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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.

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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 $f^2$.

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  • $\begingroup$ So I can see that kd(x,y) >= d(f(f(x)),f(f(y))), but I need kd(f(x),f(y)) >= d(f(f(x)),f(f(y))) to say that f has a fixed point. I still don't understand how to get there. $\endgroup$ – ztforster May 11 '15 at 2:00
  • $\begingroup$ @ztforster: $f^2\big(f(x_0)\big)=f\big(f^2(x_0)\big)=f(x_0)$. $\endgroup$ – Brian M. Scott May 11 '15 at 2:45

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