PomeD29B5BFF0A9AAlthough the mass varies, it may remain bounded. Set
the supremum of the masses of smooth strictly $\omega $-plurisubharmonic potentials. Comparing two Hermitian forms and rescaling the potentials shows that the finiteness of this supremum depends only on $X$.
Proposition 11.1.20.682711871907 If $\eta $ is another Hermitian form and $\omega \leq C\eta $ for a constant $C>0$, then
In particular, finiteness of $\overline{\mathrm v}_{\mathrm{sm}}$ is independent of the Hermitian form.
Proof. Let $u\in \mathcal H_\omega $. Since $C\eta +\mathrm{dd}^{\mathrm{c}}u\geq \omega +\mathrm{dd}^{\mathrm{c}}u>0$, the function $u/C$ belongs to $\mathcal H_\eta $. The top exterior power is monotone on positive $(1,1)$-forms: for positive Hermitian matrices $A\leq B$, all eigenvalues of $A^{-1/2}BA^{-1/2}$ are at least one, so $\det A\leq \det B$. Applying this pointwise gives
Taking the supremum over $u$ proves the inequality. Since any two Hermitian forms on the compact manifold $X$ are comparable in both directions, the last assertion follows. $\square$
The smooth supremum considered here is also the upper mass over all bounded $\omega $-plurisubharmonic potentials: proposition 13.14.15 proves this using approximation by smooth strict potentials and continuity of bounded products. In section 13.14, we then pass to unbounded truncations. By corollary 13.14.19, finiteness of this common supremum is equivalent to finiteness of the extended non-pluripolar mass of every $\omega $-plurisubharmonic potential.
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