I am having trouble with the following question:

Let $\Omega$ be a non-empty open subset of $\mathbb{C}$, and let $f$ be a continuous function on $\Omega$. Suppose that $f_1, f_2,f_3,...$ are analytic on $\Omega$, and that $$\lim_{n\rightarrow \infty} \int_{D} \vert f_n (x+ iy) - f(x + iy) \vert dxdy = 0, $$ for every closed disk $D \subset \Omega$. Show that $f$ is analytic, and that $f_n \rightarrow f$ uniformly on compact subsets of $\Omega$.


$\textbf{1.}$ I am aware that Cauchy's Theorem will ensure that $f$ analytic provided $f_n \rightarrow f$ uniformly.

$\textbf{2.}$ The limit of the integral is zero namely for $\sup_{z \in D} \vert f_n(x+iy) - f(x+iy) \vert$.

$\textbf{3.}$ Triangle Inequality. I've pushed some things around, but I don't think I am properly timing $\vert \int \vert \leq \int \vert \vert$ to see what next steps should be.

$\textbf{4.}$ Montel's Theorem and Normal families smell like they could be related.

One concern is using some circular logic that is wrapped up in the relationship of Dominated Convergence and Uniform Continuity.

Clarification, tips, resources, etc., will all be greatly appreciated.


1 Answer 1


The best way to prove this is to use the fact that analytic functions have the Mean Value Property: $m(B)|f_n(z)-f_m(z)|=|\int_B (f_n-f_m)| \leq \int_B |f_n-f_m| \leq \int_{\Omega} |f_n-f_m| $ where $B$ is any open ball around $z$ contained in $\Omega$ and $m$ is Lebesgue measure. This inequality proves that $f_n$ converges uniformly on any open ball whose closure is contained in $\Omega$. Hence $(f_n)$ is a normal family, $f$ is analytic and $f_n \to f$ uniformly on compact subsets of $\Omega$.


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