PomeProposition 15.35.12.B2DB6BD3AC4D If $\lambda _0\in \Lambda _ h(f)$, then $[\lambda _0,+\infty )\subset \Lambda _ h(f)$. Consequently, $\Lambda _ h(f)$ is empty, or equals $(a,+\infty )$ for some $a\geq 0$, or equals $[a,+\infty )$ for some $a>0$. If $b_ h(f)>0$, then $\Lambda _ h(f)=(0,+\infty )$.
15.35.11 Solvable exponents and collapse at zero threshold4B9655E85450
We return to a general Hermitian form $h$ on $X$ and a nonzero density $0\leq f\in L^1(\mathrm dV)$. Let $\Lambda _ h(f)$ be the set of exponents $\lambda >0$ for which (15.35.1) has a solution. By theorem 15.35.2, $\lambda \in \Lambda _ h(f)$ if and only if the equation has an exponential subsolution at exponent $\lambda $.
Proof. Let $u$ be a solution at $\lambda _0$, and put $s=u-\max (0,\sup _ Xu)$. Then $s\leq \min (u,0)$, so for $\lambda \geq \lambda _0$,
\begin{align*} \mathrm{MA}^{\mathrm{ext}}_ h(s) & =\mathrm e^{\lambda _0u}f\, \mathrm dV\\ & \geq \mathrm e^{\lambda _0s}f\, \mathrm dV \geq \mathrm e^{\lambda s}f\, \mathrm dV. \end{align*}
Thus $s$ is an exponential subsolution at every $\lambda \geq \lambda _0$, and theorem 15.35.2 gives a solution. Hence $\Lambda _ h(f)$ contains $[\lambda _0,+\infty )$ together with each of its points $\lambda _0$. A nonempty subset of $(0,+\infty )$ with this property is determined by its infimum $a$ and by whether $a$ belongs to it, which gives the stated forms. If $b_ h(f)>0$, corollary 15.35.3 gives $\Lambda _ h(f)=(0,+\infty )$. $\square$
Proposition 15.35.13.C791D4F48DEC Suppose $b_ h(f)=0$. Let $\lambda _ j>0$ decrease to zero, and let $u_ j$ be any solution of (15.35.1) at exponent $\lambda _ j$. Put
\[ M_ j=\sup _ Xu_ j,\qquad v_ j=u_ j-M_ j,\qquad c_ j=\mathrm e^{\lambda _ jM_ j}. \]
Then
\begin{gather*} c_ j\longrightarrow 0,\qquad \lambda _ jM_ j\longrightarrow -\infty ,\\ \| \mathrm{MA}^{\mathrm{ext}}_ h(u_ j)\| _{\mathrm{TV}} =\| \mathrm{MA}^{\mathrm{ext}}_ h(v_ j)\| _{\mathrm{TV}} \longrightarrow 0. \end{gather*}
Moreover, $\mathrm e^{\lambda _ ju_ j}\to 0$ uniformly on $X$, where the exponential is zero on the pole set, and $u_ j\to -\infty $ uniformly, that is, $\sup _ Xu_ j\to -\infty $.
Proof. Set $g_ j=\mathrm e^{\lambda _ jv_ j}f$. By lemma 5.44.5, applied to the finite non-pluripolar measure $f\, \mathrm dV$, $g_ j\to f$ in $L^1$; each $g_ j$ is nonzero, since $v_ j$ is finite almost everywhere. Since the Monge–Ampère measure is unchanged by adding constants,
\[ \mathrm{MA}^{\mathrm{ext}}_ h(v_ j)=c_ jg_ j\, \mathrm dV, \qquad \text{hence}\qquad 0<c_ j\leq b_ h(g_ j) \]
by theorem 15.28.4. The upper semicontinuity of the threshold, theorem 15.32.1, gives $\limsup _ jc_ j\leq b_ h(f)=0$. Since $0\leq g_ j\leq f$, the total masses are at most $c_ j\| f\| _{L^1}$ and tend to zero. Moreover $\lambda _ jM_ j=\log c_ j\to -\infty $, and $0\leq \mathrm e^{\lambda _ ju_ j}\leq \mathrm e^{\lambda _ jM_ j}=c_ j$. Finally, once $\lambda _ j<1$ and $\log c_ j<0$, we have $M_ j=\lambda _ j^{-1}\log c_ j\leq \log c_ j$, so $M_ j\to -\infty $. $\square$
For the density of theorem 15.35.10, solutions exist for every positive exponent, so the collapse applies as $\lambda \downarrow 0$ to every choice of solutions, in particular to the greatest ones. By contrast, the subsolutions of theorem 15.35.9 keep the mass $CV_0$ for every exponent, while the masses of the solutions above them tend to zero: on a nonclosed background, the order of two potentials does not give an order of their total Monge–Ampère masses. Whether the normalized potentials converge in capacity, and which equation a normalized cluster limit satisfies at exponent zero, are not decided by this argument. Positive endpoints of $\Lambda _ h(f)$ are governed instead by the compactness criterion below.
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