I have a function $ f $ that is analytic on the unit disc and real valued on the unit circle. I want to prove that $ f $ is real valued on the unit disk.
I know that I need to use the identity theorem, but I am unsure how. The only thing I can think to do is:
Let $ g $ be some real valued function s.t. $ g(e^{i\theta}) = f(e^{i\theta}) $ then $ f = g $, but this argument obviously isn't sound. Any help would be greatly appreciated.
Best Answer
Note: this answers a different question (assuming $f$ is holomorphic). I leave it here for now because it may have some value anyway.
I assume that $f$ is continuous on the closed disc. One approach uses Schwarz' reflection principle: The function $f$ can be extended to an entire function by defining
$$ f(z) = \overline{f(1/\overline{z})} $$
for $|z| > 1$. (The hardish part of the reflection principle is that the resulting function is indeed holomorphic also for points on the unit circle.) Now $f$ is extended to a bounded entire function (since it is bounded on the compact closed unit disc) so it must be constant.
Another approach uses the open mapping theorem. Suppose $f$ is not constant. Then the image of the open unit disc is open and bounded (since $f$ is continuous on the closed unit disc). In particular it must have a non-real boundary point. This boundary point must be attained by $f$ on the compact closed unit disc. But $f$ is an open mapping on the open unit disc so this non-real boundary point must be attained on the unit circle. Contradiction.