904042ECF361
Positive lower volume permits a small stretch away from any less singular potential, and the admissible stretch depends only on the lower volume and a mass bound. Mixing the stretched potential with a strict potential produces a strict potential lying below the original one. The same bound for mixed masses controls the change of the lower volume under a small smooth perturbation of the background. Throughout this section, $\omega $ is a fixed Hermitian form on $X$.
Lemma 13.20.1.EDD01F5B65E5 Let $u,r\in \mathrm{PSH}(X,\theta )$, with $u\leq r$, and assume $\underline{\operatorname {vol}}_\theta (u)>0$. For a Hermitian form $\Omega \geq \theta $, set
\[ b_0=1+\frac{\underline{\operatorname {vol}}_\theta (u)}{2^{n+1}n\, v_+(\Omega )}. \]
Then $\mathcal A_ b(u,r)\neq \varnothing $, so that $Q_ b(u,r)$ is defined, for $1\leq b\leq b_0$, and
\[ \lim _{b\downarrow 1}\overline{\operatorname {vol}}_\theta (Q_ b(u,r)) =\overline{\operatorname {vol}}_\theta (u). \]
Proof.
Put $K=v_+(\Omega )$. Since $0<\underline{\operatorname {vol}}_\theta (u)\leq K$, we have $1<b_0\leq 2$. After a common translation we may assume $u\leq r\leq 0$. Suppose, to reach a contradiction, that $\mathcal A_ b(u,r)=\varnothing $ for some $1<b\leq b_0$. Set
\[ u_ j=\max (u,r-j),\qquad v_ j=u+(b-1)u_ j, \qquad q_ j=P_\theta (v_ j-(b-1)r). \]
Here, as for stretching envelopes, a competitor $g$ for $q_ j$ is required to satisfy $g+(b-1)r\leq v_ j$, which is meaningful also at common poles. Off the poles the obstacle $v_ j-(b-1)r=u+(b-1)(u_ j-r)$ lies between $u-(b-1)j$ and $u$. Hence $q_ j$ is $\theta $-plurisubharmonic and $q_ j\sim u$. As $j$ increases, $q_ j$ decreases. A $\theta $-plurisubharmonic decreasing limit would belong to $\mathcal A_ b(u,r)$, so the limit is identically $-\infty $, and $\sup _ Xq_ j\to -\infty $ by theorem 5.25.2.
Let $C_ j=\{ q_ j+(b-1)r=v_ j\} $. By the contact theorem, $\mu _{q_ j}$ is carried by $C_ j$. The current $b\theta +\mathrm{dd}^{\mathrm{c}}(q_ j+(b-1)r)$ equals $(\theta +\mathrm{dd}^{\mathrm{c}}q_ j)+(b-1)(\theta +\mathrm{dd}^{\mathrm{c}}r)\geq \theta +\mathrm{dd}^{\mathrm{c}}q_ j$. Positivity of the mixed products and ordered contact in background $b\theta $ therefore give
\[ \mu _{q_ j}\leq \mathbf1_{C_ j}\bigl ((\theta +\mathrm{dd}^{\mathrm{c}}u)+(b-1)(\theta +\mathrm{dd}^{\mathrm{c}}u_ j)\bigr )^ n. \]
On $C_ j$ we have $q_ j=u+(b-1)(u_ j-r)\geq bu$, because $u_ j\geq u$ and $r\leq 0$. Fix $k>0$. For all sufficiently large $j$, $q_ j\leq -bk$ everywhere, so $C_ j\subset \{ u\leq -k\} $, and consequently
080E895C193D
13.20.2
\begin{equation} \underline{\operatorname {vol}}_\theta (u) \leq \mu _{q_ j}(X) \leq \int _{\{ u\leq -k\} }\bigl ((\theta +\mathrm{dd}^{\mathrm{c}}u)+(b-1)(\theta +\mathrm{dd}^{\mathrm{c}}u_ j)\bigr )^ n. \end{equation}
To estimate the right-hand side, note that for any $s,t\in \mathrm{PSH}(X,\theta )$ and $0\leq i\leq n$, positivity of the non-pluripolar mixed products and proposition 13.14.16 imply
\[ \begin{aligned} & \binom ni\int _ X(\theta +\mathrm{dd}^{\mathrm{c}}s)^{n-i}\wedge (\theta +\mathrm{dd}^{\mathrm{c}}t)^ i\\ & \quad \leq \binom ni\int _ X(\Omega +\mathrm{dd}^{\mathrm{c}}s)^{n-i}\wedge (\Omega +\mathrm{dd}^{\mathrm{c}}t)^ i\\ & \quad \leq 2^ n\int _ X\mathrm{MA}_\Omega ((s+t)/2)\leq 2^ nK. \end{aligned} \]
The second inequality follows by expanding $(\Omega +\mathrm{dd}^{\mathrm{c}}\frac{s+t}2)^ n$; for the possibly unbounded average, apply this expansion to its truncations at constant levels, which remain $\Omega $-plurisubharmonic. Expanding the integrand in (13.20.2), and using $(b-1)^ i\leq b-1$ for $i\geq 1$, gives
\[ \underline{\operatorname {vol}}_\theta (u) \leq \mu _ u(\{ u\leq -k\} )+n2^ nK(b-1). \]
The measure $\mu _ u$ is finite and puts no mass on the pole set of $u$, so the first term tends to zero as $k\to \infty $. By the choice of $b_0$, the remaining error is at most half of $\underline{\operatorname {vol}}_\theta (u)>0$, a contradiction. Thus $\mathcal A_ b(u,r)\neq \varnothing $ for $1<b\leq b_0$; for $b=1$ the family contains $u$, and $Q_1(u,r)=u$.
