Proposition 16.28.6547066B2F8269

Proof. The relation $u\preceq \varphi $ gives $P_{\theta ,F}[u]\leq P_{\theta ,F}[\varphi ]$. For the reverse inequality, let $a\in (0,1)$ and choose $h_ a\in \mathcal A_{1/(1-a)}(u,\varphi )$, so that $a\varphi +(1-a)h_ a\leq u$. Combining competitors below $F$ gives

Indeed, if $g_1$ and $g_2$ are competitors for $P_{\theta ,F}[\varphi ]$ and $P_{\theta ,F}[h_ a]$, then $ag_1+(1-a)g_2$ lies below $F$ and is at least as singular as $a\varphi +(1-a)h_ a\leq u$; taking the two upper envelopes preserves the inequality, by local regularization. Write $m=\min _ XF$ and $M=\max _ XF$. Translating competitors gives

Every $P_\theta [h_ a]$ has supremum zero, since it lies above $h_ a-\sup _ Xh_ a$ and below zero. The uniform $L^1$ bound for normalized potentials in theorem 5.25.2 therefore bounds the $L^1$ norms of the $P_{\theta ,F}[h_ a]$ uniformly in $a$. Letting $a\uparrow 1$ in the first inequality gives $P_{\theta ,F}[\varphi ]\leq P_{\theta ,F}[u]$ almost everywhere, and by lemma 5.24.3 the inequality holds everywhere. $\square$