# If $f$ maps a complete metric space $S$ onto $S$ and $f^2=f \circ f$ is a contraction mapping, show $f$ has a unique fixed point.

If $f$ maps a complete metric space $S$ onto $S$ and $f^2=f \circ f$ satisfies the fixed point theorem given below, show $f$ has a unique fixed point.

The following is the fixed point theorem:

If f is a mapping on a complete metric space S into S such that $d(f(x),f(y)) \leq kd(x,y)$ for all $x,y \in S$ with $0<k<1$, then the mapping has a unique fixed point.

HINT: You know that $f^2$ has a unique fixed point, say $x_0$. Show that $f(x_0)$ is also a fixed point of $f^2$.
• @ztforster: $f^2\big(f(x_0)\big)=f\big(f^2(x_0)\big)=f(x_0)$. – Brian M. Scott May 11 '15 at 2:45