Let $f \in H(B(0,1))\cap C(\overline{B(0,1)})$. Assume $f(\overline{B(0,1)}) \subset B(0,1)$. Show that $f(z)$ has only one fixed point in $B(0,1)$.
I tried to use Rouché's Theorem to show that $f(z)-z$ and $z$ have the same number of zeros inside $B(0,1)$. However, $\partial B(0,1)$ is not contained in $B(0,1)$, which causes some trouble when applying the theorem. How can I address this problem?
Suppose $a$ and $b$ are distinct fixed points of $f$. There exists $\epsilon>0$ such that $(1+\epsilon) a$ and $(1+\epsilon) b$ are both in $B(0,1)$. Also, since $f(\overline {B(0,1)})$ is a compact subset of $B(0,1)$ we can choose $\epsilon$ so small that $(1+\epsilon)f(z) \in B(0,1)$ for all $z$. Now consider the function $g$ defined by $g(z)=(1+\epsilon) f(\frac 1 {1+\epsilon} z)$. Note that $(1+\epsilon) a$ and $(1+\epsilon) b$ are fixed points for $g$ in $B(0,1)$. The difficulty you had in dealing with $f$ has vanished because $g$ is holomorphic in $B(0,1+\epsilon)$.