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Let $f(z)$ be an entire function s.t

$\forall z ; f(z+i)=f(z), f(z+1)=f(z)$

Prove: $f$ is constant

So we need will show that $f(z)$ is bounded.

How should I approach this?

Community
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gbox
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    Hint: what can you say about the function on $[0,1]\times[0,1]$ (with the usual identification $\mathbb C\simeq \mathbb R^2$)? How does this extend to $\mathbb C$? – Taladris Jul 07 '17 at 11:48

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$f$ is continuous implies that the image of $B(0,2)$ is bounded. Write $z=x+iy$, there exists $u,v$ with $|u|, |v|\leq 1$ and $x=u+n, y=v+m$, $n,m$ are integers this implies that $f(z)=f(u+n+(v+m)i)=f(u+iv)$ so $f$ is bounded.

Tsemo Aristide
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Theorem: Continuous functions attain its maximum and minimum on compact domains.

Since $A:=\{ x+iy \ : \ |x| \leq 1, \ |y| \leq 1 \}$ is compact, its image must be bounded.

For any $x+iy \in \mathbb C$, $\exists x_0,y_0$ such that $x-x_0 \in \mathbb Z$ and $y-y_0 \in \mathbb Z$ where $x_0+iy_0 \in A$.

Therefore $f(x+iy)=f(x+iy_0)=f(x_0+iy_0)$.

This clearly shows that $f$ is bounded.

ThePortakal
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  • So why cant I say, lets look at a compact domain, $f$ is continuous and therefore it is bounded and we are done? – gbox Jul 07 '17 at 12:04
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    Because it is possible that $f$ is bounded on a compact domain but not bounded on $\mathbb C$. For example $g: \mathbb R \to \mathbb R$ defined as $g(x)=x^2$ is bounded on $[-1,1]$, but it is not bounded on $\mathbb R$. – ThePortakal Jul 07 '17 at 12:10