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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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    Also interesting is the series version: http://math.stackexchange.com/questions/4603/series-converges-implies-limn-a-n-02012-03-12

4 Answers 4

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

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

Therefore,

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

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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    Great answer :)!2015-01-28