The constant entire functions

Question

Let $f$ be an entire function. Suppose that $f(z)=f(z+1)$ and $|f(z)|\leq e^{|z|}$ for all $z$. Prove that $f$ is constant.

I tried to use Cauchy estimate to show that the first derivative is zero. But it did not work as $R$ goes to infinity $|f'(z)|$ goes to infinity because of $e^{R}$ in the nominator. I have no idea about using the fact that $f(z)=f(z+1)$. Please help!

Possible duplicate of Entire function constant, where $f(z)=f(z+1)$ and $|f(z)|< e^{|z|}$. — Martin R, Jul 14 '19 at 15:27
Another one: https://math.stackexchange.com/questions/603506/entire-periodic-function-with-bounded-growth-is-constant?noredirect=1&lq=1 — Martin R, Jul 14 '19 at 16:04

score 0 · Answer 1 · answered Jul 14 '19 at 15:26

0

Start: You can use Liouville's theorem to show that $$\frac{f(z)-f(0)}{\sin(\pi z)}=c.$$

answered Jul 14 '19 at 15:26

David C. Ullrich

89,985

Shouldn't it be $\frac{f(z)-f(0)}{\sin(2\pi z)}$ , as in https://math.stackexchange.com/a/222335/42969 ? – Martin R Jul 14 '19 at 15:42
@MartinR No, in fact the $2\pi$ in that linked answer is wrong. The function $\sin(\pi z)$ has zero set exactly $\Bbb Z$. – David C. Ullrich Jul 14 '19 at 15:48
I see, you are right. (But the question is a still a duplicate.) – Martin R Jul 14 '19 at 15:50

The constant entire functions

1 Answers1