Suppose \(( f _ { n } ) _ { n \in \mathbb { N } }\) is a sequence of analytic functions on \(\mathbb { D } : = \{ z \in \mathbb { C } : | z | < 1 \}\) Show that if \(\left( f _ { n } \right)\) converges to a function \(g\colon \DD \to \CC\) uniformly on each compact subset of \(\DD\), then \(g\) is analytic on \(\DD\).
Solution
Let \[ \DD=\{z\in\CC:\abs z<1\} \] and fix \[ z_0\in\DD. \]
1. Choose \(R>0\) such that\[ \overline{B(z_0,R)}\subseteq\DD. \]Then\[ f_n\longrightarrow g \]uniformly on the circle\[ \Gamma=\{\zeta\in\CC:\abs{\zeta-z_0}=R\}. \]
Because \(\DD\) is open, there is \(R>0\) with the closed disk contained in \(\DD\). The circle \(\Gamma\) is compact, so the hypothesis of uniform convergence on every compact subset applies to \(\Gamma\).
2. For every \(z\) with\[ \abs{z-z_0}<R, \]one has\[ g(z) = \frac{1}{2\pi i} \int_\Gamma \frac{g(\zeta)}{\zeta-z}\,d\zeta. \]
Fix such a \(z\). Cauchy’s integral formula gives, for every \(n\), \[ f_n(z) = \frac{1}{2\pi i} \int_\Gamma \frac{f_n(\zeta)}{\zeta-z}\,d\zeta. \] Since \(f_n(z)\to g(z)\) pointwise and step 1 gives uniform convergence on \(\Gamma\), we have \[ \begin{aligned} &\left| \int_\Gamma \frac{f_n(\zeta)-g(\zeta)}{\zeta-z}\,d\zeta \right|\\ &\qquad\leq \frac{\operatorname{length}(\Gamma)} {\operatorname{dist}(z,\Gamma)} \sup_{\zeta\in\Gamma}\abs{f_n(\zeta)-g(\zeta)} \longrightarrow0. \end{aligned} \] Passing to the limit in Cauchy’s formula gives the displayed identity.
3. The function\[ G(z) = \frac{1}{2\pi i} \int_\Gamma \frac{g(\zeta)}{\zeta-z}\,d\zeta \]is holomorphic on\[ B(z_0,R). \]
Fix \[ z\in B(z_0,R). \] For \(h\) sufficiently small, \[ \begin{aligned} \frac{G(z+h)-G(z)}h &= \frac{1}{2\pi i} \int_\Gamma \frac{g(\zeta)} {(\zeta-z-h)(\zeta-z)} \,d\zeta. \end{aligned} \] As \(h\to0\), the integrand converges uniformly on \(\Gamma\) to \[ \frac{g(\zeta)}{(\zeta-z)^2}, \] because \(z\) has positive distance from \(\Gamma\). Therefore \[ G'(z) = \frac{1}{2\pi i} \int_\Gamma \frac{g(\zeta)}{(\zeta-z)^2}\,d\zeta. \] Thus the complex derivative exists at every point of the disk, so \(G\) is holomorphic there.
4. The function \(g\) is holomorphic on a neighborhood of \(z_0\).
Step 2 says exactly that \[ g(z)=G(z) \] for every \[ z\in B(z_0,R). \] Step 3 shows that \(G\) is holomorphic on this disk. Hence \(g\) is holomorphic there.
5. The function \(g\) is analytic on all of \(\DD\).
The point \(z_0\in\DD\) was arbitrary. By step 4, every point of \(\DD\) has a neighborhood on which \(g\) is holomorphic. Therefore \(g\) is holomorphic, hence analytic, on \(\DD\).
6. Q.E.D.
Step 5 is the required conclusion.