Classes of Entire Analytic Functions of Unbounded Type on Banach Spaces

Abstract

In this paper we investigate analytic functions of unbounded type on a complex infinite dimensional Banach space $X.$ The main question is: under which conditions is there an analytic function of unbounded type on X such that its Taylor polynomials are in prescribed subspaces of polynomials? We obtain some sufficient conditions for a function f to be of unbounded type and show that there are various subalgebras of polynomials that support analytic functions of unbounded type. In particular, some examples of symmetric analytic functions of unbounded type are constructed.

1. Introduction and Preliminaries

Let X be an infinite dimensional complex Banach space. A function $P:X\to \mathbb{C}$ is an n-homogeneous polynomial if there exists a symmetric n-linear map $B_P$ defined on the Cartesian power $X^n$ to $\mathbb{C}$ such that $P\left(x\right)=B_P(x,\dots ,x).$ The mapping $B_P$ is called n-linear form associated with P and is necessary and unique because of the well-known polarization formula (see, e.g., p. 4 [1]). The Banach space of all continuous n-homogeneous polynomials on X with respect to the norm

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

is denoted by $\mathcal{P}{(}^{n}X).$ For us the following version of the polarization inequality is important (p. 8 [1]).

$$|B_P(x_1^{n_1},\dots ,x_k^{n_k})| \le \frac{n_1!\cdots n_k!m^m}{n_1^{n_1}\cdots n_k^{n_k}m!}\parallel P\parallel \parallel x_1{\parallel }^{n_1}\cdots {\parallel x_k\parallel }^{n_k},$$

(1)

where $n_1,\dots ,n_k$ are positive integers with $n_1+\cdots +n_k=m.$

A continuous function $f:X\to \mathbb{C}$ is said to be an entire analytic function (or just entire function) if its restriction on any finite dimensional subspace is analytic. The space of all entire analytic functions on X is denoted by $H\left(X\right).$ For every entire function f there exists a sequence of continuous n-homogeneous polynomials (so-called Taylor polynomials) such that

$$f\left(x\right)=\sum _{n=0}^{\infty }{f}_n\left(x\right)$$

(2)

and the series converges for every $x\in X.$ Here $f_0=f\left(0\right)$ is a constant. The Taylor series expansion (2) uniformly converges on the open ball $rB$ centered at zero with radius $r={\varrho }_0\left(f\right),$ where

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

The radius $r={\varrho }_0\left(f\right)$ is called the radius of uniform convergence of f or the radius of boundedness of f because the ball $rB$ is the largest open ball at zero such that f is bounded on it. If ${\varrho }_0\left(f\right)=\infty ,$ then f is bounded on all bounded subsets of X and is called a function of bounded type. The set of all functions of bounded type on X is denoted by $H_b\left(X\right).$

It is known that $H_b\left(X\right)$ is a Fréchet algebra with respect to the topology of uniform convergence on bounded sets. Algebras $H_b\left(X\right)$ were considered first in [2,3] and studied by many authors for various Banach spaces $X.$

A polynomial P on X is called a polynomial of finite type if it is a finite sum of finite products of linear functionals and constants. We denote by ${\mathcal{A}}_n\left(X\right)$ the smallest closed subalgebra of $H_b\left(X\right)$ containing the space of all n-homogeneous polynomials ${\mathcal{P}}_n\left(X\right).$ In particular, ${\mathcal{A}}_1\left(X\right)$ is the closure of the space of finite type polynomials in $H_b\left(X\right).$ In the general case, ${\mathcal{A}}_n\left(X\right)\ne {\mathcal{A}}_k\left(X\right)$ if $n\ne k.$ For example, it is so for $X={\ell }_1$ (see [4] for details).

We say that a function f is an entire function of unbounded type if $f\in H\left(X\right)\setminus H_b\left(X\right).$ It is known that for given weak*-null sequences ${\varphi }_n\in X^{*},$ $\parallel \varphi \parallel =1$, which always exists (see p. 157 [5]) the function

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

(3)

is an entire function of unbounded type on $X.$ It is easy to see that the Taylor polynomials $f_n={\varphi }_n^n$ of f are polynomials of finite type. In this paper we consider the following natural question.

Question 1.

