Suppose $b_ h(f)>0$. The threshold subsolutions with supremum zero form a family that is stable under finite maxima and under $L^1$ limits. Its upper envelope is therefore again a normalized threshold subsolution, and it dominates all the others.
Proposition 15.28.6.D42D2088B421 Suppose $b=b_ h(f)>0$. The family
\[ \mathcal S_ f=\{ s\in \mathrm{PSH}(X,h):\sup _ Xs=0, \mathrm{MA}^{\mathrm{ext}}_ h(s)\geq bf\, \mathrm{d}V\} \]
is nonempty and compact in $L^1(X)$. Its upper envelope
\[ \Phi _ f=\left(\sup _{s\in \mathcal S_ f}s\right)^* \]
belongs to $\mathcal S_ f$ and is therefore its greatest member. Moreover, let $u_\lambda $ be the greatest exponential solutions of corollary 15.35.3 and $v_\lambda =u_\lambda -\sup _ Xu_\lambda $. Every $L^1$ limit of a sequence $v_{\lambda _ j}$ with $\lambda _ j\downarrow 0$ belongs to $\mathcal S_ f$, and hence lies below $\Phi _ f$.
Proof.
The family is nonempty by theorem 15.28.4. By theorem 5.25.2, an $L^1$ limit of members of $\mathcal S_ f$ is $h$-plurisubharmonic with supremum zero, and lemma 15.28.3, with $g_ j=g=bf$, shows that it satisfies the measure inequality. Thus $\mathcal S_ f$ is a closed subset of the compact set of normalized $h$-plurisubharmonic functions, hence compact. It is also stable under finite maxima: if $s_1,s_2\in \mathcal S_ f$, then $\max (s_1,s_2)$ has supremum zero, and proposition 13.14.6, applied on the sets where the maximum equals $s_1$ or $s_2$, gives $\mathrm{MA}^{\mathrm{ext}}_ h(\max (s_1,s_2))\geq bf\, \mathrm{d}V$.
By Choquet's lemma there is a sequence $(s_ i)_{i\geq 1}$ in $\mathcal S_ f$ with $(\sup _ is_ i)^*=\Phi _ f$. Fix $s_0\in \mathcal S_ f$. The finite maxima $t_ N=\max (s_0,s_1,\ldots ,s_ N)$ belong to $\mathcal S_ f$, increase with $N$, and satisfy $s_0\leq t_ N\leq 0$; their regularized limit is $\Phi _ f$. Hence $t_ N\to \Phi _ f$ almost everywhere and, by dominated convergence, in $L^1$. Lemma 15.28.3 gives $\mathrm{MA}^{\mathrm{ext}}_ h(\Phi _ f)\geq bf\, \mathrm{d}V$, and $s_0\leq \Phi _ f\leq 0$ forces $\sup _ X\Phi _ f=0$. Thus $\Phi _ f\in \mathcal S_ f$, and by construction it dominates every member of $\mathcal S_ f$.
Finally, suppose $v_{\lambda _ j}\to v$ in $L^1$ with $\lambda _ j\downarrow 0$. The measures $\mathrm{MA}^{\mathrm{ext}}_ h(v_{\lambda _ j})=\mathrm{e}^{\lambda _ ju_{\lambda _ j}}f\, \mathrm{d}V$ have densities converging in $L^1$ to $bf$, by corollary 15.35.3. Applying lemma 15.28.3 to these densities and potentials shows that $v\in \mathcal S_ f$.
$\square$
The proposition gives a greatest normalized threshold subsolution, but not the equation $\mathrm{MA}^{\mathrm{ext}}_ h(\Phi _ f)=bf\, \mathrm{d}V$; in particular, it does not show that $\Phi _ f$ has finite mass.
Proposition 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,
\[ b_ h(f)\geq \frac{v_-(h)}{\int _ Xf\, \mathrm{d}V}. \]
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.
$\square$
Pome
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