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${f_n(z)}$ 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?

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Hint: Cauchy's formula gives a representation of derivatives that plays nicely with uniform convergence. –  Miha Habič Nov 23 '12 at 14:28
    
You need to show more. You first need to show that $f$ is differentiable, otherwise the conclusion doesn't necessarily make sense. –  kahen Nov 23 '12 at 14:34

1 Answer 1

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 $D$.

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