# $f\in L^1(0,\infty)$ monotone, show $\lim_{x\rightarrow \infty} xf(x) = 0$ [duplicate]

Here is the solution:

First $f$ is monotone and integrable on $(0,\infty)$, wolg we can assume that $f>0$ and approaches $0$ as $x$ goes to infinity. Observe that $$xf(2x) \leq \int_x^{2x} f(t)$$ since when $t\in [x,2x]$ we have $f(t) \geq f(2x)$ because $f$ is decreasing. And since $f$ is integrable, we have $$\lim_{x\rightarrow \infty} \int_x^{2x} f(t) = 0,$$ thus $$\lim_{x\rightarrow \infty} 2xf(2x) = 0.$$

## marked as duplicate by user147263, Daniel Fischer, Norbert, Peter Woolfitt, HakimJul 8 '14 at 22:11

• The negation of "$\lim_{x\rightarrow\infty} xf(x)=0$" is not "$\lim_{x\rightarrow\infty} xf(x)>0$". – David Mitra Jul 8 '14 at 15:28
• Your proof looks OK. – TZakrevskiy Jul 8 '14 at 15:28
• $x$ is monotone increasing and $f$ will have to be monotone decreasing. The product $xf(x)$ needn't be monotone. You can't claim that for all $x\ge x_0$, $xf(x) \ge\epsilon$. One can say at best that there's a sequence $\{x_n\}$ such that $x_nf(x_n) \ge\epsilon$ as $n\to\infty$. – InTransit Jul 8 '14 at 15:29
• See this, though. – David Mitra Jul 8 '14 at 15:30
• This argument is by contradiction, not contraposition. – Alex Schiff Jul 8 '14 at 15:37