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On THIS site, Alexandre used Fourier transform to solve the problem.

If we use Fourier transform, how to define it to ensure any entire function has a FT?

Classical FT is defined by $$ \mathcal{F}[f] = F(\xi) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty}f(z)\mathrm{e}^{-\mathrm{i} \xi z} \mathrm{d} z. $$ This only work for $f \in L^1(\mathbb{R})$. (If improved, it can work for $f \in L^2(\mathbb{R})$.)

I know $\mathcal{F}[\mathrm{e}^{sz}] = \sqrt{2 \pi} \delta(\xi - \mathrm{i}s)$, but I'm not sure about a general definition.

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  • $\begingroup$ Look for Fourier transform of tempered distributions, for example in Hormanders volume I. $\endgroup$
    – Peter Michor
    Commented Dec 2, 2012 at 9:26

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There are two questions.

  1. For the specific functional equation considered in this question On equation f(z+1)-f(z)=f'(z), the formula I gave covers all entire solutions. I added the references there.

  2. On the general question about "Fourier transform" of entire functions or functions on the real line which are not in $L^p$. One usually replaces Fourier transform with various versions of Laplace transform. There are many versions, for various problems. I recommend Hormander, Analysis of differential operators..., Chap. 9, or the paper MR0199747 Ljubič, Ju. I.; Tkačenko, V. A. Theory and certain applications of the local Laplace transform. (Russian) Mat. Sb. (N.S.) 70 (112) 1966 416–437. There is an English translation in Math USSR Sbornik. There is also a nice little book by Carleman of Fourier transform (in French).

Edit. See also On linear independence of exponentials for an example how Laplace transform of entire functions is used.

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  • $\begingroup$ Yet I'm doing this in $\mathbb{C}$, by that, FT and LT are equivalent. Thus, if we can define LT as your state, then, we can do FT as well. $\endgroup$
    – Lwins
    Commented Dec 11, 2012 at 7:03
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The Fourier Transform $\mathcal{F}$ is at first defined on the Schwartz space $\mathcal{S}$ and is a linear isomorphism there. As always, there is a dual operator $\mathcal{F}^\prime$ that is an isomorphism on the dual space $\mathcal{S}^\prime$ of tempered distributions.

This operator $\mathcal{F}^\prime$ is the extension you are looking for, as $\mathcal{S}$ and also more general functions (for example polynomials) can be regarded as subset of $\mathcal{S}^\prime$.

However, as far as I know, not ANY entire function has a Fourier transform, only those that also lie in $\mathcal{S}^\prime$.

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There is a definition of Fourier transforms for distributions, not just tempered distributions. The Fourier transform of a test function is an entire function of exponential growth, and the Fourier transforms of distributions are defined by duality. The Fourier transforms of distributions are known as analytic functionals. You may find an exposition of this topic in the monograph by Gelfand and Shilov.

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  • $\begingroup$ Mmm.. Then, is a entire function $f$ have a Fourier transform $F$ (perhaps $F$ is a distribution) for sure? If not, then the answer for [THIS][1] is incomplete. [1]: mathoverflow.net/questions/114875 $\endgroup$
    – Lwins
    Commented Dec 2, 2012 at 11:33
  • $\begingroup$ Any distribution f (and therefore any locally integrable function defined on the reals) has a Fourier transform F, which is defined as an analytic functional. In general F is not a distribution. Analytic functionals act on test functions which are entire, for instance in your example $\delta(\xi-is)$ can be defined only for analytic test functions. $\endgroup$
    – Michael Renardy
    Commented Dec 2, 2012 at 15:02
  • $\begingroup$ Your statement about defining the Fourier transform by duality does not make sense... As you pointed out yourself, the fourier transform of a function in D is not in D again, but in S. Hence taking the dual operator gives again only an operator from S' to D'. $\endgroup$
    – Matthias Ludewig
    Commented Dec 2, 2012 at 15:12
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    $\begingroup$ The Fourier transforms of functions in D lie in a certain space of analytic functions, which is called Z. So the Fourier transform maps D to Z and vice versa. By duality, the Fourier transform then maps D' to Z' and Z' to D'. Your function f is in D', and its Fourier transform is in Z'. If you want more details, read Gelfand and Shilov as I suggested in the first place. Then you can tell me if I am making sense. $\endgroup$
    – Michael Renardy
    Commented Dec 2, 2012 at 17:13
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    $\begingroup$ I don't understand the last comment. The Fourier transform of any distribution of compact support, in particular any test function in D, is an entire function. It is holomorphic everywhere, not just outside a strip. $\endgroup$
    – Michael Renardy
    Commented Dec 3, 2012 at 2:34

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