# Problem of the Week #47 - April 22nd, 2013

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#### Chris L T521

##### Well-known member
Staff member
Here's this week's problem.

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Problem: Suppose $f$ is a non-vanishing continuous function on $\overline{\mathbb{D}}$ that is holomorphic in $\mathbb{D}$, where $\mathbb{D}=\{z\in\mathbb{C}:|z|\leq 1\}$ is the unit disc. Prove that if$$|f(z)|=1\quad\text{whenever }|z|=1,$$
then $f$ is constant.

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Hint:
Extend $f$ to all of $\mathbb{C}$ by $f(z)=1/\overline{f(1/\bar{z})}$ whenever $|z|>1$, and argue as in the Schwarz reflection principle.

#### Chris L T521

##### Well-known member
Staff member
No one answered this week's problem. You can find my solution below.

Proof: Let
$$g(z)=\left\{\begin{array}{ll}f(z) & |z|\leq 1\\1/\overline{f(1/\overline{z})} & |z|\geq 1.\end{array}\right.$$
$g(z)$ is well defined because $|z|=1\implies\dfrac{1}{\overline{z}}=z$ and
$$\frac{1}{\overline{f(1/\overline{z})}}=\frac{1}{\overline{f(z)}}=f(z)$$
due to the assumption that $|f(z)|=1$ when $|z|=1$. The function $g(z)$ is continuous on all of $\mathbb{C}$ and is holomorphic outside the boundary of $\mathbb{D}$ because $f$ is holomorphic on $\mathbb{D}$ and continuous on $\overline{\mathbb{D}}$ and $g(z)$ locally on $\mathbb{C}\backslash\overline{\mathbb{D}}$ is a convergent series from its definition. By the continuity of $g(z)$ on $\mathbb{D}$, for any bounded solid triangle $G$ in $\mathbb{C}$ whose boundary is $T$ and any $0<\epsilon<1$, if $T_{\epsilon}=\partial(G\backslash\{1-\epsilon<|z|<1+\epsilon\})$, then
$$\lim\limits_{\epsilon\to 0}\int_{T_{\epsilon}}g(z)\,dz=\int_Tg(z)\,dz.$$
Since $g$ is holomorphic on $\mathbb{C}\backslash\overline{\mathbb{D}}$, we have
$$\int_{T_{\epsilon}}g(z)=0 \,\,\text{for}\,\,0<\epsilon<1.$$
Therefore,
$$\int_T g(z)\,dz=0.$$
By Morera's theorem, $g(z)$ is holomorphic on all of $\mathbb{C}$. Since $f(z)$ is uniformly bounded on $\overline{\mathbb{D}}$, it follows that $g(z)$ is uniformly bounded and by Liouville's theorem, $g(z)$ must be constant. Therefore, $f$ is constant.$\hspace{.25in}\blacksquare$

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