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I have found that Dirac delta function $\delta (x)\in H^{s}(\mathbb{R}), \forall s<-\frac{1}{2}$, and Heaviside function $\in H^{s}(\mathbb{R}) , \forall s<\frac{1}{2}$;

Also i want to use the following formula for Cauchy principal value $p.v\left(\frac{1}{x}\right)$: $\hat{H}(\xi)=\frac{1}{2}\left ( \delta(\xi)-\frac{i}{\pi}p.v\left(\frac{1}{\xi}\right) \right )$ in order to find to which $H^{s}(\mathbb{R})$ does $p.v\left(\frac{1}{x}\right)$ belong?

How can I derive it using this formula? Or what is another way to find to which $H^{s}(\mathbb{R})$ does $p.v\left(\frac{1}{x}\right)$ belong?

Thanks a lot.

Ahaan S. Rungta
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Lilly
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1 Answers1

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You can just compute the Fourier transform of $p.v.\ 1/x$ directly: $$\int_{\epsilon<|x|<\epsilon^{-1}} \frac{1}{x} e^{-ikx} = -\pi \operatorname{sign} k \tag1$$ Indeed, the contribution of $\cos kx$ is zero, and the improper integral of $\frac{1}{x}\sin kx$ is a well-known one.

The function $(1+k^2)^{s/2}(-\pi \operatorname{sign} k )$ is in $L^2(\mathbb R)$ if and only if $s<-1/2$. Hence, $p.v.\ 1/x$ is in $H^s$ if and only if $s<-1/2$.

Using the relation with Heaviside function $H$ amounts to something similar (after all, (1) is very close to a Heaviside function), but there is an added complication because $H$ does not have zero mean. This is why you get a delta function in $\widehat{H}$.

user90090
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  • Thank you very much. And how can I find to which $H_{loc}^{s} p.v.(\frac{1}{x})$ belongs ? – Lilly Aug 21 '13 at 08:00
  • @Lilly Same values of $s$. The membership in $H^s_{\rm loc}$ is determined by integrability of $|k|^s \hat f(k)$ in a neighborhood of infinity (because the part of Fourier transform supported on a bounded interval corresponds to a real analytic function). – user90090 Aug 21 '13 at 17:01
  • Thanks a lot!If I understood you right in order to check to which $H_{loc}^{s}$ function $f$ belongs I need to check if $\int \left |k \right |^{s}\hat{f}(k)dk< \infty$? But in this case we get $-\pi \int \left |k \right |^{s} sign(k) dk$ and this integral converges for $s<-1$, and not for$s<-\frac{1}{2}$. Can you tell me what is my mistake, please? – Lilly Aug 21 '13 at 20:05