Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 11 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 $\{x_i\}$.

share|cite|improve this answer
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 '15 at 0:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.