Analysis Problem of the Day 1

The following appears as Problem 8 on the UCLA Fall 2014 Analysis Qual:

Problem 8. Let $f: \mathbb{C} \to \mathbb{C}$ be an entire function. Show that

$|f(z)| \leq C e^{a|z|}, \quad z \in \mathbb{C},$

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

$|f^{(n)}(0)| \leq M^{n+1}, \quad n=0,1,...$

for some constant $M.$

Solution: Suppose $|f(z)| \leq C e^{a|z|}$ for some constants $C,a>0.$ Then, by Cauchy’s estimates,

$|f^{n}(0)| \leq \frac{n! \sup_{|z| \leq R} |f(z)|}{R^n} \leq \frac{n! C e^{aR}}{R^n}$

for all $R>0.$ Setting $R=n$ and using the crude bound $n! \leq n^n,$ we get that

$|f^{(n)}(0)| \leq C e^{an} = C (e^a)^n,$

so setting $M=\max(C,a)$ yields the desired claim. Conversely, we use the locally uniformly convergent power series

$f(z) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} z^n,$

which yields

$|f(z)| \leq \sum_{n=0}^\infty \frac{|f^{(n)}(0)|}{n!} |z|^n \leq M \sum_{n=0}^\infty \frac{1}{n!} M^n|z|^n = M e^{M|z|},$

so $C=a=M$ suffices.