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Suppose f is entire and $f(z)=f(z+1)=f(z+\pi)$. Does this imply $f$ is constant?

I want to prove that it is constant.I see that it is enough to consider the value of $f(z)$ in between the lines $z=1$ and $z=-1$. Clearly $f$ does not have a pole at $\infty$ (gonig to $\infty$ along the real line). I only need to show that it does not have essential singularity at $\infty$. But I cannot proceed further.Also I am not using the second condition. Any help is highly appreciated. Also I am not sure if the answer is yes.

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    The function $e^{2\pi iz}$ is non-constant and has period $1$ so you definitely need the second condition. Using the fact that $\pi$ is irrational, can you prove that $f$ is constant on the real line? How would that help you? – frakbak Jul 25 '16 at 13:56
  • Related: http://math.stackexchange.com/questions/775718/a-real-continuous-periodic-function-with-two-incommensurate-periods-is-constant. – Martin R Jul 25 '16 at 14:07

2 Answers2

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Yes it is constant. To see that, you have to consider the additive subgroup of $\mathbb{R}$ generated by $1$ and $\pi$. As it turns out, this subgroup has no smallest positive element: If $x$ were a smallest element, then we take the least $n \in \mathbb{N}$ such that $nx>1$ or $nx>\pi$. Such an $n$ exists, because if not, then $\pi$ would be rational. Then $nx -1$ or $nx - \pi$ would be smaller than $x$.

Thus the group has no smallest element, and is hence dense in $\mathbb{R}$. Therefore $f$ is constant on each horizontal line, and hence globally constant, by a simple application of the Cauchy-Riemann equations.

Oles Wohnzimmer
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  • @2015 fixed that to absolute value. – Oles Wohnzimmer Jul 25 '16 at 14:21
  • I don't understand the statement 'Such an $n$ exists, because if not, then $\pi$ would be rational'. Also why $nx-\pi$ is smaller than $x$? –  Jul 25 '16 at 14:39
  • @2015 If no such n existed then we had that $nx=1$ and $mx=\pi$ and hence $\pi=\frac{m}{n}$. As $n$ was the least $n$ such that $nx > \pi$ by definition $nx - \pi<x$, as else the n would not be minimal – Oles Wohnzimmer Jul 25 '16 at 14:40
  • @2015 I don't understand your question. – Oles Wohnzimmer Jul 25 '16 at 14:45
  • The negation of the statement'there exist a least $n$ such that $nx\gneq 1$ or $nx\lneq \pi$' is 'for all $m$ and $n$, $nx \leq 1$ and $mx \geq \pi$'? –  Jul 25 '16 at 14:50
  • Anyway thank you! The statement which I am enquiring follows from Archimedian property. –  Jul 25 '16 at 15:00
  • @2015 It would perhaps help here to remember (or to know...) that a non-trivial subgroup of $;\Bbb R;$ is either cyclic (and it thus contains a minimal positive element) or else it is dense in $;\Bbb R;$ . – DonAntonio Jul 25 '16 at 15:28
  • @DonAntonio I think here the subgroup is not cyclic as $\pi$ is not rational and $1$ is rational. In his argument I don't see the importance of $\pi$. –  Jul 25 '16 at 15:40
  • @2015 It seems to be that's the case, indeed. I think that's what Oles remarked: if there's a minimal positive element then the subgroup is cyclic and etc. – DonAntonio Jul 25 '16 at 15:42
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Take $g(z)=f(z+1)-f(z)$ and a sequence $z_n=\frac{1}{n}$ then $g(1/n)=f(\frac{1}{n}+1)-f(\frac{1}{n})=0$ means $g(z)$ is zero in the $nbd.$ of $z=0$. Use Identity theorem to get $g(z)\equiv 0$ which implies $f(z)$ is constant.

Nitin Uniyal
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  • $g(z)\equiv 0$ implies $f(z+1)=f(z)$. So we get nothing new! As the comment by frakbak suggested you need to use the second condition also. –  Jul 27 '16 at 04:44