Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|cite|improve this question
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

1 Answer 1

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.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.