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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.

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  • $\begingroup$ You need to assume $f$ is non constant. $\endgroup$
    – zhw.
    Oct 18, 2017 at 18:36

1 Answer 1

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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.

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