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Suppose that $(X,d)$ is a complete metric space and $f:X \rightarrow X$ is such that $f\circ f$ is a strong contraction. Must $f$ have a fixed point?

So, it is given that $f\circ f$ is a strong contraction. Therefore,

$$d\left(f\circ f(x),f\circ f(y)\right)\leq cd(x,y)$$ for $0<c<1$ and also there exist an $x_0 \in X$ such that $f\circ f(x_0)=x_0$. I want to know whether $f$ has a fixed point or not and please give some hint to prove it.

Thanks!

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    $\begingroup$ Hint: fixed points of $f\circ f$ are unique. Show that both $x_0$ and $f(x_0)$ are fixed points of $f\circ f$. $\endgroup$ Apr 2, 2016 at 17:34

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Hint

By hypothesis $f\circ f$ has unique fixed point $x_0$. We have $f\circ f\circ f(x_0)=f(x_0)$ hence $f(x_0)$ is a fixed point for $f\circ f$. Can you take it from here?

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