Is there a continuous function such that $f(f(x)) = g(x)$? 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.
 A: Yes.  Note first that $g^n(x)\to\infty$ as $n\to\infty$ for any $x$, since otherwise $g^n(x)$ would approach some finite limit which would then be a fixed point of $g$ by continuity.  Similarly, $g^n(x)\to-\infty$ as $n\to-\infty$.  Now let $a_n=g^n(0)$ for each $n\in\mathbb{Z}$, and observe that each element of $\mathbb{R}$ is in exactly one of the intervals $I_n=[a_n,a_{n+1})$.  Moreover, $g$ maps $I_n$ to $I_{n+1}$ by an order-preserving bijection for each $n$.
Now the idea is that we can make $g$ conjugate to $x\mapsto x+1$ by mapping each $I_n$ to $[n,n+1)$ in a manner compatible with $g$.  Explicitly, define $h:\mathbb{R}\to\mathbb{R}$ to map $I_0$ to $[0,1)$ by some particular order-preserving bijection $h_0$ (say, the linear one) and map $I_n$ to $[n,n+1)$ for each other $n\in\mathbb{Z}$ by $h(x)=h_0(g^{-n}(x))+n$.  We then see that $h$ is an order-preserving bijection and $g(x)=h^{-1}(h(x)+1)$.  So, if we define $f(x)=h^{-1}(h(x)+1/2)$, then $f$ is continuous and $f^2=g$.
More generally, we can define $g^r(x)=h^{-1}(h(x)+r)$ for all $r\in\mathbb{R}$ and these will satisfy $g^1=g$, $g^r\circ g^s=g^{r+s}$ for all $r$ and $s$.  These "fractional iterates" are not uniquely defined though, since they depend on the choice of $h_0$.
