Suppose $f$ is a non-vanishing continuous function on $\bar{\mathbb{D}}$ that is
holomorphic in $\mathbb{D}$. Prove that if
$$|f(z)|=1~~~\text{whenever}~~~|z|=1$$
then $f$ is constant.
I have proved this by showing that the function
$$F(z)=\left\{\begin{array}{cc}f(z)&\text{when}~~|z|\leq1\\ 1/\bar{f}(\bar{z})&\text{otherwise}\end{array}\right.$$
is bounded and entire.
Is there any other more elegant way to do this problem, because my method is turning out to be too gruesome for this beautiful problem.
Thanks in advance!
Best Answer
According the Maximum Modulus Principle we have $\vert f(0)\vert\leq 1$.
But since $f$ does not vanish, the function $z\mapsto1/f(z)$ satisfies the same conditions as $f$ and consequently we have also $\left\vert \frac{1}{f(0)}\right\vert\leq 1$, or equivalently $\vert f(0)\vert\geq 1$.
Combining these two inequalities we get $\vert f(0)\vert = 1$, and the Maximum Modulus Principle implies in this case that $f$ is constant since it attains its maximum inside $\Bbb{D}$.