Existence of entire function that yields periodicity 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. $$\forall x,y: g(1,y)=g(0,y) \text{ and }g(x,1)=g(x,0).$$
 A: *

*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.


*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.


*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 $f(z)=ce^{-2\pi iz}$.
