Show that if \(f\) is holomorphic on \(\DD\) and continuous on \(\bar\DD\) with \(f(\bar \DD) \subseteq \DD\), then \(f\) has a unique fixed point in \(\DD\).
Note: this is subtle because \(\DD\) is not compact!
Solution
Continuous images of compact sets are compact, so \(f(\bar\DD)\) is a compact subset of \(\DD\) and thus contained in some \(\DD_r(0)\) with \(0<r<1\). On this disc, \[ \abs{f(z)} = \abs{f(z) - z + z} < \abs{z} .\] By Rouché, \(f(z)-z\) and \(z\) have the same number of zeros, which is one. This holds for any \(r'\) with \(r<r'<1\), and thus holds on \(\DD\).