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, Hakim Jul 8 '14 at 22:11

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  • $\begingroup$ The negation of "$\lim_{x\rightarrow\infty} xf(x)=0$" is not "$\lim_{x\rightarrow\infty} xf(x)>0$". $\endgroup$ – David Mitra Jul 8 '14 at 15:28
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    $\begingroup$ Your proof looks OK. $\endgroup$ – TZakrevskiy Jul 8 '14 at 15:28
  • $\begingroup$ $x$ 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 $n\to\infty$. $\endgroup$ – InTransit Jul 8 '14 at 15:29
  • $\begingroup$ See this, though. $\endgroup$ – David Mitra Jul 8 '14 at 15:30
  • $\begingroup$ This argument is by contradiction, not contraposition. $\endgroup$ – Alex Schiff Jul 8 '14 at 15:37