If ${f_n}$ is a uniformly bounded sequence of holomorphic functions in $\Omega$ such that ${f_n(z)}$ converges for every $z \in \Omega$, is the convergence is uniform on every compact subset of $\Omega$?
I'm trying to prove using Dominated Convergence Theorem with Cauchy's Integral Formula. For any compact $K \subset \Omega$, let $\gamma = \partial K$. Then for any $z \in K$, by the CIF,
$$|f_n(z) – f_m(z)| \leq \frac{1}{2\pi i}\int_{\gamma}\left|\frac{f_n(w)-f_m(w)}{w-z}\right|\:dz.$$
Now since we are talking in compact sets, $|f_n(z)| \leq B \in L^1$, so by applying Dominated Convergence Theorem, and by taking the limit as $m \to \infty$,
$$|f_n(z)-f(z)| \leq \frac{1}{2\pi i}\int_{\gamma}\left|\frac{f_n(w)-f(w)}{w-z}\right|\:dz.$$
But by the pointwise convergence, I am trying to claim that RHS can be made sufficiently small for large enough $n$. Am I on the right track?
Best Answer
Here is a powerful convergence theorem.