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Let $(E):\ \sum_{k=0}^n P_k(z)g(z+k)=0$ with the $P_k\in\mathbb C[z]$ a finite differences equation in $\mathbb C$.

Is it true that every any entire solution $g$ of (E) of exponential type $<\pi$ is an exponential polynomial?

Thanks in advance

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Do you mean $g(z+i)$? – abx Mar 11 '14 at 18:35
Is my edition correct? – Alexandre Eremenko Mar 12 '14 at 4:46
It's best to avoid using $i$ as a summation index in contexts such as this where $i$ might be needed as a square root of $-1$ (e.g. one natural thing to try is the Laplace transform, whose inversion involves complex exponentials). Better to sum over $j$, $k$, or $m$. – Noam D. Elkies Mar 12 '14 at 5:04
Alexandre > Thanks your edit was correct. Following Noam, I changed the summation index $i$ by $k$. – joaopa Mar 12 '14 at 6:18
up vote 7 down vote accepted

The answer is no. Function $g(z)=(e^z-1)/z$ is entire, of exponential type $1$, is not an exponential polynomial, but it satisfies the equation $$(z+2)g(z+2)-(1+e)(z+1)g(z+1)+ezg(z)=0.$$ Ref. J-P Bezivin et F. Gramain, Solutiuons entieres d'un systeme d'equations aux differences, Ann. Inst. Fourier, Grenoble, 43 (1993) 792-814.

Notice that you can modify this example to achieve arbitrarily small exponential type.

Edit. There exists also such equations with entire solutions of smaller growth, of order less than one. This paper has explicit examples with order $1/3$ and $1/5$: K. Ishizaki and N. Yanagihara, Wiman-Valiron method for difference equations, Nagoya Math. J., 175 (2004) 75-102.

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Beautiful counter-example. Thanks a lot – joaopa Mar 12 '14 at 18:54

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