Let $f(z)$ be analytic on $|z|\le 1$ and satisfy $|f(z)|<1$ if $|z|=1$. Show that $f$ has a unique fixed point on $|z|<1$, namely a unique point $z_0$ in the disk $|z|<1$ such that $f(z_0)=z_0$.
My solution:
Since $f$ is an analytic, hence continuous function from the closed disk to itself (Maximum modulus principle), then $f$ has at least one fixed point according to the Brouwer fixed-point theorem. Now it leaves us to check that there is only one fixed point.
Whereas, we have known that if $f:\mathbb{D}\rightarrow\mathbb{D}$ is analytic with two distinct fixed points then $f$ is identity. contradicting the condition that $|f(z)|<1$ if $|z|=1$. And we are done.
My question is, can we avoid the Brouwer fixed-point theorem here? Thank you.
Let $g(z)=f(z)-z$. By the hypothesis and Rouche’s theorem, $g$ has the same number of zeros as $-z$ in the disk $|z|<1$, which is one.