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Fall 2015 - Problem 8

periodic functions

Assume that $f\p{z}$ is an entire function that is $2\pi$ -periodic in the sense that $f\p{z + 2\pi} = f\p{z}$ , and

\abs{f\p{x + iy}} \leq Ce^{\alpha\abs{y}},

for some $C > 0$ , where $0 < \alpha < 1$ . Prove that $f$ is constant.

Solution.

Since $f$ is $2\pi$ -periodic, we may write $f\p{z} = g\p{e^{iz}}$ , where $g\p{z} = f\p{\frac{\log{z}}{i}}$ . Indeed, by periodicity, $g$ is well-defined regardless of the branch chosen, and it is holomorphic except at the origin. When $\abs{z} < 1$ , we get by assumption that

\abs{g\p{z}} \leq Ce^{\alpha\abs{\log\abs{z}}} = Ce^{-\alpha\log\abs{z}} = C\abs{z}^{-\alpha},

and because $0 < \alpha < 1$ , we see that $\abs{zg\p{z}} \to 0$ , so $g$ extends analytically to the origin, i.e., $g$ is entire. On the other hand, if $\abs{z} > 1$ , $\abs{g\p{z}} \leq C\abs{z}^\alpha$ . Hence, by the Cauchy estimates,

\abs{g^{\p{n}}\p{z}} \leq \frac{CR^\alpha n!}{R^n} = \frac{Cn!}{R^{n-\alpha}} \xrightarrow{R\to\infty} 0,

since $n - \alpha \geq 1 - \alpha > 0$ . Thus, $g$ is constant, which implies that $f$ is constant as well.