Suppose that $f(z)$ is complex analytic on $|z| \leq 1$ and satisfies
$|f(z)| < 1$ for $|z|=1$.
(a) Prove that the equation $f(z)=z$ has exactly one root
(counting multiplicities) in $|z|<1$.
(b) Prove that if $|z_0| \leq 1$, then the sequence $z_n$ defined recursively
by $z_n= f(z_{n-1}) , n=1,2,…$, converges to the fixed point of $f$.
I was able to prove (a) using Rouche's theorem, but (b) stumps me.
I know that (b) is true for analytic fuctions such that $f(0)=0$ or
$|f'(z)|<1$ on the disc, neither of which are necessarily true in general.
The farthest I was able to get was $|f(z)-z^*|<\frac{1}{1-|z*|}|z-z^*|$,
where $z^*$ is the fixed point of $f$, but
$\frac{1}{1-|z^*|}>1$, so I don't think this helps me.
Can someone please point me in the right direction?
Best Answer
One can reduce it to the case of $f(0)=0$ (i.e. $0$ is the fixed point) by making a suitable linear fractional transformation of the disk. Namely, if $z^*$ is the fixed point, apply the above argument to $L \circ f \circ L^{-1}$ where $L$ sends $z^* \to 0$ and is a LFT.