Suppose \(f_k: \Omega\to \CC\) is a sequence of differentiable functions converging locally uniformly to \(f:\Omega\to \CC\). Show that
- \(f\) is continuous,
- \(f\) is differentiable,
- \(\ts{f_k'}\to f'\) locally uniformly.
Thus if \(f(z) = \sum{k\geq 0} c_k (z-z_0)^k\) is a power series, since \(S_N\to f\) locally uniformly, \(f\) can be differentiated term-by-term within its radius of convergence.
Solution
That \(f\) is continuous is a local question: fixing a point \(z_0\), take a closed disc \(\DD+z_0\) about \(z_0\). By local uniform convergence \(f_k\to f\) uniformly on \(\DD+z_0\), and differentiable \(\implies\) continuous. So each \(f_k\) is continuous, making \(f\) continuous on \(\DD+z_0\) by the uniform limit theorem.
That \(f\) is differentiable is again a local question: fix \(z\) and write \(\gamma \da \bar{\DD + z}\) as the boundary of the disc about \(z\). Define \(g_k(\xi) \da {f_k\over \xi-z}\), so \(g_k \to {f \over \xi-a}\) locally uniformly. Now apply Cauchy’s integral formula at \(z\): \[ f(z) &= \lim_k f_k(z) \\ &= \lim_k {1\over 2\pi i}\int_\gamma {f_k(\xi) \over \xi - z}\dxi \\ &= \lim_k {1\over 2\pi i}\int_\gamma g_k(\xi)\dxi\\ &= {1\over 2\pi i}\int_\gamma \lim_k g_k(\xi)\dxi \\ &= {1\over 2\pi i}\int_\gamma g(\xi)\dxi\\ &= {1\over 2\pi i}\int_\gamma {f(\xi) \over \xi - z} \dxi ,\] where we’ve used uniform convergence on \(\gamma\) to commute the limit and integral. So \(f\) has an integral representation, making it differentiable.
That \(f_k'\to f'\): \[ \lim_k f_k'(z) &= \lim_k {1\over 2\pi i}\int_\gamma {f_k(\xi) \over (\xi - z)^2 }\dxi\\ &= {1\over 2\pi i}\int_\gamma \lim_k {f_k(\xi) \over (\xi - z)^2}\dxi\\ &= {1\over 2\pi i}\int_\gamma {f(\xi) \over (\xi - z)^2}\dxi\\ &= f'(z) .\]
That the convergence is locally uniform: first consider what happens on an closed discs \(K = D\) with \(\gamma \da \bd{D}\). Then for \(z\in D\), \[ \abs{f'(z) - f_k'(z) } &= \abs{{1\over 2\pi i} \int_{\gamma} {f(\xi) - f_k(\xi) \over (\xi - z)^2}\dxi }\\ &\leq {1\over 2\pi}\int_{\gamma} {\abs{f(\xi) - f_k(\xi) } \over \abs{\xi - z}^2} \dxi\\ &\leq {1\over 2\pi}\int_{\gamma } { \sup_{\xi \in \gamma} \abs{f(\xi) - f_k(\xi) } \over r^2 } \dxi\\ &= {1\over 2\pi} { \sup_{\xi \in \gamma } \abs{f(\xi) - f_k(\xi) } \over r^2} \cdot {2\pi r} \\ &= { \sup_{\xi \in \gamma } \abs{f(\xi) - f_k(\xi) }}/r .\] Since \(\gamma\) is compact, using locally uniform convergence of \(f_k\to f\), there exists an \(n_0\) such that \(n\geq n_0\) bounds this \(\sup\) by \(\eps\). For \(K\) arbitrary, cover \(K\) by discs \(D_z\) for every \(z\in K\) and extract a finite cover \(\ts{D_{z_k}}_{k\leq N}\). Produce \(n_0, n_1,\cdots, n_N\) as in the above argument, and take \(n\da \max\ts{n_k}_{k\leq N}\) to obtain uniform convergence on every \(D_{z_k}\) and thus on \(K\).