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I know $\lim_{x\rightarrow 0} x\log x＝０$ can be proved by using l'Hospital, but I heard that this statement can also be shown by using dominated convergence theorem. Hint says we use the function $f（t,x）＝1[x,1]（t）x/t$. Could you give me the process of applying DCT to this problem? Thank you in advance.

Bernard
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Pont
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2 Answers2

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Note that for $0<x<1$ $$|x\log x| = \left|\int_x^1\frac{x}{t} \, {\rm d}t\right| \leq \int_x^1\frac{x}{t^{3/2}} \, {\rm d}t = 2\sqrt{x}(1-\sqrt{x}) \, .$$

Diger
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  • How can I proceed from here? Is integral used here is lubesgue's or Riemann's ? Thank you! – Pont Jul 01 '20 at 03:29
  • I'm not sure what you mean "proceed from here". I think you are done. Just take the limit $x\rightarrow 0$. Even though I considered it as an ordinary Riemann-Integral, I presume it can be considered as either type of integral for fixed $x$. – Diger Jul 01 '20 at 10:54
  • Thank you I understood.. Where did you use lubesgue dominated convergence theorem? – Pont Jul 01 '20 at 11:09
  • There is only 1 inequality. – Diger Jul 01 '20 at 11:20
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$\int_0^{1} I_{(x,1)} (t) \frac x tdt \to \int_0^{1} dt=0$ by DCT because $I_{(x,1)} (t) \frac x t \to 0$ as $x \to 0+$ for every $t$ and $0 \leq I_{(x,1)} (t) \frac x t \leq 1$. The constant function $1$ is integrable on $[0,1]$ so DCT is applicable. Of course, $\int_0^{1} I_{(x,1)} (t) \frac x tdt = -x \ln x$ so we get $x \ln x \to 0$ as $x \to 0+$

Kavi Rama Murthy
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  • Thank you ! Sorry,but what is a sequence of measurable function here?What is a character parametrizing the sequence? x? – Pont Jun 28 '20 at 13:41
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    Proving that $x \ln x \to 0$ is equivalent to proving that $x_n \ln x_n \to 0$ for every sequence $x_n \to 0$. So the sequence of functions can be taken as $f_n(t)=I_{(x_n,1)} (t) \frac {x_n} t$. @bellow – Kavi Rama Murthy Jun 28 '20 at 23:13
  • Do you say given limit is equivalent to lim[n→∞]∮[0,1]（-f（t,xn）dt ? – Pont Jun 29 '20 at 13:26
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    @bellow Yes, that is correct. – Kavi Rama Murthy Jun 29 '20 at 23:15
  • Thank you! Why we us indicator function? What is the merit（or necessarity） to use indicator function ?And, in your answer, why ’of course' ∮I（x,1）（t）x/tdt＝-xinx? – Pont Jun 30 '20 at 03:27
  • Sorry,but is ∮ here a lubesgue ones? or Riemann one? The same thing?thank you – Pont Jul 01 '20 at 03:28
  • On $[x,1]$ we are integrating a continuous function. The function is both Riemann integrable and Lebesgue integrable and its Riemann integral is equal to its Lebesgue integral. – Kavi Rama Murthy Jul 01 '20 at 05:11
  • we integrate on [0,1], how can I explain the value integrated on {0,1] is same as on [xn,1]? – Pont Jul 01 '20 at 11:51
  • All integrals are actually over $(0,1)$ but f or fixed $x$ the function we are integrating is $0$ for $t <x$ so it is enough to integrate from $x$ to $1$. @bellow – Kavi Rama Murthy Jul 01 '20 at 12:04