On the unit ball, part 1 controls the $L^2$ norm. Outside it, $|x|^\alpha\geq1$, so
$$
\int_{|x|>1}|u|^2\leq\int|x|^\alpha|u|^2.
$$
Hence $\|u\|_2\leq C\|u\|_\Sigma$, proving that the <linear map>[embedding] $\Sigma\hookrightarrow L^2$ is <continuous function>[continuous].

For compactness, let $(u_n)$ be bounded in $\Sigma$. On each ball, it is bounded in $H^1$, so the <Rellich-Kondrachov compactness theorem> gives a subsequence convergent in local $L^2$. The tail estimate
$$
\int_{|x|>R}|u_n|^2
\leq R^{-\alpha}\int|x|^\alpha|u_n|^2
$$
is uniform in $n$ and tends to zero as $R\to\infty$. A <diagonal argument> therefore gives convergence in all of $L^2$. This is the <compact embedding of a confining-potential energy space>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 154 / 1 / 1 / 2 / Solution

The embedding $\Sigma_V\hookrightarrow L^2(\mathbb R^d)$ is compact. The <Rellich-Kondrachov compactness theorem> gives compactness on each fixed ball, while
$$
\int_{|x|>R}|u|^2
\leq\left(\inf_{|x|>R}V(x)\right)^{-1}
\int_{\mathbb R^d}V|u|^2
$$
makes the tails uniformly small.

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