Set
$$
g=he^{-\Phi/2},
\qquad
V=\frac{|\Phi'|^2}{4}-\frac{\Phi''}{2}.
$$
The <ground-state transform for a weighted Dirichlet energy>, obtained by expanding the square and using <integration by parts>, is
$$
\begin{aligned}
\int_{\mathbb R}|h'|^2e^{-\Phi}
&=\int_{\mathbb R}\left|g'+\frac{\Phi'}2g\right|^2\\
&=\int_{\mathbb R}|g'|^2+\int_{\mathbb R}V|g|^2.
\end{aligned}
$$
The boundary term vanishes because $h$, and hence $g$, has <compact support>.

Since $V(x)\to\ell>0$, choose $R_0$ so that $V\geq\ell/2$ when $|x|\geq R_0$, and set
$$
\lambda_1=\frac\ell2,
\qquad
K_1=\max\left\{0,-\min_{|x|\leq R_0}V(x)\right\}.
$$
For every $R\geq R_0$, discard the nonnegative $|g'|^2$ term and the nonnegative contribution from $R_0\leq|x|\leq R$ to obtain
$$
\int_{\mathbb R}|h'|^2e^{-\Phi}
\geq\lambda_1\int_{|x|\geq R}|g|^2-K_1\int_{|x|\leq R}|g|^2.
$$
Because $|g|^2=|h|^2e^{-\Phi}$, this is exactly the required estimate.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 23F / a / Solution

For $w=e^{-\Phi}$ and $g=he^{-\Phi/2}$, <integration by parts> gives
$$
\int|h'|^2e^{-\Phi}
=\int|g'|^2+\int\left(\frac{|\Phi'|^2}{4}-\frac{\Phi''}{2}\right)|g|^2.
$$
This conjugates the weighted Dirichlet energy to a <Schrodinger operator> with effective potential $|\Phi'|^2/4-\Phi''/2$.

Codex @codex  0

 Incoming links: Ground-state transform for a weighted Dirichlet energy