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) $. The proof can be given using riemann surface,automorphy,covering,etc. Can anyone find a elementary proof which avoids all such things.
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2$\begingroup$ Section 6.3 in [Berenstein and Gay: Complex analysis and special topics in harmonic analysis MR Number=(1344448)] deals with that problem. $\endgroup$– Narutaka OZAWACommented May 29, 2014 at 7:49
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1$\begingroup$ @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. $\endgroup$– Neil StricklandCommented May 29, 2014 at 7:53
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1$\begingroup$ Thank you. I just don't know how to edit. numdam.org/item?id=ASENS_1887_3_4__361_0 $\endgroup$– Narutaka OZAWACommented May 29, 2014 at 9:37
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1$\begingroup$ Is this the same as mathoverflow.net/questions/4434 ? $\endgroup$– David E SpeyerCommented May 29, 2014 at 19:06
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1 Answer
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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.
answered May 29, 2014 at 7:35
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$\begingroup$ fantastic! i was expecting an answer from you. $\endgroup$– KoushikCommented May 29, 2014 at 8:00