# A sequence of holomorphic functions in D(0,1)

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 $D(0,1)$?

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.
• Why not just go directly to $e^{-f_n}$ is uniformly bounded? – zhw. Jul 16 '18 at 23:46
• @zhw. Actually I have a better reason than I did when I wrote the first version. Saw a cute way to handle the technicalities at the end; for that to work with the $e^{-f_n}$ version I need to know that $|e^{-f_n}|\ge c>0$ on compact sets. – David C. Ullrich Jul 16 '18 at 23:58
• @Rachel.L Seriously? $g$ is holomorphic in a disk and has no zero, so it has a holomorphic logarithm... – David C. Ullrich Jul 24 '18 at 20:57