Minimize $\|u\|_\Sigma^2$ subject to $\|u\|_2=1$. A minimizing sequence is bounded in $\Sigma$, and part 2 supplies a subsequence converging strongly in $L^2$ and weakly in $\Sigma$. The constraint survives the strong convergence, while <weak lower semicontinuity> of the norm shows that the limit $\psi$ attains the minimum. Since $\||\psi|\|_2=\|\psi\|_2$ and $|\nabla|\psi||\leq|\nabla\psi|$ <almost everywhere>, we may take $\psi\geq0$.

The <Lagrange multiplier> equation is
$$
\int\nabla v\mathbin{\cdot}\nabla\psi+\int|x|^\alpha v\psi
=\lambda\int v\psi
\qquad(v\in\Sigma),
$$
where testing with $v=\psi$ shows that $\lambda=\|\psi\|_\Sigma^2>0$. Thus
$$
(-\Delta+|x|^\alpha)\psi=\lambda\psi
\quad\text{in }\mathcal D'(\mathbb R^d).
$$
This is the <ground-state eigenfunction of a confining Schrödinger operator>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 154 / 1 / 1 / 5 / Solution

The minimum of
$$
\|\nabla u\|_2^2+\|V^{1/2}u\|_2^2
$$
subject to $\|u\|_2=1$ is attained because $\Sigma_V$ embeds compactly into $L^2$. Replacing a minimizer by its <absolute value function>[absolute value] does not increase its energy. The <Euler-Lagrange equation> therefore produces a nonnegative eigenfunction of $-\Delta+V$ for its lowest eigenvalue.

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 Incoming links: Ground-state eigenfunction of a confining Schrödinger operator