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Find all functions $f:\mathbb C\to\mathbb C$ such that $f(z+1)-f(z)$ is entire.

I am curious about this, because an algebraic analog states that if $f:\mathbb Z\to\mathbb N$ is such that $f(m+1)-f(m)$ agrees with a polynomial in $m$ for $m\gg0$ then $f(m)$ agrees with a polynomial in $m$ for $m\gg0$.

Matemáticos Chibchas
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    If $f(z)=\sin 2\pi z$, then $f(z+1)-f(z)$ is a polynomial but $f$ is not. :-) – Jyrki Lahtonen Jul 04 '15 at 13:16
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    For the existence, this post is useful http://mathoverflow.net/questions/4434/elementary-solutions-to-fz1-fz-gz-in-entire-functions. – mathcounterexamples.net Jul 04 '15 at 13:24
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    Indeed, if $e$ is entire and $p$ is an arbitrary $1$-periodic function, define $f(x)=e(x)+p(x)$, and then $f(x+1)-f(x) = e(x+1)-e(x)$ clearly is entire. However I have no idea if you get all such functions that way. – celtschk Jul 04 '15 at 13:24
  • @JyrkiLahtonen You are right. My mistake for omitting the technicalities. Edited now. – Matemáticos Chibchas Jul 04 '15 at 13:27
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    Perhaps a trifle, but just in case: $f$ needn't be entire. Take $f(z)=\bar z$. – ajotatxe Jul 04 '15 at 13:28
  • well $f$ needn't be continuous anywhere. If $\Delta(f)(z) = f(z+1)-f(z)$ and $A$ is a domain like ${z \mid 0 \le \Re(z) < 1}$, then the map $f \mapsto (\Delta(f),f\mid_A)$ is a bijection from $\Bbb C^\Bbb C$ to $\Bbb C^\Bbb C \times \Bbb C^A$. So for any arbitrary entire $\Delta(f)$ you can pick an arbitrarily ugly $f\mid_A$ and you recover $f$ from those. – mercio Jul 04 '15 at 13:31
  • A concrete example of a nowhere continuous function fulfilling the condition: The product of the Dirichlet functions of the real and the imaginary part. Clearly $f(x+1)=f(x)$, therefore $f(x+1)-f(x)=0$ is entire. – celtschk Jul 04 '15 at 13:38
  • Related : http://math.stackexchange.com/questions/1174799/existence-of-an-holomorphic-function/1195779#1195779 –  Jul 04 '15 at 13:55

1 Answers1

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The set of all functions $f:\mathbb C\to\mathbb C$ such that $f(z+1)−f(z)$ is entire is exactly the set of funcions $f(z)=e(z)+p(z)$ where $e$ is entire and $p:\mathbb C\to\mathbb C$ is $1$-periodic.

Proof:

Be $g(z)=f(z+1)-f(z)$. Since $g$ is entire, according to this Mathoverflow post (linked to by mathcounterexamples.net in the comments) there exists an entire function $e$ such that $g(z) = e(z+1) - e(z)$.

Now define $p(z) = f(z) - e(z)$. Now $p(z+1) - p(z) = (f(z+1) - f(z)) - (e(z+1) - e(z)) = g(z) - g(z) = 0$, therefore $p(z)$ is $1$-periodic.

Therefore every function $f$ with the proposed property can be written as the sum of an entire function $e$ and a $1$-periodic function $p$.

On the other hand, if $e$ is an arbitrary entire function and $p$ is an arbitrary $1$-periodic function, then one easily verifies that $f(z)=e(z)+p(z)$ has the desired property.

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