Derivative of Uniform Limit of Analytic Functions

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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:

\(\ds \map {f'_n} w\)

\(=\)

\(\ds \frac 1 {2 \pi i} \int_{\partial \map {D_r} w} \frac {\map {f_n} z} {\paren {z - w}^2} \rd z\)

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 $\paren 1$

$\blacksquare$

Also see

• Uniform Limit of Analytic Functions is Analytic