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This question here shows the first part, ie. $f(f(x))$ has unique fixed point $\implies$ $f(x)$ has a unique fixed point.

This is for a general metric space $(X,d)$, $f: X \to X$.

I want to prove the reverse implication that $f(x)$ has unique fixed point $\implies$ $f(f(x))$ has a unique fixed point.

I have tried to show that $g: X \to X$, $g(x) = f(f(x))$ is a contraction but I don't think this will help because $X$ is not necessarily complete.

Zarathustra
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    Is this true? over $\mathbb R$, we could have $f(x)=-x$. Then the only fixed point of $f(x)$ is $0$ but $f\circ f(x)=x$ has infinitely many fixed points. – lulu Jan 21 '20 at 16:26
  • That does seem right... The question reads "prove that $f$ has a unique fixed point if and only if $g$ has a unique fixed point" (where g is defined as above) but could be mistaken? – Zarathustra Jan 21 '20 at 16:40
  • It is not iff. It is clear that any fixed point of $f$ is a fixed point of $f \circ f$, but not the other way around as the example $f(x) = -x$ shows. – copper.hat Jan 21 '20 at 16:42
  • Yes - that is a good example. I suppose there must be a mistake in the question I am looking at. Thank you. – Zarathustra Jan 21 '20 at 16:43
  • Is there maybe a condition on $f$ such as that it is a weak contraction ($d(f(x),f(y)) < d(x,y)$ for all $x \neq y$)? – Daniel Fischer Jan 21 '20 at 16:58
  • The question specifies no conditions on $f$ unfortunately. Do you have a condition in mind which would help? – Zarathustra Jan 21 '20 at 17:01
  • That $f$ is a weak contraction for example. Then $f\circ f$ also is one, and hence it can have at most one fixed point. – Daniel Fischer Jan 21 '20 at 17:07
  • Doesn't that require that $X$ is complete? – Zarathustra Jan 21 '20 at 17:08
  • The uniqueness of fixed points (that there is at most one) doesn't require completeness. The existence of at least one fixed point of $f\circ f$ is here given by the hypothesis that $f$ has a fixed point. Generally, for a weak contraction, completeness is not enough to guarantee the existence of a fixed point, consider $X = [1,+\infty)$ and $f(x) = x + \frac{1}{x}$. Then $f(x) > x$ for all $x$, but since $f'(x) = 1 - \frac{1}{x^2} < 1$ for all $x$, $f$ is a weak contraction. – Daniel Fischer Jan 21 '20 at 18:26

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From the comments, the if and only statement is not true.

The example to disprove this statement is $f(x)=-x$ as the only fixed point of $f$ is $x=0$ by $f(f(x)) = -(-x)$ has infinitely many fixed points.

Thank you to @lulu.

Zarathustra
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