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Fall 2011 - Problem 12

Cauchy estimates

A holomorphic function $\func{f}{\C}{\C}$ is said to be of exponential type if there are constants $c_1, c_2 > 0$ such that

\abs{f\p{z}} \leq c_1e^{c_2\abs{z}} \quad\text{for all}\quad z \in \C.

Show that $f$ is of exponential type if and only of $f'$ is of exponential type.

Solution.

" $\implies$ "

Suppose $f$ is of exponential type with constants $c_1, c_2$ . Let $z \in \C$ and $C\p{z, R}$ the circle of radius $R$ centered at $z$ . We get the Cauchy estimate

\begin{aligned} \abs{f'\p{z}} &= \abs{\frac{1}{2\pi i} \int_{C\p{z,R}} \frac{f\p{\zeta}}{\p{\zeta - z}^2} \,\diff\zeta} \\ &\leq \frac{1}{2\pi} \cdot 2\pi R \cdot \sup_{\zeta \in C\p{z,R}} \frac{c_1e^{c_2\abs{\zeta}}}{R^2} \\ &\leq \frac{c_1}{R}e^{c_2\abs{z}+c_2R} \\ &= \p{\frac{c_1}{R}e^{c_2R}}e^{c_2\abs{z}}. \end{aligned}

Since $R > 0$ is a fixed constant, $f'$ is of exponential type.

" $\impliedby$ "

Suppose $f'$ is of exponential type with constants $c_1, c_2$ and let $z \in \C$ . Consider the segment $\gamma$ parametrized by $t \mapsto tz$ , so that by the fundamental theorem of calculus,

f\p{z} = f\p{0} + \int_{\gamma} f'\p{z} \,\diff{z}.

Then

\begin{aligned} \abs{f\p{z}} &\leq \abs{f\p{0}} + \int_\gamma \abs{f'\p{z}} \,\diff\abs{z} \\ &\leq \abs{f\p{0}} + \abs{z}c_1e^{c_2\abs{z}} \\ &\leq \abs{f\p{0}}e^{\p{c_2+1}\abs{z}} + c_1e^{\p{c_2+1}\abs{z}} \\ &= \p{\abs{f\p{0}} + c_1}e^{\p{c_1+1}\abs{z}}, \end{aligned}

since $e^x \geq x$ and $e^x \geq 1$ for $x \geq 0$ . Hence, $f$ is of exponential type as well.