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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.

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Well, for example $f(z) = 1$ satisfies the hypothesis, but the Schwarz Lemma does not apply to $f(z)/z = 1/z$. – GEdgar Mar 1 '12 at 19:39
up vote 3 down vote accepted

Fixed point theorem should suffice, but here's a slightly more complex-analysis-style proof:

By Rouche's theorem, we note that $|f(z)|<1=|z|$ on $\{|z|=1\}$ therefore $N_{f(z)-z}(D)=N_z(D)=1$, so we've proven something slightly stronger; that there exists a unique $z_0 \in D$ such that $f(z_0)-z_0=0$ therefore $f(z_0)=z_0$

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$|f|$ has maximum on the boundary, hence it maps $D$ to $D$, and you can use Brouwer fixed point theorem.

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