Let $\omega \subset \mathbb C$ be a simple connected set and $\,f:\omega \to A$ is an analytic function where $A \subset \omega$ is compact. Show that $f$ has an unique fixed point.

I think we can reduce the problem by helping from Riemann open mapping theorem to unit disk but I can't prove the existence of fixed point.

  • $\begingroup$ I don't think you need the open mapping theorem... $\endgroup$ – ireallydonknow Nov 27 '14 at 14:05
  • $\begingroup$ if i don't need open mapping then how can i solve this problem $\endgroup$ – erfan soheil Nov 27 '14 at 14:10
  • $\begingroup$ I assume you take open to be included in the definition of simply connected. Otherwise you could take $\omega=A$ and the identity map, which has many fixed points. On the other hand if $\omega$ is open and $A$ is compact then $A\not=\omega$, hence $f$ cannot be the identity. $\endgroup$ – Mirko Nov 27 '14 at 14:58
  • $\begingroup$ $\omega $ is not open because $\omega = f^{-1}(A)$ and why we can take $\omega = A$ ? $\endgroup$ – erfan soheil Nov 27 '14 at 15:53
  • $\begingroup$ Analytic functions are defined on open sets. $\endgroup$ – Yiorgos S. Smyrlis Nov 27 '14 at 16:29

I imagine that the domain $\omega$ of $f$ is open.

If $f$ is constant, then it has a unique fixed point and we are done. Assume that $f$ is not constant, and hence an open map (cf. Open Mapping Theorem).

Let $K=\overline{f(\omega)}$. Then $K\subset A\subset \omega$, and as $A$ is compact, so is $K$. Thus $$ \omega\supset K\supset f(\omega)\supset f(K)\supset \cdots\supset f^{n}(\omega)\supset f^{n}(K)\supset\cdots, $$
and hence $$\bigcap_{n\in\mathbb N}f^{n}(\omega)=\bigcap_{n\in\mathbb N}f^{n}(K)=L.$$ Clearly, $L$ is compact and non-empty, as the intersection of a decreasing sequence of non-empty compact sets.

Next, define the sequence $f_1=f$, $\,f_2=f\circ f$, and $f_{n+1}=f\circ f_n$.
Clearly, $\{\,f_n\}_{n\in\mathbb N}$ is a sequence of bounded holomorphic functions, and by virtue of Montel's Theorem, there is a subsequence $\{\,f_{k_n}\}$ converging uniformly in the compact subsets of $\omega$ to say $g\in\mathcal H(\omega)$. Also, as the image of $f^n$ is $f^n(\omega)$, then the image of $g$ is a subset of $L$. We shall show that $g(\omega)=L$, which implies that $g$ is constant (otherwise $g(\omega)$ would be an open set). If $z_0\in L$, then, $$ z_0\in f^{n+1}(\omega)=f^n\big(\,f(\omega)\big)\subset f^n(K), $$ for every $n\in\mathbb N$, and hence there exists a $w_n\in K$, such that $z_0=f^n(w_n).$ But $\{w_{k_n}\}_{n\in\mathbb N}\subset K$ has a convergent subsequence $\{w_{\ell_n}\}_{n\in\mathbb N}$, with $w_{\ell_n}\to w_0$. Now we have that $$ z_0=f^{\ell_n}(w_{\ell_n})\to g(w_0). $$ Indeed, $g$ is constant. Let $g(z)=z_0$, for all $z\in\omega$. Then $$ f(z_0)=f\big(g(z_0)\big)=\lim_{n\to\infty} f\big(\,f^{k_n}(z_0)\big) =\lim_{n\to\infty} f^{k_n}\big(\,f(z_0)\big)=g\big(f(z)\big)=z_0. $$ Thus $z_0$ is a fixed point for $f$. If $f$ had a second fixed point, say $z_1$, then $$ z_1=f(z_1)=f^2(z_1)=\cdots=f^{k_n}(z_1)\to g(z_1)=z_0. $$ Indeed, $f$ has a unique fixed point.

Note. The domain $\omega$ does not have to be simply connected.

  • $\begingroup$ Thank you very much. This answer in in any book? $\endgroup$ – erfan soheil Nov 27 '14 at 16:34
  • 1
    $\begingroup$ I do not think so. I've been teaching CV for years, and I have not seen this problem in any book. $\endgroup$ – Yiorgos S. Smyrlis Nov 27 '14 at 16:56
  • $\begingroup$ Thank you mr. your way is ccomplete and logical $\endgroup$ – erfan soheil Nov 27 '14 at 17:06
  • $\begingroup$ I have two questions about your answer, 1- where did you use connnectivity of $\omega$ 2- i saw montel's theorem and i didn'n find any relation whit answer, whould you tell me how you realized that $\{f_n\}$ has a converges subsequense $\endgroup$ – erfan soheil Dec 8 '14 at 9:58
  • $\begingroup$ As $\omega$ is connected and $g$ not open, then $g$ is constant. If $\omega$ was not connected this would not be true. $\{f_n\}$ possesses a convergent subsequence, uniformly in every compact subset of $\omega$. This is due to the fact that $\{f_n\}$ is uniformly bounded. $\endgroup$ – Yiorgos S. Smyrlis Dec 8 '14 at 10:32

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