Let
$$
\overline h_R=\frac{\int_{-R}^Rh(y)e^{-\Phi(y)}\,dy}{\int_{-R}^Re^{-\Phi(y)}\,dy},
\qquad
q=h-\overline h_R.
$$
Then $q'=h'$ and $q$ has zero weighted mean. Part (b) gives
$$
\|q\|_\infty^2\leq C_R^2\int_{-R}^R|h'|^2e^{-\Phi}.
$$
Therefore, with $Z_R=\int_{-R}^Re^{-\Phi}$,
$$
\int_{-R}^R|h-\overline h_R|^2e^{-\Phi}
\leq Z_RC_R^2\int_{-R}^R|h'|^2e^{-\Phi}.
$$
Thus the compact-interval <Weighted Poincare inequality> holds with
$$
\boxed{\lambda_R=(Z_RC_R^2)^{-1}>0}.
$$

Solution

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

{c}

For a <probability density function> $w$ on a domain, a weighted Poincare inequality has the form
$$
\int|h'|^2w\geq\lambda\int|h-\mathbb E_w h|^2w.
$$
The optimal $\lambda$ is the <spectral gap> of the associated weighted diffusion operator.

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 Incoming links: Weighted Poincare inequality