I'm stuck with this problem from Stein-Shakarchi:
Let $f$ be non-constant and holomorphic in an open set containing the closed unit disc.
a) Show that if $|f(z)| = 1$ whenever $|z| = 1$, then the image of $f$ contains the unit disc.
b) If $|f(z)| \geq 1$ whenever $|z| = 1$ and there exists $z_{0} \in D(0,1) $ such that $|f(z_{0})| < 1$, then the image of $f$ contains the unit disc.
Any idea ?
Best Answer
a) is a special case of b), since the maximum modulus principle implies $\lvert f(z)\rvert < 1$ for all $z$ in the open unit disk when $f$ is non-constant.
Consider $h \colon \mathbb{D}\to \mathbb{C}$,
$$h(w) = \frac{1}{2\pi i} \int_{\partial \mathbb{D}} \frac{f'(z)}{f(z)-w}\,dz.$$
By the residue theorem (in the form of the argument principle), $h(w)$ is the number of times the value $w$ is attained by $f$ in the open unit disk (counting multiplicities). In particular, $h$ is integer-valued. On the other hand, since the integrand depends continuously on $w$, $h$ is continuous on $\mathbb{D}$. It follows that $h$ is constant, and by assumption $h(f(z_0)) \geqslant 1$, so $h(w) \geqslant 1$ for all $w\in \mathbb{D}$, which means $w\in f(\mathbb{D})$ for all $w\in \mathbb{D}$.