Problem
I have a function $f: \mathbb{R}^+ \to \mathbb{R}^+$, which is twice continuously differentiable and satisfies the following property: $$ f''(x) > 0 \implies f'(x) < 0, \quad \forall x. $$ I call it hump-shaped because this property essentially guarantees that it is either fully concave, or it is concave for a while and decreasing afterwards. (This is useful because it guarantees at most two crossings with increasing, convex functions.)
I am trying to prove that the same property holds for the (possibly weighted) running integral of this function, which is defined as $$ F(x) = \int_0^1 w(s) f(sx) \mathrm{d}s, $$ with $w(s) > 0 \;\forall s \in [0, 1]$. That is, I want to show that
$$ F''(x) > 0 \implies F'(x) < 0, \quad\forall x. $$
I have a hunch that it is true, and could not find a counterexample, but I have not managed to prove it yet. I would appreciate any help (or counterexamples, or additional assumptions I might need).
Where I'm stuck
I essentially have to show that $$ \int_0^1 s^2 w(s) f''(sx) \mathrm{d}s > 0 \implies \int_0^1 s w(s) f'(sx) \mathrm{d}s < 0 $$ It is trivial if $f$ is decreasing everywhere, or concave everywhere, but seems tricky when it switches at some point. In particular, the left hand side integral puts more weight on the "later" parts of the function, where $f$ might become convex, while the right hand side integral puts relatively more weight on the earlier parts, where $f$ might be increasing.