Pome69BB1D568698Proposition 15.28.7.69BB1D568698 Let $0\leq f\in L^1(\mathrm{d}V)$ have positive integral. If $f\in L^ p(\mathrm{d}V)$ for some $p>1$, then $b_ h(f)=c_ h(f)$. In general,
If $h$ is Kähler, then $b_ h(f)=\int _ Xh^ n/\int _ Xf\, \mathrm{d}V$.
Proof. Suppose that $f\in L^ p$ with $p>1$, and let $v$ be a continuous solution of $\mathrm{MA}_ h(v)=c_ h(f)f\, \mathrm{d}V$. It is a subsolution with coefficient $c_ h(f)$, so theorem 15.28.4 gives $c_ h(f)\leq b_ h(f)$. Conversely, lemma 15.27.1, applied to a threshold subsolution, gives $b_ h(f)\leq c_ h(f)$.
For general $f$, the continuous solution for $f_ k$ has total mass $c_ h(f_ k)\int _ Xf_ k\, \mathrm{d}V$, which is at least $v_-(h)$ by the definition of the lower mass. Letting $k\to \infty $ and using monotone convergence gives the lower bound for $b_ h(f)$. If $h$ is Kähler, Stokes' theorem for bounded potentials gives instead the identity $c_ h(f_ k)\int _ Xf_ k\, \mathrm{d}V=\int _ Xh^ n$, and the same limit proves the last assertion. $\square$
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