Lemma 6.17.47EFB7FFA3AB8

Proof. After shrinking the neighborhood we may subtract a constant from $v$ and assume $v\leq 0$; this changes neither the ideals nor the integrability of $\mathrm e^{m(v-u)}$. Put $p=\lambda '/\lambda >1$, and note that $m\geq \lambda p/(p-1)>\lambda $. Let $f\in \mathcal I(\lambda 'v)_ x$, defined on a small neighborhood on which $|f|^2\leq C$ for a constant $C$. All the following inequalities are taken outside the pole sets of $u$ and $v$, which have measure zero. Where $u\geq pv$, we have $\lambda u\geq \lambda 'v$ and hence $|f|^2\mathrm e^{-\lambda u}\leq |f|^2\mathrm e^{-\lambda 'v}$. Where $u<pv$, since $m-\lambda >0$, $v\leq 0$ and $(m-\lambda )p-m=m(p-1)-\lambda p\geq 0$,

that is, $-\lambda u\leq m(v-u)$; hence $|f|^2\mathrm e^{-\lambda u}\leq C\mathrm e^{m(v-u)}$. Both bounds are integrable near $x$, so $f\in \mathcal I(\lambda u)_ x$. $\square$