Let ${\mathcal{P}}_0{(}^{n}X)$ be subspaces of $\mathcal{P}{(}^{n}X),$ $n\in \mathbb{N}.$ Under which conditions is there a function $f={\sum }_{n=0}^{\infty }{f}_n\in H\left(X\right)\setminus H_b\left(X\right)$ such that $f_n\in {\mathcal{P}}_0{(}^{n}X)$?

In Section 2 we obtain some general results on entire functions of unbounded type. In particular, we show that if X is such that ${\mathcal{A}}_m\left(X\right)\ne {\mathcal{A}}_{m-1}\left(X\right)$ for some $m>1,$ then there exists a function $f\in H\left(X\right)\setminus H_b\left(X\right)$ such that all its Taylor polynomials are in ${\mathcal{A}}_m\left(X\right).$ Also, we establish some sufficient conditions for a function f to be in $H\left(X\right)\setminus H_b\left(X\right).$ In Section 3 the obtained results are applied to construct examples of symmetric analytic functions of unbounded type on ${\ell }_p,$ $1\le p\le \infty .$ These examples can be considered as generalizations of the example, constructed in [6] on ${\ell }_1.$ In addition, the paper contains some discussions and open questions.

For details on analytic functions of bounded type we refer the reader to [1,5,7]. Entire analytic functions of unbounded type were investigated in [8,9,10]. Symmetric analytic functions on Banach spaces were studied in [11,12,13,14,15,16,17,18].

2. General Results

Evidently, the set of entire functions of unbounded type is not a linear space. Moreover, the following example shows that the product of two functions of unbounded type is not necessarily a function of unbounded type because there are invertible analytic functions of unbounded type.

Example 1.

Let $X={\ell }_p$ or $c_0$ for $1\le p<\infty$ and ${\varphi }_n$ be the coordinate functionals multiplied by $-1$

$${\varphi }_n:x=(x_1,x_2,\dots ,x_n,\dots )\mapsto -x_n.$$

So the function f of the form (3) is of unbounded type. We set

$$g\left(x\right)=e^{f\left(x\right)}.$$

Clearly, $g\in H\left(X\right)\setminus H_b\left(X\right)$ because it is unbounded on the bounded set

$$z_j=(\underset{j}{\underbrace{0,\dots ,0,2}},0,\dots ),sinceg\left(z_{2j}\right)=e^{2^{2j}}.$$

On the other hand,

$$\frac{1}{g\left(z_{2j+1}\right)}=e^{-f\left(z_{2j+1}\right)}=e^{2^{2j+1}}$$

and so $1/g\in H\left(X\right)\setminus H_b\left(X\right).$ However, the product of g and $1/g$ equals $1,$ which is bounded.

Note that $H\left(X\right)\setminus H_b\left(X\right)$ surprisedly contains infinite dimension linear subspaces and subalgebras without the zero vector [19].

The following proposition shows that the set of entire functions of unbounded type has some kind of ideal property.

Proposition 1.

Let

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

and $P_n\in \mathcal{P}{(}^{n}X),$ $0<c\le \parallel P_n\parallel \le C<\infty$ for some constants $c,C>0,$ and $n\in \mathbb{N}.$ Then

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

Proof. 

Let us show that g is well-defined on $X.$ Note first that since $f\left(x\right)$ is well-defined on $X,$ the series

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

is convergent on $X.$ Indeed, for every fixed $x\in X$ the function $\gamma \left(t\right):=f\left(tx\right),$ $t\in \mathbb{C}$ is in $H\left(\mathbb{C}\right)$ and so the power series

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

is absolutely convergent. It is true, in particular, for $t=1.$

For every $x\in X$

$$\left|g\left(x\right)\right|\le \sum _{n=1}^{\infty }|g_n\left(x\right)|=\sum _{n=1}^{\infty }|P_n\left(x\right)f_n\left(x\right)|\le C\sum _{n=1}^{\infty }\left|f_n\left(x\right)\right|<\infty .$$

To prove that g is analytic but not of bounded type we need to show that the radius of boundedness ${\varrho }_0\left(g\right)$ of g at zero satisfies $0<{\varrho }_0\left(g\right)<\infty .$ One can check that

$$\frac{1}{{\varrho }_0\left(g\right)}=\underset{n\to \infty }{lim\; sup}\parallel g_n{\parallel }^{1/n}=\underset{n\to \infty }{lim\; sup}{\parallel P_n{f}_n\parallel }^{1/2n}.$$

