It seems from what I read that the Heaviside function is in $H^{s}(\mathbb{R})$ for all $s<\frac{1}{2}$. See for example To which Sobolev local space Dirac delta function belongs to?.
It seems it makes no sense for two reasons. The first is that is not localized in any way (that is not even squared-integrable on $\mathbb{R}$)
Furthermore, if we follow the procedure, one way to decide if a functions $f$ is in $H^s(\mathbb{R})$ is to check if $(1+\omega^2)^{s/2}\hat{f}(\omega)$ is in $L^2(\mathbb{R})$.
We have that $$\hat{H}(\omega)= \pi \delta(\omega)- \mathrm{PV}\left(\frac{i}{\omega}\right).$$ So it is not true that $(1+\omega^2)^{s/2}\hat{H}(\omega)$ is in $L^2(\mathbb{R})$ for all $s<1/2$ because of the Dirac Delta function.
It seems like people will check the condition on the signum function instead, as its Fourier transform is $$ \widehat{{\text{sign}}}(\omega)=-\mathrm{PV}\left(\frac{2i}{\omega}\right). $$ Thus it checks the condition if and only if $s<1/2$.
My question: what do people mean by saying that the Heaviside function is in $H^{s}(\mathbb{R})$ for all $s<1/2$? Maybe is that those spaces admit discontinuous functions? Indeed, $H^{1/2}(\mathbb{R})$ is a space where each function can be made continuous. So that would mean that this property is lost as soon as $s$ is less than 1/2? Is there a way to prove this?
