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Original Problem

If $f$ is an entire function such that $$ f(z+1)-f(z)=f'(z) $$ for all $z$. Is there a non-trivial solution? ($f(z)=az+b$ is trivial)

And here is something uncertainty

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.

Daniele Tampieri
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Lwins
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1 Answers1

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Linear functional equations can be solved with Fourier transform. Let $\lambda_k$ be the roots of the equation $e^\lambda-1=\lambda$. There are infinitely many such roots. Then

$$f(z)=\sum_k a_ke^{\lambda_k z}$$

is a solution. Here the sum can be finite, and $a_k$ arbitrary, or the sum can be infinite, and $a_k$ tend to zero with such speed that the series converges.

The trivial solution is covered, if you interpret the answer correctly. The equation $e^\lambda-1=\lambda$ has a DOUBLE root at $0$. For the case of a double root one includes not only the term $e^{\lambda z}$ but also the term $ze^{\lambda z}$. Similarly for multiple roots $z^ke^{\lambda z}$. So the root $\lambda=0$ exactly covers the solution $az+b$.

EDIT: Actually all entire solutions can be represented in this way, for the proof I refer to Gelfond, Calculus of finite differences, MR0342890, Chap. 5 Sect 7, Thm II and Corollaries. He gives the equation $f'(z)=f(z-1)$ as an example, but your equation is treated similarly.

EDIT2: The problem can be generalized as follows: Let $\omega$ be a distribution with bounded support. Equation $f\star w=0$, where $\star$ is the convolution, is called a convolution equation, and its solutions are called mean-periodic functions (fonctions moyenne-periodiques). The case we are discussing is $\omega(x)=\delta(x+1)-\delta(x)-\delta'(x)$ where $\delta$ is the delta function. The theory of mean periodic functions was created by Delsarte and Schwartz in 1940-s. Schwartz has a general result that all mean-periodic functions can be obtained as limit of exponential sums, like in this problem.

Alexandre Eremenko
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    Cool ! But seems trivial solution f = z is not covered by the answer... – Alexander Chervov Nov 29 '12 at 13:00
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    And yet the "trivial" solution $z$ is not obtained in this way. – Gerald Edgar Nov 29 '12 at 13:03
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    If $\lambda$ is a double root of $e^\lambda-1-\lambda$, then we should include $z e^{\lambda z}$ as well. This recovers the trivial solution, since $0$ is a double root. – Henry Cohn Nov 29 '12 at 13:27
  • $\lambda=0$ is the only double root. The other ones are simple. – Denis Serre Nov 29 '12 at 14:38
  • Excellent, but how to define "Fourier Transform" to ensure that any possible entire function have its FT? Classical FT is only defined in $L^2(\mathbb{R})$. – Lwins Dec 02 '12 at 07:11
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    I do not know a way to define FT for ALL entire functions. And I suppose there is no reasonable way. FT can be defined for functions on the real line which satisfy $|f(x)|=O(e^{\epsilon x})$ for every $\epsilon>0$ which is enough for most applications. Speaking of that functional equation, I did not claim that I described ALL entire solutions. But this was not asked:-) – Alexandre Eremenko Dec 02 '12 at 15:12
  • Regarding Lwins.Gafields last comment: the lesson I take from this beautiful and enlightening answer is that whatever its scope, Fourier Transform provides a valuable "heuristic" (I think I'm using the term correctly according to Hoyle) for arriving at correct solutions. ("Heuristic" and "enlightening" harking back to this exchange: http://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics) – Todd Trimble Dec 03 '12 at 06:35
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    We have $-(\lambda +1)e^{-(\lambda+1)}=-e^{-1}$, so the $\lambda$ are determined by the multiple branches of Lambert's $W$-function, namely: $\lambda = -1-W(-e^{-1})$. – Thomas Feb 17 '17 at 23:38