Let g(z) be an entire function of a complex variable z. Does there exist an entire function f(z) such that f(z+1)-f(z)=g(z)? As I learned several years back, the answer to this [is apparently 'yes'](http://www.math.niu.edu/~rusin/known-math/01_incoming/coho_func), but I have not felt satisfied with the proof because it goes beyond my expertise.


I tried to find f using the power series expansion of g, for that works when g is a polynomial. But the results of partial inversions kept diverging. Representing g as an integral via Cauchy's formula, and doing inversion inside the integration led to similar problems. Perhaps I am overly optimistic, but a question this elementary should have equally elementary solution. Is there such a solution? If not, is there a reason to expect that no simple and elementary solution should exist?



Elementary solutions to f(z+1)-f(z)=g(z) in entire functions
It took me some time to find a solution that satisfies both requirements:


a) If should be based on the power series expansion


b) It should use no tools heavier than contour integration.


So, let $g(z)=\sum a\_ k z^k$. We know that $a\_ k$ decay faster than any geometric progression. We want analytic functions $F\_ k(z)$ such that $F\_ k(z+1)-F\_ k(z)=z^k$ and $F\_ k(z)$ grows not faster than a geometric progression as a function of $k$ in every disk. Then $f=\sum a\_ k F\_ k$ is what we want. So, just choose any odd multiples $r\_ k\in(k,k+2\pi)$ of $\pi$ and put
$$
F\_ k(z)=\frac{k!}{2\pi i}\oint\_ {|z|=r\_ k}\frac {e^{z\zeta}}{e^\zeta-1}\frac{d\zeta}{\zeta^{k+1}}
$$  

The key is that $|e^\zeta-1|\ge \frac 12$ on the circle and $r\_ k^k\ge k!$, so $|F\_ k(z)|\le 2e^{|z|(k+2\pi)}$ on the plane.


