Show that if $f(x)$ does not have a fixed point then $f(f(x))$ also does not have a fixed point

Question

Show that if $f(x)$ does not have a fixed point then $f(f(x))$ also does not have a fixed point.

Update:

$f$ is continous

I tried proving this by contradicting $f(f(x))$ has an unique fixed point, prove $f(x)$ has an unique fixed point too. but no luck, any help is highly appreciated.

Answer 1

If $f$ is a continuous function from an interval $I \subseteq \Bbb R$ into itself without fixed point then

• either $f(x) < x$ for all $x \in I$
• or $f(x) > x$ for all $x \in I$.

In both cases, $f \circ f$ has no fixed point.

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Without continuity the assertion is false: $$ f(x) = \begin{cases} 1 & \text{ if $x = 0$}\\ -x & \text{ if $x \ne 0$}\\ \end{cases} $$ has no fixed point, but $$ f(f(x)) = \begin{cases} -1 & \text{ if $x = 0$}\\ x & \text{ if $x \ne 0$}\\ \end{cases} $$ has infinitely many.

If the domain is not an interval then the assertion is also false, an example is $$ f: D \to D, x \mapsto -x \, . $$ with $D = (-1, 0) \cup (0, 1)$.

Answer 2

Suppose that $f(f(p))=p$ for some $p$. Then, $f(f(f(p)))=f(p)$ so $f(p)$ is a fixed point of $f(f(x))$ as well. If you can prove that $f$ is such that, if $f(f(x))$ admits a fixed point, then this point is unique, then it follows that $f(p)=p$ and $f$ has a fixed point. For example, if $f(f(x))$ is a Banach contraction of a complete metric space, then the claim obviously holds.