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.
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 $\widehat{H}$.