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. $$\forall x,y: g(1,y)=g(0,y) \text{ and }g(x,1)=g(x,0).$$
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Sign up to join this communityI 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. $$\forall x,y: g(1,y)=g(0,y) \text{ and }g(x,1)=g(x,0).$$
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 $f(z)=ce^{-2\pi iz}$.