Suppose \(f: \HH\union \RR\to \CC\) satisfies the following:
\(f(i) = i\)
\(f\) is continuous
\(f\) is analytic on \(\HH\)
\(f(z) \in \RR \iff z\in \RR\).
Show that \(f(\HH)\) is a dense subset of \(\HH\).
Solution
Ideas:
If an entire function doesn’t have dense image, it’s constant by Liouville using the proof idea of Casorati-Weierstrass.
Conjugate \(f\) by \(T:\HH\to \DD\) where \(T(z) = {z-i\over z+i}\), then \(\tilde f(0) = 0\)
Use that \(T(\RR) = S^1\), so \(\abs{\tilde f(z)} = 1\) when \(\abs{z} = 1\).
Schwarz reflection applies to \(\tilde f\) to define an entire function – if \(f\) isn’t dense, then the extension of \(\tilde f\) isn’t dense…?
No clue how to use \(f(i) = i\), although it implies \(\tilde f(0) = 0\) and Schwarz applies.