Intersection of all Sobolev spaces with negative order

Question

Call $H^{s}$ the usual $L^2$-based Sobolev spaces on, say, a closed manifold, for $s \in \mathbb R$. The intersection $\bigcap _{s<0} H^s $ contains $L^2$. Is this intersection equal to $L^2$?

Thank you.

Answer 1

No. Consider $f \in \cap_{s < 0}H^s(\mathbb{R})$ defined by $\hat f(t) = (1+t^2)^{-1/4}$. This function is not in $L^2$.