Corollary 15.35.3.B2672237F244 Suppose $b=b_ h(f)>0$. Then (15.35.1) has a greatest solution $u_\lambda $ for every $\lambda >0$. If $v\in \mathrm{PSH}(X,h)$ satisfies $\sup _ Xv=0$ and $\mathrm{MA}^{\mathrm{ext}}_ h(v)\geq bf\, \mathrm{d}V$, then
\[ u_\lambda \geq v+\lambda ^{-1}\log b. \]
Put
\[ M_\lambda =\sup _ Xu_\lambda ,\qquad v_\lambda =u_\lambda -M_\lambda ,\qquad c_\lambda =\mathrm{e}^{\lambda M_\lambda }. \]
Then, as $\lambda \downarrow 0$,
\[ c_\lambda \longrightarrow b_ h(f),\qquad \mathrm{e}^{\lambda u_\lambda }f\longrightarrow b_ h(f)f \quad \text{in }L^1(\mathrm{d}V). \]
Proof.
By theorem 15.28.4, a function $v$ as in the statement exists. Since $v\leq 0$, the function $s=v+\lambda ^{-1}\log b$ satisfies
\[ \mathrm{MA}^{\mathrm{ext}}_ h(s)\geq bf\, \mathrm{d}V \geq \mathrm{e}^{\lambda s}f\, \mathrm{d}V, \]
so it is an exponential subsolution. Theorem 15.35.2 gives the greatest solution $u_\lambda $ and the bound $u_\lambda \geq s$; in particular, $c_\lambda \geq b$.
Set $g_\lambda =\mathrm{e}^{\lambda v_\lambda }f$. By lemma 5.44.5, applied to the finite measure $f\, \mathrm{d}V$, we have $g_\lambda \to f$ in $L^1$ as $\lambda \downarrow 0$. Each $g_\lambda $ is nonzero, since $v_\lambda $ is finite almost everywhere. The equation for $u_\lambda $ reads $\mathrm{MA}^{\mathrm{ext}}_ h(v_\lambda )=c_\lambda g_\lambda \, \mathrm{d}V$, so theorem 15.28.4, applied to the density $g_\lambda $, gives
\[ c_\lambda \leq b_ h(g_\lambda ). \]
The upper semicontinuity of the threshold (theorem 15.32.1) therefore gives
\[ b\leq \liminf _{\lambda \downarrow 0}c_\lambda \leq \limsup _{\lambda \downarrow 0}c_\lambda \leq b_ h(f)=b. \]
This proves $c_\lambda \to b$. Since $\mathrm{e}^{\lambda u_\lambda }f=c_\lambda g_\lambda $ and $g_\lambda \to f$ in $L^1$, it also proves $\mathrm{e}^{\lambda u_\lambda }f\to bf$ in $L^1$.
$\square$
Pome
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