3

I have to show that by defining $$\langle u, f\rangle=\lim_{\varepsilon\rightarrow 0}\int_{-\infty}^{-\varepsilon}+\int_{\varepsilon}^{\infty}\frac{f(x)}{x}dx$$

with $f\in\mathcal{D}(\mathbb{R})$, we have $u\in\mathcal{D}'(\mathbb{R})$ and find $u'$.

The tip is to use the mean value theorem. I did the second part with $u'$ but my problem is to show that it is well defined.

So I got $$\int_{-\infty}^{-\varepsilon}\frac{f(x)}{x}dx=\int_{\varepsilon}^{\infty}\frac{f(-x)}{x}dx$$

so

$$\int_{-\infty}^{-\varepsilon}+\int_{\varepsilon}^{\infty}\frac{f(x)}{x}dx=\int_{\varepsilon}^{\infty} \frac{f(x)+f(-x)}{x}dx$$

It would be great if I got $\frac{f(x)-f(-x)}{x}$ because then

$$\frac{f(x)-f(-x)}{x}=2\frac{f(x)-f(-x)}{x-(-x)}=2f'(y)$$

for some $y\in(-x,x)$. But now how can I apply the mean value theorem to show that it is well defined?

rom
  • 831
  • 1
    Sign error: $$\int_{-\infty}^{-\varepsilon}\frac{f(x)}{x}dx= -\int_{\varepsilon}^{\infty} \frac{f(-x)}{x}dx$$ Just a remark: another way to approach this construction is to begin with the distribution $\log|x|$ and take its derivative. –  Jan 15 '13 at 03:03
  • @rom As the integral domain and sup $|f'|$ are bounded you are done. – Vobo Jan 15 '13 at 07:01
  • @5PM with the change of variables $x=-y$ we have $dx=-dy$ so I think that I have the sign correctly. Initially I got the minus but then I couldn't find $u'$. – rom Jan 15 '13 at 12:08
  • @Vobo I don't quite see it, how do I make $f'$ to appear. – rom Jan 15 '13 at 12:10
  • @rom You also flipped the limits of integration; that changes the sign. –  Jan 15 '13 at 12:12
  • Oh! right! Then I got $u'$ wrongly. – rom Jan 15 '13 at 13:02
  • @rom Your integrand is bounded independently of $\varepsilon>0$ as $f'$ is, so what is your problem? – Vobo Jan 15 '13 at 21:19

0 Answers0