3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ prolly not..... $\endgroup$ – mathworker21 May 31 at 13:03
1
$\begingroup$

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 $L^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.