Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 23F (Analysis of Functions R)

SetThe ground-state transform for a weighted Dirichlet energy, obtained by expanding the square and using integration by parts, isThe boundary term vanishes because $h$, and hence $g$, has compact support.

Since $V(x)\to \ell >0$, choose $R_0$ so that $V\ge \ell /2$ when $|x|\ge R_0$, and setFor every $R\ge R_0$, discard the nonnegative $|g^{\prime }{|}^2$ term and the nonnegative contribution from $R_0\le |x|\le R$ to obtainBecause $|g{|}^2=|h{|}^2{e}^{-\Phi }$, this is exactly the required estimate.

LetThe fundamental theorem of calculus and the Cauchy-Schwarz inequality giveThus the one-half Hölder seminorm is at most $m_R^{-1/2}{E}_R(h{)}^{1/2}$.

Writing $Z_R={\int }_{-R}^{R}w>0$ and using the zero weighted mean, for each $x$ we haveSince $|x-y|\le 2R$, the preceding estimate yieldsConsequently the required inequality holds, for example, with

LetThen $q^{\prime }=h^{\prime }$ and $q$ has zero weighted mean. Part (b) givesTherefore, with $Z_R={\int }_{-R}^R{e}^{-\Phi }$,Thus the compact-interval Weighted Poincare inequality holds with

Subtracting the global weighted mean from $h$ leaves its derivative unchanged, so it suffices to assumeWrite $w=e^{-\Phi }$, choose $R\ge R_0$, and putThe weighted mean of $h$ on $[-R,R]$ isBy the Cauchy-Schwarz inequality,Part (c), applied on $[-R,R]$, therefore gives for $E={\int }_{\mathbb{R}}|h^{\prime }{|}^{2}w$Part (a) gives independently

Choose ${ϵ}_0>0$ so small that ${ϵ}_0{K}_1\le 1/2$, and then choose $R$ large enough that $q/Z\le {ϵ}_0{\lambda }_1/2$. Multiply the last inequality by $ϵ={ϵ}_0{\lambda }_R$ and add it to the local inequality. This yieldswhere $c_R=\frac{{\lambda }_R}{2}\min \{1,{ϵ}_0{\lambda }_1\}>0$. Hence, with ${\lambda }_0=c_R/(1+ϵ)$,