Let $\{f_n\}$ be a sequence of holomorphic functions in the open unit disk $D(0,1)$ such that $Re f_n \geq 0$.
How can I show that either $|f_n(z)| \rightarrow \infty$ as $ n \rightarrow \infty$ for all $z \in D(0,1)$ or $\{f_n\}$ has a subsequence which converges uniformly on every compact subset of $D(0,1)$?
Best Answer
It follows easily from the version of Harnak's inequality here that if $K$ is a compact subset of $\Bbb D=D(0,1)$ there exist $c_K$ and $C_K$ such that $$c_Ku(0)\le u(z)\le C_Ku(0)\quad(z\in K)$$for every positive harmonic function $u$ in $\Bbb D$.
So if $\Re f_n(0)\to\infty$ then $|f_n(z)|\to\infty$ uniformly on compact sets. Otoh if $\Re f_n(0)\not\to\infty$ then restricting to a subsequence we can assume that $\Re f_n(0)$ is bounded; now the inequality above shows that $e^{f_n}$ is uniformly bounded on compact sets...
Heh, a slick way to get around unpleasantness with the logarithm for the punchline: So again passing to a subequence we can assume that $e^{f_n}\to g$ uniformly on compact sets. Since $\left|e^{f_n}\right|\ge1$ it follows that $g$ has no zero, so $g=e^f$. How to get from there to $f_n\to f$: Since normal convergence implies normal convergence of the derivatives we have $f_n'e^{f_n}\to f'e^f$, and now since $\left|e^{f_n}\right|\ge1$ it follows that $f_n'\to f'$ uniformly on compact sets. So if we choose our logarithms so that $f_n(0)\to f(0)$ it follows that $f_n\to f$ uniformly on compact sets.