${f_n(z)}$ be a sequence of analytic function in a domain $D$ such that $f_n\to f$ uniformly in $D$. Then $f_n’\to f'$ uniformly in $D$.
How can I show that the above statement is true/false?
${f_n(z)}$ be a sequence of analytic function in a domain $D$ such that $f_n\to f$ uniformly in $D$. Then $f_n’\to f'$ uniformly in $D$.
How can I show that the above statement is true/false?
In general it's not true. Consider
$$f_n(z) := \frac{1}{n} \cdot z^n$$
where $z \in D:=B(0,1):= \{z \in \mathbb{C}; |z|<1\}$. Then
$$|f_n(z)| \leq \frac{1}{n} \to 0 \quad (n \to \infty)$$
i.e. $f_n \to 0$ uniformly in $D$. But:
$$|f_n'(z)| = |z^{n-1}|=|z|^{n-1}$$
does not converge uniformly (to $0$) in $B(0,1)$.
The following statement is true: Let $f_n$ holomorphic. If $f_n \to f$ uniformly, then $f$ is holomorphic and $f_n' \to f'$ uniformly on compact subsets of $D$.
Hints : take $f_n(z) = \frac{z^n}{n^2}$