Suppose $f$ is a non-vanishing continous function on $\overline{D(0,1)} $ and holomorphic on ${D(0,1)} $ such that $$|f(z) | = 1$$ whenever $$|z | = 1$$
Then I have to prove that f is constant.
We can extend $f$ to all $\mathbb{C}$ by setting $$f(z) = \frac{1}{\overline{f(\frac{1}{\bar{z}})}}$$ and the resulting function is holomorphic on ${D(0,1)} \ $, $\mathbb{C} – \overline{D(0,1)}$ and continous on $\partial D(0,1)$.
But how can we say that the resulting function is holomorphic in $z \in \partial D(0,1)$ ?
Best Answer
To prove that $f$ is identically constant, you'd better employ the maximum principle, according to which the maximum of $|f|$ on $\overline{D}$ equals 1. But $f$ is non-vanishing on $\overline{D}$, hence $\frac{1}{f}$ is holomorphic on $D$ while $\bigl|\frac{1}{f}\bigr|=1$ on $\partial{D}$. By the same maximum principle now follows that the minimum of $|f|$ on $\overline{D}$ also equals 1. Hence $|f|=1$ on $\overline{D}$, whence readily follows the desired result.