Problem 7

Exercise 9.7 (Compact convergence in the plane)   Let $\Omega\subseteq\mathbb{C}$ be open and suppose $\{f_n\}_{n=1}^\infty$ is a sequence of holomorphic functions over $\Omega$ converging uniformly to $f:\Omega\to\mathbb{C}$. For any $\delta>0$, the $f'_n\to f'$ uniformly on the set

$$K_\delta := \{z\in \Omega \;\vert\; \overline{B_\delta(z)}\subseteq \Omega\}$$ (9.38)

Proof. Note that if $z\in K_\delta$, then $\overline{B_{\delta/2}(z)} \subseteq K_{\delta/2}$, because if $x\in \overline{B_{\delta/2}(z)}$, then $\overline{B_{\delta/2}(x)}\subseteq \overline{B_\delta(z)} \subseteq \Omega$, so $x\in K_{\delta/2}$.

For any $z\in K_\delta$, the Cauchy Integral Formula can be applied on a circle of radius $r=\delta/2$ around $z$ to find the derivative sequence in terms of the original sequence. For any $f_n$, we have

$$f'_n(z) = \frac{1}{2\pi i} \int_{C_r(z)} \frac{f'_n(\zeta)}{\zeta-z}$$ (9.39)

It can be shown that

$$\int_{C_r(z)} \frac{f'_n(\zeta)}{\zeta-z} = \int_{C_r(z)}\frac{f_n(\zeta)}{(\zeta-z)^2}$$ (9.40)

Then

$$\vert f'_n(z)-f'(z)\vert =\frac{1}{2\pi} \left\vert \int_{C_r(z)}\frac{f_n(\zeta)}{(\zeta-z)^2} -\int_{C_r(z)}\frac{f(\zeta)}{(\zeta-z)^2} \right\vert$$ (9.41)

$$\leq\frac{1}{2\pi} \int_{C_r(z)}\frac{\vert f_n(\zeta)-f(\zeta)\vert}{\vert\zeta-z\vert^2}$$ (9.42)

$$\leq \frac{1}{2\pi r^2} \int_{C_r(z)}\vert f_n(\zeta)-f(\zeta)\vert$$ (9.43)

$$\leq \frac{1}{2\pi r^2} \Vert f_n-f\Vert _{C_r(z)}\mu(C_r(z))$$ (9.44)

$$\leq \frac{1}{r}\Vert f_n-f\Vert _{K_{\delta/2}}$$ (9.45)

$$\leq \frac{1}{r}\Vert f_n-f\Vert _{\Omega}$$ (9.46)

Letting $n\to\infty$ shows $f'_n\to f'$ uniformly on $K_\delta$. $\qedsymbol$