Theorem 15.28.4CC13A9B1D927

Proof. If $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq af\, \mathrm{d}V$ with $a>0$, then also $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq af_ k\, \mathrm{d}V$ for every $k$, and lemma 15.27.1 gives $a\leq c_ h(f_ k)$. Letting $k\to \infty $ gives $a\leq b_ h(f)$. If $b_ h(f)=0$, the coefficient zero is attained by the zero potential.

Suppose now that $b=b_ h(f)>0$, and let $u_ k$ be continuous solutions of

By theorem 5.25.2, a subsequence $u_{k_ j}$ converges in $L^1$ to some $u\in \mathrm{PSH}(X,h)$ with $\sup _ Xu=0$. The densities $c_ h(f_{k_ j})f_{k_ j}$ are dominated by $c_ h(f_1)f$ and converge to $bf$ almost everywhere, hence in $L^1$. Lemma 15.28.3 now gives $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq bf\, \mathrm{d}V$, so the maximum is attained at $a=b$. $\square$