I'm having difficulty with the following exercise in Ahlfors' text, on page 227.
Prove that in any region $\Omega$ the family of analytic functions with positive real part is normal. Under what added condition is it locally bounded? Hint: Consider the functions $e^{-f}$.
Here is what I've tried:
I will start with a remark:
Apparently, Ahlfors wants us to show that the family is "normal in the classical sense". That is, every sequence of functions in the family has a subsequence which converges uniformly on compact subsets or tends uniformly to $\infty$ on compact subsets. In order to see why this is the right definition, consider the sequence $f_n(z)=n$. It is contained in the family but has no appropriate subsequence (in the sense of definition 2 in the text with $S=\mathbb C$).
Now, to the attempt itself:
Let $\Omega \subset \mathbb C$ be a fixed region, and consider the family $$\mathfrak F=\{f: \Omega \to \mathbb C | f \text{ is analytic and } \Re(f) >0 \}. $$
We would like to show that $\mathfrak F$ is normal in the classical sense. Following the hint, we examine the family $$ \mathfrak G=\{e^{-f}:f \in \mathfrak F \}.$$
$\mathfrak G$ is locally bounded (since $|e^{-f}|=e^{- \Re (f)}<1$ for every $f \in \mathfrak F$), thus it is normal with respect to $\mathbb C$ (theorem 15), and obviously, it is normal in the classical sense as well.
Let $\{ f_n \}$ be a sequence in $\mathfrak F$, and consider the sequence $\{ g_n \}=\{e^{-f_n} \}$ in $\mathfrak G$. According to normality it has a convergent subsequence $\{ g_{n_k} \}=\{e^{-f_{n_k}} \}$ which converges uniformly on compact subsets of $\Omega$ to some function $g$ (which is analytic by Weierstrass' theorem).
Since each $\{ g_{n_k} \}$ is nonvanishing, the limit function $g$ is either identically zero, or non vanishing as well (Hurwitz's theorem). In the former case it is easy to show that the subsequence $\{ f_{n_k} \}$, obtained by the same indices, tends to $\infty$ uniformly on compact sets. Hence, we will assume from now that $g(z) \neq 0$ for all $z \in \Omega$.
Up until now, I was trying to show that the subsequence $\{ f_{n_k} \}$ works in all cases, but sadly, this is not the case. Consider the sequence $f_n(z) \equiv 1+2 \pi i (-1)^n \in \mathfrak F$. In that case $g_n(z)=e^{-1}$, and an admissible subsequence is $g_{n_k}=g_k=e^{-1}$. However, $f_{n_k}=1+2 \pi i (-1)^k$ diverges everywhere.
Can anyone please help me finish this proof? Or maybe give me some hints?
Thanks!
Best Answer
It would be much simpler to prove the result considering the family obtained by composing with the Möbius transformation $$z \mapsto \frac{z-1}{z+1}$$ that maps the right half plane biholomorphically to the unit disk.
But well, let's look at what we got from considering $e^{-f}$. Without loss of generality, we can assume that the entire sequence $e^{-f_n}$ converges compactly to a nonzero function $g$.
As you observed, that does not yet guarantee that the sequence $f_n$ itself converges compactly to a holomorphic function. So let's fix some $z_0 \in \Omega$ and consider the sequence $f_n(z_0)$. Either the sequence converges to $\infty$, or we can extract a subsequence converging to a complex number.
Consider first the case where we can extract a subsequence converging to a complex number. Without loss of generality, assume the entire sequence converges to $w_0 \in \mathbb{C}$. In a neighbourhood of $e^{-w_0}$, there is a branch of the logarithm with $\log e^{-w_0} = -w_0$ defined.
Then $f_n$ converges uniformly to $\log g$ in a neighbourhood of $z_0$.
If $f_n(z_0) \to \infty$, then, taking a branch of the logarithm in a neighbourhood of $g(z_0)$, we obtain a sequence $k_n$ of integers with $\lvert k_n\rvert \to \infty$ and $f_n(z_0) - 2\pi i k_n \to \log g(z_0)$. Thus the sequence $f_n - 2\pi i k_n$ converges uniformly to a holomorphic function in a neighbourhood of $z_0$, and since $\lvert k_n\rvert \to \infty$, the sequence $f_n$ itself converges uniformly to $\infty$ in a neighbourhood of $z_0$.
It remains to see that the uniform convergence to either a holomorphic function or $\infty$ extends (as locally uniform convergence) to all of $\Omega$.
Let $A = \{z \in \Omega : f_n(z) \to \infty\}$ and $B = \{z \in \Omega : f_n(z) \text{ is bounded}\}$ and $C = \Omega \setminus (A\cup B)$.
The argument above shows that all, $A$, $B$ and $C$ are open, and they are disjoint. Since $\Omega$ is connected, we have $\Omega = A$, $\Omega = B$, or $\Omega = C$. By having extracted the subsequence converging (to $\infty$ or $w_0$) at $z_0$, we have arranged that $z_0 \notin C$, hence $C = \varnothing$, so $\Omega = A$ if $f_n(z_0) \to \infty$, and $\Omega = B$ if $f_n(z_0) \to w_0$.