Put $h_0=Q_{b_0}(u,r)$ and, for $1<b\leq b_0$, $t=(b-1)/(b_0-1)\in (0,1]$. The inequality $h_0\leq b_0u-(b_0-1)r$ shows that $(1-t)u+th_0$ is a competitor for $Q_ b(u,r)$, and $Q_ b(u,r)\leq bu-(b-1)r\leq u$. Hence
\[ (1-t)u+th_0\leq Q_ b(u,r)\leq u. \]
Addition, homogeneity and volume monotonicity give
\[ (1-t)^ n\overline{\operatorname {vol}}_\theta (u) \leq \overline{\operatorname {vol}}_\theta (Q_ b(u,r)) \leq \overline{\operatorname {vol}}_\theta (u). \]
Letting $b\downarrow 1$ proves the limit. Undoing the common translation changes neither the volumes nor the nonemptiness of the competitor families.
$\square$
Corollary 13.20.3.D51B82B18D30 If $u\in \mathrm{PSH}(X,\theta )$ has positive lower volume, then there are $w\in \mathrm{PSH}(X,\theta )$ and $\varepsilon >0$ with
\[ w\leq u,\qquad \theta +\mathrm{dd}^{\mathrm{c}}w\geq \varepsilon \omega . \]
Proof.
Translate $u$ so that $u\leq V_\theta $. By bigness, there are $\rho \in \mathrm{PSH}(X,\theta )$ and $\delta >0$ with $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq \delta \omega $; after a translation, $\rho \leq V_\theta $. By lemma 13.20.1 with $r=V_\theta $, there is $b>1$ for which $h=Q_ b(u,V_\theta )$ is defined. Put
\[ w=b^{-1}h+(1-b^{-1})\rho . \]
Since $b^{-1}h+(1-b^{-1})V_\theta \leq u$ and $\rho \leq V_\theta $, we have $w\leq u$; moreover $\theta +\mathrm{dd}^{\mathrm{c}}w\geq (1-b^{-1})\delta \omega $. Undoing the translation of $u$ by the same translation of $w$ gives the assertion with $\varepsilon =(1-b^{-1})\delta $.
$\square$
Proposition 13.20.4.0729BE10D92E Assume in addition that $\theta \geq 0$. If $u\in \mathrm{PSH}(X,\theta )$ and $\underline{\operatorname {vol}}_\theta (u)>0$, then $P_{\varepsilon \theta }(u)\not\equiv -\infty $ for some $\varepsilon \in (0,1)$.
Proof.
By corollary 13.20.3, choose $w\leq u$ and $\delta >0$ with $\theta +\mathrm{dd}^{\mathrm{c}}w\geq \delta \omega $. Choose $C>0$ with $\theta \leq C\omega $, and $0<t<\min (1,\delta /C)$. Then
\[ (1-t)\theta +\mathrm{dd}^{\mathrm{c}}w\geq (\delta -tC)\omega \geq 0. \]
Thus $w$ is a competitor for $P_{(1-t)\theta }(u)$, and $\varepsilon =1-t$ has the required property.
