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

$(2): \quad \map {D_r} w \subseteq \map {D_{2 r} } a \subseteq U$
 $\ds \map {f'_n} w$ $=$ $\ds \frac 1 {2 \pi i} \int_{\partial \map {D_r} w} \frac {\map {f_n} z} {\paren {z - w}^2} \rd z$ and: $\ds \map {f'} w$ $=$ $\ds \frac 1 {2 \pi i} \int_{\partial \map {D_r} w} \frac {\map f z} {\paren {z - w}^2} \rd z$

Therefore:

 $\ds \cmod {\map {f'_n} w - \map {f'} w}$ $=$ $\ds \frac 1 {2 \pi} \cmod {\int_{\partial \map {D_r} w} \frac {\map {f_n} z - \map f z} {\paren {z - w}^2} \rd z}$ $\ds$ $\le$ $\ds \frac {2 \pi r} {2 \pi} \sup_{z \mathop \in \partial \map {D_r} w} \frac {\cmod {\map {f_n} z - \map f z} } {r^2}$ Estimation Lemma for Contour Integrals $\ds$ $=$ $\ds \frac 1 r \sup_{z \mathop \in \partial \map {D_r} w} \cmod {\map {f_n} z - \map f z}$ $\ds$ $\le$ $\ds \frac 1 r \sup_{z \mathop \in \map {D_{2 r} } a} \cmod {\map {f_n} z - \map f z}$ by $(2)$

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

 $\ds \sup_{w \mathop \in \map {D_r} a} \cmod {\map {f'_n} w - \map {f'} w}$ $\le$ $\ds \frac 1 r \sup_{z \mathop \in \map {D_{2 r} } a} \cmod {\map {f_n} z - \map f z}$ $\ds$ $\to$ $\ds 0$ as $n \to \infty$ by $\paren 1$

$\blacksquare$