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Suppose $f:\Bbb{C}\rightarrow \Bbb{C}$ is holomorphic, $f(z+1)=f(z)$ for all $z \in \Bbb{C}$ and $$|f(z)|\leq \exp\left(\frac{2}{\pi}|z|\right).$$ Is $f(z)$ a constant in $\Bbb{C}$?

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up vote 3 down vote accepted

By periodicity of $f$ the following function is well defined and holomorphic on $\mathbb{C}^{\ast}$:

$$ h(z) = f\left(\frac{\log(z)}{2\pi i}\right) $$

Considering all branches of $\log$ the modulus of $h$ can be estimated by

$$ |h(z)| = \left|f\left(\frac{\log(z)}{2\pi i}\right) \right| \leq \min_{k \in \mathbb{Z}} \exp \left( \left| \frac{\log(z)}{ \pi^2 i} + \frac{2 k}{\pi} \right| \right) \leq \exp \left( \frac{1}{\pi} + \frac{|\log|z| \,|}{ \pi^2} \right). $$

For $|z| \leq 1$ this implies

$$ | z \, h(z) | \leq e^{1/\pi} |z|^{1-1/\pi^2} $$

which shows that $z\,h(z)$ is bounded near $z=0$ and hence extends to an entire function. Since $z\,h(z)$ vanishes at $z=0$ it follows that $h(z)$ is itself entire.

Similarly for $|z| \geq 1$ we get

$$ |z \, h(z^{-1})| \leq e^{1/\pi} |z|^{1-1/\pi^2} $$

and therefore $h(z^{-1})$ is also entire. Then $h(z)$ is a bounded entire function so it must be constant. then $f$ itself is constant as well.

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I think you can use Liouville theroem. Since $f$ is an entire function, then $g=f\exp(2z/\pi)$ is also an entire function. Easily we can see that $g$ is a constant in $\Bbb C.$ So we can get $f(z)=c\exp(2z/\pi)$, however, we have $f(z)=f(z+1)$, so $c=0,$ and as a result $f(z)=0$ for all $z\in \mathbb{C}$

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Why would $g$ be constant or even bounded? Also, $f(z)=1$ for all $z \in \mathbb{C}$ does satisfy the required bound and this would not be possible? – WimC Apr 18 '13 at 11:33
I am so sorry that my idea is wrong.However,today I read a proof of this problem.It says that a period function can be expanded by trigonometric series,like polynomials.Like the proof in the first analysis provided by WimC,estimate the max|f(z)| when z=r,then you can get an contradiction if the expansion is not a constant.I hope this idea can help you. – HSLiu Apr 20 '13 at 11:46

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