Let $f$ be a function which is continuous on the closed unit disc and analytic on the open disc.

Question

Let $f$ be a function which is continuous on the closed unit disc and analytic on the open disc. Assume that $|f(z)| = 1$ whenever $|z| = 1$. Show that the function $f$ can be extended meromorphically to the whole complex plane with at most finitely many poles, i.e., there is a meromorphic function $F(z)$ defined in $\mathbb{C}$ such that $F(z) = f(z)$ for $z \in D(0, 1)$

Answer 1

First of all, since $|f(z)|=1$ on $\partial\mathbb{B}$ and the function is continuous, there exists an $r<1$ such that $1>|z|>r\rightarrow f(z)\neq 0$. By the identity principle, we obtain that the function must have a finite number of zeros in $\mathbb{B}$.

We now turn to proving the extension. There are tree ways of proving this:

1. By direct riflection: $$F(z):=\frac{1}{\overline{f(\overline{z}^{-1})}}$$ It is simply a matter of calculations (hint: use morera theorem) to prove that $F(z)$ is meromorphic.
2. Applying Cayley Transform, and considering the function $$\hat{f}(z):=\frac{f\left(\frac{z-i}{z+i}\right)+1}{1-f\left(\frac{z-i}{z+i}\right)}\\ \mathbb{H}\to\mathbb{H}$$ Once we are here, we note that the function must have only a finite number of real values that are sent to $\infty$. Using the reflection principle for meromorphic functions, we obtain the assertion for $\hat{f}$. Applying the inverse cayley transform we obtain the assertion.
3. Since the function has only a finite number of zeros, it has a finite Blaschke factorization:

$$f(z)=g(z)z^m\prod_{k}\frac{|a_k|}{a_k}\frac{a_k-z}{1-\overline{a_k}z}$$

$g$ can be extended by reflection to an holomorphic function on all of $\mathbb{C}$ (by a reflection similar to the one in (1)), and the blaschke product is already meromorphic on all of $\mathbb{C}$. Since it has a finite number of factors, their product has only a finite number of poles.

Side note: It is interesting to note that for $F(z)$, $\infty$ is at most a pole:

1. Follows easily from the fact that $0$ is at most a zero of finite order of $f$
2. Follows from the fact that if $f$ is non constant (and that it extends to a meromorphic function on $\mathbb{C}$), $f-1$ is at most a zero of finite order, and so $\hat{f}$ has at most a pole in $\infty$. The assertion for $F$ follows applying the inverse cayley transform
3. We note that, by reflection, $g(z)=\frac{1}{\overline{g(\overline{z}^{-1})}}$ and so it has at most a pole in $\infty$. The assertion then follows by the observation that the finite blaschke product has at most a pole in $\infty$

The only entire functions that have $\infty$ as a pole are the rational functions. Thus, we have proved that $f$ has to be rational.