$f : [0,\infty] \to \Bbb R $ is monotone and $\displaystyle∫^∞_0f(x)\,dx$ converges.

Note: we also proved before $\lim_{x→∞}f(x)=0$

Show that even $\lim_{x→∞}xf(x)=0$


  • 2
    $\begingroup$ It reminds me of the theorem on infinite series : If $a_{n}$ is a positive and decreasing sequence and $\sum a_{n}$ is convergent then $\lim_{n \to \infty}na_{n} = 0$. This can be used to show that $\sum (1/n)$ is divergent. $\endgroup$ – Paramanand Singh Nov 10 '13 at 9:34

Begin by noting that it's clear that $f$ is monotone decreasing.

Proceed by contrapositive: Suppose that there exists an $\epsilon > 0$ such that there are arbitrarily large $x$ for which

$$xf(x) > \epsilon$$

Without any loss of generality, $\epsilon = 2$. Choose a strictly increasing sequence $x_n$ converging to infinity such that the above inequality holds; in fact, we choose the sequence such that for every $n$,

$$x_n > 2x_{n - 1}$$

Now since $f$ is monotone decreasing, we have the estimate

$$\int_{x_{n - 1}}^{x_n}f \ge \int_{x_{n - 1}}^{x_n} f(x_n) = (x_n - x_{n - 1}) f(x_n)$$

But $x_n - x_{n - 1} > \frac{1}{2}x_n$ by construction, and

$$\int_{x_{n - 1}}^{x_n} \ge \frac{1}{2} x_n f(x_n) \ge 1$$

Now sum over $n$, and conclude that $f$ is not integrable.

  • $\begingroup$ why it is monotone decreasing? $\endgroup$ – SundayCat Nov 10 '13 at 8:19
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    $\begingroup$ @sundaycat It's monotone by assumption. Can it be increasing? $\endgroup$ – user61527 Nov 10 '13 at 8:20
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    $\begingroup$ @sundaycat No. If it's increasing and there is some $x_0$ for which $f(x_0) \ne 0$, then $$\int_0^{\infty} f(x) dx \ge \int_{x_0}^{\infty} f(x) dx \ge f(x_0) \int_{x_0}^{\infty} dx = \infty$$ $\endgroup$ – user61527 Nov 10 '13 at 8:27
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    $\begingroup$ $f$ could be negative, increasing, and still have a finite integral, though. $\endgroup$ – detnvvp Nov 10 '13 at 8:40
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    $\begingroup$ In line two of your answer, where you write "infinitely many $x$", I think you mean "arbitrarily large $x$". (Also, this is just a matter of taste, but: why take $\epsilon = 2$? Strictly speaking it is a loss of generality. What you mean is that the argument can be carried through the same way in the general case...but then why not just do so?) $\endgroup$ – Pete L. Clark Nov 12 '13 at 6:30

As JLA began in this answer, there exists some $M >0$ such that for $x>M$ either $f(x)\geq 0$ or $f(x)\leq 0.$ WLOG assume the former. Then by the convergence of $\displaystyle\int_{0}^{\infty}f(x)dx,$ we must have that $f$ is monotone decreasing for sufficiently large $x.$ Then for large enough $x,$ $f(x)$ is non-negative and monotone decreasing and hence by the integral test, $\displaystyle \sum_{n=1}^{\infty}f(n)$ converges. Let $K >0$ be such that for $x>K$ we have $f(x)>0$ and $f$ is monotone decreasing.

Since $\displaystyle \sum_{n=1}^{\infty}f(n)$ converges and $0\leq f(n+1)\leq f(n)$ for $n > K$, by the Abel-Olivier-Pringsheim condition (that is, the theorem that Paramanand Singh notes in his first comment), we have $\displaystyle \lim_{n \to \infty} nf(n)=0$

Claim: $\displaystyle \lim_{x \to \infty} xf(x) =0$

It suffices to show that if $(a_n)$ is any sequence such that $a_n \to \infty$ then $a_nf(a_n)\to 0.$ Suppose not. Then there exists $\varepsilon_{0}>0$ such that for all $n >K$ we have $a_nf(a_n)\geq \varepsilon_{0}.$

For each $m \in \mathbb{N}$ set $n_m = \lfloor a_{m}\rfloor$ so that we have $n_m \leq a_{m}<n_m +1.$ For $m >K$ we have $f(a_m) \leq f(n_m)$ (as $f$ is monotone decreasing) and $\varepsilon_{0} \leq a_{m}f(a_m)$.

Since $nf(n) \to 0$ and $f(n) \to 0$ there exists $r \in \mathbb{N}$ such that for all $n \geq r$ we have $|nf(n)+ f(n)|< \dfrac{\varepsilon_{0}}{2}$

For $m \geq \max\{r, K\}$ consider:

$\dfrac{\varepsilon_{0}}{a_m}\leq f(a_m)\leq f(n_m)$

$\Rightarrow \varepsilon_{0} < a_mf(n_m) < (n_m +1)f(n_m)< \dfrac{\varepsilon_{0}}{2},$

a contradiction.

Therefore we must have $a_nf(a_n) \to 0$ and since $(a_n)$ is arbitrary it follows that $\displaystyle \lim_{x \to \infty}xf(x)=0$


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