For which negative values of $s$ is $\delta\in W^{2,s}(\mathbb{R}^{n})$?

Question

Recall that $$\delta\in W^{2,s}(\mathbb{R}^{n})\iff\int_{\mathbb{R}^{n}}(1+|\xi|^{2})^{s}|\langle\hat{\delta},\varphi\rangle|^{2}d\xi<\infty,\qquad\varphi\in\mathcal{S}(\mathbb{R}^{n}).$$

We have

$$\begin{aligned} \int_{\mathbb{R}^{n}}(1+|\xi|^{2})^{s}|\langle\hat{\delta},\varphi\rangle|^{2}d\xi&=\int_{\mathbb{R}^{n}}(1+|\xi|^{2})^{s}|\langle\delta,\hat{\varphi}\rangle|^{2}d\xi \\ &=\int_{\mathbb{R}^{n}}(1+|\xi|^{2})^{s}|\hat{\varphi}(0)|^{2}d\xi \\ &\le\int_{\mathbb{R}^{n}}(1+|\xi|^{2})^{s}d\xi \end{aligned}$$

Then we are just integrating a polynomial here, which is convergent for arbitrarily many negative $s$, no?

Answer 1

The integral $\int_{\mathbb{R}^{n}}(1+|\xi|^{2})^{s}d\xi$ converges if and only if $2s<-n$, that is $s<-n/2$. This determines the range of $s$ for which the delta-function is in the Sobolev space of order $s$.

As a notational remark: $W^{s,2}$ is the more common way of ordering the exponents: order of derivative first, exponent of integrability second (compare to $C^{k,\alpha}$ where the order of derivative also comes first). The space is also often denoted by $H^s$.