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Jan 10, 2013 at 14:31 comment added Pietro Majer (or, also: if $f^n$ is a contraction on a complete metric space, it has a unique fixed point $x$; but $f(x)$ is a fixed point of $f^n$ too, so by the uniqueness $x=f(x)$.)
Sep 18, 2012 at 19:48 comment added jbc It is a trivial remark (not due to me) that if a mapping $ f $ on a set $ X $ (no metric required) has a unique fixed point, then this is also a fixed point for any mapping which commutes with $f$. This implies immediately that a mapping on a complete metric space has a fixed point whenever some iterate is a contraction. No continuity assumption is required.
Jul 19, 2010 at 15:40 vote accept user3014
Jul 18, 2010 at 22:37 history answered Pietro Majer CC BY-SA 2.5