There is no nonconstant entire function $f$ such that $f(z+1)=f(z)$ and $f(z+i)=f(z)$

Question

Claim: there is no entire non-constant function $f$ such that $f(z+1)=f(z)$ and $f(z+i)=f(z), \forall z\in \mathbb{C}.$

May I verify if my proof is valid? Or is there a better way to approach this problem? Thank you.

Suppose there exists such $f.$ Since $f$ is continuous, $f$ is bounded on some compact set $S:=\{a+bi \in \mathbb{C} : a,b \in [0,1]\}.$

Given $z=x+iy \in \mathbb{C},\ f(x+iy)=f(a+bi+ \lfloor {x}\rfloor + i\lfloor {y}\rfloor)=...=f(a+ib),$ where $a,b \in [0,1].$

Hence, $f$ is bounded in $\mathbb{C}$ and Liouville's theorem follows.

Why is $f$ constant on $S$? Every entire holomorphic function is bounded on $S$. [You have your ingredients in the wrong order.] — Daniel Fischer, Aug 28 '14 at 17:50
Thanks for pointing out the mistakes. So how should I proceed? — Alexy Vincenzo, Aug 28 '14 at 17:51
Remove the $= c$ from your last line, move the last line up, and deduce that $f$ is bounded on $\mathbb{C}$, then Liouville applies. — Daniel Fischer, Aug 28 '14 at 17:58
An aside: Such function are studied as doubly periodic functions. As you showed, they cannot be entire. But they can be meromorphic in the plane. — GEdgar, Aug 28 '14 at 18:25

score 1 · Answer 1 · answered Aug 28 '14 at 19:21

Also another way of thinking in different context, when you see your function $f$ (assumed non-constant) is doubly periodic, think about the lattice $L$ generated by $1, i.$ Your $f$ induces a holomorphic map from $g: R=\mathbb{C}/L \to \mathbb{C}.$ Now, since $R$ is compact (is a doughnut with nothing stuffed in), by maximum modulus theorem, $g$ has to attain its maximum on the boundary of $R$ which is empty!

There is no nonconstant entire function $f$ such that $f(z+1)=f(z)$ and $f(z+i)=f(z)$

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