Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
1  
Do you mean $g(z+i)$? –  abx Mar 11 at 18:35
    
Is my edition correct? –  Alexandre Eremenko Mar 12 at 4:46
1  
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 at 5:04
    
Alexandre > Thanks your edit was correct. Following Noam, I changed the summation index $i$ by $k$. –  joaopa Mar 12 at 6:18

1 Answer 1

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.

share|improve this answer
    
Beautiful counter-example. Thanks a lot –  joaopa Mar 12 at 18:54

Your Answer

 
discard

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.