Proposition 16.28.6249C69232C604

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 $b\geq 1$. $\square$