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Given a function $ g $ entire on the whole complex plane $ C $, it is possible to find an entire function $f $ such that $ f(z+1) -f(z)=g(z) $. The proof can be given using riemann surface,automorphy,covering,etc. Can anyone find a elementary proof which avoids all such things.

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Section 6.3 in [Berenstein and Gay: Complex analysis and special topics in harmonic analysis MR Number=(1344448)] deals with that problem. – Narutaka OZAWA May 29 '14 at 7:49
@NarutakaOZAWA: I liked the reference in the earlier comment that you just deleted. Did you just replace it because it was very old and in French? I think it would be nice for you to give both references. – Neil Strickland May 29 '14 at 7:53
Thank you. I just don't know how to edit. – Narutaka OZAWA May 29 '14 at 9:37
Is this the same as ? – David Speyer May 29 '14 at 19:06
up vote 7 down vote accepted

Let $L$ be your difference operator: $(Lf)(z)=f(z+1)-f(z)$. Consider these polynomials $$P_n(z)=\frac{1}{n!}z(z-1)\ldots(z-n+1),\quad n=0,1,2,\ldots.$$ Simple computation shows that $LP_n=P_{n-1}$. Polynomials $P_n$ make a basis in the space of all polynmials, because there is one polynomial of each degree. This allows you to find a solution of any equation with polynomial RHS. Then perform a limit process. For the details see any book under the title Calculus of finite differences. For example, by N\"orlund or by Gelfond.

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fantastic! i was expecting an answer from you. – Koushik May 29 '14 at 8:00

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