So

$$\underset{n\to \infty }{lim\; sup}{c}^{1/2n}\parallel f_n{\parallel }^{1/2n}\le \frac{1}{{\varrho }_0\left(g\right)}\le \underset{n\to \infty }{lim\; sup}{C}^{1/2n}{\parallel f_n\parallel }^{1/2n}.$$

That is, ${\varrho }_0\left(g\right)=\sqrt{{\varrho }_0\left(f\right)}$ and so $g\in H\left(X\right)\setminus H_b\left(X\right).$ □

Corollary 1.

Let X be such that ${\mathcal{A}}_m\left(X\right)\ne {\mathcal{A}}_{m-1}\left(X\right)$ for some $m>1.$ Then there exists a function $f\in H\left(X\right)\setminus H_b\left(X\right)$ such that all $f_n,$ $n\ge m$ are in ${\mathcal{A}}_m\left(X\right).$

Theorem 1.

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 $m\in \mathbb{N}$ such 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 $n>k.$ Then

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

Proof. 

Note first that

$${\varrho }_0\left(g\right)=\frac{1}{{lim\; sup}_{n\to \infty }{\parallel P_n\parallel }^{1/n}}=\frac{1}{c}$$

and so $0<{\varrho }_0\left(g\right)<\infty .$ Thus g is locally bounded and if it is well-defined on $X,$ then it belongs to $H\left(X\right)\setminus H_b\left(X\right).$

Let us show that g is well-defined on $X.$ Let $x\in X$ and $z\in \Omega$ such that $\parallel x-z\parallel <1/2c.$ Let $y=x-z.$ Then

$$g\left(x\right)=g(z+y)=\sum _{n=1}^{\infty }\sum _{k\le m}{B}_{P_n}(\underset{k}{\underbrace{z,\dots ,z}},\underset{n-k}{\underbrace{y,\dots ,y}}).$$

For every $0\le k\le m$ we set

$$g^{\left[k\right]}\left(y\right)=\sum _{n=1}^{\infty }{B}_{P_n}(\underset{k}{\underbrace{z,\dots ,z}},\underset{n-k}{\underbrace{y,\dots ,y}}).$$

For the fixed $z,$ $g^{\left[k\right]}\left(y\right)$ is an entire function of y and its j-homogeneous Taylor polynomial is

$$g_j^{\left[k\right]}\left(y\right)=B_{P_{k+j}}(\underset{k}{\underbrace{z,\dots ,z}},\underset{j}{\underbrace{y,\dots ,y}}).$$

Taking into account inequality (1) we have

$$\parallel g_j^{\left[k\right]}\parallel =\underset{\parallel y\parallel \le 1}{sup}|B_{P_{k+j}}(\underset{k}{\underbrace{z,\dots ,z}},\underset{j}{\underbrace{y,\dots ,y}})|\le \frac{{k!j!\parallel z\parallel }^k{(k+j)}^{(k+j)}}{k^k{j}^j(k+j)!}\parallel P_{k+j}\parallel .$$

So, using the Stirling asymptotic formula for $j!/j^j,$ we have

$$\begin{array}{cc}\hfill & \underset{j\to \infty }{lim\; sup}{\parallel g_j^{\left[k\right]}\parallel }^{1/j}=\underset{j\to \infty }{lim\; sup}{\left(\frac{{k!j!\parallel z\parallel }^k{(k+j)}^{(k+j)}}{k^k{j}^j(k+j)!}\parallel P_{k+j}\parallel \right)}^{1/j}\hfill \\ \hfill & =\underset{j\to \infty }{lim\; sup}{\left(\frac{e^{(j+k)}\sqrt{2\pi j}}{e^j\sqrt{2\pi (j+k)}}\parallel P_{k+j}\parallel \right)}^{1/j}=\underset{j\to \infty }{lim\; sup}(\parallel P_{k+j}{\parallel )}^{1/j}.\hfill \end{array}$$

