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

Asked Jul 08 '14 at 15:18

Active Jul 08 '14 at 20:22

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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 $$\lim_{x\rightarrow \infty} 2xf(2x) = 0.$$

edited Jul 08 '14 at 20:22

asked Jul 08 '14 at 15:18

Xiao

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The negation of "$\lim_{x\rightarrow\infty} xf(x)=0$" is not "$\lim_{x\rightarrow\infty} xf(x)>0$". – David Mitra Jul 08 '14 at 15:28
1

Your proof looks OK. – TZakrevskiy Jul 08 '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 08 '14 at 15:29
See this, though. – David Mitra Jul 08 '14 at 15:30
This argument is by contradiction, not contraposition. – Alex Schiff Jul 08 '14 at 15:37
Thank you guys for the reply! I remembered how I worked on this problem a few months ago. The result comes directly from the fact that $$xf(2x) \leq \int_x^{2x} f(t).$$ Now I will try to fix the hole in my contradiction proof. – Xiao Jul 08 '14 at 15:56
could u please show the fact in detail... – Toeplitz Jul 08 '14 at 19:06

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

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