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



Entire solutions of finite difference equations
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.


