I am preparing for qualifying exams, and this is a question from the Penn State Qualifying Exam for Fall 2015. It is stated as follows
Let $\epsilon > 0$ and let $f$ be holomorphic (analytic) on the disk $S = \{z \in \mathbb{C} \ | \ |z| < 1 + \epsilon \}$. Suppose that $f(z)$ is real valued whenever $|z| = 1$. Prove that $f$ is constant.
I have tried a few things in showing this to be true, but I keep finding holes in my logic. My largest issue is that the portion of the domain in which $f$ is real valued is not open, and so I'm not able to use many of the theorems I otherwise feel would be helpful (Open Mapping Theorem, Cauchy-Riemann Equations, Maximum Modulus, etc.).
Most of my attacks towards this problem have centered around showing things for the unit disk $\mathbb{D}$ instead of $S$, because I figure if I can show that $f$ is constant on $\mathbb{D}$, then I can use the Identity Theorem to extend it to $S$. However, since I don't know anything about the specific values that $f$ obtains, I can't use Schwarz' Lemma either.
I played with the idea, also, of suggesting that, if we consider the closure of $\mathbb{D}$, then $f(\partial{\mathbb{D}}) \in \mathbb{R}$. Since $\partial\mathbb{D}$ is closed and bounded and $f$ is analytic, it's image should also be bounded. That would put, for some $M \in \mathbb{R}$, $f(\partial{\mathbb{D}}) \in [-M,M]$, which is bounded and obtains a maximum (since $f$ is continuous and real valued here). I know that this set isn't open, but there is a Corollary in my text (Complex Analysis – Freitag) that says the following:
"If $K$ is a compact subset of the domain $D$ and $f:D \rightarrow \mathbb{C}$ is analytic, then the restriction of $f|K$ being a continuous function has a maximal modulus on $K$. By the Maximum Modulus Principle, we can moreover affirm that the maximal modulus value is necessarily taken on the boundary of $\mathbb{D}$."
The proof of that statement follows from the Open Mapping Theorem. Would it be appropriate to use that in this case, then? If $K = \partial \mathbb{D}$, even though it's technically not an open set, can I say that $f$ obtains its maximum on $\mathbb{D}$, state that it's constant, and then extend this to $S$ using the Identity Theorem? My issue here is that the constant itself isn't actually in $\mathbb{D}$,
I appreciate any help for this problem (or any tips for showing that complex functions will be constant, as these types of problems show up often).
Edit: My function $f$ is not necessarily entire, so Liouville's Theorem does not apply
Best Answer
By the Schwarz reflection principle, the formula $f(z)=\overline{f(1/\overline{z})}$ for $|z|>1$ extends $f$ analytically to the entire complex plane. Then by definition of the extension, $f(\mathbb{C})\subset f(\overline{\mathbb{D}}) \cup \overline{f(\overline{\mathbb{D}})}$, which is bounded by compactness.