I have a model where $f(x)$ is a decreasing, convex function from mapping $[0,\infty)$ to $[0,1]$. How do I interpret a condition on $f(x)$ that the function $\frac{f(x)}{f'(x)}$ is also decreasing in $x$?
Is there a family of functions that I need to consider in this case?
On
I will assume that the function is $C^2$ and $f'(x)\neq 0$, also those are just thoughts and obervations I had when seeing your inequality. If a function is decreasing, you have $f'(x)\leq 0$. If you know that $f(x)/f'(x)$ is decreasing as well, you have:
$$
\frac{d}{dx}\frac{f(x)}{f'(x)}=1-\frac{f(x)f''(x)}{f'(x)^2} \leq 0
$$
Rearranging the equation gives you:
$$
f'(x)^2\leq {f(x)f''(x)}
$$
Integrating both sides over any bounded set $B \subset \mathbb{R}_+$ and using Cauchy-Schwarz leads to:
$$
\int_B f'(x)^2 \leq \int_B f(x)^2 \int_B f''(x)^2
$$
which is a nice interpolation inequality. Since $|f(x)|\leq 1$, you can expand a little further to get a Poincare-Type estimate:
$$
||f'(x)||_{L^2(B)}\leq C(B)||f''(x)||_{L^2(B)}
$$
where $C(B)$ is a depending only on $B$. Those results can be helpful if you are looking for a family of functions in the space of $BV$ (Bounded variation) und in a Sobolev Space(e.g. $W^{1,2}(B)$). Both of those spaces have really nice compact properties.
On another note, those types of questions often seem to involve the subdifferential, which statisfies the estimate:
$$
f'(x)(y-x)=\nabla f(x)(y-x)\leq f(y)-f(x)
$$
Convexity, monoticity and information about the derivative suggest me that you could try to find more results by using those methods.
Yes, we can interpret it with help of the (family of) logarithms
$$\frac{f(x)}{f'(x)} = \left(\frac{f'(x)}{f(x)}\right)^{-1}=\left(\frac{d\{\log(f(x))\}}{dx}\right)^{-1}$$
Which is due to the famous logarithmic derivative which you can derive with the chain rule if you want to.
The multiplicative inverse of something decreasing must be increasing.
So the derivative of the logarithm of the function is increasing.