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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 $W^{k,2}(\mathbb R^n) \to \mathcal M$?

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  • $\begingroup$ $k \geq 0$ 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
    Commented 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
    Commented Nov 8, 2021 at 16:52
  • $\begingroup$ Related post: math.stackexchange.com/questions/4033589/… $\endgroup$
    – RaffaeleScandone
    Commented 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
    Commented Nov 8, 2021 at 20:31
  • $\begingroup$ @AndréHenriques Yes, it does. You are welcome. $\endgroup$
    – RaffaeleScandone
    Commented Nov 8, 2021 at 23:03

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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.

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