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