# Continuous Extension of a Bounded Holomorphic Function on the unit disk?

Let $f$ be a bounded holomorphic function on the upper-half plane. Is it true that such a map always admits an (unique) extension to real line, so that $f$ is continuous on the closure of the upper half plane? I feel like the answer to this question is "no," since a lot of theorems concerning bounded holomorphic functions start out with, "Assuming that $f$ is continuous on the closure," but I'm not having much luck with either a proof or a simple counterexample.

If the statement is false, what if we further assume that $f$ is periodic with period 1? That is, if $f$ is a bounded holomorphic function on the upper half plane satisfying $f(z+1) = f(z)$?

-
I think (at least for your first question) you could take a function which is only holomorphic on the unit disc and use a Möbius transformation to map it to the upper half plane. –  Sam Aug 20 '11 at 21:32
@Sam - I'm guessing you are giving me a counterexample to the first one. Is there a function that is both bounded and only holomorphic on the unit disc? –  Braindead Aug 20 '11 at 21:37
@Sam - Wait I don't think I understand your comment anymore... would you mind clarifying? –  Braindead Aug 20 '11 at 21:44
I didn't read the magical word "bounded"... But in this case: If you develop the function on the unit disc into a series expansion about $z=0$ and $f$ is bounded, then (if I remember correctly - which I'm not sure of... ;-( ) the radius of convergence must be strictly bigger than $1$. Which would show that one can extend $f$. By composing with a Möbius transformation this would also imply an answer to your question. (However to repeat: I'm really only at a beginner's level in complex analysis, so don't hold me accountable if it's complete rubbish, pls. =) ) –  Sam Aug 20 '11 at 21:50
@Sam: For a simple example of a bounded analytic function on the disk whose Maclaurin series has radius of convergence exactly $1$, consider $f(z)=\sum\limits_{k=0}^\infty\frac{z^k}{k^2}$. This example is also continuous on the closed disk, however. In my answer I mention the example of Blaschke products with infinitely many zeros that do not have such continuous extensions. –  Jonas Meyer Aug 20 '11 at 23:54

Summary: The answer to the first question is no. See Robert Israel's answer for the periodic case.

A bounded analytic function $f$ on the open upper-half plane induces a bounded analytic function $\tilde f$ on the open unit disk $D$ via the Cayley transform, such that if $f$ has a continuous extension to $\mathbb R$, then $\tilde f$ has a continuous extension to $\partial D\setminus\{1\}$. However, using Blaschke products, one can give examples of bounded analytic functions on the open unit disk that have no such extension. All bounded analytic functions have nontangential limiting values a.e. on the boundary, and in the case of Blaschke products the boundary functions that come from these limits have modulus $1$ a.e. On the other hand, the set of zeros of a Blaschke product can be chosen to contain the boundary in its closure, so any continuous extension would have to vanish on the boundary, and hence can't exist. See also Myke's related question on the case of the disk.

Robert Israel's answer covers the periodic case, but here is a general fact about bounded analytic periodic functions that might be useful. On page 183 of Gamelin's Complex analysis it is shown under a weaker hypothesis that such $f$ has a pointwise absolutely convergent series expansion in the open half-plane, $f(z)=\sum\limits_{k=0}^\infty a_ke^{2\pi i kz}$, converging uniformly on each half-plane $\mathrm{Im}\ z\geq \varepsilon>0$.

You can find much more in Garnett's Bounded analytic functions. For example, a necessary and sufficient condition for a sequence $(z_n)_n$ to be the set of zeros of some nonconstant bounded analytic function on the open half plane is that $$\sum_n\frac{\mathrm{Im}\ z_n}{1+|z_n|^2}<\infty,$$ as seen on page 53 of Garnett. So the set of zeros of a nonconstant bounded analytic function can be chosen to contain $\mathbb R$ in its closure, implying that if there were a continuous extension it would have to vanish on $\mathbb R$. But in Section II.4 it is shown that the boundary function $f:\mathbb R\to\mathbb C$ of a nonconstant analytic function with continuous extension to the real line must satisfy $$\int_{\mathbb R}\frac{\log|f(t)|}{1+t^2}dt>-\infty,$$ implying that $f$ is nonvanishing off a set of measure $0$.

-
Thank you so much for the detailed answer! –  Braindead Aug 21 '11 at 15:24

I assume the $f(z+1) = f(z)$ refers to a function on the upper half plane, as others have remarked. If $g$ is a bounded analytic function on the open unit disk, $f(z) = g(\exp(2 \pi i z))$ is a bounded analytic function on the open upper half plane which also satisfies $f(z+1) = f(z)$. So take a bounded analytic function on the open unit disk that has no contininuous extension to the closed disk ...

-
You are right! That's so simple. I'd love to accept both your answer and Jonas's answer, but unfortunately I don't think I can do that... –  Braindead Aug 21 '11 at 15:24