16.28.60 Relative envelope classesE04112C46C03

Proof. If $u\geq \varphi -C$, then $h=\varphi -bC$ belongs to $\mathcal A_ b(u,\varphi )$ for each $b$; hence every $u\sim \varphi $ belongs to the class. More generally, if $u'\geq u-C$ and $h\in \mathcal A_ b(u,\varphi )$, then $h-bC\in \mathcal A_ b(u',\varphi )$. Since the condition $u\preceq \varphi $ is also unchanged by bounded changes of $u$, the class is stable under such changes. A convex combination of two competitors is a competitor for the same convex combination of the two potentials, which proves convexity. Finally, a competitor for $u$ is a competitor for any larger $v$, and the assumption $v\preceq \varphi $ gives the remaining condition.

If $\varphi '\in \mathrm{PSH}(X,\theta )$ and $\varphi '\sim \varphi $, choose $C\geq 0$ with $|\varphi '-\varphi |\leq C$ off their common pole set. For $h\in \mathcal A_ b(u,\varphi )$, the function $h-(b-1)C$ belongs to $\mathcal A_ b(u,\varphi ')$: the required inequality holds off the poles, and it extends everywhere by lemma 5.24.3. The reverse implication follows by interchanging the references, and $u\preceq \varphi $ is equivalent to $u\preceq \varphi '$.

For the last assertion, note that $h\mapsto h-f$ is a bijection from $\mathrm{PSH}(X,\theta )$ onto $\mathrm{PSH}(X,\theta +\mathrm{dd}^{\mathrm{c}}f)$. It preserves the order $\preceq $ and transforms each stretching inequality into

This identity also holds at poles, because $f$ is smooth and finite. Thus the corresponding families of competitors are in bijection for every $b\geq 1$. $\square$

Proof. Fix $b,s\geq 1$ and put $h=Q_{bs}(u,\varphi )$. Its defining inequality gives, off the pole sets,

The left-hand side is $\theta $-plurisubharmonic, so it is at most $Q_ b(u,\varphi )$. Hence $h\in \mathcal A_ s(Q_ b(u,\varphi ),\varphi )$ for every $s$. Also $Q_ b(u,\varphi )\leq u\leq \varphi $, so $Q_ b(u,\varphi )\preceq \varphi $. The inequalities between convex combinations extend across the poles by lemma 5.24.3, which proves the first assertion.

For the second assertion, choose a constant $C$ such that $u-C\leq \varphi $ and $w-C\leq \varphi $. Since $Q_ b(u-C,w-C)=Q_ b(u,w)-C$ and class membership is unchanged by translation, we may assume $u,w\leq \varphi $. Because $u\leq w\leq \varphi $, the obstacle for $Q_ b(u,w)$ lies above that for $Q_ b(u,\varphi )$ and below $u$. Thus

Since $Q_ b(u,\varphi )\in \mathcal E_\varphi $ by the first assertion, the stability under increase proved in proposition 16.28.62 gives the conclusion. $\square$

Proof. Adding a common constant to all $u_ i$, we may assume $u_ i\leq \varphi $. By lemma 16.28.63, proposition 16.28.62,

Each summand is at most $\varphi $. For every $i$, the defining inequality of $Q_ N(u_ i,\varphi )$ therefore gives

These inequalities between convex combinations also hold at common poles. Thus $h$ is a competitor for $p=P_\theta (\min _ i u_ i)$, so $p$ is not identically $-\infty $, and $h\leq p\leq u_1\preceq \varphi $. Stability under increase gives $p\in \mathcal E_\varphi $, and undoing the translation preserves this conclusion. $\square$

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. $\square$