"Consider the set
$T = \{ \alpha ∈ C|∃ a, b ∈ Z :\alpha = a + bi \}$
Let $g$ be an entire function which satisfies that
$g(z + \alpha) = g(z)$
for all $z ∈ \mathbb{C}$ and all $\alpha ∈ T$.
Prove that $g$ is bounded on the set $ U= \{\beta ∈ C|∃ x ∈ [0, 1], y ∈
[0, 1] :\beta = x + yi\}$.
Prove that $g$ is constant on $\mathbb{C}$."
I think I have a outline of a proof:
(1) Prove U is compact (closed and bounded)
(2) Use (1) to prove $g$ is bounded on $U$ (Every entire function on a compact subset of $\mathbb{C}$ is bounded)
(3) Then, as $g$ is entire by definition and is bounded, I can use Liouville's Theorem to show $g$ is constant (although I feel this only proves $g$ is bounded on $U$ and not on all of $\mathbb{C}$)
However, I'm stuck on step (1). I mean, obviously $U$ is closed and bounded, but I have no idea how to prove that. And then I'm not confident with (3). Is Liouville's Theorem enough to prove $g$ is constant everywhere?
Best Answer
In order to prove (1), your idea is correct: $U$ is closed and bounded. It is clear that $U$ is a subset of the closed ball centered at $0$ with radiues $\sqrt2$; therefore, $U$ is bounded. On the other if a sequence $(z_n)_{n\in\mathbb N}$ of elements of $U$ converges to $z\in\mathbb C$, then both the real and the imaginary part of each $z_n$ is in $[0,1]$ and therefore the same thing happens to $z$; so $z\in U$. This proes that $U$ is closed.
And you are right about (3) too. Since $g|_U$ is bounded and $g(z+a)=g(z)$ for each $a\in T$, then $g$ is bounded. Therefore, by Liouvile's theorem, $g$ is constant.