Choose a <space of test functions>[test function] $\chi$ that equals one on the unit ball and vanishes outside the ball of radius two. Since
$$
\operatorname{div}(x\chi)=d\chi+x\mathbin{\cdot}\nabla\chi,
$$
<integration by parts> gives
$$
d\int\chi u^2
=-\int(x\mathbin{\cdot}\nabla\chi)u^2-2\int\chi u\,x\mathbin{\cdot}\nabla u.
$$
The first term on the right is supported where $1\leq|x|\leq2$, where $u^2\leq |x|^\alpha u^2$. The <Cauchy-Schwarz inequality> and <Young inequality> bound the second term by
$$
2\left|\int\chi u\,x\mathbin{\cdot}\nabla u\right|
\leq\frac d2\int\chi u^2+C\int|\nabla u|^2.
$$
After absorbing the first integral,
$$
\int_{|x|\leq1}|u|^2
\leq C\left(\int|\nabla u|^2+\int|x|^\alpha|u|^2\right)
=C\|u\|_\Sigma^2.
$$

This estimate also proves completeness. Indeed, a <Cauchy sequence> in $\Sigma$ is Cauchy in $H^1$ locally, while its gradients and the functions $|x|^{\alpha/2}u_n$ converge in $L^2$. The limits agree locally with a function $u$, so $u\in\Sigma$ and $u_n\to u$ in the energy norm. Thus $\Sigma$ is a <Hilbert space>; it is the <confining-potential energy space> for $V(x)=|x|^\alpha$.

Solution

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

{title2=$\Sigma_V$}

For a nonnegative potential $V(x)\to\infty$ as $|x|\to\infty$, the confining-potential energy space is
$$
\Sigma_V=\{u:\nabla u\in L^2,\ V^{1/2}u\in L^2\},
\qquad
\|u\|_{\Sigma_V}^2=\|\nabla u\|_2^2+\|V^{1/2}u\|_2^2.
$$
A local <Poincare inequality> controls the missing $L^2$ term, so this energy norm makes $\Sigma_V$ a <Hilbert space>.

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 Incoming links: Confining-potential energy space