The dual space of entire functions is known. It is the space of functions analytic around infinity with non-constant term, $\mathcal{O}^\infty_0$. The action of $F\in \mathcal{O}^\infty_0$ on an entire function $\phi$ is defined by $$ \langle f,\phi\rangle = \frac{1}{2\pi i}\oint F(z) \phi(z) dz, $$ where the integral is performed on a circle of big enough radius.

Sometimes these are described as functionals of compact support. This is because for any integrable real function $f$ with compact support and any entire function $\phi$ we can define the pairing $$\langle f,\phi\rangle = \int_{-\infty}^\infty f(x) \phi(x) dx. $$ Then this functional can be included in $\mathcal{O}^\infty_0$ by defining $$ F(z) = \int_{-\infty}^\infty \frac{f(x)}{x-z} d x .$$ The function $F$ is in $\mathcal{O}^\infty_0$ and behaves in the expected way. So in a sense the support of an analytic functional is the set of points on which it is not analytic.

I couldn't find anything on the dual space of entire functions of exponential type. Since this is a smaller space, its dual has to be bigger. Clearly if $\phi$ is of exponential type then the map $$ \phi \mapsto \int_{-\infty}^\infty \phi(x) e^{-x^2} dx $$ defines a continuous functional and clearly it does not have compact support.

So is this space known? Is there a description of it as a functional space?

  • $\begingroup$ Ah, yes that's true. It is not a priori clear since it has to respect the topology of the space of functions of exponential type. But I still think it is the case that the space has to be bigger than the dual of entire functions. $\endgroup$ – tst Dec 7 '17 at 18:36
  • 1
    $\begingroup$ On second thoughts, it is actually clear that if $F$ was continuous on entire functions, then it's also continuous on exp type when restricted (and with the new topology), which is what you were saying. In any event, there are two topologies involved. $\endgroup$ – Christian Remling Dec 7 '17 at 19:30
  • $\begingroup$ Whenever you have ''natural spaces'' $X \subseteq Y$ with their "natural topologies", the inclusion should be continuous (in many cases this follows from the closed graph theorem). $\endgroup$ – Jochen Wengenroth Dec 8 '17 at 12:35
  • $\begingroup$ Before asking about the dual, you have to specify the topology on the space of entire functions of exponential type. $\endgroup$ – Alexandre Eremenko Feb 1 '18 at 2:20
  • $\begingroup$ I assumed the standard topology on the space, see en.wikipedia.org/wiki/Exponential_type $\endgroup$ – tst Feb 1 '18 at 4:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.