Assume that a function $f$ is continuous on the closed unit disk, analytic on the open unit disk and $|f(z)| \le 1$ when $|z|=1$. Show that $f$ has at least one fixed point on the unit closed disk and that if $f$ has no fixed points on $|z|=1$ then $f$ has exactly one fixed point inside the disk.
I know that if $f$ has no fixed points on the unit circle, we could use Rouché's Theorem to conclude that $f(z)-z$ has exaclty one zero on the unit disk. But I am having some trouble with the first part. If $f$ has no fixed points on the closed disk, then $g(z)=\frac{1}{f(z)-z}$ is analytic. I tried to apply the Maximum Modulus Principle on $g$, but i got nowhere, since we don't have an equality condition for $f$ on $|z|=1$. Any hints?
Best Answer
One idea that is used for the 2D version of Brouwer's Theorem is as follows.
Assume $f$ has no fixed point on the closed disc. For each $x$ define $r$ to be the point on the boundary determined by a half-line from $x$ to $f(x)$. Then $r$ is a 'retraction' i.e. a continuous function from the disc to its boundary which fixes every point on the boundary.
If you already know that there is no such retraction then you are finished.