If a function $f$ is holomorphic on the closed unit disk centered at the origin and is real valued whenever $|z| = 1$, then $f$ is constant.

Question

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

Answer 1

One can get this from the Schwarz Reflection Principle. That only works for the disk, or other special domains. It's easy to give an argument that works for any bounded domain:

The imaginary part is harmonic and vanishes on the boundary, so the imaginary part is $0$. So $f$ is real-valued. Hence $f$ is constant (Cauchy-Riemann equations, Open Mapping Theorem, who knows what else).

Or: Apply Maximum Modulus to $e^{if}$ and $e^{-if}$.

Answer 2

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.