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?


  • $\begingroup$ I would prefer to consider the family $\mathfrak{C} = \left\{\frac{f-1}{f+1} : f \in \mathfrak{F} \right\}$. Can you see why? $\endgroup$ – Daniel Fischer Sep 16 '13 at 22:54
  • $\begingroup$ @DanielFischer is it because of its simpler inverse function? $\endgroup$ – user1337 Sep 16 '13 at 22:56
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    $\begingroup$ Simpler? It has an inverse. $\endgroup$ – Daniel Fischer Sep 16 '13 at 22:58
  • $\begingroup$ Yuck. I could certainly organize the proof better, but the argument itself is messy enough. I certainly don't feel like cleaning the proof up tonight. Hope you can understand it nevertheless. $\endgroup$ – Daniel Fischer Sep 16 '13 at 23:37
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    $\begingroup$ Yes. Since Möbius transformations are automorphisms of $\widehat{\mathbb{C}}$, composition with Möbius transformations conserves compact convergence. $\endgroup$ – Daniel Fischer Sep 17 '13 at 0:00

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 $f_n(z_0) \to w_0$.

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  • $\begingroup$ Thanks, do you have any idea how to attack the other part of the question. (how can I make the family locally bounded as well). I could remove all sequences that tend to $\infty$ uniformly on compact sets, but that seems too artificial... $\endgroup$ – user1337 Sep 17 '13 at 10:34
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    $\begingroup$ For the local boundedness, the only criterion I see is that the family is locally bounded if and only if there is a point $z_0 \in \Omega$ such that $\{f(z_0) : f \in \mathfrak{F}\}$ is bounded. Of course, the family of all holomorphic functions with positive real parts cannot be locally bounded, since it contains all constant functions with positive real part. $\endgroup$ – Daniel Fischer Sep 17 '13 at 10:51

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.

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