TITLE: What is an exact characterization for the functions $f$ such that $xf'(x) \leq 2f(x)$?
QUESTION [5 upvotes]: What is an exact characterization for the functions $f$ such that $xf'(x) \leq 2f(x)$?
I know, for instance, that the inequality holds for all functions $f(x) = c_0 + c_1x + c_2x^2$, with $c_0, c_1, c_2 \geq 0$. But it does not hold for $f(x) = x^2 - x$ or $f(x) = e^x$. I'm interested in the broadest possible class of functions that it holds for.
Note 1: For my application, we can also assume that $f(x), f'(x), f''(x) \geq 0$ for all $x \geq 0$ (i.e., it is positive, non-decreasing, and convex over $[0, \infty)$). 
Note 2: I am interested in a slightly more general solution, $xf'(x) \leq bf(x)$, where $b \geq 1$ is a constant. But I should be able to generalize any solution for the $b = 2$ case.

REPLY [4 votes]: I'm not sure what you mean by "characterization".  For $x>0$, this should hold whenever $x^{-2} f(x)$ is a nonincreasing function, i.e. $f(x) = x^2 g(x)$ for any nonincreasing differentiable function $g$.
To derive this, write the given inequality as
$$
x\,f'(x) - 2\,f(x) \;\leq\; 0
$$
Assuming $x> 0$, we can multiply through by the integrating factor $x^{-3}$ to get
$$
x^{-2}f'(x) \,-\, 2x^{-3}f(x) \;\leq\; 0,
$$
which can be written
$$
\frac{d}{dx}\biggl[x^{-2} f(x) \biggr] \;\leq\; 0.
$$

REPLY [2 votes]: For $x > 0$, the condition is equivalent to $\frac{d}{dx} \frac{f(x)}{x^2} \le 0$ while for $x = 0$, it simply says $f(0) \ge 0$. Therefore these are the differentiable functions on $[0,\infty)$ that satisfy 
$$f(0) \ge 0, \quad x \mapsto x^{-2}f(x) \quad \text{ is non-increasing.}
$$
The generalization to the case $xf'(x) \le b f(x)$ is now obvious.

REPLY [2 votes]: The inequality yields (Let's say you have $b$ instead of $2$)
For $x>0$,  $xf'(x)-bf(x)\leq 0\Rightarrow x^bf'(x)-bx^{b-1}f(x)\leq 0$  
$\Rightarrow \frac{x^bf'(x)-bx^{b-1}f(x)}{x^{2b}}\leq 0 \Rightarrow (\frac{f(x)}{x^b})'\leq 0 $
This means that the function $g(x)=\frac{f(x)}{x^b}$ is strictly decreasing.
I think this is what you want.