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
Note that by Uniform Limit of Analytic Functions is Analytic, $f$ is analytic.

Let $a\in U$.

Let $D$ be a disk of radius $r$ about $a$, contained in $U$.

We have Cauchy's Integral Formula for Derivatives


 * $\ds \map { {f_n}'} a = \frac 1 {2 \pi i} \int_{\partial D} \frac {\map {f_n} z} {\paren {z - a}^2} \rd z$

and:


 * $\ds \map {f'} a = \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^2} \rd z$

Therefore:

Now the $f_n$ tend uniformly to $f$, and we can bound $r$ away from zero.

It follows that ${f_n}' \to f'$ uniformly in each compact disk contained in $U$.

Also see

 * Uniform Limit of Analytic Functions is Analytic