It is easy to check (c.f. Lemma 2 [6]) that

$$\underset{j\to \infty }{lim\; sup}\parallel P_{k+j}{\parallel }^{1/j}=\underset{j\to \infty }{lim\; sup}{\parallel P_j\parallel }^{1/j}.$$

So the radius of boundedness of $g^{\left[k\right]}$ is equal to $1/c.$ Thus $g^{\left[k\right]}$ is defined at y because $\parallel y\parallel <1/2c.$ Since

$$g\left(x\right)=g(z+y)=\sum _{k=0}^m{g}^{\left[k\right]}\left(y\right),$$

g is well-defined at $x.$ □

3. Symmetric Analytic Functions of Unbounded Type

Let X be a complex Banach space and $\mathcal{S}$ be a semigroup of isometric operators on $X.$ A polynomial $P\in \mathcal{P}\left(X\right)$ is said to be $\mathcal{S}$-symmetric if it is invariant with respect to the action of $\mathcal{S},$ that is, $P\left(\sigma \right(x\left)\right)=P\left(x\right)$ for every $\sigma \in \mathcal{S}.$ $\mathcal{S}$-symmetric polynomials from the general point of view were considered in [12,13].

Symmetric polynomials on ${\ell }_p,$ $1\le p<\infty$ can be defined as $\mathcal{S}$-symmetric polynomials if $\mathcal{S}$ is the group of permutation of the standard basis vectors in ${\ell }_p.$ Due to [20] we know that polynomials

$$F_k\left(x\right)=\sum _{i=1}^{\infty }{x}_i^k,k=\lceil p\rceil ,\lceil p\rceil +1\cdots$$

form an algebraic basis in the algebra of all symmetric polynomials on ${\ell }_p$ (here $\lceil p\rceil$ is the smallest integer that is greater than or equal to $p),$ $x=(x_1,x_2,\dots ,x_n,\dots )\in {\ell }_p.$ For the case ${\ell }_1$ we can use, also, another algebraic basis

$$G_k\left(x\right)=\sum _{n_1<n_2<\cdots <n_k}{x}_{n_1}\dots x_{n_k}.$$

In [15] it is proved that $\parallel G_n\parallel =1/n!.$ Let $P_n=n!G_n.$ It is easy to check that polynomials $P_n$ satisfy the condition of Theorem 1 if $\Omega$ is the subspace $c_{00}$ of all finite sequences in ${\ell }_1.$ In [6] there is a direct proof that the following function

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

belongs to $H\left({\ell }_1\right)\setminus H_b\left({\ell }_1\right).$ We will prove it for more general situation. For a given positive integer number s we denote

$$G_k^{\left(s\right)}\left(x\right)=\sum _{n_1<n_2<\cdots <n_k}{x}_{n_1}^s\dots x_{n_k}^s.$$

Theorem 2.

If $s\ge p,$ then $G_k^{\left(s\right)}\in \mathcal{P}{(}^{k+s}{\ell }_p)$ and polynomials

$$P_n^{\left(s\right)}\left(x\right)=\frac{G_n^{\left(s\right)}\left(x\right)}{\parallel G_n^{\left(s\right)}\parallel }$$

satisfy the condition of Theorem 1 for $\Omega =c_{00}.$ In particular,

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

Proof. 

If $s\ge p,$ then for every $x\in {\ell }_p,$ $\parallel x\parallel \le 1$ we have

$$|G_k^{\left(s\right)}\left(x\right)|\le {\left(\sum _{n=1}^{\infty }{\parallel x_n\parallel }^s\right)}^k\le {\parallel x\parallel }^{ps}\le 1.$$

So $\parallel G_k^{\left(s\right)}\parallel \le 1$ for every $k.$ Thus $G_k^{\left(s\right)}$ is continuous and well-defined on ${\ell }_p.$

Let $z\in c_{00},$ $z=(z_1,\dots ,z_m,0,\dots ,0).$ Then, for $n>k>sm$

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

because it is a linear span of elements

$$\sum _{j_1<\cdots <j_{n-k}}{z}_1^{s_1}\cdots z_m^{s_m}\underset{k-s_1-\cdots -s_m}{\underbrace{0\cdots 0}}{y}_{j_1}\cdots y_{j_{n-k}}=0,$$

where $0\le s_1+\cdots +s_n\le s.$ So $P_n^{\left(s\right)}$ satisfies the condition of Theorem 1 for $\Omega =c_{00}.$ □

