# Which entire functions $h(z)$ can be written as $h(z)=f(z+1)-f(z)$ for some entire function $f$? [duplicate]

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Question: For which entire functions $h(z)$ does there exist an entire function $f(z)$ such that $h(z)=f(z+1)-f(z)$?

What I have tried:

Suppose that $f:\mathbb{C}\to\mathbb{C}$ is an entire function, and let $\displaystyle f(z)=\sum_{n=0}^\infty a_nz^n$ be its Taylor series expansion. Then $$\displaystyle f(z+1)=\sum_{n=0}^\infty a_n\sum_{i=0}^n\binom{n}{i}z^i=\sum_{n=0}^\infty z^n\left[\sum_{j=n}^\infty a_j\binom{j}{n}\right].$$

Therefore $$f(z+1)-f(z)=\sum_{n=0}^\infty z^n\left[\sum_{j=n+1}^\infty a_j\binom{j}{n}\right].$$

For $\{a_n\}\in\mathbb{C}^\infty$, define $c_n=\displaystyle\sum_{j=n+1}^\infty a_j\binom{j}{n}$, if this sequence converges. If $\{a_n\}$ is a sequence for which each $c_n$ converges, then define $\Pi(\{a_n\})=\{c_n\}$.

Lets let $\mathscr{H}$ denote the collection of all sequences of complex numbers $\{a_n\}$ such that $\Pi(\{a_n\})$ is well-defined. Let $\mathscr{H}_e$ denote the collection of sequences such that $\displaystyle\sum_{n=0}^\infty a_nz^n$ has infinite radius of convergence. It is not hard to see from the above work that $\mathscr{H}_e\subset\mathscr{H}$. Let $\mathscr{H}_0\subset\mathscr{H}_e$ be the collection of finite sequences (ie those corresponding to polynomials).

I know that $\Pi:\mathscr{H}_0\to\mathscr{H}_0$ is surjective. I want to know what $\Pi(\mathscr{H}_e)$ is (I now know that $\Pi:\mathscr{H}_e\to\mathscr{H}_e$ is not surjective as noted in my comment below).

Bonus question: If we mod out $\mathscr{H}$ by the relation $\{a_n\}\sim\{b_n\}$ if and only if $a_k=b_k$ for each $k>0$, then $\Pi:\mathscr{H}_0\to\mathscr{H}_0$ is injective. Is $\Pi:\mathscr{H}\to\mathscr{H}$ injective when modded out similarly?

## marked as duplicate by mercio, Shailesh, Community♦Mar 24 '17 at 13:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Sometimes called "indefinite summation". See also mathoverflow.net/a/43967/454 mathoverflow.net/q/42484/454 mathoverflow.net/q/41011/454 math.stackexchange.com/q/1710/442 – GEdgar Aug 24 '16 at 14:44
• I don't think that $\Pi$ is well-defined on ${\cal H}$ (sums defining $c_n$ need not converge). so not sure how to answer the various questions. – H. H. Rugh Aug 24 '16 at 14:46
• @H.H.Rugh True, but $\Pi$ is well-defined on $\mathscr{H}_e$, which is really what my question is asking about. I will fix it. – Trevor Richards Aug 24 '16 at 14:54
• OK, then there are answers. – H. H. Rugh Aug 24 '16 at 14:59
• @H.H.Rugh I just updated the question to avoid issues of well-definedness. Thanks for the interest. – Trevor Richards Aug 24 '16 at 15:08

## 1 Answer

Partial answers: Already mentioned but there is a large kernel of period-1 functions. So addressing the question when taking modulo those:

$h_a(z)=e^{az}$ admits $f(z)=(e^a-1)^{-1} e^{az}$ as solution to $\Delta f(z)=f(z+1)-f(z)=h_a(z)$ for $e^a\neq 1$. When $e^a=1$ or more generally when you look at any entire function $h$ for which $h(z+1)=h(z)$ then $f(z)=zg(z)$ satisfies: $$f(z+1)-f(z) = (z+1)g(z+1)-zg(z)=(z+1)g(z)-zg(z)=g(z)$$ So the kernel of $\Delta$ is in the image of $\Delta$ as well. In particular, any linear combination of exponentials is in the image of $\Delta$. But the fact that the solution blows up as $e^a$ tends to 1 is again an indication of non-surjectivity ...

Conjecture (due to the above mentioned blow-up): $h(z)=\int_0^1 e^{2\pi i a z} da=\frac{e^{2\pi i z}-1}{2\pi i z}$ is not in the image of $\Delta$.