 Solution

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 ${\lambda }_0=c_R/(1+ϵ)$,