The family of analytic functions with positive real part is normal

Question

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!

Answer 1

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 $f_n(z_0) \to w_0$.

Answer 2

This is an attempt to supplement Daniel Fischer's answer. I hope it provides useful information (if it is in fact correct - I have my doubts).

I started off with the same observation about the Fractional Linear Map $w \to \frac{w-1}{w+1}$.

Define as above a new family $\mathfrak{G}$ by $g(z) = \frac{f(z)-1}{f(z)+1}$.
This family is normal in the "non-classical" sense (Ahlfor's Definition #2, p.220), since each function $g$ takes values in the unit disk.

As a locally bounded family, the family of derivatives $g'$ of functions $g \in \mathfrak{G}$ is itself a locally bounded family. Differentiating, I hope that we get $$ g'(z) = 2 \frac{f'(z)}{(1+f(z))^2} $$ Conveniently we get $$ \frac{2|f'(z)|}{1 + |f(z)|^2} \le \frac{4|f'(z)|}{|1+f(z)|^2} = 2 |g'(z)| $$ Now apply Marty's Theorem (Ahlfors Theorem 17 p.226) to get that $\mathfrak{F}$ is normal in the classical sense.

I also have no suggestions for the second part of the question except to decree that there is a point in the domain on which $\mathfrak{F}$ is bounded.

Answer 3

I recommend Hurwitz Theorem to approach this:

If the function are analytic and $\neq 0$ in a region $\Omega$, and if $f_n(z)$ converges to $f(z)$, uniformly on every compact subset of $\Omega$, then $f(z)$ is either identically zero or never equal to zero in $\Omega$. (From Ahlfors' book)

Apply this to $g_n(z)=e^{-f_n(z)}$. Since is locally bounded, it is normal, i.e. there exists some subsequence $n_k$ such that $g_{n_k}(z)\rightarrow g(z)$ uniformly on compact subsets. Easy to know that $g_n(z)\neq 0$ and are analytic, then by the theorem above, $g(z)\equiv 0$ or $g(z)$ never equal to zero. When $g(z)\equiv 0$, the real part of $f_{n_k}(z)$ goes to $\infty$. In the latter case, we can find $f(z)=\log g(z)$ and $h_k(z)$ such that $h_k(z)=f_{n_k}(z)+2\pi a_k i\rightarrow f(z)$, where $a_k\in\mathbb{Z}$. Then $f_{n_k}(z)\rightarrow \infty$ if $a_k\rightarrow\infty$. Otherwise, we can find a subsequence of $a_k$ such that $a_{k_l}$ converges, then $f_{n_{k_l}}(z)$ converges. Since $a_k$ constant over $\Omega$, $f_{n_{k_l}}(z)$ converges on compact subsets

Second question, to make $f(z)$ locally bounded is to avoid the case $g(z)\equiv 0$ or $a_k\rightarrow\infty$. This forces the existence of $z$ such that $\{f(z): f\in$} is bounded.

Note: This is also my homework problem and I read Daniel's answer, finding his branch step very confusing. Happen to find out "Hurwitz" and think it really helpful! Anyway, thanks Daniel for giving me some ideas and perspective.