Pome93CEF832AEA293CEF832AEA2
Throughout the rest of the chapter, $h$ denotes a Hermitian form on $X$, used as the background, and densities are taken with respect to the fixed smooth positive probability volume form $\mathrm{d}V$; the sections that use another background say so explicitly. No singularity type is prescribed: the unknown potentials range over all of $\mathrm{PSH}(X,h)$. Constants belong to $\mathrm{PSH}(X,h)$, and every $u\in \mathrm{PSH}(X,h)$ is bounded above, so $u\preceq v$ for every bounded $v\in \mathrm{PSH}(X,h)$.
The masses of bounded potentials are measured by the lower and upper masses of definition 13.14.14,
where $v$ ranges over the bounded functions in $\mathrm{PSH}(X,h)$. The form $h$ has positive lower mass if $v_-(h)>0$. If $h$ is Kähler, Stokes' theorem gives $v_-(h)=v_+(h)=\int _ Xh^ n$; on a non-Kähler manifold the lower mass can vanish (theorem 12.19.1).
For a nonzero nonnegative $q\in L^ p(\mathrm{d}V)$ with $p>1$, theorem 14.19.2 gives a constant $c>0$ and a continuous $u\in \mathrm{PSH}(X,h)$ with $\mathrm{MA}_ h(u)=cq\, \mathrm{d}V$, and proposition 14.19.34 shows that $c$ is determined by $q$. We call $c_ h(q)=c$ the normalization scalar of $q$, and for the measure $\nu =q\, \mathrm{d}V$ we also write $c_ h(\nu )=c_ h(q)$. We say that a finite measure $\nu $ admits a positive scaled subsolution if there are $u\in \mathrm{PSH}(X,h)$ and a constant $a>0$ such that $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq a\nu $.
If $\nu $ admits a solution of $\mathrm{MA}^{\mathrm{ext}}_ h(u)=c\nu $ with $c>0$, then it admits in particular a positive scaled subsolution. We show that every nonzero finite non-pluripolar measure admits such a solution if and only if $h$ has positive lower mass. The key estimate is a consequence of strict domination: if $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq aq\, \mathrm{d}V$, then $a\leq c_ h(q)$. When $v_-(h)=0$, there are smooth probability densities with arbitrarily small normalization scalars, and a weighted sum of such densities, with scalars decreasing fast enough, is a single probability density admitting no positive scaled subsolution.
Lemma 15.27.1.CE5CA1071A3C Let $0\leq q\in L^ p(\mathrm{d}V)$, $p>1$, be nonzero. If $u\in \mathrm{PSH}(X,h)$ and $a>0$ satisfy
then $a\leq c_ h(q)$. Consequently, for nonzero nonnegative bounded densities $q,r$,
Proof. Let $v$ be a continuous solution of $\mathrm{MA}_ h(v)=c_ h(q)q\, \mathrm{d}V$, and suppose, to reach a contradiction, that $a>c_ h(q)$. Then $\tau =c_ h(q)/a<1$ and, for every real $A$,
The translate $v+A$ is bounded, so $P_{h/2}(v+A)\geq \inf _ Xv+A$ is not identically $-\infty $, and $u\preceq v+A$. Theorem 15.2.2, with $\delta =1/2$ and $c=\tau $, therefore gives $v+A\geq u$ for every real $A$. Letting $A\to -\infty $ at a point where $u$ is finite gives a contradiction. Hence $a\leq c_ h(q)$.
