Are all continuous linear operators on the space of entire functions "simple"? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T13:09:24Z http://mathoverflow.net/feeds/question/64097 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64097/are-all-continuous-linear-operators-on-the-space-of-entire-functions-simple Are all continuous linear operators on the space of entire functions "simple"? Ricky Demer 2011-05-06T08:39:10Z 2011-05-06T09:38:21Z <p>Let $\langle \operatorname{Ent},+,\cdot \rangle$ be the (complex) vector space of entire functions. <br> For all members $n$ of <code>$\{1,2,3,...\}$</code>, define $||\cdot ||_n : \operatorname{Ent} \to \mathbb{R}$ by <code>$||f||_n = \operatorname{sup}(\{|f(z)| : |z|\leq n\})$</code>. <br> <code>$\big\langle \operatorname{Ent},+,\cdot,\{||.||_n : n\in \{1,2,3,...\}\} \big\rangle$</code> is a Frechet space.</p> <p><br></p> <p>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 <br><br> $(i) \quad (L_1(f))(z) = g(z)\cdot f(z)$ <br><br> $(ii) \quad (L_2(f))(z) = f(g(z))$ <br><br> $(iii) \quad (L_3(f))(z) = f'(z)$ <br><br> $(iv) \quad (L_4(f))(z) = \displaystyle\int_0^z f$ <br><br> are all continuous and linear. <br><br><br> Let $S$ be the set of all functions obtainable by the above. <br> Let $\mathbf{L}$ be continuous operator algebra on $\operatorname{Ent}$. <br> Let $T$ be the closure of $S$ as a sub-algebra of $\mathbf{L}$.</p> <p><br></p> <p>Does $\:$ $T = \mathbf{L}$ $\:$ ? <br> If no, is $T$ dense in $\mathbf{L}$? (uniform operator topology) <br> If no again, is $T$ dense in $\mathbf{L}$ in some weaker topology?</p> http://mathoverflow.net/questions/64097/are-all-continuous-linear-operators-on-the-space-of-entire-functions-simple/64101#64101 Answer by Neil Strickland for Are all continuous linear operators on the space of entire functions "simple"? Neil Strickland 2011-05-06T09:38:21Z 2011-05-06T09:38:21Z <p>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:</p> <pre><code> \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)}}, } </code></pre> <p>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 $\text{Ent}$. </p>