From Theorem 3 it follows that there exists an entire function of unbounded type $g\left(x\right)$ on ${\ell }_p$ such that its Taylor polynomials $P_n^{\left(s\right)}$ are symmetric. From [4] it follows, also that $P_n^{\left(s\right)}\in {\mathcal{A}}_{sn}\left(X\right)$ for $s\ge p.$ Using similar arguments like in Theorem 3, it is possible to construct examples of separately symmetric and entire block-symmetric functions on a finite Cartesian power of ${\ell }_p,$ $1\le p<\infty .$ For the definition and properties of separately symmetric analytic functions we refer the reader to [21]. Block symmetric polynomials on a Cartesian power of ${\ell }_p$ were studied in [22,23]. Here we consider the case of ${\ell }_{\infty }.$

Each vector x in ${\ell }_{\infty }$ is a bounded number sequence $x=(x_1,x_2,\dots ,x_n,\dots ).$ So we can consider symmetric polynomials on ${\ell }_{\infty },$ that is, invariant polynomials with respect to all permutations $\sigma :\mathbb{N}\to \mathbb{N}$

$$(x_1,x_2,\dots ,x_n,\dots )\mapsto (x_{\sigma \left(1\right)},x_{\sigma \left(2\right)},\dots ,x_{\sigma \left(n\right)},\dots ).$$

In [17] it is proved that only constants are symmetric polynomials on ${\ell }_{\infty }.$ However, ${\ell }_{\infty }$ admits polynomials that are invariant with respect to the subgroup of finite permutations. Such polynomials are called finitely symmetric. A permutation $\sigma :\mathbb{N}\to \mathbb{N}$ is finite if there is $m\in \mathbb{N}$ such that $\sigma \left(n\right)=n$ for every $n>m.$ The following example shows that there exists a finitely symmetric entire function of unbounded type on ${\ell }_{\infty }.$

Example 2.

Let us fix a presentation of the set of positive integers as a disjoint union

$$\mathbb{N}=\coprod _{i=1}^{\infty }{N}_i,\left|N_i\right|=\infty .$$

Let ${\mathcal{U}}_i$ be a free ultrafilter on $N_i$ for every $i\in \mathbb{N}.$ We denote by

$${\varphi }_i\left(x\right):=\underset{{\mathcal{U}}_i}{lim}{x}_n,x=(x_1,x_2,\dots ,x_n,\dots )\in {\ell }_{\infty }.$$

Set

$$Q_1\left(x\right)={\varphi }_1\left(x\right)-{\varphi }_2\left(x\right),\dots ,Q_k\left(x\right)=\prod _{i<j\le k}({\varphi }_i\left(x\right)-{\varphi }_j\left(x\right)),\dots .$$

It is clear that $Q_k$ are nontrivial continuous n-homogeneous polynomials for

$$n=\left(\genfrac{}{}{0pt}{}{k}{2}\right)=\frac{k(k-1)}{2},k\in \mathbb{N}.$$

Let us define

$$P_n\left(x\right)=\left\{\begin{array}{ccc}\frac{Q_k\left(x\right)}{\parallel Q_k\parallel }& if& n=\left(\genfrac{}{}{0pt}{}{k}{2}\right)\\ 0& & otherwise.\end{array}\right.$$

Let

$${\Omega }_m=\left\{x\in {\ell }_{\infty }:x_j\in \{c_1,c_2,\dots ,c_m\}\right\}$$

for some fixed constants $c_1,c_2,\dots ,c_m$ and

$$\Omega =\bigcup _{m\in \mathbb{N}}{\Omega }_m.$$

In other words, if $x\in \Omega ,$ then the function $x\left(n\right)=x_n$ has just a finite number of different values. By the definition of $P_n$ one can check that if $x\in {\Omega }_m,$ then

$$P_n\left(x\right)=0forn>\left(\genfrac{}{}{0pt}{}{m}{2}\right)=\frac{m(m-1)}{2}.$$

Since Ω is dense in ${\ell }_{\infty },$ polynomials $P_n$ satisfy the condition of Theorem 1. In addition, polynomials $P_n,$ $n\in \mathbb{N}$ are finitely symmetric because all functionals are evidently so.

