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Can we define the Cauchy principal value of $\frac{1}{|x|}$? If yes, how does it act on a smooth function with a compact support? Is it defined as $$\text{p.v.} \frac{1}{|x|}(\psi) = \lim_{\epsilon \to 0+}\int_{\mathbb{R}\setminus[-\epsilon, \epsilon]}\frac{\psi(x)}{|x|}dx $$ for $\psi(x) \in C^\infty_0(\mathbb{R})$? It seems the above integral does not converge for $\psi \in C^\infty_0(\mathbb{R})$ if $\psi(0) \neq 0$. Do we have other ways to make $\frac{1}{|x|}$ a distribution on $C_0^\infty(\mathbb{R})$?

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3 Answers 3

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I wouldn't call it principal value but rather finite part. It can be defined as the distributional derivative of $\operatorname{sign}(x)\ln|x|$: $$ \langle \operatorname{fp}\frac{1}{|x|}, \psi(x) \rangle := \langle (\operatorname{sign}(x)\ln|x|)', \psi(x) \rangle = - \langle \operatorname{sign}(x)\ln|x|, \psi'(x) \rangle \\ = -\lim_{\epsilon\to 0} \int_{|x|>\epsilon} \operatorname{sign}(x) \, \ln|x| \, \psi'(x) \, dx \\ = \lim_{\epsilon\to 0} \left( \int_{-\infty}^{-\epsilon} \ln|x| \, \psi'(x) \, dx - \int_{\epsilon}^{\infty} \ln|x| \, \psi'(x) \, dx \right) \\ = \lim_{\epsilon\to 0} \left( \left[ \ln|x| \, \psi(x) \right]_{-\infty}^{-\epsilon} - \int_{-\infty}^{-\epsilon} \frac{1}{x} \, \psi(x) \, dx - \left[ \ln|x| \, \psi(x) \right]_{\epsilon}^{\infty} + \int_{\epsilon}^{\infty} \frac{1}{x} \, \psi(x) \, dx \right) \\ = \lim_{\epsilon\to 0} \left( 2 \psi(0) \ln|\epsilon| + \int_{|x|>\epsilon} \frac{1}{|x|} \, \psi(x) \, dx \right) . $$

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  • $\begingroup$ This is one possible distribution. Another would be $$\int_{-\infty}^\infty \frac{\psi(x)-\psi(0)\xi_{[-1,1]}(x)}{|x|}\,dx=\int_{|x|\le 1}\frac{\psi(x)-\psi(0)}{|x|}\,dx+\int_{|x|\ge 1}\frac{\psi(x)}{|x|}\,dx$$which is commonly used. $\endgroup$
    – Mark Viola
    Commented Jun 16, 2023 at 17:28
  • $\begingroup$ @MarkViola. I guess that they only differ by some multiple of $\delta.$ $\endgroup$
    – md2perpe
    Commented Jun 16, 2023 at 18:28
  • $\begingroup$ @MarkViola. Actually I find that your definition gives exactly the same distribution as mine. $\endgroup$
    – md2perpe
    Commented Jun 16, 2023 at 19:12
  • $\begingroup$ Hi. That is most curious. Which distribution? The one with $r=1$? $\endgroup$
    – Mark Viola
    Commented Jun 16, 2023 at 19:14
  • $\begingroup$ Ah. Yes. The one with $r=1$ $\endgroup$
    – Mark Viola
    Commented Jun 16, 2023 at 19:17
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In Exercise $13$ of THIS, Terrance Tao asks the reader to show that a distributional interpretation of $1/|x|$ can be defined for any $r>0$ as the functional $\lambda_r$

$$\langle \phi, \lambda_r \rangle=\int_{|x|\le r}\frac{\psi(x)-\psi(0)}{|x|}\,dx+\int_{|x|\ge r}\frac{\psi(x)}{|x|}\,dx$$

It is interesting to note that the difference between two such functionals differs by a constant time the Dirac Delta distribution. To see this note that

$$\begin{align} \langle \psi, \lambda_r\rangle -\langle \psi, \lambda_s\rangle&=\phi(0)\int_{-\infty}^\infty \frac{\xi_{[-s,s]}(x)-\xi_{[-r,r]}(x)}{|x|}\,dx\\\\ &=2\phi(0)\log(s/r) \end{align}$$

from which we see that in distribution we have

$$\lambda_r-\lambda_s=2\log(s/r)\delta(x)$$

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Depending what properties we need the "regularization/extension" to have, the answer to the question can be "yes" or "no". Yes, as in the other answers, we can write formulas that extend integration-against-$1/|x|$ to all test functions (or Schwartz functions). On the other hand, if we require that the extension still be even, and homogeneous of the same degree (however we define it) as integration-against-$1/|x|$, then this is provably impossible.

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