# Proving that $f$ has a fixed point on the closed unit disk

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?

• Brower fixed point theorem Commented Nov 15, 2019 at 18:41
• @user124910 I know this theorem, but since it's not mentioned in the book, I guess there is a direct proof. Commented Nov 15, 2019 at 18:42
• @CélioAugusto I believe user124910 is suggesting you look at a proof of the Brouwer fixed point theorem to get ideas. Commented Nov 15, 2019 at 18:55

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.
• How do you prove $r$ is a smooth function in terms of differential geometry? Commented Jan 11 at 13:18