Derivative of Uniform Limit of Analytic Functions

Theorem

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

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

Let $\left\langle{f_n}\right\rangle$ converge locally uniformly to $f$ on $U$.

Then the sequence $\left\langle{f_n'}\right\rangle_{n \mathop \in \N}$ 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 $U$.

$\displaystyle f'_n \left({a}\right) = \frac 1 {2 \pi i} \int_{\partial D} \frac {f_n \left({z}\right)} {\left({z - a}\right)^2} \, \mathrm d z$

and:

$\displaystyle f'\left({a}\right) = \frac 1 {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)}{\left({z - a}\right)^2} \, \mathrm d z$

Therefore:

 $\displaystyle \left\vert{f'_n \left({a}\right) - f' \left({a}\right) }\right\vert$ $=$ $\displaystyle \frac 1 {2 \pi} \left\vert{ \int_{\partial D} \frac {f_n \left({z}\right) - f \left({z}\right)} {\left({z - a}\right)^2} \, \mathrm d z }\right\vert$ $\displaystyle$ $\le$ $\displaystyle \frac r {r^2} \sup_{z \mathop \in \partial D} \left\vert{ f_n \left({z}\right) - f \left({z}\right) }\right\vert$ Estimation Lemma for complex integrals $\displaystyle$ $=$ $\displaystyle \frac 1 r \sup_{z \mathop \in \partial D} \left\vert{ f_n \left({z}\right) - f \left({z}\right) }\right\vert$

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 $U$.

$\blacksquare$