# A sequence of holomorphic functions in open connected set

Let $\{f_n\}$ be a sequence of holomorphic functions on an open connected subset $\Omega$ of $\mathbb{C}$. Suppose $\mathrm{Re} f_n(z)$ is positive for all $n$ and $z \in \Omega$. Show that $\{f_n\}$ has a subsequence either converges uniformly to a holomorphic function on compact subsets of $\Omega$ or converges uniformly to $\infty$ on compact subsets of $\Omega$.

I tried to show it using Harnack's Inequality, but I have no idea how to show $|f_{n_k}(z)| \rightarrow \infty$ uniformly on compact sets of $\Omega$(I showed only pointwise). Also, I am not sure where I need to use connectedness of $\Omega$ as I can find a counterexample if it's not connected.

## 1 Answer

Let $g_n=e^{-f_n}$. Then $|e^{-f_n}| \leq 1$ so there is a subseqeunce $g_{n_{k}}$ converging uniformly on compact sets. I claim that this subsequence $f_{n_{k}}$ is locally bounded. As an easy consequence of Rouche's Theorem it follows that $g\equiv 0$ or $g$ has no zeros. [Connectedness is needed here. See details below]. In the first case $Re f_{n_{k}}\to \infty$ which implies $|f_{n_{k}}|\to \infty$ and there is nothing to prove. In the second case each point has a neighborhood $N$ such that there is a holomorphic logarithm on $g(N)$. This shows that $f_{n_{k}}$ is bounded on $N$. Since locally bounded seqeunces of holomorphic functions form a normal family, it follows that $f_{n_{k}}$ has a further subseqeuence which converges uniformly on compact subsets.

Lemma Suppose $\{h_n\}$ is a sequence of holomorphic functions on a connected open set with no zeros. Suppose $h_n \to h$ uniformly on compact subsets. Then either $h \equiv 0$ or $h$ has no zeros.

Proof: suppose, if possible, $h(c)=0$ for some $c$ but $h$ is not identically zero. Since the domain is connected, the zeros of $h$ do not have any limit point in it. Hence, if $r>0$ is small enough, there are no zeros of $h$ on $\{z:|z-c|=r\}$. By continuity, $|h|$ attains a minimum at some point on this circle and this minimum is positive. Hence there exists $\delta >0$ such that $|h(z)|>\delta$ for all $z$ on this circle. Now $\{z:|z-c|\leq r\}$ is a compact set and $h_n \to h$ uniformly on this compact set. Hence $|h_n(z)-h(z)| <\delta$ for all $z$ in this disk for $n$ sufficiently large. In particular $|h_n(z)-h(z)| <|h(z)|$ for all $z$ with $|z-c|=r$. By Rouche's Theorem this implies that $h_n$ and $h$ have the same number of zeros in the disk $\{z:|z-c|\leq r\}$. This is a contradiction because $h_n$ has no zeros and $h(c)=0$.

• How can I apply Rouche's Theorem here? – Rachel.L Aug 16 '18 at 1:35
• @Rachel.L I have added the proof of this fact to my answer now. Let me know if you need any additional details. – Kavi Rama Murthy Aug 16 '18 at 5:20