Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|cite|improve this question
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

1 Answer 1

up vote 1 down vote accepted

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:

    author={Luecking, D. H.},
    author={Rubel, L. A.},
    title={Complex analysis},
    note={A functional analysis approach},
    place={New York},
    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 $\text{Ent}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.