Let $f(z)$ be an analytic function on $D=\{z : |z|\leq 1\}$. $f(z) < 1$ if $|z|=1$. How to show that there exists $z_0 \in D$ such that $f(z_0)=z_0$. I try to define $f(z)/z$ and use Schwarz Lemma but is not successful.
Edit: Hypothesis is changed to $f(z) < 1$ if $|z|=1$. I try the following. If $f$ is constant, then conclusion is true. Suppose that $f$ is not constant and $f(z_0)\neq z_0$ for all $z_0\in D$. Then $g(z)=\frac{1}{f(z)-z}$ is analytic. If I can show that $g$ is bounded, then we are done. But it seems that $g$ is not bounded.