15.35.14 Positive endpoints and one-sided limits44A2CB44D7CB

Proof. Set $q=\min (f,1)$, a nonzero bounded density. By theorem 14.19.3, there is a continuous $w\in \mathrm{PSH}(X,h)$ with $\mathrm{MA}_ h(w)=\mathrm e^{\tau w}q\, \mathrm dV$. Put $t_\tau =w+\max (0,-\inf _ Xw)$. Then $t_\tau \geq 0$ and $t_\tau \geq w$, so for $\lambda \geq \tau $,

Let $u$ be an exponential subsolution at an exponent $\lambda \geq \tau $, and let $\epsilon >0$. On $\{ t_\tau <u-\epsilon \} $,

The bounded function $t_\tau $ satisfies $P_{h/2}(t_\tau )\not\equiv -\infty $, and $u-\epsilon \preceq t_\tau $. Thus theorem 15.2.2 gives $t_\tau \geq u-\epsilon $, and letting $\epsilon \downarrow 0$ gives $t_\tau \geq u$.

For the order estimate, put $\delta =K_\tau (\lambda _2-\lambda _1)/\lambda _2$. By the first part, $U_{\lambda _1}\leq t_\tau \leq K_\tau $, so $(\lambda _2-\lambda _1)U_{\lambda _1}\leq \lambda _2\delta $, that is, $\lambda _1U_{\lambda _1}\geq \lambda _2(U_{\lambda _1}-\delta )$. Since the Monge–Ampère measure is unchanged by adding constants,

Thus $U_{\lambda _1}-\delta $ is an exponential subsolution at $\lambda _2$, and the maximality of $U_{\lambda _2}$ gives $U_{\lambda _2}\geq U_{\lambda _1}-\delta $. Finally, $1-x^{-1}\leq \log x$ for $x\geq 1$; with $x=\lambda _2/\lambda _1$ this gives $\delta \leq K_\tau \log (\lambda _2/\lambda _1)$, which proves the monotonicity. $\square$

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 $\mathrm e^{aM_ j}\| f\| _{L^1}$. $\square$

Proof. Let $K=K_ a$ be the constant of the lemma with $\tau =a$, and let $\lambda _ j>a$ decrease to $a$. By the lemma, the functions

decrease and satisfy $V_ j\geq U_ a$. Their limit $V$ therefore belongs to $\mathrm{PSH}(X,h)$ and satisfies $V\geq U_ a$. Dominated convergence between the integrable bounds $U_ a$ and $V_1$ gives $V_ j\to V$ in $L^1$, and lemma 5.29.19 gives convergence in capacity. Since the added constants tend to zero, $U_{\lambda _ j}\to V$ in both senses.

The nonpositive functions $U_{\lambda _ j}-K$ satisfy

and $\mathrm e^{\lambda _ jK}f\to \mathrm e^{aK}f$ in $L^1$. Theorem 15.34.2 therefore gives $\mathrm{MA}^{\mathrm{ext}}_ h(V)=\mathrm e^{aV}f\, \mathrm dV$, together with total-variation convergence of the measures $\mathrm{MA}^{\mathrm{ext}}_ h(U_{\lambda _ j})$. Hence $V$ is a solution at $a$, so $V\leq U_ a$ by maximality, while $V\geq U_ a$ by construction. Thus $V=U_ a$. Since every sequence of exponents converging to $a$ from above has a decreasing subsequence, this proves the assertions. $\square$

Proof. Choose $\tau \in \Lambda _ h(f)$ with $0<\tau <a$; the upper-interval property gives such a choice. Put $K=K_\tau $ from lemma 15.35.15 and

This family is increasing. For a sequence $\lambda _ j\uparrow a$ with $\tau \leq \lambda _ j<a$, its members lie between the integrable potentials $V_\tau $ and $V_ a$. The upper-envelope construction in section 5.40 gives

Indeed cofinality and the order give the same pointwise supremum for every such sequence. Regularization changes it only on a pluripolar set. Dominated convergence between the integrable bounds gives $L^1$ convergence to $V^-_ a$, and lemma 5.29.19 gives convergence in capacity. Since the added constants converge, set

Then $U_{\lambda _ j}\to L_ a$ in both topologies and $L_ a\leq U_ a$; the latter inequality survives regularization because $V_ a$ is upper semicontinuous.

The potentials $U_{\lambda _ j}-K$ and their limit $L_ a-K$ are nonpositive. Apply theorem 15.34.2 with exponents tending to $a>0$ and densities $\mathrm e^{\lambda _ jK}f\to \mathrm e^{aK}f$ in $L^1$. It proves the exact equation and total-variation convergence.

Every sequence approaching $a$ from the left has a cofinal increasing subsequence. If potential convergence in either topology, or total-variation convergence of the equation measures, failed, a sequence staying outside a fixed neighbourhood of the proposed limit would therefore have an increasing subsequence, contrary to the preceding argument. This proves the full left limit and measure convergence. Repeating the construction with another $\tau $ gives the same $L^1$ limit; canonical quasi-plurisubharmonic representatives that agree almost everywhere agree everywhere. Thus the auxiliary choice does not change $L_ a$. $\square$

Proof. Fix $a_0<\alpha <\beta $, choose an admitted $0<\tau <\alpha $, and use the corrected family $V_\lambda $ from the preceding proof on $[\alpha ,\beta ]$. Its integrals

form a finite bounded increasing real function. For $a\in (\alpha ,\beta ]$, the left-limit proposition gives

If $L_ a\ne U_ a$, this quantity is positive: the integrable nonnegative difference cannot vanish almost everywhere without equality of the canonical potentials. For any finite ordered collection of such points, the sum of their jumps is at most $I(\beta )-I(\alpha )$. Consequently there are only finitely many jumps at least $1/m$ for each positive integer $m$, and at most countably many positive jumps. A countable cover by intervals with rational endpoints inside $(a_0,+\infty )$ proves the assertion about $D$.

If $a\notin D$, the left limit is $U_ a$, and proposition 15.35.17 supplies the same right limit in both topologies and for the equation measures. Splitting an arbitrary approaching sequence into its two sides proves the stated two-sided convergence. $\square$