Function satisfying (sort of) differential inequality: $|f'(x) x| \le Cf(x)$ During my research I stumbled upon some inequality which I tried to tackle using cutoff functions. Ideally, I would need a $C^1$ (at least) function $f:[\delta, 2\delta] \rightarrow [0,1]$, with $\delta>0$, such that $f(\delta)=1, f(2\delta)=0$ and, most importantly,
$$
|f'(x) x| \le Cf(x) \, \, \forall x\in(\delta, 2\delta), C>0.
$$
It would also be nice if the function was decreasing (since we could use $f'(x)\le 0$ and simplify the inequality) , but this is not required.
Nevertheless, I somehow struggle to find an explicit function satisfying those conditions, mostly due to the fact that $|f(x)| \rightarrow 0$ as $x\rightarrow 2\delta$, implying $f'(x) \rightarrow 0$ as well for $x\rightarrow 2\delta$. All examples I had in mind seem to fail because (for them):
$$
\lim\limits_{x \rightarrow 2\delta}\frac{|f'(x)x|}{|f(x)|} = \infty.
$$
This led me to believe that maybe such functions cannot exist. Then again, I can't see why that would be true. The question is then to either

*

*Provide an example of such function, or a hint towards such example;

*Prove that such function cannot exist (or, once again, hint towards a proof).

 A: Such a function does not exist, and it suffices to assume that $f$ is differentiable, the continuity of the derivative is not needed.
If $f$ satisfies $|f'(x) x| \le Cf(x)$ on some interval $[a, b] \subset (0, \infty)$ then
$$
 \left( x^C f(x)\right)' = C x^{C-1}f(x) + x^C f'(x) \ge C x^{C-1}f(x) - x^C C \frac{f(x)}{x} = 0 \, , 
$$
so that $x \mapsto  x^C f(x)$ is non-decreasing.
It follows that if $f(a) > 0$ then $f(x) > 0$ on the entire interval $[a, b]$.
In particular, $f(\delta)=1$ and $f(2\delta)=0$ is not possible.

How did I c0me up with this? If we first assume that $f(x) > 0$ on some interval $[\delta, b]$ then the given inequality can be written as
$$
 -\frac Cx \le \frac{f'(x)}{f(x)} \le \frac Cx
$$
and integration from $\delta $ to $x$ gives
$$
 -C (\log x - \log \delta) \le \log(f(x)) \le C (\log x - \log \delta)
$$
or
$$
 \left(\frac{\delta}{x}\right)^C \le f(x) \le \left(\frac{x}{\delta}\right)^C \, .
$$
The left inequality shows that $f$ is bounded below by some strictly positive function. Writing the lower bound as
$$
 x^C f(x) \ge \delta^C
$$
suggest to consider $ x^C f(x)$ in the first place, and that makes the initial assumption $f(x) > 0$ obsolete.
