# If monotone decreasing and $\int_0^\infty f(x)dx <\infty$ then $\lim_{x\to\infty} xf(x)=0.$

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.

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 $\{x_i\}$.

• 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 :)! – MathMan Jan 28 '15 at 0:48

For every $$c>0$$, there exists $$R$$ such that for $$x>0, |\int_R^xf(x)dx| and there exists $$R'$$ such that $$x>R'$$ implies that $$f(x)<{c\over {2R}}$$. Let $$x>sup(R,R')$$, we have $$c/2\geq \int_R^xf(x)dx\geq (x-R)f(x)$$ since $$f$$ decreases and is positive. We deduce that $$xf(x)\leq c/2+Rf(x)\leq c$$.

For all $$x>0$$, in the interval $$[\frac{1}{2}x ; x]$$ the minimum of $$f$$ is $$f(x)$$, since $$f$$ is decreasing. Thus, $$\frac{1}{2} x f(x) = \int_{\frac{1}{2}x}^{x} f(x) \mathrm dt \le \int_{\frac{1}{2}x}^{x} f(t) \mathrm dt \le \int_{\frac{1}{2}x}^{+ \infty} f(t) \mathrm dt \to 0$$ as $$x \to + \infty$$. Thus, by comparison, $$\lim_{x \to + \infty} \frac{1}{2}x f(x) = 0$$ which is equivalent to $$\lim_{x \to + \infty} x f(x) = 0$$

You can solve this using BCT. First, note that for the integral to exist, we must have

$$\lim_{x \rightarrow \infty} f(x) = 0$$

Now we know that the original limit is of indeterminate form. Now from here, since the improper integral converges, we know for sure that $$f(x) < \frac{1}{x^a}$$ where $$a > 1$$. This is by BCT since the improper integral of $$\frac{1}{x}$$ doesn't converge. Thus we know that $$xf(x) < x(\frac{1}{x^a})$$