1

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?

Samiha Belmor
  • 27
  • Mainly to get compactness you need to prevent oscillations and parts of the function leaving at infinity ... The first is obtained by regularity. The second is always verified on bounded domains, but you need weights to prevent this happening in the whole space – LL 3.14 Feb 03 '22 at 00:09

2 Answers2

1

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...

paul garrett
  • 52,465
  • yeah you're right, I have been searching, and I came across a similar question that was answered before here https://math.stackexchange.com/questions/929738/is-it-true-that-l2-is-compactly-embedded-in-w1-2-0-ast?rq=1 , the user said that "Pretty much nothing is compactly embedded on $\mathbb{R^n}$, unless you use a weight decaying at infinity,
    Does that mean I may find a compact embedding in $L^2$-weighted Lebesgue space?     – Samiha Belmor Jan 27 '22 at 13:46
  • @SamihaBelmor, as far as I know, one needs both a smoothness constraint and a growth constraint. So, for example, with the "quantum harmonic oscillator" $-\Delta+|x|^2$, and spaces defined analogously to Sobolev spaces, the inclusions are compact and so on... – paul garrett Jan 28 '22 at 22:10
0

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.

timur
  • 16,599