1
$\begingroup$

I have tried to use the Lebesgue Dominated Convergence Theorem to evaluate: $$\lim_{n\rightarrow \infty} \int_{(0,1]} f_n \;d\mu $$ with $f_n(x)=\dfrac{n\sqrt{x}}{1+n^2x^2}$ and $f_n(x)=\dfrac{n\;x\log(x)}{1+n^2x^2}$.

The problem is I can't find the function $g$ such that $\lvert f_n(x) \rvert \leq g(x)$ with $g\in L^1$. $g(x)=\dfrac{\lvert \log(x) \rvert}{x}$ and $g(x)=\dfrac{1}{x^{3/2}}$ is the best I got and these functions are not Lebesgue integrable.

$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

For $|f_n(x)| = \left|\dfrac{nx \log (x)}{1+n^2x^2}\right| \leq 2|\log x| = -2\log x, x \in (0,1]$. Take $g(x) = - 2 \log x$.

For $|f_n(x)| = \left|\dfrac{n\sqrt{x}}{1+n^2x^2}\right| = \left|\dfrac{nx}{1+n^2x^2}\right|\cdot \dfrac{1}{\sqrt{x}} \leq \dfrac{1}{2\sqrt{x}}$. Take $g(x) = \dfrac{1}{2\sqrt{x}}$.

$\endgroup$
3
  • $\begingroup$ I think it requires some justification that $\left|\frac{nx}{1+n^2x^2}\right|\le\frac12$. $\endgroup$
    – Math1000
    Dec 26, 2014 at 18:51
  • 3
    $\begingroup$ $1+n^2x^2 -2nx = (1-nx)^2 \geq 0 \to 1+n^2x^2 \geq 2nx$. $\endgroup$
    – DeepSea
    Dec 26, 2014 at 18:52
  • $\begingroup$ Very nice argument. I was trying to find the maximum of $\frac{nx}{1+n^2x^2}$ on $(0,1]$ and it got really ugly. $\endgroup$
    – Math1000
    Dec 26, 2014 at 19:01
0
$\begingroup$

For $|f_{n}(x)|=|\dfrac{nx\log(x)}{1+n^{2}x^{2}}|\leq |\dfrac{\log(x)}{2}|\leq |\log(x)|$. So for $g(x) = \log(x)$. Is that right??

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .