$\int_C|f_n(z)||dz|\leq M$ for each $f_n$. Prove $\{f_n\}$ has a subsequence converging uniformly on compact subsets of $D$.

Question: Let $\{f_n\}$ be a sequence of analytic functions in a domain $D\subseteq\mathbb{C}$, and suppose there exists a positive constant $M$ such that $\int_C|f_n(z)||dz|\leq M$ for each $f_n$ and for every circle $C$ lying in $D$. Prove that $\{f_n\}$ has a subsequence converging uniformly on compact subsets of $D$.

Thoughts: I was thinking this was maybe somewhat similar to infinite product stuff where if you want to show that an infinite product converges absolutely on every compact set then you sum that the sum of the log of the stuff we're taking the infinite product of is bounded, but I don't think thats right. My next though was maybe talking about the derivatives of $f_n$, and using Cauchy to bound each derivative and if we can bound each derivative by a constant, then we can get $f_n$ to converge to some polynomial... but that wasn't coming out to anything. So, I wasn't sure where to go from there. Any help would be greatly appreciated! Maybe max mod?

Solution 1:

Hints: If $K$ is a compact subset of $D$ then there exists $r>0$ such that for each point $z$ of $K$ the disk $D(z,r)$ is contained in $D$. Analytic functions have mean value property: $f(z)$ is the average of $f$ over any circle around $z$ contained in $D$. It follows from these that the given family $(f_n)$ is bounded on compact sets. Hence, it is normal family, so there is a subsequence which converges uniformly on compact subsets.