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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?

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

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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.

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  • $\begingroup$ How do you prove $r$ is a smooth function in terms of differential geometry? $\endgroup$
    – Oopsilon
    Jan 11 at 13:18

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