If $q\leq r$, a continuous solution of $\mathrm{MA}_ h(w)=c_ h(r)r\, \mathrm{d}V$ satisfies $\mathrm{MA}_ h(w)\geq c_ h(r)q\, \mathrm{d}V$, so the first assertion gives $c_ h(r)\leq c_ h(q)$. Finally, the equation $\mathrm{MA}_ h(v)=c_ h(q)q\, \mathrm{d}V$ can be written as $\mathrm{MA}_ h(v)=t^{-1}c_ h(q)\, (tq)\, \mathrm{d}V$, and uniqueness of the scalar gives $c_ h(tq)=t^{-1}c_ h(q)$. $\square$
The proof uses only that $v$ is bounded. Hence, more generally, if a nonnegative density $g$ satisfies $\mathrm{MA}_ h(v)=c\, g\, \mathrm{d}V$ for some bounded $v\in \mathrm{PSH}(X,h)$ and some $c>0$, then every $u\in \mathrm{PSH}(X,h)$ and $a>0$ with $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq ag\, \mathrm{d}V$ satisfy $a\leq c$.
Lemma 15.27.2.366F6832AB46 Fix $p>1$. If nonzero nonnegative densities $q_ j,q\in L^ p(\mathrm{d}V)$ satisfy $q_ j\to q$ in $L^ p$, then $c_ h(q_ j)\to c_ h(q)$.
Proof. We first show that the scalars $c_ h(q_ j)$ lie in a compact subset of $(0,+\infty )$. Fix a Gauduchon metric $\gamma $ on $X$ (definition 11.4.1; in dimension one, any Hermitian metric). Pairing the equation for $q_ j$ with $\gamma ^{n-1}$, as in proposition 15.14.2 with $\theta =h$ and barrier $\rho =0$, gives
Since $q_ j^{1/n}\to q^{1/n}$ in $L^ n$, the integrals on the left converge to the corresponding integral for $q$, which is positive. Hence the scalars are bounded above. The lower-density estimate of theorem 14.19.19 gives $c_ h(q_ j)\| q_ j\| _{L^ p}\geq c_ p>0$, so they are also bounded away from zero.
Consequently, the normalized continuous solutions $u_ j$ of $\mathrm{MA}_ h(u_ j)=c_ h(q_ j)q_ j\, \mathrm{d}V$ have uniformly bounded $L^ p$ densities. By theorem 14.19.21, every subsequence has a further subsequence along which $u_ j$ converges uniformly to some $u$ and $c_ h(q_ j)$ converges to some $c>0$. Continuity of bounded Monge–Ampère products under uniform convergence gives $\mathrm{MA}_ h(u)=cq\, \mathrm{d}V$, and uniqueness of the scalar gives $c=c_ h(q)$. Since every subsequence of $(c_ h(q_ j))$ has a further subsequence converging to $c_ h(q)$, the whole sequence converges to $c_ h(q)$. $\square$
Theorem 15.27.3.C7AA7377C144 Suppose $v_-(h)=0$. There is a probability measure $\mu =f\, \mathrm{d}V$, with $f\in L^1(\mathrm{d}V)$ and $f\geq m$ almost everywhere for a constant $m>0$, such that no $u\in \mathrm{PSH}(X,h)$ and $a>0$ satisfy
In particular, the equation $\mathrm{MA}^{\mathrm{ext}}_ h(u)=c\mu $ has no solution $u\in \mathrm{PSH}(X,h)$ for any $c>0$; neither a singularity type nor finiteness of the mass of $u$ is imposed.
Proof. By theorem 12.15.4, theorem 12.2.3, $v_-(h)$ is the infimum of the normalization scalars $c_ h(\nu )$ over smooth strictly positive probability measures $\nu $. Since $v_-(h)=0$, we can choose such measures $\nu _ j=f_ j\, \mathrm{d}V$ with
Set
By monotone convergence, $\int _ Xf\, \mathrm{d}V=1$, so $f\in L^1$. Moreover $f\geq \tfrac 12\min _ X f_1=:m>0$, and $\mu $ is non-pluripolar, being absolutely continuous with respect to $\mathrm{d}V$.
