Sufficient conditions to have $f' = O(f(x)/x)$.

Question

Suppose $f$ a nonnegative real-valued function, non-decreasing, $O(x^m)$ for some $m \in \mathbb{Z}_{\geqslant 0}$ and $C^1$, with $f'$ being monotonic and nonnegative. Are this sufficient conditions to have $f'(x) = O(f(x)/x)$ as $x\to +\infty$?

I've took these hypotheses from known cases, like $\sqrt[3]{x}$, $\sqrt{x}\log{x}$, $x/{\log{x}}$, etc., but I was unable to prove this in general, not even assuming $f = O(x)$ and $f'$ bounded.

Answer 1

One such condition is that $f$ be continuous and differentiable everywhere and that $f'$ be locally integrable and have regular variation at infinity.

Let $\alpha \in \mathbb{R}$ A function $g$ is said to have regular variation of index $\alpha$ at infinity if for all $y > 0$,

$$\lim_{x \to + \infty} \frac{g(xy)}{g(x)} = y^\alpha.$$

It is said to have regular variation (at infinity) if it has regular variation of index $\alpha$ for some $\alpha \in \mathbb{R}$. Functions with regular variation of index $0$ are also called functions with slow variation.

For instance, rational functions have regular variation (of integer index); logarithms have slow variation ; $x \mapsto x^\alpha$ has regular variation of index $\alpha$; products of functions with regular variation have regular variation...

Then the result follows from Karamata's theorem.

Note that regular variation often include a measurability condition, which is always true for a derivative. I think we can do without the "locally integrable" condition here, but I don't have the Bingham-Goldie-Teugels with me to check.

Answer 2

[OP changed the question since this answer was posted]

Have you tried $f(x)=e^x$? It's a smooth nonnegative nondecreasing function with $f'(x)=e^x$ which is monotonic and nonnegative, but $$e^x\ne O(e^x/x)$$