Derivative of Uniform Limit of Analytic Functions

Theorem
Let $U$ be an open subset of $\C$.

Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence of analytic functions $f_n : U \to \C$.

Let $\sequence {f_n}$ converge locally uniformly to $f$ on $U$.

Let $f'$ denote the derivative of $f$.

Then the sequence $\sequence { {f_n}'}_{n \mathop \in \N}$ converges locally uniformly to $f'$.

Proof
Let $a \in U$.

By definition of locally uniform convergence, there exists an open disk $\map {D_{2 r} } a \subseteq U$ such that $f_n$ converges uniformly to $f$ on $\map {D_{2 r} } a$.

That is:
 * $\ds (1): \quad \lim_{n \mathop \to \infty} \sup_{z \mathop \in \map {D_{2 r} } a} \cmod {\map {f_n} z - \map f z} = 0$

We shall show that:
 * $\ds \lim_{n \mathop \to \infty} \sup_{w \mathop \in \map {D_r} a} \cmod {\map {f'_n} w - \map {f'} w} = 0$

Let $w \in \map {D_r} a$.

By Triangle Inequality for Complex Numbers:
 * $(2): \quad \map {D_r} w \subseteq \map {D_{2 r} } a \subseteq U$

Thus by Cauchy's Integral Formula for Derivatives:

Therefore:

Since $w \in \map {D_r} a$ was arbitrary, we have:

Also see

 * Uniform Limit of Analytic Functions is Analytic