Proposition 15.35.169E30269224CD

Proof. Suppose first that $a\in \Lambda _ h(f)$, let $u_ a$ be a solution at $a$, and put $s=u_ a-\max (0,\sup _ Xu_ a)$. As in the proof of proposition 15.35.12, $s$ is an exponential subsolution at every $\lambda \geq a$, so $U_\lambda \geq s$. Hence the greatest solutions at any exponents $\lambda _ j\downarrow a$ have suprema at least $\sup _ Xs>-\infty $.

Conversely, let $\lambda _ j$ and $u_ j$ be as in the statement. The lemma with $\tau =a$ gives the common upper bound $u_ j\leq K_ a$. Write $M_ j=\sup _ Xu_ j$. After extraction, $M_ j\to M\in \mathbb R$ and, by theorem 5.25.2, $u_ j-M_ j\to v$ in $L^1$ for some $v\in \mathrm{PSH}(X,h)$. Then $u_ j\to u=v+M$ in $L^1$ and, after a further extraction, almost everywhere. Since $a\leq \lambda _ j\leq \lambda _1$, $u_ j\leq K_ a$ and $K_ a\geq 0$, the densities $\mathrm e^{\lambda _ ju_ j}f$ are bounded by $\mathrm e^{\lambda _1K_ a}f$, and dominated convergence gives

Lemma 15.28.3 now gives $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq \mathrm e^{au}f\, \mathrm dV$. Thus $u$ is an exponential subsolution at $a$, and theorem 15.35.2 gives a solution at $a$.

Finally, suppose $a\notin \Lambda _ h(f)$, and let $u_ j$ be solutions at exponents $\lambda _ j\downarrow a$. If $\sup _ Xu_ j$ did not tend to $-\infty $, a subsequence would satisfy the condition just shown to imply $a\in \Lambda _ h(f)$. Hence $M_ j\to -\infty $. Once $M_ j<0$, since $\lambda _ j\geq a>0$,

and the total mass $\int _ X\mathrm e^{\lambda _ ju_ j}f\, \mathrm dV$ is at most $\mathrm e^{aM_ j}\| f\| _{L^1}$. $\square$