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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.

Savannah
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1 Answers1

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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 $V \subset \subset L^{2}(\mathbb{R}^{d})$.

Claudio Moneo
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    Could you elaborate on the steps after applying R-K compactness theorem? Why do we find such functions and how are you concluding in the final step? – Savannah Jun 19 '20 at 11:38
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    @Savannah This is a characterization of compactness in metric spaces, I have added it to my answer. Basically this is "For every collection of sets covering the compact $A$ we find a finite collection of sets that already covers $A$". Please ask if you need further explanation :) – Claudio Moneo Jun 19 '20 at 15:44
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    For the last inequality I guess you are trying to consider $|u-f_j|{L^2(\mathbb{R}^d)} <|u-f_j|{L^2(B^c_N)}+|u-f_j|{L^2(B_N)}<|u-f_j|{L^2(B^c_N)}+ \frac{\epsilon}{2}<|f_j|_{L^2(B^c_N)}+\epsilon $, how do you finish from here? And where are you using the fact that we are in Weighted L^2 space, it seems like your arguments will work for $H^1(\mathbb{R}^d)$ as well, which cannot be true. – Savannah Jun 20 '20 at 03:14
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    @Savannah The $f_{j}$ are $\in L^{2}(B_{N})$ and thus zero outside of $B_{N}$. That is why the last inequality works. To your second point: Such $N$ only exists because we are in the weighted space! Otherwise, if we are in $H^{1}(\mathbb{R}^{d})$, consider a sequence of bump functions travelling to infinity - we will never be able to find a suitable $N$ – Claudio Moneo Jun 20 '20 at 06:23
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    Which property of weighted space are you using and is there any particular benefit of the weight $|x|^2$ or it is true for any positive function $\phi$? – Savannah Jun 20 '20 at 06:47
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    @Savannah I use the fact that $x^2 \rightarrow \infty$. I have added the precise argument to my answer – Claudio Moneo Jun 20 '20 at 07:19