$\square$
For a reference potential $\varphi $, the positive-volume model condition reads
F65E6B4224EA
13.20.5
\begin{equation} P_\theta [\varphi ]=\varphi ,\qquad \underline{\operatorname {vol}}_\theta (\varphi )>0. \end{equation}
When $\theta \geq 0$, proposition 13.20.4 shows that (13.20.5) implies the smaller-background model condition (16.28.67) of subsection 16.28.66. If moreover $v_-(\theta )>0$, the converse holds. Indeed, if $P_{\varepsilon \theta }(\varphi )\not\equiv -\infty $, then a translate of the same competitor shows that $P_{\varepsilon \theta }(\psi )\not\equiv -\infty $ for every $\psi \sim \varphi $ in $\mathrm{PSH}(X,\theta )$, and lemma 16.28.69 gives $\int _ X\mathrm{MA}_\theta (\psi )\geq (1-\varepsilon )^ nv_-(\theta )$. Hence
\[ \underline{\operatorname {vol}}_\theta (\varphi ) \geq (1-\varepsilon )^ n v_-(\theta )>0. \]
Without the lower bound $v_-(\theta )>0$, the same lemma still shows that each potential equivalent to $\varphi $ has positive mass, but not that these masses are bounded away from zero.
A small smooth increase of the background changes the lower volume by at most a linear error, and a strict potential lying below $u$ gives a multiplicative comparison for both volumes.
Proposition 13.20.6.F4C607C7C503 There is $C>0$, depending only on $\theta $ and $\omega $, such that, for every $u\in \mathrm{PSH}(X,\theta )$, every $\varepsilon \in [0,1]$ and every smooth real $(1,1)$-form $\eta $ with $\theta \leq \eta \leq \theta +\varepsilon \omega $,
\[ 0\leq \underline{\operatorname {vol}}_\eta (u) -\underline{\operatorname {vol}}_\theta (u)\leq C\varepsilon . \]
If $\underline{\operatorname {vol}}_\theta (u)>0$, there is also $\delta >0$, depending on $u$, such that, for every $\varepsilon \geq 0$ and every smooth $\eta $ with $\theta \leq \eta \leq \theta +\varepsilon \omega $, both singularity volumes satisfy
\[ \operatorname {vol}_\theta (u) \leq \operatorname {vol}_\eta (u) \leq (1+\varepsilon /\delta )^ n\operatorname {vol}_\theta (u). \]
Proof.
Choose a Hermitian form $\Omega $ with $\Omega \geq \theta $ and $\Omega \geq \omega $, and set $C=n2^ n v_+(\Omega )$. For every $v\in \mathrm{PSH}(X,\theta )$ and $1\leq k\leq n$, the mixed-mass estimate in the proof of lemma 13.20.1, applied to $s=v$ and $t=0$, together with $\omega \leq \Omega $, gives
\[ \binom nk\int _ X(\theta +\mathrm{dd}^{\mathrm{c}}v)^{n-k}\wedge \omega ^ k \leq 2^ n v_+(\Omega ). \]
Expanding in the smooth positive perturbation and using $\varepsilon ^ k\leq \varepsilon $ yields
\[ \int _ X\mathrm{MA}_\eta (v) \leq \int _ X\mathrm{MA}_{\theta +\varepsilon \omega }(v) \leq \int _ X\mathrm{MA}_\theta (v)+C\varepsilon . \]
Each $v\sim u$ in $\mathrm{PSH}(X,\theta )$ is also a competitor for the lower volume in background $\eta $. Taking the infimum over these $v$ proves the upper bound for $\underline{\operatorname {vol}}_\eta (u)-\underline{\operatorname {vol}}_\theta (u)$. The lower bound is background monotonicity (theorem 13.17.1).
Now assume that $u$ has positive lower volume, and choose $w\leq u$ and $\delta >0$ with $\theta +\mathrm{dd}^{\mathrm{c}}w\geq \delta \omega $ by corollary 13.20.3. The case $\varepsilon =0$ is immediate. For $\varepsilon >0$ put $t=\varepsilon /\delta $. The potential $tw$ belongs to $\mathrm{PSH}(X,t(\theta -\delta \omega ))$, and the form $\eta +t(\theta -\delta \omega )$ is big, since adding $\mathrm{dd}^{\mathrm{c}}((1+t)w)$ to it gives a form at least $(1+t)\delta \omega -t\delta \omega =\delta \omega $. Addition and monotonicity give
\[ \begin{split} \operatorname {vol}_\eta (u) & \leq \operatorname {vol}_{\eta +t(\theta -\delta \omega )}(u+tw)\\ & \leq \operatorname {vol}_{(1+t)\theta }((1+t)u) =(1+t)^ n\operatorname {vol}_\theta (u). \end{split} \]
Here we used $\eta +t(\theta -\delta \omega )\leq (1+t)\theta $, which holds because $\eta \leq \theta +\varepsilon \omega $, and $u+tw\leq (1+t)u$. The lower bound is again background monotonicity.
$\square$
Pome
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