 Solution

For real $s$, the Sobolev space $H^s(\mathbb{R})$ consists of those tempered distributions $u$ whose Fourier transform is represented by a locally square-integrable function and for whichThis is the Sobolev norm used here; inserting the fixed factor $(2\pi {)}^{-1}$ gives an equivalent convention.

For the constant function $c$, Fourier inversion at zero givesA nonzero Dirac delta distribution cannot be represented by a locally square-integrable function. Indeed, if $G$ represented $2\pi c{\delta }_0$, choose a test function $\rho$ with $\rho (0)=1$, and put ${\rho }_{\epsilon }(\xi )=\rho (\xi /\epsilon )$. On a fixed compact interval containing their supports, Cauchy-Schwarz inequality givesThus $c=0$. Every Sobolev space requires local square integrability of the Fourier transform, regardless of how negative $s$ is. ThereforeThe zero constant function belongs to every Sobolev space.