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"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?

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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.

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  • $\begingroup$ I'm struggling to see how g being bounded on U and g(z+a) = g(z) proves g is bounded. Is it because z + a can be written as z' + b where b is an element of U? Even then, I don't completely understand why that would prove g is bounded $\endgroup$ – Dino Nov 21 '18 at 17:32
  • $\begingroup$ Let $z$ be any complex number. Then$$z=z-\lfloor\operatorname{Re}z\rfloor-\lfloor\operatorname{Im}z\rfloor i+\lfloor\operatorname{Re}z\rfloor+\lfloor\operatorname{Im}z\rfloor i$$and therefore$$g(z)=g\bigl(z-\lfloor\operatorname{Re}z\rfloor-\lfloor\operatorname{Im}z\rfloor i\bigr),$$since $\lfloor\operatorname{Re}z\rfloor+\lfloor\operatorname{Im}z\rfloor i\in T$. But $z-\lfloor\operatorname{Re}z\rfloor-\lfloor\operatorname{Im}z\rfloor i\in U$ and so $\bigl\lvert g(z)\bigr\rvert\leqslant\sup\lvert g\rvert(U)$. $\endgroup$ – José Carlos Santos Nov 21 '18 at 18:12

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