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.

  • $\begingroup$ Prove that $x^{2}f''(x) >\epsilon$ for $x \geq M$ implies $-xf'(x)>\epsilon$ for $x>M$. $\endgroup$ Commented Nov 7, 2021 at 6:44
  • $\begingroup$ 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. $-xf'(x)>\epsilon$) $\endgroup$
    – Jong
    Commented Nov 7, 2021 at 6:58
  • $\begingroup$ Integrate from $x$ to $\infty$. $\endgroup$ Commented Nov 7, 2021 at 7:21

1 Answer 1


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.


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