Conjecture on the elasticity of a positive, bounded, and decreasing function. Consider a function $f$ defined on $[0,a]$ where $a>1$ and $a$ is possibly $+\infty$. Assume that it is positive, bounded, decreasing, with $\lim_{x\to a}f(x)=0$. Also assume that $f$ is smooth. I conjecture that the elasticity of $f$ (in absolute value) is strictly increasing on $[0,a]$. By absolute value of the elasticity, I mean $−d\ln(f)/d\ln(x)$ or, equivalently, $-x \cdot [f'(x)/f(x)]$.
I would like to know (1) if the conjecture is indeed true; (2) if it is true, how to prove it; and (3) if the conjecture is not true, what is the minimal amount of additional restrictions on the function $f$ that would make the conjecture true.
Something I know is that by defining a function piecewise, it is easy to construct examples of functions that are positive, bounded, decreasing, with a limit of 0 in $a$, but that do not have an increasing elasticity. For instance: $f(x)=2-x$ on $[0,1]$ and then $1/x$ on $[1,\infty]$. But this function is not smooth.
Thank you!
 A: What you are really asking is that $$-\frac{xf'(x)}{f(x)}$$ is strictly increasing, which would mean that $$-\frac{f'(x)f(x) + xf''(x)f(x) - xf'(x)^2                 }{(f(x))^2} > 0,$$ which is equivalent to $$\frac{f'(x)}{f(x)} < \frac{1}{x} + \frac{f''(x)}{f'(x)}, (\ast)$$ because $f'(x) < 0$. Note that you might not have strict equality all the time (sometimes equal to zero).
Thinking that you had already tried $x^{-p}$, I tried each of these: $f(x) = e^{-\lambda x^n}$ and $f(x) = \frac{1}{1+\lambda x^n}$, for $\lambda > 0$ and $n \in \mathbb{N}$ and they both worked. ... This idea led to nothing.

So, we now look at the general case. Suppose that $f$ is strictly decreasing and twice differentiable (this is where your example breaks down). Write $f(x) = e^{g(x)}$, where $g(x) = \operatorname{ln}(f(x))$, then $g$ is decreasing in the same was as $f$ since the exponential function is strictly increasing. So, we see that the elasticity is
$$x|g'(x)|,$$ which is positive. This is why the $C^0$ (continuous but not smooth) counter example works because for $x < 1$, $x|g'(x)| = x$ is increasing and for $x > 1$, $x|g'(x)| = 1/x$ is decreasing. 
So, setting $h(x) = g'(x)$, we are trying to investigate the existence of a nice function $h$ that is negative and $x|h(x)|$ is increasing and $g(x) = \int_0^{x}|h(t)|dt \to g(0)$ as $x \to \infty$.
This, however, is asking too much. Set 
$$ f(x) = e^{\tan^{-1}(x)}.$$ We then see that $f$ is a counter-example:
$$ \frac{\frac{d}{dx}\operatorname{tan^{-1}(x)}}{\frac{d}{dx} \operatorname{ln}(x)} = \frac{\frac{1}{1+x^2}}{\frac{1}{x}}  = \frac{x}{1+x^2}.$$
You can check that this is not increasing. See: Graph.
A: If $y=f(x)$, then defining $u=\ln x$ and $v=\ln y$ we have $v=\ln f(\exp u)$; call this function $v=g(u)$. Then ${\mathrm d\ln f(x)}/{\mathrm d\ln x}={\mathrm dv}/{\mathrm du}=g'$.
Now we have $g:(-\infty,\ln a)\to\mathbb R$ is bounded above, decreasing, smooth, and has limit $\lim_{u\to\ln a} g(u)=-\infty$. This is certainly not sufficient to ensure that $|g'|$ is strictly increasing, as you can most easily convince yourself by trying to sketch a decreasing graph with the specified asymptotes and finding that nothing forces you to always increase the slope.
For an explicit example in the $a=\infty$ case, take $g(u)=-\exp(u+\sin u)$, i.e. $f(x)=\exp(-\exp(\ln x+\sin\ln x))=\exp(-x\exp\sin\log x)$.
Here's a slight variation using $\frac1{2\pi}\sin(2\pi u)$ instead of $\sin u$, which looks more interesting when plotted.
$v=g(u)$:

$y=f(x)$:

