I was recenty reading about the weighted Lebesgue spaces and came accross an exercise that asks to prove that $H^1(\mathbb{R}^d) \cap L^2(\mathbb{R}^d,|x|^2\,dx)$ is compactly embedded in $L^2(\mathbb{R}^d)$. Where $H^1(\mathbb{R}^d)$ is the usual Sobolev space and $L^2(\mathbb{R}^d,|x|^2\,dx)$ is the weighted Lebesgue space containing functions $f$ for which $\int_{\mathbb{R}^d} |f|^2\,|x|^2\,dx< \infty$
The compact embedding theorems I know work on a bounded domain, not sure how to prove this one.
Best Answer
Assume $V\subset W$ is a bounded set, where $W$ is your weighted space.
Now note that $\forall \epsilon>0$ we find some $N>0$ such that for all $u \in V$ $$\int_{B_{N}^{c}} u^{2} < \epsilon/2$$
Assume this is not the case. Then for some $\epsilon>0$ for every $N>0$ there is some $u \in V$ such that $||u||_{L^{2}(B_{N}^{c})}\ge \epsilon$. But then $||u||_{W} \ge ||u(x) \cdot x^2||_{L^{2}(B_{N}^{c})} \ge N^2 \epsilon$ which contradicts boundedness of $V$.
On $B_{N}$ (the ball of radius $N$ in $\mathbb{R}^{d}$), we can apply the normal Rellich-Kondrachov to ensure the compact embedding to $L^{2}(B_{N})$.
Now recall that compactness of a set $A \subset X$ in a metric space $X$ means that $\forall \epsilon >0$ we find $x_{1},...,x_{k=k(\epsilon)} \in X$ such that $A \subset \cup_{j=1}^{k} B_{\epsilon}(x_{j})$ .
This means that $\forall \epsilon>0$ we find functions $f_{1},...,f_{k(\epsilon)} \in L^{2}(B_{N})$ such that $\forall u \in V$ we have some $f_{j}$ such that $$||u-f_{j}||_{L^{2}(B_{N}^{c})} < \epsilon/2$$ But $||u||_{L^{2}(B_{N}^{c})} < \epsilon/2$ anyway, so we even have $$||u-f_{j}||_{L^{2}(\mathbb{R}^{d})} < \epsilon$$ which implies the compact embedding $V \subset \subset L^{2}(\mathbb{R}^{d})$.