x_1\right).$$

Let $H_0=\{f_{\gamma }:\gamma \in Ho{l}_{\mathbb{C}}\left(0\right)\}\subset Hol\left(0\right).$ Then

$$\mathcal{A}\left(f_{\gamma }\right)\left(x\right)=\sum _{n=0}^{\infty }{\gamma }_n{x}_{n+1}^n.$$

We claim that $\mathcal{A}$ is a quasi-extension operator. Indeed, let $\left|z\right|<{\varrho }_0\left(\gamma \right).$ Then

$$\parallel (\underset{k}{\underbrace{z,\dots ,z}},0,0,\dots )\parallel =\left|z\right|<{\varrho }_0\left(f_{\gamma }\right),$$

and

$$\mathcal{A}\left(f_{\gamma }\right)(\underset{k}{\underbrace{z,\dots ,z}},0,0,\dots )=\sum _{n=0}^k{\gamma }_n{z}^n\to \gamma \left(z\right)=f_{\gamma }(z,0,0,\dots )ask\to \infty .$$

5. Conclusions

In the paper, we constructed the most general analogs of the weighted backward shift operator for (possible non-separable) Banach spaces, admitting a Schauder decomposition. We proved that under some natural conditions, such operators are topologically transitive. Next, we observed that properties of the backward shift are similar to conditions of Theorem 3. Using this observation, we proposed a definition of generators of analytic functions of the unbounded type and showed that the backward shift satisfies the definition. By Corollary 2, if X is a non-reflexive quotient of a von Neumann algebra, it does not support a generator of analytic functions of the unbounded type. So, we obtained precisely the conditions prohibiting the existence of topologically transitive operators as in [15]. We found conditions implying that multiplication by a constant generator of analytic functions of the unbounded type is topologically transitive. In addition, applying the obtained results, we showed that the backward shift has some kind of extension property that we call a quasi-extension. We considered quasi-extension operators in general and proposed some examples.