Let $w\in L^n(\mathbb{R}^n)$, ($n\geq 3$), and for $\tau\in\mathbb{C}$, $Im(\tau)\neq 0$, let $R_{\tau}=(-\Delta-\tau)^{-1}$ be the resolvent of the Laplacian. I need to show that $T:=wR_{\tau}w$ is a compact operator on $L^2(\mathbb{R}^n)$. The estimates $L^{2n/(n+2)}\rightarrow L^{2n/(n-2)}$ for the resolvent of Laplacian makes $T$ a bounded operator on $L^2$. The paper i`m studying on gives the following hints:
i) suppose $w$ bounded and compactly supported, the general case follows by a density argument,
ii) $H^2$ embeds compactly inside $L^{2n/(n-2)}$ within any ball of finite radius,
iii) outside the support of $w$, we have the bound
$$|R_{\tau}wg(x)|\lesssim_{\tau}\Vert g\Vert_2|x|^{\frac{1-n}{2}}$$
iv) finish the proof by expressing $wR_{\tau}w$ as norm-limit of compact operator on $L^2$.
I'm not able to work out a proof, particularly it's not clear to me how to (and why) use the pointwise estimates outside $supp(w)$. I'm grateful for any help, thank you.
This is what I think is happening:
Assume that $w\in L^\infty$ and $\mathrm{supp}\,w\subset B$ for some ball $B\subset\mathbb{R}^n$. Take $g\in L^2$. Then we have $$ \|wg\|_{L^2}\leq\|w\|_{L^\infty}\|g\|_{L^2}. $$ Hence by the basic property of the resolvent, we have $R_\tau wg\in H^2$ and $$ \|R_\tau wg\|_{H^2}\leq C\|w\|_{L^\infty}\|g\|_{L^2}. $$ With the restriction operator $r:H^2(\mathbb{R}^n)\to H^2(B)$, and $p=\frac{2n}{n-2}$, the operator $$ rR_\tau w:L^2(\mathbb{R}^n)\to L^p(B), $$ is compact, which means that $$ wrR_\tau w:L^2(\mathbb{R}^n)\to L^2(\mathbb{R}^n), $$ is compact. Supposing that $B$ is centred at the origin and its radius is $R$, and using the decay estimate, we get $$ \|(1-r)R_\tau wg\|_{L^p} \lesssim R^{-\frac12}\|g\|_{L^2} . $$ This means that $wR_\tau w$ can be approximated in norm by a compact operator.