Suppose $f(z)$ is analytic in the closed unit disc and $$|f(z)|<1 \quad \text{for} \quad |z|=1$$ Show that $f(z)$ has one and only one fixed point; that is, there exist a unique point $z_0$ in the unit disc such that $f(z_0)=z_0$.

Should I approach this using proof by contradiction? Assume there exist more that one fixed point?

## Best Answer

Define $g(z) = -z$ and $h(z) = f(z) + g(z) = f(z) - z$.

Note that $h(z) - g(z) = f(z)$, and so $|h(z) - g(z)| = |f(z)| < 1$ for all $z$ such that $|z| = 1$. Note also that since $g(z) = -z$, then $|g(z)| = |-z| = |z| = 1$ for all such $z$.

So we have $|h(z) - g(z)| < |g(z)|$ for all $z$ such that $|z| = 1$.

Therefore, by RouchÃ©'s Theorem, the functions $g(z)$ and $g(z) + (h(z) - g(z)) = h(z)$ have the same number of roots in the interior of $|z| = 1$, i.e., on the set $D = \{z : |z| < 1\}$.

Since $g(z)$ clearly has only one root in $D$ (and actually one root everywhere; it's $z=0$), then this means $h(z)=0$ also has only one solution in $D$. And since $h(z) = f(z) -z$, then we see that $f(z) -z=0$ has only one solution in $D$, i.e., $f$ has only one fixed point in $D$.