$L^{2}(\mathbb{R^n})$ is compactly embedded in which space?

Question

As answered in this question $L^2$ compactness of embedding, we have $L^2(\Omega)$ is indeed compactly embedded in $H^{-1}(\Omega)$ (here $\Omega$ is bounded domain in $R^n$), and accounting for the answer of this questions subset of $H^1(\mathbb{R}^d)$ is compactly embedded in $L^2(\mathbb{R}^d)$

Now, what about $L^2(\mathbb{R^n}$)? does the embedding still hold $( L^2(\mathbb{R^n}) \subset \subset H^{-1}(\mathbb{R^n}))$? I am wondering if there's any space that $L^2(\mathbb{R^n}$) is compactly embedded into?

Answer 1

I am not sure if this is what you were looking for but $L^2(\mathbb{R}^n)$ is compactly embedded into $L^2(\mathbb{R}^n)$ equipped with its weak topology. In general the unit ball in any Hilbert space is compact with respect to the weak topology.

Answer 2

No, on $\mathbb R^n$, the inclusions of $L^2$ Sobolev spaces $H^{s+\varepsilon} \to H^s$ (for $\varepsilon>0$) are not compact, no matter how large $\varepsilon>0$ may be.

This can be seen easily on the spectral side: the operator (multiplication-by) $(1+|x|^2)^{-\varepsilon/2}$ has purely continuous spectrum...