Analytic function defined in the punctured unit disk real in the unit circle then $f(z)=\overline{f(1/\overline{z})}$

This is a follow up to a question I asked a few days ago. I initially thought I understood the solution to the problem, but there's something I can't quite grasp:

Let $f$ be analytic in the set $\{ x \in \mathbb{C} : 0<|z|<1\}$. If $f$ is real in the unit circle $\{z\in \mathbb{C}: |z|=1\}$, and has a continuous extension to the unit circle, then show that: $$f(z)=\overline{f\left ( \frac{1}{\overline{z}} \right )} \ \forall z\in \mathbb{C}$$

My approach was as follows: Let $\phi: \mathbb{C}\setminus\{-i\} \to \mathbb{C}\setminus \{1\}$ be given by $$\phi(z)=\frac{z-i}{z+i}$$

It is easy to prove that $\phi$ maps the upper half plane into the unit disk, this is $\phi (\mathbb{C}^+)= D$ and that it maps the real axis into the unit circle, i.e. $|\phi(x)|=1$ for real $x$. Note that $\phi^{-1}:\mathbb{C}\setminus \{1\} \to \mathbb{C}\setminus\{-i\}$ is given by: $$\phi^{-1}(z)=i\frac{z+1}{1-z}$$ Also note that $$\overline{\phi^{-1}(z)}=\phi^{-1}\left ( \frac{1}{\overline{z}}\right )\tag1$$

Now, by the preceding remarks, the composition $g=f\circ \phi : \overline{\mathbb{C}^+}\setminus\{i\} \to \mathbb{C}$ is continuous in its domain and analytic in $\mathbb{C}^+ \setminus\{i\}$. Further, since $g(\mathbb{R})\subset \mathbb{R}$, we can apply the Schwarz reflection principle to extend $g$ to $\mathbb{C}\setminus\{i,-i\}$, that is the function $h:\mathbb{C}\setminus\{i,-i\} \to \mathbb{C}$ given by:

$$h(z)=\begin{cases}g(z)=(f\circ \phi)(z) & \textrm{if} \ z \in \overline{\mathbb{C}^+}\setminus\{i\} \\\overline{g(\overline{z})}=\overline{(f\circ \phi)(\overline {z})} & \textrm{if} \ z \in \overline{\mathbb{C}^-}\setminus\{-i\} \end{cases}\tag2$$

is analytic in $\mathbb{C}\setminus\{i,-i\}$ and verifies $h(z)=\overline{h(\overline{z})}$ for all $z \in \mathbb{C}\setminus \{i,-i\}$.

Now, if we take $z\neq i,-i$, from $(1)$ and the fact that $h(z)=\overline{h(\overline{z})}$ we get that:

$$f(z)=(f\circ \phi)(\phi^{-1}(z))=h(\phi^{-1}(z))=\overline{h(\overline{\phi^{-1}(z)})}=\overline{h\left ( \phi^{-1}\left ( \frac{1}{\overline{z}}\right ) \right )}=\overline{f\left ( \frac{1}{\overline{z}} \right )}\tag3$$

Which would complete the proof (the cases $z=i,-i$ are easily handled separately since the function is real in the unit circle).

My question is: In $(3)$, why does $h\circ \phi^{-1}=f$ (this is used in the second and final equalities). If $\phi^{-1}(z) \in \overline{\mathbb{C}^+}\setminus\{i\}$, then in is clear from $(2)$ that $h(\phi^{-1}(z))=(f\circ \phi)(\phi^{-1}(z))=f(z)$, since this is the definition of $h$ in that region, but what if $\phi^{-1}(z) \in \overline{\mathbb{C}^-}\setminus\{-i\}$? Using $(2)$ and some other identities we already know I can get:

$$h(\phi^{-1}(z))=\overline{(f\circ \phi)(\overline {\phi^{-1}z})}=\overline{(f\circ \phi)\left ( \phi^{-1}\left ( \frac{1}{\overline{z}} \right )\right )}=\overline{f\left ( \frac{1}{\overline{z}}\right )}$$

Alternatively:

$$h(\phi^{-1}(z))=\overline{h(\overline{\phi^{-1}(z)})}=\overline{h\left ( \phi^{-1}\left (\frac{1}{\overline z} \right ) \right )}$$

Both paths seemingly lead nowhere, and the argument becomes almost circular.

Any help?

• @kilimanjaro What do you mean? I explained below that I don't think that step is correct due to the piecewise definition of $h$; i.e. $h(\phi^{-1}(z))$ depends on whether $\phi^{-1}(z)$ is in the lower/upper half-plane. – Reveillark Sep 23 '15 at 16:50
• What are you taking the domain of $f$ to be? – Michael Albanese Jan 26 '16 at 7:05