Let $T : H^1(\mathbb{R}^d) \to L^2(\mathbb{R}^d)$ be the operator such that $Tu(x)=\frac{u(x)}{1+|x|^2}$. (here $H^1$ denotes the Sobolev space)
Question : show that $T$ is compact.
My attempt is this one : let $(u_n)$ be a bounded sequence in $H^1(\mathbb{R}^d)$ I want to show that there is a subsequence $(u_{\phi (n)})$ converging strongly in $L^2(\mathbb{R}^d)$. All I know (and want to use) is that $H^1(\mathbb{R}^d)$ continuously embbed in $L^2$ in every compact. But here the result has to be shown not only for compacts !
How can I do that ?
Thank you.
Since $C_0^\infty(R^d)$ is dense in $H^1(R^d)$, it suffices to show this for $C_0^\infty(R^d)$. Denote $B_i$ the centered ball in $R^d$ with center $0$ and radius $i$. Suppose $\{u_m\}$ is a bounded sequence in $C_0^\infty(R^d)$, say norm is less equal to $1$. Consider their restriction on $B_i$. Since $T:H^1(B_i)\rightarrow L^2(B_i)$ is compact, we find a subsequence $\{u^i_m\}$ such that for there exists $N_i$ such that for $m,n> N_i$ \begin{equation} \|u^i_m-u^i_n\|_{L^2(B_i)}\leq \frac{1}{2^i}. \end{equation} In particular for $i+1$-th step, the sequence can be chosen from the sequence which was used for the $i$-th step, i.e., $\{u^{i+1}_m\}\subset\{u^i_m\}$. Now let $v_i:=u^i_i$. Then \begin{align} \|(v_m-v_n)/(1+|\cdot|^2)\|_{L^2(R^d)}&=\|(v_m-v_n)/(1+|\cdot|^2)\|_{L^2(B_i)}+\|(v_m-v_n)/(1+|\cdot|^2)\|_{L^2(R^d/B_i)}\\ &=:L_1+L_2. \end{align} Since $\{v_i\}$ is bounded in $C_0^\infty(R^d)$ by norm $1$, for each $\varepsilon>0$ $L_2$ can be estimated by $2\|(1+|\cdot|^2)\|_{L^2(R^d/B_i)}<\varepsilon$ for all sufficiently large $i$. Due to the construction of $\{v_i\}$, $L_1$ can be arbitrary small for all enough large $m,n$, then the claim is proved.