I have the following question:

Does there exist an entire function $f(z)$ where $z=x+iy$ such that

$$g(x,y) =e^{-2\pi y^2}f(z)$$

is periodic in both $x$ and $y$ direction, i.e. $$\forall x,y: g(1,y)=g(0,y) \text{ and }g(x,1)=g(x,0).$$

  • 2
    $\begingroup$ $g(1, y) = g(0, y)$ doesn't sound like periodicity. May be you've meant $g(x+1, y) = g(x, y)$? If so, then I believe the following should be a promising approach: from your periodicity prove an a priory bound that $f$ is entire of order $2$ with periodic zeroes. This means that $f$ is a finite product of shifts of Weierstrass sigma functions times $\exp(az^2+bz+c)$, which then should be easy to check whether it works or not. $\endgroup$ – Aleksei Kulikov Jun 27 at 17:17
  1. If you correct your definition to the correct definition of periodicity, $g(x,y+1)=g(x,y)$, for all $x,y$, then the answer is no (except when $f=0$). Indeed, let $z=x+iy$, and assuming $g$ is periodic with respect to $y$, we obtain $$f(z+i)=g(x,y+1)e^{-2\pi(y+1)^2}=g(x,y)e^{-2\pi y^2}e^{-4\pi y-2\pi}=f(z)e^{-4\pi y-2\pi},$$ and this holds identically in $z$. Therefore $f(z+i)/f(z)=\exp(-4\pi y-2\pi)$ is not an analytic function anywhere, since it is real, contradiction.

  2. On the other hand, if one takes your conditions literally as you wrote them (and don't call them "periodicity"), then such a function exists: namely $f(z)=\exp(-2\pi iz)$. It satisfies $f(z+1)=f(z)$ and $f(z+i)=e^{2\pi}f(z)$, therefore it satisfies your conditions.

  3. Moreover, this $f$ is unique, up to a constant factor. Indeed, your first conditon is equivalent to $f(z+1)=f(z)$ therefore $f$ has period one, and has Fourier expansion $$f(z)=\sum_{-\infty}^\infty c_ne^{2\pi inz}=g(e^{2\pi i z}),$$ where $g$ is analytic in $C^*$. And your second condition means that $g(e^{-2\pi} w)=e^{2\pi}g(w)$ which easily implies that all $c_j$ except $c_{-1}$ are zero. This gives $f(z)=ce^{-2\pi iz}$.

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  • 1
    $\begingroup$ Technically one needs to exclude the degenerate case $f=0$ in your analysis of interpretation 1. $\endgroup$ – Terry Tao 2 days ago
  • $\begingroup$ @Terry Tao: Thanks. I corrected. $\endgroup$ – Alexandre Eremenko 2 days ago

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