# Are all continuous linear operators on the space of entire functions “simple”?

Let $\langle \operatorname{Ent},+,\cdot \rangle$ be the (complex) vector space of entire functions.
For all members $n$ of $\{1,2,3,...\}$, define $||\cdot ||_n : \operatorname{Ent} \to \mathbb{R}$ by $||f||_n = \operatorname{sup}(\{|f(z)| : |z|\leq n\})$.
$\big\langle \operatorname{Ent},+,\cdot,\{||.||_n : n\in \{1,2,3,...\}\} \big\rangle$ is a Frechet space.

For all complex numbers $z_0$ and members $g$ of $\operatorname{Ent}$, the operators $L_1,...,L_4 : \operatorname{Ent} \to \operatorname{Ent}$ defined by

$(i) \quad (L_1(f))(z) = g(z)\cdot f(z)$

$(ii) \quad (L_2(f))(z) = f(g(z))$

$(iii) \quad (L_3(f))(z) = f'(z)$

$(iv) \quad (L_4(f))(z) = \displaystyle\int_0^z f$

are all continuous and linear.

Let $S$ be the set of all functions obtainable by the above.
Let $\mathbf{L}$ be continuous operator algebra on $\operatorname{Ent}$.
Let $T$ be the closure of $S$ as a sub-algebra of $\mathbf{L}$.

Does $\:$ $T = \mathbf{L}$ $\:$ ?
If no, is $T$ dense in $\mathbf{L}$? (uniform operator topology)
If no again, is $T$ dense in $\mathbf{L}$ in some weaker topology?

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It seems to me that you forget the following operators: $$(L_6(f))(z)=f\circ g(z),$$ where $g\in$Ent is given. –  Denis Serre May 6 '11 at 9:02
Why would you expect this? You mean probably $S$ being the algebra generated by the operators? $S$ will seperate points using a Taylor expansion in $0$, but how would you get $f(g(z))$ or $g(f(z))$, e.g. start $f( \alpha z)$ for some $\alpha \in \mathbb{C}$. –  Marc Palm May 6 '11 at 9:02
@Denis, that cuts it down to $L_1,...,L_4$. (as I just edited to reflect) –  Ricky Demer May 6 '11 at 9:15
What about restricting your functions to a compact subset of the reals and Fourier transforming it? –  Marc Palm May 6 '11 at 9:18

Suppose you have a function $u(z)$ that is defined and holomorphic for $|z|>R$, with $u(z)\to 0$ as $z\to\infty$. You can then define $L_u:\text{Ent}\to\mathbb{C}$ by $L_u(f)=\oint_C f(z)u(z)\\,dz$, where $C$ is a circle around the origin of radius $R'>R$. This only depends on the germ of $u$ at $\infty$. I think one can show that this gives an isomorphism from the space of germs to the dual of $\text{Ent}$. There are theorems of this type in the following book:

 \bib{MR745622}{book}{
author={Luecking, D. H.},
author={Rubel, L. A.},
title={Complex analysis},
series={Universitext},
note={A functional analysis approach},
publisher={Springer-Verlag},
place={New York},
date={1984},
pages={vii+176},
isbn={0-387-90993-1},
review={\MR{745622 (86d:30002)}},
}


However, I am not 100% sure if they apply to the whole plane or just to bounded subsets. Anyway, for a holomorphic function $u(z,w)$ with suitable domain we can now define an operator $T_u:\text{Ent}\to\text{Ent}$ by $T_u(f)(z)=\oint_C u(z,w)f(w)\\,dw$, and using the description of $\text{Ent}^*$ it should follow that this gives all possible continuous operators on $\text{Ent}$.

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