Limit property of second derivative of bounded function.

Suppose $$f:\mathbb R \rightarrow \mathbb R$$ is twice continuously differentiable, bounded and monotone, and $$\lim_{x\rightarrow \infty} xf'(x)= 0$$. I am trying to show this implies $$\lim\inf_{x\rightarrow \infty} x^{2}f''(x)\leq 0$$ or come up with a counterexample.

• Prove that $x^{2}f''(x) >\epsilon$ for $x \geq M$ implies $-xf'(x)>\epsilon$ for $x>M$. Commented Nov 7, 2021 at 6:44
• Could you provide a hint of how to do this? I've tried to prove the original statement by showing that $x^2f''(x)>\epsilon$ for $x>M$ implies $f'(x) > -\epsilon/x + \epsilon/M + f'(M)$ but am not sure how to make the jump to the statement you suggest (i.e. $-xf'(x)>\epsilon$)
– Jong
Commented Nov 7, 2021 at 6:58
• Integrate from $x$ to $\infty$. Commented Nov 7, 2021 at 7:21

As it was suggested in a previous comment, assume by contradiction that $$t^2f(t)\geq\varepsilon$$ for all $$t\geq M$$. Divide by $$t^2$$ the inequality, integrate between $$y$$ and $$x$$ (both bigger than $$M$$) and multiply everything by $$y$$ so that you obtain
$$yf^\prime(y)\geq y\biggl(\frac{xf^\prime(x)+\varepsilon}{x}\biggr)-\varepsilon.$$
Choose $$x$$ so big (and $$y$$ bigger) that $$|f^\prime(x)x|\leq\varepsilon/2$$ and in this way you get a contradiction.