prove the existence of a fixed point

Question

Prove that every holomorphic function $f$ on the closed disk $\overline{D}(0,1)$ with $|f(z)|<1$ when $z\in \overline{D}(0,1)$ has at least one fixed point in $D(0,1)$.

My attempt:
Since $f$ is holomorphic on $\mathbb{D}$, $f$ is either constant or attains its maximum on boundary.
If $f$ is constant, we finish.
If $f$ is not constant, then $f$ attains its maximum on $\partial\mathbb{D}$.
According to Maximum modulus principle, we have $$|f(z)|\leq |f(z_0)|\quad\forall z\in\mathbb{D}\quad (\text{for some }|z_0|=1).$$ I don't know how to continue. I haven't used the hypothesis $|f(z)|<1\quad\forall z\in\overline{D}(0,1)$.
Could someone have any idea how to solve this problem?

Answer 1

Consider $g(z)=-z$. Then, in the unit circle, $|f(z)|<1=|g(z)|$.

By Rouche's Theorem, $f(z)+g(z)=-z+f(z)$ has the same number of roots as $-z$ on $\overline{D}(0,1)$. But the roots of $-z+f(z)$ are the fixed points of $f$ and $-z$ has one root in $\overline{D}(0,1)$, so not only does $f$ have one fixed point, it is unique!

Answer 2

Argument using Contraction Mapping Theorem:

There is a point $z_0$ in the closed disk at which $|f|$ attains its maximum. If $r=|f(z_0)|$ then $r<1$ and $|f(z)| \le r$ for all $z$. By Cauchy's estimates (Ref. Rudin's RCA) we get $|f'(z)| \le r$ for all $z$ and this makes $f$ a contraction mapping of the closed unit disk. Hence, there is a unique fixed point.

[$|f(u)-f(v)|=|\int_{[u,v]} f(w)dw|\leq r|u-v|$ where $[u,v]$ is the line segment from $u$ to $v$].