Normal family of complex functions

Question

Anyone can give me a hint for solve this problem?

Show if $\mathcal{F} \subseteq \textit{H}(\Omega)$ is a normal family, then $\mathcal{F}'= \{f' | f \in \mathcal{F}\}$ is normal. ($\Omega \subset \mathbb{C}$)

I really appreciate it any hint.

Answer 1

Recall the Weierstrass convergence theorem: If $(f_n)\subset\text{Hol}(\Omega)$ and $f_n\to f$ locally uniformly (that is, uniformly over each compact set) then $f\in\text{Hol}(\Omega)$ and moreover $f_n^{(k)}\to f^{(k)}$ locally uniformly.

Let $\mathcal{F}\subset\text{Hol}(\Omega)$ be a normal family. By definition, this means that if $(f_n)\subset\mathcal{F}$ then $(f_n)$ has a subsequence $(f_{n_k})$ such that $f_{n_k}\to f$ locally uniformly, where $f\in\text{Hol}(\Omega)$. Now Let $\mathcal{F}'$ be the family you described and let $(f_n')$ be a sequence in $\mathcal{F}'$. Since $(f_n)\subset\mathcal{F}$, let $(f_{n_k})$ be the locally-uniformly-convergent subsequence of $(f_n)$ and let $f$ be the limit function. Then by Weierstrass's theorem, $(f_{n_k}')$ converges to $f'$ locally uniformly and $f'\in\text{Hol}(\Omega)$, since $f$ is analytic.

The converse is not true: Let $\Omega=\mathbb{C}$ and $\mathcal{F}=\{f_n:n\in\mathbb{N}\}$ with $f_n:\mathbb{C}\to\mathbb{C}$ and $f_n(z)=z+n$. Obviously, $\mathcal{F}'=\{h\}$, where $h$ denotes the constant function $1$, therefore it is a normal family. But there exists a sequence of $\mathcal{F}$ (actually, every sequence has this property) that has no locally uniformly convergent subsequence: indeed, suppose that $(f_{m_k})\subset\mathcal{F}$ converges locally uniformly to $f\in\text{Hol}(\mathbb{C})$. Then $f(0)=\lim_{k\to\infty}f_{m_k}(0)=\lim_{k\to\infty}m_k=\infty$, which is impossible, since $f$ admits values in $\mathbb{C}$.