To show a function is analytic

Question

Let $G\subset\mathbb C$ be open and connected, and function $h$ is analytic on $G$. $\{f_n(z)\}$ is a sequence of analytic functions on $G$ for which $\lim_{n\rightarrow \infty}f_n(z)$ exists for any $z$ in $G$. Define $f(z)=\lim_{n\rightarrow \infty}{f_n(z)}$. Suppose that $|f'_n(z)|\leq|h(z)|$ for any $z\in G$. Prove that $f\in H(G)$.

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My idea: first to show $f(z)$ is continuous then use Morera's Theorem to show $f$ is analytic. But I have no idea of $h$. How to use $|f'_n(z)|\leq|h(z)|$ for any $z\in G$? What does it imply?

Answer 1

Hint: Use Morera's theorem for $f_n$. The details are below.

For any compact $K$ of $G$ the inequality $|f'_n(z)|<|h(z)|$ shows that $f'_n$ are uniformly bounded on $K$. Therefore the family $\{f'_n\}$ is normal (Montel's theorem). Therefore there is a subsequence $f'_{n_k}$ that converges uniformly. Hence $f_{n_k}$ converges uniformly (to $f$). Now your plan is applicable. $f$ is continuous for being the uniform limit of continuous function (analytic actually). And Morera is satisfied for $f$ because it is satisfied for $f_{n_k}$ and they converge uniformly to $f$.