2
$\begingroup$

$\DeclareMathOperator{\vp}{v.p.}$ We define $\vp \frac 1x \in \mathcal D'(\mathbb R)$ (the principal value of $\frac 1x$) as

$$\left\langle \vp \frac 1x, \varphi \right \rangle = \lim_{\varepsilon \to 0_+} \int_{|x| > \varepsilon} \frac{\varphi(x)}{x} \ dx = \vp \int_\mathbb R \frac{\varphi(x)}x \ \ \forall \varphi \in \mathcal D(\mathbb R)$$

Now I am required to show that $\displaystyle \frac{d}{dx}\ln |x| = \vp \frac 1x$

This is the solution, which I do not understand;

As $\ln |x| \in L^1_{loc} (\mathbb R) \subset \mathcal D'(\mathbb R)$, for all $\varphi \in \mathcal D(\mathbb R)$ we have

$$\langle d / dx \ln |x|, \varphi \rangle = - \langle \ln |x|, \varphi '\rangle = - \int_\mathbb R \ln|x|\varphi'(x)\ dx =$$$$= - \lim_{\varepsilon \to 0_+, R \to \infty} \int_{-R}^{-\varepsilon} \ln|x| \varphi'(x) \ dx + \int_{\varepsilon}^R \ln|x| \varphi'(x) \ dx = \vp \int_\mathbb R \frac{\varphi(x)}x$$$$= \langle \vp \frac 1x, \varphi \rangle$$

And the only comment to the above calculations is : We have made an integration by parts, remembering that $\text{spt }\varphi$ is compact and $\lim_{\varepsilon \to 0} (\varphi(\varepsilon) - \varphi(-\varepsilon))\ln \varepsilon = 0$

Can somebody explain to me what is going on here? I do not understand this solution nor the comment :-/

$\endgroup$
4
$\begingroup$

First of all note that $\ln |x|$ is integrable near $0$ so that for any $\phi \in \mathcal{D}'$ the integral $$\int_{\mathbb R} \ln |x| \phi'(x) \, dx$$ is defined, and that $$\int_{\mathbb R} \ln |x| \phi'(x) \, dx = \lim_{\epsilon \to 0^+} \int_{|x| \ge \epsilon} \ln|x| \phi'(x) \, dx.$$ Now, $$\int_{|x| \ge \epsilon} \ln|x| \phi'(x) \, dx = \int_{-\infty}^{-\epsilon} \ln(-x) \phi'(x) \, dx + \int_{\epsilon}^\infty \ln x \phi'(x) \, dx$$ so that (taking into account that $\phi$ is compactly supported means the limits at infinity do not contribute to the boundary terms) an integration by parts on each integral separately gives \begin{align*}\int_{|x| \ge \epsilon} \ln|x| \phi'(x) \, dx &= \ln(\epsilon) \phi(-\epsilon) - \int_{-\infty}^{-\epsilon} \frac 1x \phi(x) \, dx - \ln(\epsilon) \phi(\epsilon) - \int_\epsilon^\infty \frac 1x \phi(x) \, dx\\ &= -\ln(\epsilon) [\phi(\epsilon) - \phi(-\epsilon)] - \int_{|x| \ge \epsilon} \frac 1x \phi(x) \, dx.\end{align*}

As for the first term, note that $$ \lim_{\epsilon \to 0^+} \ln(\epsilon) [\phi(\epsilon) - \phi(-\epsilon)] = \lim_{\epsilon \to 0^+} 2\epsilon \ln(\epsilon) \cdot \frac{\phi(\epsilon) - \phi(-\epsilon)}{2\epsilon} = 0 \cdot \phi'(0) = 0.$$ Thus $$ \lim_{\epsilon \to 0^+} \int_{|x| \ge \epsilon} \ln|x| \phi'(x) \, dx = - \lim_{\epsilon \to 0^+} \int_{|x| \ge \epsilon} \frac 1x \phi(x) \, dx.$$ That is, $$\int_{\mathbb R} \ln |x| \phi'(x) \, dx = - \lim_{\epsilon \to 0^+} \int_{|x| \ge \epsilon} \frac 1x \phi(x) \, dx.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ well thank you very much! It seems all clear. In case I'll have any doubts I'll ask :) $\endgroup$ – Ant Jan 29 '15 at 16:49

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.