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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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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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  • $\begingroup$ I understand the first inequality, but I don't see how you got the second one. $\endgroup$ – Galois Mar 12 '12 at 20:47
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    $\begingroup$ 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. $\endgroup$ – William DeMeo Mar 12 '12 at 20:59
  • $\begingroup$ Yes its clear thanks $\endgroup$ – Galois Mar 15 '12 at 5:42
  • $\begingroup$ Great answer :)! $\endgroup$ – MathMan Jan 28 '15 at 0:48
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For every $c>0$, there exists $R$ such that for $x>0, |\int_R^xf(x)dx|<c/2$ 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 $xf(x)\leq c/2+Rf(x)\leq c$.

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

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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 $xf(x) < x(\frac{1}{x^a})$

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