I have the complex-valued function

\begin{equation} f(z) = \frac{1}{1 + a(1 - z)}, \quad 0 < a < 1, ~ |z| \le 1. \end{equation}

For $|z| \le 1$ we have $|f(z)| \le 1$. Further, $z = f(z)$ has a single root in $|z| \le 1$, namely $z = 1$.

Define the sequence $z_n = f(z_{n - 1}), ~ n = 1,2,\ldots$ and set $|z_0| \le 1$. I want to show that $\lim_{n \to \infty} z_n = 1$.

Could anyone point me to some references? This should be easy, but I am not well-versed in complex-analysis nor in fixed-point iterations.


HINT.- First at all you have to prove that $$\left|\frac{a}{(1+a(1-z))(1+a(1-w))}\right|\le K\lt 1$$ for $f(z)$ be a contraction, i.e. $|f(z)-f(w)|\le K|z-w|$. The corresponding fixed point in the complet metric space $|z|\le 1$ is the limit of iterated $fofof....fof(z_0)$ for any arbitrary $z_0$ in the unit ball $|z|\le 1$.

Since $z_1=f(z_0),z_2=f(z_1),....,z_n=f(z_{n-1})$ one has $$|z_2-z_1|\le K|z_1-z_0|\\|z_3-z_2|\le K|z_2-z_1|\le K^2|z_1-z_0|\\.....\\|z_{n+1}-z_{n}|\le K^n|z_1-z_0|$$ It follows the Cauchy sequence $\{z_n\}$ because $$|z_{n+p}-z_n|\le |z_{n+p}-z_{n+p-1}|+|z_{n+p-1}-z_{n+p-2}|+.....+|z_{n+1}-z_n|$$ $$|z_{n+p}-z_n|\le (K^{p-1}+K^{p-2}+.....+K+1)K^n|z_1-z_0| \le\frac{K^n}{1-K}|z_1-z_0|$$ (we know that $\sum_{k=0}^{k=\infty}K^i=\frac{1}{1-K}$ and $K^n$ tends to $0$).

Hence $\{z_n\}$ has a limit $z$ and, because $f$ is continuous, if $z_n\to z$ then $f(z_n)\to f(z)$.

On the other hand, the uniqueness of this fixed point is clear because if $z$ and $w$ were distinct fixed points then one has the contradiction $$|z-w|\le K|z-w|\lt |z-w|$$ We finish with the fixed point $z=1$ having in hand and because of the uniqueness.

  • $\begingroup$ @Ritz: The proof of $K\lt 1$ is not difficult! $\endgroup$ – Piquito Sep 12 '16 at 13:06
  • $\begingroup$ Thanks for this. I am still working through it. For the first step, aren't we trying to show that for $w,z$ with $|w|,|z| \le 1$ and $w \neq z$ that $| \ldots | \le K < 1$ instead of $| \ldots | = K < 1$? In the first case, I find that $K = a$. $\endgroup$ – Ritz Sep 12 '16 at 13:19
  • $\begingroup$ Tha's right. It was a typo (Take the maximun). $\endgroup$ – Piquito Sep 12 '16 at 13:41

$\displaystyle g(z):=\frac{1}{1+a(1-z)}-1$

$\displaystyle \frac{g(z)}{g'(\Re(z))}=-(1-z)\frac{(1+a(1-\Re(z)))^2}{1+a(1-z)}$

Newton-Fourier method (https://en.wikipedia.org/wiki/Newton%27s_method below)

for $g(z)=0$: $\displaystyle \,\,z_{n+1}=z_n-\frac{g(z_n)}{g'(\Re(z_n))}$

Therefore $z=1$ or $\displaystyle \Re(z)=1+\frac{1}{a}$ and with $|z|\leq 1$ follows $z=1$.


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