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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 ?

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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 $w\in \mathbb{D}$.

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  • $\begingroup$ $h(w)$ 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. What is the meaning of this sentence? Could you please tell more details? Thanks! $\endgroup$ – Lorence Apr 16 '16 at 8:17
  • $\begingroup$ @Lorence On the one hand, from the integral representation, we can read off that $h$ is continuous (the integrand is a continuous function on $\partial \mathbb{D}\times \mathbb{D}$ since $\lvert f(z) \rvert \geqslant 1 > \lvert w\rvert$ for $z\in \partial \mathbb{D}$ and $w \in \mathbb{D}$). On the other hand, from the residue theorem (argument principle) we know that $h$ is integer-valued, namely, $h(w)$ counts how often $f$ attains the value $w$ (counting multiplicities) in the unit disk. Since a continuous integer-valued function on a connected space is constant, it follows that $h$ is … $\endgroup$ – Daniel Fischer Apr 16 '16 at 12:43
  • $\begingroup$ constant, i.e. all values from the unit disk are attained equally often (counting multiplicities, e.g. for $f(z) = z^2$ the value $0$ is attained at only one point, but with multiplicity $2$, while all other values are attained at two points, with multiplicity $1$ at each point). By assumption, there is a $z_0\in \mathbb{D}$ with $\lvert f(z_0)\rvert < 1$, so we know that at least one value is attained at least once. It follows that all values in the unit disk are attained at least once, i.e. $\mathbb{D}\subseteq f(\mathbb{D})$. $\endgroup$ – Daniel Fischer Apr 16 '16 at 12:43

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