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Assume that $f(z)$ is analytic on the open unit disk and continuous on the boundary $|z|=1$ with $|f(z)|=1$ if $|z|=1$. Show that $f$ is a rational function.


My attempt:

Since $f$ maps the boundary of the unit disk to the boundary of the unit disk, we can compose $f$ with some conformal map $\varphi$, such that $F\colon=\varphi^{-1}\circ f\circ\varphi$ maps the upper half-plane to the upper half-plane and preserves the real axis. For example, we may set $$ \varphi(z)=\frac{z-i}{z+i} .$$ Therefore, we can extend $F$ to an entire function with $F(z)=\overline{F(\bar z)}$ if $z$ lies in the lower half-plane.

Now it suffices to show that $\infty$ is a pole then we invoke that every meromorphic function on $\overline{\mathbb{C}}$ is a rational function to conclude. But I am stuck... Any help?


Edit:

Can we claim that $$\lim_{z\to\infty}F(z)=\lim_{z\to x_0}z=x_0$$ where $x_0\in\mathbb R\cup\{\infty\}$ by tracing the images of the point which goes to infinity?

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Since $f$ does not vanish on $|z|=1$ it follows that it has at most finitely many zeros. Let $c_1,c_2,...,c_N$ be the zeros counted according to multiplicities. Consider $g(z)=\prod_{j \leq N} \frac {z-c_j} {1-\overset {-} {c_j}z }$. Observe that $|g(z)|=1$ when $|z|=1$. Verify that $h(z)=\frac {f{(z)}} {g(z)}$ is analytic on the open unit disk (ignoring removable singularities) and continuous on the closed disk with no zeros. Apply MMP to $h$ and $\frac 1 h$ to conclude that $|h|$ is a constant. This implies that $h$ is a constant and completes the proof.

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    $\begingroup$ Technically, $h$ as written is meromorphic, with removable singularities. $\endgroup$
    – Arthur
    Jul 30 '19 at 7:49
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    $\begingroup$ It should be $g(z)=\prod_{j \leq N} \frac {z-c_j} {1-\bar{c_j}z }$ (Blaschke product) if you take the approach. $\endgroup$
    – Martin R
    Jul 30 '19 at 8:04
  • $\begingroup$ @MartinR Thank you very much. I was so careless. I indeed wanted $z-c_j$ in the numerator. $\endgroup$ Jul 30 '19 at 8:08
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Yes, your approach works. Just define $F=\varphi^{-1}\circ f\circ\varphi$ on the extended upper half-plane $$\overline H = \{ z: \operatorname{Im} z \ge 0 \} \cup \{ \infty \} \subset \hat{\Bbb C} \,. $$ That is possible because $f$ is continuous on $\overline{\Bbb D}$. Then extend $F$ to the lower half-plane via the Schwarz reflection principle, as you did. Now $F$ is meromorphic in $\hat{\Bbb C}$ and therefore a rational function.

Alternatively extend $f$ directly to a meromorphic function on $\Bbb C$ via reflection at the unit circle: $$ g(z) = \begin{cases} f(z) & \text{ if } |z| \le 1 \\ \frac{1}{\overline{f(1/\bar z)}} &\text{ if } 1 < |z| < \infty . \end{cases} $$ Then $\lim_{z \to \infty} g(z) = \frac{1}{\overline{f(0)}}$ so that $g$ has a removable singularity or a pole at $\infty$, and therefore is a rational function.

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    $\begingroup$ @Bach: $g$ is continuous, because $f(z) \cdot \overline{f(z)} = |f(z)|^2 = 1$ on the unit circle. $\endgroup$
    – Martin R
    Jul 30 '19 at 8:09

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