Can every increasing negative function be expressed as a product with these properties? 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 $C^1$ functions $\lambda:(0,1] \to (-\infty,0)$, $g:(0,1] \to (0,1]$ such that
$$
h=\lambda g,  \,\,\,\,\,\lambda'>0, \,\,\,\,\,g'<0 \tag{1}
$$
Note that $h'=\lambda' g+\lambda g'>0$,
since both summands are positive. Thus, $h'>0$ is a necessary condition for $h$ to be expressible as in $(1)$.
Is it sufficient? Can every $h$ with $h'>0$ be expressed in this way?

The motivation for this convoluted question comes from trying to analyses when the solution to a certain minimization  problem is convex. (a bit too long to describe here).

An equivalent reformulation: (thanks to Alex Jones).
Take any $g: (0,1] \to (0,1]$ with $g' < 0$. Then, we must have $\lambda = h/g < 0$. Since $g > 0$, the condition $\lambda' > 0$ is equivalent to $h'/h < g'/g$.
So, the question boils down to this: Does there exist a function $g:(0,1] \to (0,1]$, with $g' < 0$, such that  $h'/h < g'/g$?
Since the quantity $g'/g$ is invariant under positive scaling of $g$ ($g \to c g$ for $c >0$), it suffices to search for bounded $g$.
Note that if $h$ is bounded, then by taking $g=-h$ we get $h'/h = g'/g$ instead of $h'/h < g'/g$. So we are "nearly there".

You can always choose $g(x) = 1-e^{h(x)}$. Then $0 < g(x) < 1$ and $g$ is decreasing.
Finally, $\lambda = h/g$ is strictly increasing because
$$
 \lambda(x) = \frac{h(x)}{1-e^{h(x)}} = \phi(h(x))
$$
where
$$
 \phi(u) = \frac{u}{1-e^u}
$$
is strictly increasing on $(-\infty, 0)$.
