# Entire doubly-periodic function is constant [duplicate]

I want to show that a function $f$ entire s.t.

f(z+1)=f(z)=f(z+i)

for all $z$, must be constant.

I think what I need to do is show that the function is bounded in some region of the complex plane, and then by periodicity it must be true that the function is bounded everywhere, and hence by Liouville must be constant. Can someone help me do this? Thanks!

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• That's doubly periodic. The distinction is crucial. – zhw. Aug 27 '16 at 22:39
• Note that $f(\mathhbb{C}) = f(\{ a + bi : a, b \in [0, 1] \})$, and that's a compact set, so $f$ is bounded. – AJY Aug 27 '16 at 23:22

Let $z\in C, z=u+vi, u,v\in R$, you can write $u=a+n, v=b+m, 0\leq a,b\leq 1$, $f(z)=f(a+n+(b+m)i)=f(a+ib)$, this implies that $f$ is bounded since its image is contained in the image of the compact space $\{a+ib, a,b\leq 1\}$. We deduce that $f$ is constant.

Consider the closed disk: $D= \{ z : |z| \leq 2 \}$. Since it is closed and bounded, it must be compact.

Extreme Value Theorem: Continuous functions attain maximum and minimum values on compact domains.

So, $f$ attains its minimum $m$ and maximum $M$ on $D$.

Let $z= a + ib$ for $a,b \in \mathbb R$, and $w = \big(a-\lfloor a \rfloor \big) + i \big(b-\lfloor b \rfloor \big)$.

Observe that $w \in D$ and $f(z) = f(w)$. So, for any $z \in \mathbb C$, we have $m \leq f(z) \leq M$.

Therefore $f$ is bounded.