Fixed factor model for analytic features on the device disc

Suppose that $f(z)$ is intricate analytic on $|z| \leq 1$ and also pleases $|f(z)| < 1$ for $|z|=1$.

(a) Prove that the formula $f(z)=z$ has specifically one origin (checking multiplicities) in $|z|<1$.

(b) Prove that if $|z_0| \leq 1$, after that the series $z_n$ specified recursively by $z_n= f(z_{n-1}) , n=1,2,...$, merges to the set factor of $f$.

I had the ability to confirm (a) making use of Rouche is theorem, yet (b) stumps me. I recognize that (b) holds true for analytic fuctions such that $f(0)=0$ or $|f'(z)|<1$ on the disc, neither of which are always real as a whole. The farthest I had the ability to get was $|f(z)-z^*|<\frac{1}{1-|z*|}|z-z^*|$, where $z^*$ is the set factor of $f$, yet $\frac{1}{1-|z^*|}>1$, so I do not assume this aids me. Can a person please factor me in the appropriate instructions?

2019-12-02 02:53:24
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Answers: 1

One can lower it to the instance of $f(0)=0$ (i.e. $0$ is the set factor) by making an ideal straight fractional makeover of the disk. Particularly, if $z^*$ is the set factor, use the above argument to $L \circ f \circ L^{-1}$ where $L$ sends out $z^* \to 0$ and also is a LFT.

2019-12-03 04:20:55