3
$\begingroup$

Consider the fractional Sobolev space $$ W^{k,2}(\mathbb R^n):=\big\{f \in \mathcal S'\,\big|\,(1+\|\xi\|)^k\hat f(\xi)\in L^2(\mathbb R^n)\big\} $$ for some $k\in\mathbb R$, and let $\mathcal M$ denote the space of Lebesgue-measurable functions on $\mathbb R^n$ (equivalence classes of functions, where two functions are deemed equivalent if they differ on a set of measure zero), equipped with the topology of local convergence in measure.

For which values of $k\in\mathbb R$ do we have $W^{k,2}(\mathbb R^n)\subset \mathcal M$?

More formally, for which $k\in\mathbb R$ does the identity map $C^\infty_c(\mathbb R^n)\hookrightarrow \mathcal M$ extend by continuity to a map $W^{k,2}(\mathbb R^n) \to \mathcal M$?

Improve this question
$\endgroup$
5
  • $\begingroup$ $k \geq 0$ seems to be a necessary condition, because for $k<0$ elements of $W^{k,2}$ are not measurable functions. For any $k \geq0$, convergence of a sequence in $W^{k,2}$ at least implies convergence in $L^2$. But convergence in $L^2$ implies convergence globally (and thus locally) in measure (en.wikipedia.org/wiki/Convergence_in_measure#Properties). Since $L^2$ is first countable, sequential and usual continuity are equivalent. So the embedding should be continuous for all $k \geq 0$. $\endgroup$
    – almosteverywhere
    Nov 8, 2021 at 16:44
  • $\begingroup$ @almosteverywhere "for $𝑘<0$ elements of $𝑊^{𝑘,2}$ are not measurable functions" That's exactly my question. Can you explain why this is the case? $\endgroup$
    – André Henriques
    Nov 8, 2021 at 16:52
  • $\begingroup$ Related post: math.stackexchange.com/questions/4033589/… $\endgroup$
    – RaffaeleScandone
    Nov 8, 2021 at 19:17
  • $\begingroup$ @RaffaeleScandone: Thank you. If I understand things correctly, the post you linked seems to answer my question. $\endgroup$
    – André Henriques
    Nov 8, 2021 at 20:31
  • $\begingroup$ @AndréHenriques Yes, it does. You are welcome. $\endgroup$
    – RaffaeleScandone
    Nov 8, 2021 at 23:03

1 Answer 1

Reset to default
0
$\begingroup$

My question has been answered in this MathStackexchange post: https://math.stackexchange.com/questions/4033589/sobolev-space-with-negative-index

For every $k<0$, there exist a measure $\mu_k$ which is singular with respect to Lebesgue measure, and such that $\mu_k\in W^{k,2}(\mathbb R^n)$.

So $W^{k,2}(\mathbb R^n)$ does not embed into the space of Lebesgue-measurable functions.

Thank you Raffaele Scandone.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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