I am looking for hints for this old prelim exam question.

Let $\mathcal{F}$ be a normal family of entire functions, and let $$A_n=\sup\left\{|a_n|:f(z)=\sum_{j=0}^{\infty}a_jz^j,f\in\mathcal{F}\right\}$$ for $n=0,1,2,\ldots.$ Show that $\sum_{n=0}^{\infty}A_nr^n<\infty$ for all $r>0.$

My first step is to recognize that $A_n=\frac{1}{n!}\sup_{f\in\mathcal{F}}|f^{(n)}(0)|.$ For each $n,$ there exists a sequence of functions $(f_{n,k})_{k=1}^{\infty}$ in $\mathcal{F}$ such that $$\lim_{k\to\infty}\frac{\left|f_{n,k}^{(n)}(0)\right|}{n!}=A_n.$$ Now since $\mathcal{F}$ is a normal family, the sequence $(f_{n,k})_{k=1}^{\infty}$ admits a subsequence $\left(f_{n,k_j}\right)_{j=1}^{\infty}$ such that there exists a function $g_n\colon \mathbb{C}\to\mathbb{C}$ such that $f_{n,k_j}\xrightarrow[j\to\infty]{}g_n$ uniformly on compact sets. Then it can be shown that $g_n$ is entire for each $n,$ and $f_{n,k_j}^{(m)}\xrightarrow[j\to\infty]{}g_n^{(m)}$ uniformly on compact sets for all orders $m\in \mathbb{N},$ in particular $f_{n,k_j}^{(n)}\xrightarrow[j\to\infty]{}g_n^{(n)}$ uniformly on compact sets. It follows that $\frac{\left|g_n^{(n)}(0)\right|}{n!}=A_n.$ My hope is to use the functions $g_n$ find an entire function $h$ with the property that $h^{(n)}=g_n^{(n)}$ for all orders $n,$ but don't see how to do this. Does this seem to be a good strategy, or is there a better approach I am missing? I feel that I am not making enough use of the fact that $\mathcal{F}$ consists of entire functions - is there a way to use this fact here?


1 Answer 1


$\cal F$ is uniformly bounded on compact sets, so that $$ M_R = \sup \{ |f(z)| : |z| \le R, f \in \cal F \} $$ is finite for every $R> 0$. Using Cauchy's integral formula, this gives a uniform upper bound for $|a_n|$, and therefore an upper bound for $A_n$.

$A_n \le \frac{M_R}{R^n}$.

Using this bound, you can prove the convergence of $\sum_{n=0}^{\infty}A_nr^n$ for all $r > 0$, for example with the root test, or by comparison with the geometric series.

Choose $R= 2r$. Then $A_n r^n \le \frac{M_R}{2^n}$.


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