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Fall 2014 - Problem 8

entire functions

Let $\func{f}{\C}{\C}$ be an entire function. Show that

\abs{f\p{z}} \leq Ce^{a\abs{z}},\quad z\in \C,

for some constants $C$ and $a$ if and only if we have

\abs{f^{\p{n}}\p{0}} \leq M^{n+1},\quad n = 0, 1, \ldots,

for some constant $M$ .

Solution.

" $\implies$ "

By the Cauchy estimates, we get for any $n > 0$ that

\begin{aligned} \abs{f^{\p{n}}\p{0}} &\leq \frac{n!}{2\pi} \int_{\partial B\p{0,n}} \frac{\abs{f\p{\zeta}}}{\abs{\zeta}^{n+1}} \,\diff\abs{\zeta} \\ &\leq \frac{n!}{n^n} Ce^{an} \\ &\leq Ae^{an} \end{aligned}

by Stirling's approximation. Increasing $A$ if necessary, we may assume that $A > 1$ so that $\abs{f^{\p{n}}\p{0}} \leq \p{Ae^a}^{n+1}$ for all $n \geq 0$ .

" $\impliedby$ "

Writing $f\p{z} = \sum_{n=0}^\infty a_nz^n$ , we see

\abs{a_n} = \abs{\frac{f^{\p{n}}\p{0}}{n!}} \leq \frac{M^{n+1}}{n!}.

Thus,

\abs{f\p{z}} \leq \sum_{n=0}^\infty \frac{M^{n+1}}{n!}\abs{z}^n = Me^{M\abs{z}}.