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Spring 2014 - Problem 7

entire functions

Characterize all entire functions $f$ with $\abs{f\p{z}} > 0$ for $\abs{z}$ large and

\limsup_{z\to\infty}\,\frac{\abs{\log\,\abs{f\p{z}}}}{\abs{z}} < \infty.

Solution.

By definition, there exist $R, M > 0$ such that

\begin{aligned} \frac{\abs{\log\,\abs{f\p{z}}}}{\abs{z}} \leq M &\implies \abs{\log\,\abs{f\p{z}}} \leq M\abs{z} \\ &\implies \abs{f\p{z}} \leq e^{M\abs{z}} \end{aligned}

for $\abs{z} \geq R$ . Since $\cl{B\p{0, R}}$ is compact, it follows that $f$ is entire of order $1$ . Conversely, if $f$ is entire of order $1$ , then $\abs{f\p{z}} \leq Ce^{a\abs{z}}$ for all $z \geq 1$ . Thus,

\frac{\abs{\log{\abs{f\p{z}}}}}{\abs{z}} \leq \frac{\log{C} + a\abs{z}}{\abs{z}} \implies \limsup_{z\to\infty}\,\frac{\abs{\log{\abs{f\p{z}}}}}{\abs{z}} \leq a < \infty,

so all entire functions satisfying this property are entire of order $1$ . Since $\abs{f\p{z}} > 0$ for $\abs{z}$ large, $f$ only has finitely many zeroes, so by Hadamard's theorem, $f$ has the form

f\p{z} = e^{az+b} \prod_{n=1}^N \p{z - z_n},

where the $z_n$ are the zeroes of $f$ .