4. Discussion, Conclusions and Open Questions

As a result, we can say that various classes of polynomials on Banach spaces support entire analytic functions of unbounded type. However, does an entire symmetric analytic function of unbounded type exist on $L_{\infty }[0;1]$? A function on $L_{\infty }[0;1]$ is symmetric if it is invariant with respect to measuring and preserving automorphisms of the interval $[0;1].$ Polynomials

$$R_n\left(x\right)={\int }_{[0;1]}{\left(x\left(t\right)\right)}^{n}dt,x\left(t\right)\in L_{\infty }[0;1],n\in \mathbb{N}$$

form an algebraic basis in the space of all symmetric polynomials on $L_{\infty }[0;1]$ [18]. It would be interesting, also, to construct an entire supersymmetric function of unbounded type on ${\ell }_1\left({\mathbb{Z}}_0\right),$ where ${\mathbb{Z}}_0=\mathbb{Z}\setminus \left\{0\right\}.$ Let us recall that a polynomial on ${\ell }_1\left({\mathbb{Z}}_0\right)$ is supersymmetric if it is an algebraic combination of polynomials

$$T_k\left(x\right)=\sum _{n=1}^{+\infty }{x}_n^k-\sum _{n=-1}^{-\infty }{x}_n^k,x=(\dots ,x_{-n},\dots ,x_{-1},x_1,\cdots ,x_n,\dots )\in {\ell }_1\left({\mathbb{Z}}_0\right)$$

(see [24]).

In a more general case, let $\mathbf{P}=\left(P_n\right),$ $n\in \mathbb{N}$ be a sequence of algebraically independent polynomials on a Banach space $X.$ Let us denote by $H_{b\mathbf{P}}\left(X\right)$ the minimal subalgebra of $H_b\left(X\right)$ containing all polynomials in $\mathbf{P}.$ Algebras $H_{b\mathbf{P}}\left(X\right)$ were studied in [16,25]. It is natural to ask:

Question 2.

Under which conditions on $\mathbf{P}$ does there exist an entire functions f of unbounded type on X such that its Taylor polynomials are in $H_{b\mathbf{P}}\left(X\right)$?

Note that algebras of symmetric (block-symmetric, supersymmetric) analytic functions of bounded types are partial cases of $H_{b\mathbf{P}}\left(X\right).$ Let us consider the following example.

Example 3.

Let $X={\ell }_p$ for $1\le p\le \infty$ and

$$P_n\left(x\right)=x_n^n,x=(x_1,\dots ,x_n,\dots )\in {\ell }_p.$$

If $p<\infty ,$ then polynomials in $H_{b\mathbf{P}}\left({\ell }_p\right)$ support a function of unbounded type, for example, $f\left(x\right)={\sum }_{n=1}^{\infty }{P}_n\left(x\right).$ However, if $p=\infty ,$ the function f is no longer defined on the whole space.

Question 3.

Does there exist an entire function of unbounded type on ${\ell }_{\infty }$ with Taylor polynomials in $H_{b\mathbf{P}}\left({\ell }_p\right),$ where $\mathbf{P}$ is as in Example 3?

Let us make a note about products of functions of unbounded type. In [19] it is proved that if f is a function of unbounded type on X of the form (3) and P is a nonzero continuous polynomial on $X,$ then $Pf\in H\left(X\right)\setminus H_b\left(X\right).$ Example 1 shows that the product of two functions of unbounded type is not necessarily of unbounded type.

Question 4.

Does a nonzero function $f\in H_b\left(X\right)$ and $g\in H\left(X\right)\setminus H_b\left(X\right)$ exist such that $fg\in H_b\left(X\right)$?

By the following example we can see that the answer to this question is positive for the real case but we do not know the answer for the complex case. Analytic functions of unbounded type on real Banach spaces were studied in [26].

Example 4.

Let f be a function of unbounded type defined by (3) on a real Banach space $X.$ We set

$$g\left(x\right)=e^{{\left(f\left(x\right)\right)}^2}.$$

Clearly $g\in H\left(X\right)\setminus H_b\left(X\right).$ However,

$$\frac{1}{g}=e^{-{\left(f\left(x\right)\right)}^2}$$

has a bounded range and so is of bounded type.