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Let $f:\mathbb{R}_+ \to \mathbb{R}_+$ be a monotone decreasing function defined on the positive real numbers with $$\int_0^\infty f(x)dx <\infty.$$ Show that $$\lim_{x\to\infty} xf(x)=0.$$

This is my proof: Suppose not. Then there is $\varepsilon$ such that for any $M>0$ there exists $x\geq M$ such that $xf(x)\geq \varepsilon$. So we can construct a sequence $(x_n)$ such that $x_n \to \infty $ and $x_n f(x_n ) \geq \varepsilon$. So $$\frac{\varepsilon}{x_n}\leq f(x_n) \implies \sum_{n\in\mathbb{N}}\frac{\varepsilon}{x_n} \leq \sum_{n\in\mathbb{N}} f(x_n) \leq \int_0^1 f(x)dx.$$ So we get a contradiction. I feal like I have the correct idea but some details are wrong. Any help would be appreciated.

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This might be relevant. – Nick Strehlke Mar 12 '12 at 20:15
Also interesting is the series version:… – dls Mar 12 '12 at 20:59

1 Answer 1

up vote 9 down vote accepted

Notice that, since $f$ is monotone decreasing, you have for each $x$,

$$0\leq f(x) (x - \frac{x}{2}) \leq \int_{\frac{x}{2}}^{x} f(t) \, dt$$


$$0\leq xf(x) \leq 2\int_{\frac{x}{2}}^{x} f(t) \, dt$$

The right hand side goes to zero since the integral converges.

Added: You should convince yourself that the last sentence is true. You could do this by writing the integral as a sum of terms of the form $\int_{x_i/2}^{x_i} f(t) \,dt$, for an appropriate sequence $\{x_i\}$.

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I understand the first inequality, but I don't see how you got the second one. – Galois Mar 12 '12 at 20:47
The very first inequality is $0\leq f(x)(x−x/2)$. I assume that's the one you understand. The second inequality holds because the minimum value $f$ takes on the interval $[\frac{x}{2},x]$ is $f(x)$, since it's monotonically decreasing. Let me know if it's still not clear. – William DeMeo Mar 12 '12 at 20:59
Yes its clear thanks – Galois Mar 15 '12 at 5:42
Great answer :)! – Wanderer Jan 28 at 0:48

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