# The Schwarz Reflection Principle for a circle

I'm working on the following exercise (not homework) from Ahlfors' text:

" If $f(z)$ is analytic in $|z| \leq 1$ and satisfies $|f| = 1$ on $|z| = 1$, show that $f(z)$ is rational."

I already know about the reflection principle for the case of a half plane, so I tried using the "Cayley transform" $$T (\zeta)=\frac{\zeta-i}{\zeta+i}$$ Which maps the closed upper half plane onto the closed unit disk with $1$ removed.

I defined $$g(\zeta)=(T^{-1} \circ f \circ T)(\zeta)=i\frac{1+f \left( \frac{\zeta-i}{\zeta+i} \right)}{1-f \left( \frac{\zeta-i}{\zeta+i} \right)},$$ And tried to apply the reflection principle in the book. $g$ is indeed analytic in the upper half plane, but for $\zeta \in \mathbb R$, I'm afraid that $g$ might get infinite (because on the boundary, $f$ takes values on the unit circle). If so, it will not be continuous and not even real, and the reflection principle is not applicable.

Am I missing something here? After all Ahlfors does mention in the text a generalized reflection principle for arbitrary circles $C,C'$.

Thanks

## 2 Answers

In the given situation, we can proceed directly. The reflection in the unit circle is given by

$$\rho(z) = \overline{z}^{-1},$$

so by setting

$$g(z) = \frac{1}{\overline{f(\overline{z}^{-1})}},$$

we obtain a function $g$ that is meromorphic in the outside of the unit disk. Since $f$ can have only finitely many zeros in $\mathbb{D}$, $g$ has only finitely man poles in $\hat{\mathbb{C}} \setminus \overline{\mathbb{D}}$,

and since $\lvert f(z)\rvert = 1$ for $\lvert z\rvert = 1$, the function

$$h(z) = \begin{cases}f(z) &, \lvert z\rvert \leqslant 1\\ g(z) &, \lvert z\rvert > 1\end{cases}$$

is continuous (outside the poles, none of which lies on $\partial \mathbb{D}$), and holomorphic outside $\partial \mathbb{D} \cup \{\text{poles}\}$. By a small modification of Morera's theorem (you can map each arc on the circle to the real axis by a Möbius transformation), it is meromorphic on all of $\hat{\mathbb{C}}$, hence rational.

You can also use the Cayley transform as you started with, if $f$ is not constant, then $f$ can take the value $1$ only finitely often on $\partial\mathbb{D}$, and $g = T^{-1}\circ f \circ T$ has only finitely many poles on $\mathbb{R}$, and either a pole or a removable singularity in $\infty$, on each interval between two poles, you can apply the ultra-classic reflection principle to see that $g$ can be extended by reflection to a meromorphic function on $\hat{\mathbb{C}} \setminus \{\text{poles}\}$, hence is rational.

• Thanks Daniel. About the other proof: how can you tell that the singularity at $\infty$ is either removable or a pole? – user1337 Jul 22 '13 at 23:08
• Which "other"? If the direct reflection in the circle, in a neighbourhood of $\infty$, you have - modulo two conjugations - that $1/g(z)$ is $f(z)$ in a neighbourhood of $0$, hence if $f(0) \neq 0$, you have a removable singularity, and a pole (whose order is the multiplicity of the zero of $f$ in $0$) if $f(0) = 0$. If the Cayley-transform, $f$ is analytic in a neighbourhood of $1$ by hypothesis, hence you get a pole in $\infty$ if $f(1) = 1$, and a removable singularity otherwise. – Daniel Fischer Jul 22 '13 at 23:16

First note that the hypothesis implies that $f$ has only a finite number of zeros in the unit disk $\mathbb{D}$, say $\alpha_1, \dots, \alpha_n$. Consider now the function $$B(z):=\prod_{j=1}^n \frac{z-\alpha_j}{1-\overline{\alpha_j}z}.$$ This is a finite Blaschke product and $|B(z)| =1$ for all $z \in \partial \mathbb{D}$. Since $B$ has the same zeros as $f$, it follows that both $f/B$ and $B/f$ are analytic in $\mathbb{D}$ and continuous on $\overline{\mathbb{D}}$. By the maximum principle applied to both quotients, we deduce that $|f/B|=1$ everywhere in $\mathbb{D}$, so that $f/B$ is a unimodular constant $\lambda$, by the open mapping theorem. Therefore $f= \lambda B$, a rational function.