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Problem Statement

Find all entire functions such that $f(it)=f(it+i)=f(it+i\sqrt{2})$ for all real $t$.

My approach:

Naturally a constant function satisfies the above conditions and thus I thought about trying to prove that any function that satisfies the above conditions must be a constant function.

Liouville's theorem sounds relevant since the function is entire but I can't see a way of proving it must be bounded.

An exercise of a similar nature goes like this:

"prove that if $f(z)$ is entire and doubly periodic, meaning it has $2$ linearly independent (over the reals) periods, then $f(z)$ is a constant"

I do know how to prove the above and it is proven by Liouville's theorem. But I can't seem to replice the proof that $f(z)$ is bounded in my exercise.

Please don't give me a full solution, hints will great :)

Thank you so much in advance!

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  • $\begingroup$ it would either have to be not differentiable or a constant. $\endgroup$ – Jacob Claassen Jun 12 '17 at 13:45
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    $\begingroup$ Identity theorem. But it's also useful to prove that a continuous function $g \colon \mathbb{R}\to \mathbb{C}$ with periods $1$ and $\sqrt{2}$ is constant. $\endgroup$ – Daniel Fischer Jun 12 '17 at 13:46
  • $\begingroup$ Though he doesn't want full solutions, I guess he wants full hints. ;-) $\endgroup$ – Professor Vector Jun 12 '17 at 13:47
  • $\begingroup$ See for instance math.stackexchange.com/questions/136665/… or math.stackexchange.com/questions/90177/… $\endgroup$ – LutzL Jun 12 '17 at 13:52
  • $\begingroup$ Thanks, I understand now :) $\endgroup$ – zokomoko Jun 12 '17 at 19:08

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