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?



Are all continuous linear operators on the space of entire functions "simple"?
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 $\text{Ent}$. 