Suppose that (15.27.4) held for some $u$ and $a>0$. Since $\mu \geq 2^{-j}\nu _ j$, we would have
for every $j$, and lemma 15.27.1 would give $a2^{-j}\leq c_ h(\nu _ j)\leq 2^{-2j}$. Thus $a\leq 2^{-j}$ for every $j$, contradicting $a>0$. $\square$
For the reference $0$, definition 16.28.61 says that $u\in \mathrm{PSH}(X,h)$ belongs to the relative envelope class $\mathcal E_{\mathrm{env}}(X,h,0)$ if, for every $b\geq 1$, some $g\in \mathrm{PSH}(X,h)$ satisfies $g\leq bu$.
Corollary 15.27.5.810CC86832BF The following conditions are equivalent:
$v_-(h)>0$.
For every nonzero finite non-pluripolar measure $\mu $ there are $c>0$ and $u\in \mathcal E_{\mathrm{env}}(X,h,0)$ such that $\mathrm{MA}^{\mathrm{ext}}_ h(u)=c\mu $.
For every probability density $f\in L^1(\mathrm{d}V)$ bounded below by a positive constant, there are $u\in \mathrm{PSH}(X,h)$ and $a>0$ such that $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq af\, \mathrm{d}V$.
Proof. The Hermitian form $h$ is semipositive and big, and the reference $0$ satisfies (16.28.67), since $P_ h[0]=0$ and $P_{\varepsilon h}(0)=0$ for every $0<\varepsilon <1$. Hence, if $v_-(h)>0$, theorem 15.37.4, applied with $\theta =h$ and $\varphi =0$, gives (2). Condition (2) implies (3), because a finite measure that is absolutely continuous with respect to $\mathrm{d}V$ is non-pluripolar. Finally, if $v_-(h)=0$, the density of theorem 15.27.3 violates (3); thus (3) implies (1). $\square$
Corollary 15.27.6.05B92E5A4684 Assume $v_-(h)=0$, and let $\mu =f\, \mathrm{d}V$ be the measure of theorem 15.27.3. Then $f\notin L^ p(\mathrm{d}V)$ for every $p>1$. Moreover, for every probability density $g\in L^1(\mathrm{d}V)$ and every $0<t<1$, the probability density
admits no positive scaled subsolution, although $\| g_ t-g\| _{L^1}\leq 2t$. In particular, the probability densities admitting no positive scaled subsolution are dense among all probability densities in $L^1(\mathrm{d}V)$.
Proof. If $f\in L^ p$ for some $p>1$, theorem 14.19.2 would give a continuous solution of $\mathrm{MA}_ h(u)=cf\, \mathrm{d}V$ with $c>0$, contradicting the theorem. Since $g_ t\, \mathrm{d}V\geq t\mu $, a positive scaled subsolution for $g_ t\, \mathrm{d}V$ would also give one for $\mu $. Finally, $\| g_ t-g\| _{L^1}=t\| f-g\| _{L^1}\leq 2t$. $\square$
Smooth strictly positive probability densities are also dense in $L^1(\mathrm{d}V)$, and each of them admits a smooth solution of the normalized equation. Thus, when the lower mass vanishes, solvability is not preserved under $L^1$ approximation. The obstruction comes from the way the density concentrates, not from its size: the density in theorem 15.27.3 is bounded below by a positive constant. It occurs, for instance, on the compact threefold of theorem 12.19.1, where every Hermitian form has zero lower mass.
Theorem 15.27.3 does not exclude unbounded solutions of the exponential equation $\mathrm{MA}^{\mathrm{ext}}_ h(u)=\mathrm{e}^{\lambda u}\mu $ with $\lambda >0$, because the factor $\mathrm{e}^{\lambda u}$ need not have a positive lower bound. It does exclude bounded solutions, since a bounded solution would satisfy $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq \mathrm{e}^{\lambda \inf _ Xu}\mu $. On suitable products, theorem 15.35.10 chooses the concentrating components of such a density so that its exponential equation has a solution for every positive exponent. We do not know whether this holds for every density admitting no positive scaled subsolution, nor which individual measures admit a solution of the normalized equation when $v_-(h)=0$.
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