If $f:\mathbb{R}\to\mathbb{R}$ is continuous and $f(x)\neq x$ for all $x$, must it be true that $f(f(x))\neq x$ for all $x$? Let $f: \Bbb R → \Bbb R$ be a continuous function such that $f(x)=x$ has no real solution  . Then is it true that $f(f(x))=x$ also has no real solution ?
Since $f$ is continuous, saying that $f(x) = x$ has no solution means that either $f(x) &lt; x$ for all $x$ or $f(x) &gt; x$ for all $x$. Let's assume wlog that $f(x) &gt; x$, then $f(f(x)) &gt; f(x) &gt; x$ for all $x$ so $f(f(x))=x$ has no solution either.
Let $\mbox{Id}_\mathbb{R}:\mathbb{R}\to \mathbb{R}$ the function $\mbox{Id}_\mathbb{R}(x)=x$. By hypotheses $0\notin \mbox{Im}(f-\mbox{Id}_{\mathbb{R}})$. Note that :


$\mbox{Im}(f -\mbox{id}_\mathbb{R})|_{\mbox{Im}(f)} \subset \mbox{Im}(f-\mbox{id}_{\mathbb{R}}) \implies  0\notin \, \mbox{Im}(f -\mbox{id}_\mathbb{R})|_{\mbox{Im}(f)} $
$\mbox{Im}(f -\mbox{id}_\mathbb{R})|_{\mbox{Im}(f)^C} \subset \mbox{Im}(f-\mbox{id}_{\mathbb{R}}) \implies  0\notin \, \mbox{Im}(f -\mbox{id}_\mathbb{R})|_{\mbox{Im}(f)^C} $
$\mbox{Im}(f\circ f-\mbox{id}_\mathbb{R})
\supset
\mbox{Im}(f|_{\mbox{Im}(f)} -\mbox{id}_\mathbb{R}|_{\mbox{Im}(f)})
=
\mbox{Im}(f -\mbox{id}_\mathbb{R})|_{\mbox{Im}(f)}$.
$\mbox{Im}(f\circ f-\mbox{id}_\mathbb{R})
\supset
\mbox{Im}(f|_{\mbox{Im}(f)^C} -\mbox{id}_\mathbb{R}|_{\mbox{Im}(f)^C})
=
\mbox{Im}(f -\mbox{id}_\mathbb{R})|_{\mbox{Im}(f)^C}$.
$\mbox{Im}(f\circ -\mbox{id}_\mathbb{R})= \mbox{Im}(f -\mbox{id}_\mathbb{R})|_{\mbox{Im}(f)^C}\cup \mbox{Im}(f -\mbox{id}_\mathbb{R})|_{\mbox{Im}(f)}$


Then if $f|_{\mbox{id}_\mathbb{R}} -\mbox{id}_\mathbb{R}$ no have real roots then $f\circ f-\mbox{id}_\mathbb{R}$ too no have real roots.