1
$\begingroup$

I'm trying to show that p.v.$\frac{1}{x}(\varphi):= \lim_{\epsilon \to 0^{+}}\int_{|x|> \epsilon} \frac{\varphi(x)}{x}dx, \varphi \in \mathcal{S}$ exists and defines a tempered distribution, where $\mathcal{S}$ denotes Schwartz space.

First step is I split the integral up:

p.v.$\frac{1}{x}(\varphi) = \int_{|x|<1}\frac{\varphi(x)-\varphi(0)}{x}dx + \int_{|x|>1} \frac{\varphi(x)}{x}dx$

Now I notice that the first integral on the right hand side is essentially the mean value theorem so is equal to some derivative of $\varphi$.

Question:

  • How do i deal with the second integral?
  • I know that $\int \frac{1}{x}dx$ is not integrable near the origin due to it have a singularity, so are my steps justified so far and if so why?
$\endgroup$
1
$\begingroup$

First question. \begin{align*} \int_{|x|>1}\dfrac{|\varphi(x)|}{x}dx&=\int_{|x|>1}\dfrac{|x\varphi(x)|}{x^{2}}dx\\ &\leq\sup_{x\in{\bf{R}}}|x\varphi(x)|\int_{|x|>1}\dfrac{1}{x^{2}}dx\\ &=2\sup_{x\in{\bf{R}}}|x\varphi(x)|. \end{align*} Second question. You should write $\lim_{\epsilon\rightarrow 0^{+}}\displaystyle\int_{\epsilon<|x|<1}\dfrac{\varphi(x)-\varphi(0)}{x}dx$. Now we see that \begin{align*} \chi_{\epsilon<|x|<1}\left|\dfrac{\varphi(x)-\varphi(0)}{x}\right|\leq\chi_{|x|<1}\|\varphi'\|_{L^{\infty}({\bf{R}})}, \end{align*} and $\chi_{|x|<1}\|\varphi'\|_{L^{\infty}({\bf{R}})}\in L^{1}({\bf{R}})$, so Lebesgue Dominated Convergence Theorem applies.

$\endgroup$
  • $\begingroup$ why is $\chi_{|x|<1}||\varphi'||_{L^{\infty}(\mathbb{R})} \in L^{1}(\mathbb{R}) $ $\endgroup$ – VBACODER Mar 26 '18 at 9:23
  • $\begingroup$ That $\|\varphi'\|_{L^{1}({\bf{R}})}$ is a constant, now $\displaystyle\int_{{\bf{R}}}\chi_{|x|<1}dx=v_{n}<\infty$, the volume of the unit ball. $\endgroup$ – user284331 Mar 26 '18 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.