= 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 which
$$
\boxed{\|u\|_{H^s}^2:=\int_{\mathbb R}(1+\xi^2)^s|\widehat u(\xi)|^2\,d\xi<\infty.}
$$
This 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 gives
$$
\langle\widehat c,\phi\rangle=c\int\widehat\phi(x)\,dx=2\pi c\phi(0),
\qquad \widehat c=2\pi c\delta_0.
$$
A 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 $\rho_\varepsilon(\xi)=\rho(\xi/\varepsilon)$. On a fixed <compact interval> containing their supports, <Cauchy-Schwarz inequality> gives
$$
|2\pi c|=\left|\int G\rho_\varepsilon\right|
\leq\|G\|_{L^2(K)}\sqrt\varepsilon\,\|\rho\|_2\longrightarrow0.
$$
Thus $c=0$. Every <Sobolev space> requires local square integrability of the <Fourier transform>, regardless of how negative $s$ is. Therefore
$$
\boxed{c\in H^s(\mathbb R)\text{ for some }s\in\mathbb R\iff c=0.}
$$
The zero <constant function> belongs to every <Sobolev space>.
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