TITLE: Continuous Extension of a Bounded Holomorphic Function on the unit disk?
QUESTION [7 upvotes]: 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 $f(z+1) = f(z)$?

REPLY [8 votes]: 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 ...