1
$\begingroup$

Let $f: \bar{D} \rightarrow \mathbb{C}$ be analytic function in $D$ and continues on $\bar{D}$ where $D$ denotes the open unit disk and $\bar{D}$ denotes the close unit disk. Also, it is given than for $z \in \partial D$, $|f(z)| = 1$. In other words, the boundary of the disk goes to itself. The exercise has 4 parts.

  1. Show that for $|z| < 1$, $|f(z)| < 1$, it's direct corollary of the maximum principle.
  2. If additionally, $f$ is onto on the boundary, in other words, for every $\omega \in \partial D$, there exist $z \in \partial D$ such that $f(z)=\omega$, than $f$ has a zero on the interior of the unit disk. Assume not, then by the minimum principle, $|f|$ gets its minimum on the boundary, but than $|f|$'s minimum is $1$, as well as its maximum, so $|f|$ is constant so $f$ is constant, contradiction.
  3. With the previous assumptions, show that $f$ is onto in the open disk as well. Using Rouche's theorem, let $|z_0|<1$, so on the boundary, $|z_0| < f$, so $f$ and $f-z_0$ have the same number of zeroes (with multiplicities), which is positive.
  4. Assuming additionally, that $f$ is one to one on the boundary, in other words, if $z, \omega \in \partial D$ and $f(z) = f(\omega)$ then $z = \omega$, then it's the same for the interior. I'm stuck here.

Some thoughts: if I'll show that for some $|z_0| < 1$, $f(z) - z_0$ has one zero (with multiplicities), then according to Roche's theorem, that's true for all $|z|<1$. However, it's not clear how to use the one-to-one'ness of the function on the boundary.

I'll appreciate any help.

$\endgroup$
1
  • $\begingroup$ You need to assume $f$ is non constant. $\endgroup$ – zhw. Oct 18 '17 at 18:36
1
$\begingroup$

For each $z_0\in D$ we write $N(z_0)$ for the number of zeroes of the analytic function $f(z)-z_0$ in $D$. Then from the argument principle it follows that \begin{align} N(z_0)=&\frac{1}{2\pi i}\oint_{\partial D}\frac{f'(z)}{f(z)-z_0}\mathrm dz\\ =&\frac{1}{2\pi i}\oint_{f}\frac{1}{z}\mathrm dz\\ =&\textit{winding number}(f)\\ =&\pm1. \end{align} The last equal sign upward comes from the fact that $f\colon\partial D\to\partial D$ is a homeomorphism and thus its winding number must be $\pm 1$, while as you have elucidated, $N(z_0)$ must be positive, and thus each $N(z_0)=1$, i.e., $f\colon D\to D$ is one-to-one and onto.

ps. The condition that $f$ is analytic on $\overline D$ can be loosened to that $f$ is analytic in $D$ and continuous on $\overline D$ and similar conclusion holds, for which you can see here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.