I want to prove a statement regarding the holomorphic extension of a function holomorphic on the unit disk.

Let $D \subset \mathbb{C}$ be the (open) unit disk and $f: \bar{D} \to > \mathbb{C}$ be a continous map, such that $f$ restricted to $D$ is holomorphic and $|f(z)|=1$ for all $z \in \partial D$. Show that $f$ can be extended to a function holomorphic on $\mathbb{C}$ up to finitely many isolated singularities, i.e. there are finitely many points $z_1,..., z_n \in \mathbb{C}$ and a holomorphic function $g: \mathbb{C} \setminus \{z_1,...,z_n \} \to \mathbb{C}$ such that $f=g$ on $D$.

Let $C_0=\partial D$. Then $C_0$ is the unit circle. Consider $f|_D: D \to \mathbb{C}$. Then $f|_D$ is holomorphic because $f$ is holomorphic. Let $K \subset C_0$ be a circular arc. Because $|f(z)|=1 $ for all $z \in \partial D$ it follows $f(K) \subset C_0$. Let $\sigma_0$ be the inversion with respect to $C_0$. Consider

\begin{align*} g: D \cup K \cup \sigma_0(D), \ F(z)=\begin{cases} f(z) & , \ z \in D \\ \sigma_0(f(\sigma_0(z))) & , \ z \in \sigma_0(G) \\ \sigma_0(f(\sigma_0(z))) & , \ z \in K \end{cases} \end{align*}

Then $f(z)=\sigma_0(f(\sigma_0(z)))$ for all $z \in K$. Because $f|_D$ can be extended to a function that is conitnous on $\bar{D}$ it follows from the Schwarz reflection principle that the function $g$ is holomorphic. The function $g$ has a singularity at $a \in \mathbb{C}$ if and only if $a\in \sigma_0(G)$ and $f(\sigma_0(a))=0$. Define $S:=\{z \in \sigma_0(D) \ | \ f(\sigma_0(z))=0 \}$. Then $S$ is the set of singularities of $g$. Suppose that $S$ contains infinitely many elements. Because $\sigma_0$ is bijective it follows that there are infinitely many elements $z \in D$ such that $f(z)=0$. Thus $f$ has infinitely many zeros. For every zero of order $m \geq 1$ there is a neighborhood and a holomorphic function $h$ such that $f(z)=(h(z))^m$ in that neighborhood.

But I do not see how to proceed.


$f$ has only finitely many zeros because otherwise these zeros would have an an accumulation point in the closed unit disk.

But the zeros cannot accumulate in the interior of the disk because of the identity theorem. And they can also not accumulate at a point of the boundary because $|f(z)| = 1$ on the boundary.

If $z_1, \ldots, z_n$ is the (possibly empty) list of zeros of $f$ then $g$ can be defined in $\Bbb C \setminus \{ \sigma_0(z_1), \ldots, \sigma_0(z_n) \} $ as $$ g(z) = \begin{cases} f(z) & \text{ if $|z| \le 1$ } \\ \sigma_0(f(\sigma_0(z))) & \text{ if $|z| > 1$, $f(\sigma_0(z)) \ne 0$ } \\ \end{cases} $$ As you said, $g$ is holomorphic in $\Bbb C$ save for poles at $\sigma_0(z_1), \ldots, \sigma_0(z_n)$.

In fact $g$ is a rational function because the singularity at $z = \infty$ is a pole or removable (depending on whether $f(0)$ is zero or not).

  • $\begingroup$ "But the zeros cannot accumulate in the interior of the disk because of the identity theorem. " I am not sure I see how this follows. Since $f$ has infinitely many zeros I know there is a sequence $(z_n)_{n \in \mathbb{N}}$ in $D$ such that $f(z_n)=0$ for all $n \in \mathbb{N}$. Since $(z_n)$ is contained in $\bar{D}$ it is bounded. Then it follows from the Bolzano-Weierstra\ss theorem that $(z_n)$ has a subsequence $(z_{n_k})_{k \in \mathbb{N}}$ such that $\lim_{k \to \infty} z_{n_k}=a \in \bar{D}$. Suppose that $a \in D$. Now could you please elaborate on how to use the identity theorem? $\endgroup$
    – Polymorph
    Jun 10 '20 at 20:48
  • $\begingroup$ @Polymorph: If the zeros of $f$ accumulate at $a \in D$ then the identity theorem states that $f=0$ everywhere in $D$. $\endgroup$
    – Martin R
    Jun 10 '20 at 20:58
  • $\begingroup$ @Polymorph: See for example en.wikipedia.org/wiki/Analytic_function: “If the set of zeros of an analytic function ƒ has an accumulation point inside its domain, then ƒ is zero everywhere on the connected component containing the accumulation point.” $\endgroup$
    – Martin R
    Jun 10 '20 at 21:06

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