Consider a twice continuously differentiable function $f: \mathbb{R}_+^0 \rightarrow \mathbb{R}$ with $f(0)=0$, $f'>0$, $f''<0$.
Does that function feature non-increasing elasticity, i.e. $$\frac{ \partial \frac{f'(x) x}{f(x)}}{\partial x} \leq 0\quad \forall x?$$
I am unable to prove this simple point, but also unable to find a counter example.
EDIT: I have made some progress, but it is not yet a proof. Take any $x>0$ and calculate the derivative on the LHS of the statement. The statement is true if (and only if)
$$ f'(x) x - f(x) \geq \frac{f''(x) f(x) x}{f'(x)}$$
Using a second order Taylor expansion we obtain
$$0=f(0)=f(x) - f'(x)x + 1/2 f''(x)x^2 - O(x^3).$$
Rearranging and plugging into the inequality we get
$$ f''(x)x^2/2 - O(x^3) \geq \frac{f''(x) f(x) x}{f'(x)}$$
Which (at least in a neighborhood of 0 or if $-O(x^3)\geq 0$) has a sufficient condition in
$$ f'(x) \leq 2\frac{f(x)}{x}.$$
Since $f(x)/x \geq f'(x)$ (because $f$ is concave, increasing and goes through the origin) that is true. Yet we don't know anything about what happens if $O(x^3) >0$ and we are far away from $x=0$.