Let $h:(0,1) \to (-\infty,0)$ be a $C^1$ function, with $h'>0$.
I am looking for sufficient conditions on $h$ that imply the existence of a $C^1$ decreasing* function $g:(0,1) \to (0,1)$ with $g'<0$ such that $g\circ g=\rm id$ and $h'/h < g'/g$.
Ideally, these conditions would be phrased in terms of $h$ and its derivatives, and won't involve auxiliary functions; but I am open to various sorts of options.
Comments:
If one removes the requirement $g\circ g=\rm id$, such a $g$ always exists; e.g. one can take $g(x) = 1-e^{h(x)}$. See here for details.
Even though my main goal is finding sufficient conditions on $h$, I am also interested in necessary conditions on it. (Of course, the best result would be to find a necessary and sufficient condition on $h$ that is not a tautology/trivial). I have a rather indirect argument that not every such $h$ admits a suitable $g$.
Here is how involutions $g$ arise: Choose arbitrarily a point $0<a<1$ and a decreasing function $g:[0,a]\rightarrow [a,1]$ such that $g(0)=1,g(a)=a$. Then extend the definition of $g$ to $(a,1]$ by $g(x):=g^{-1}(x)$. Then $g\circ g$ over $[0,1]$.
*Any involution $g$ which is not the identity, is necessarily strictly decreasing. Indeed, if $g(x)\ne x$, suppose that $z:=g(x)>x$; then $z>x$ and $g(z)=x<g(x)$. And similarly for the case where $g(x)<x$.
The motivation for this convoluted question comes from trying to analyse when the solution to a certain minimization problem is convex. (a bit too long to describe here).