Let $g:\mathbb{R}\to\mathbb{R}$ be strictly increasing, continuous, and satisfy $g(x) > x$ and $\lim\limits_{x\rightarrow -\infty}g(x) = -\infty $. Is there a continuous function $f :\mathbb{R} \rightarrow \mathbb{R} $ such that $f(f(x)) = g(x)$?
I have noted a few properties about a candidate $f$; one must have
(1) $f$ is injective
(2) $f$ is strictly increasing or strictly decreasing
(3) $f$ is surjective
We also have that $g$ has a continuous inverse $g^{-1}$.
The following maybe useful;
We can define an equivalence relation $\sim$ on $\mathbb{R}$ where $x \sim y$ if $y =g_k(x)$ (here $g_{m+1}(x) = g(g_{m}(x))$ , $g_{m-1}(x) = g^{-1}(g_m(x))$, $g_{0}(x) = x$ ).
One can also observe that for a candidate $f$ that if $f(a) = b$ then $a \nsim b$ as if $f(a) = g_k(a)$ then $f(f(a)) = f(g_k(a))$ or $g_1(a) = f(g_k(a))$; hence $f(g_1(a)) = g_{k+1}(a)$; thus by two-sided induction we must have for $u \in \mathbb{Z} $ that $f(g_u(a)) = g_{k+u}(a)$; hence $f(g_k(a)) = g_{2k}(a)$ but from above $f(g_k(a)) = f(f(a)) = g(a)$; hence $g(a) = g_{2k}(a)$ which is impossible as $g$ has no fixed points.
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