Proposition 10.9.10CB97B7394604

Proof. We replace the factors $\omega _ v$ of $\omega _ v^ n$ by $\omega _ u$ one at a time, using lemma 10.9.4; the errors are traces of products with fewer factors different from $\omega $, and Steps 1 and 2 bound them by multiples of $\int _ E\omega _ u^ n$. For a product $T$ of $n-1$ factors chosen among $\omega ,\omega _ u,\omega _ v$, let $\mathcal P(T)$ be the set of products obtained from $T$ by deleting one factor and inserting $\omega ^2$, or by deleting two factors and inserting $\omega ^3$. The right side of (10.9.6) has at most $n^2$ terms, each of which is an element of $\mathcal P(T)$, and $0<v-u\leq s$ on $E$. Hence lemma 10.9.4 gives

with a constant $B\geq 1$ depending only on $(X,\omega )$.

Step 1. Pure traces. For $0\leq k\leq n$ put $a_ k=\int _ E\omega _ u^ k\wedge \omega ^{n-k}\geq 0$, and apply (10.9.11) to $T=\omega _ u^ k\wedge \omega ^{n-k-1}$, $0\leq k<n$. The function $2v$ belongs to $\mathrm{PSH}(X,\omega )$, because $\omega +\mathrm{dd}^{\mathrm{c}}(2v)=2\omega _ v-\omega \geq 0$, and it is bounded. Since the products are affine in each potential and $v=\frac12\cdot 0+\frac12\cdot 2v$, we have $\omega _ v\wedge T=\frac12\omega \wedge T+\frac12\omega _{2v}\wedge T$, and $\omega _{2v}\wedge T$ is positive by lemma 10.9.3(2); hence $\int _ E\omega _ v\wedge T\geq \tfrac 12a_ k$. Moreover $\int _ E\omega _ u\wedge T=a_{k+1}$, and the elements of $\mathcal P(T)$ are among the products $\omega _ u^ h\wedge \omega ^{n-h}$ with $h\in \{ k-2,k-1,k\} $. Thus (10.9.11) gives

Choose $s_0>0$ with $s_0B(n+1)4^ n\leq 1/4$; in particular $sB\leq 1$. We prove $a_ k\leq 4a_{k+1}$ for $0\leq k<n$ by induction on $k$. If it holds for all smaller indices, then $a_ h\leq 4^{k-h}a_ k$ for $h\leq k$, so $\sum _{h=0}^ ka_ h\leq (k+1)4^ ka_ k$, and the error term in (10.9.12) is at most $s_0B(n+1)4^ na_ k\leq \tfrac 14a_ k$. Hence $\tfrac 12a_ k\leq a_{k+1}+\tfrac 14a_ k$, that is, $a_ k\leq 4a_{k+1}$. Consequently

Step 2. Mixed traces. For $j,k\geq 0$ with $j+k\leq n$ put $a_{j,k}=\int _ E\omega _ u^ j\wedge \omega _ v^ k\wedge \omega ^{n-j-k}\geq 0$, so that $a_{j,0}=a_ j$. If $k\geq 1$, apply (10.9.11) to $T=\omega _ u^ j\wedge \omega _ v^{k-1}\wedge \omega ^{n-j-k}$. Then $\int _ E\omega _ v\wedge T=a_{j,k}$ and $\int _ E\omega _ u\wedge T=a_{j+1,k-1}$, while every element of $\mathcal P(T)$ has fewer than $j+k$ factors different from $\omega $. Thus

We show by induction on $m=j+k$ that $a_{j,k}\leq C_ m\sum _{h=0}^ ma_ h$ with $C_ m$ depending only on $n$. For $k=0$ this holds with $C_ m=1$. For $k\geq 1$, since $sB\leq 1$, applying the last inequality $k$ times, which replaces the factors $\omega _ v$ one at a time by $\omega _ u$, gives

and by the induction hypothesis the last sum, which has at most $(m+1)^2$ terms, is at most a constant depending only on $n$ times $\sum _{h<m}a_ h$. Together with Step 1 this shows $a_{j,k}\leq C'_ na_ n$ for all $j,k$, where $C'_ n$ depends only on $n$.

Step 3. Conclusion. By symmetry of the products,

since the sum telescopes. Apply (10.9.11) to each of the $n$ products $T=\omega _ v^ i\wedge \omega _ u^{n-1-i}$. The resulting error terms are $sB$ times sums of at most $n^2$ traces $a_{p,q}$, hence at most a constant times $sa_ n$ by Step 2. Since $a_ n=\int _ E\omega _ u^ n$, this proves $\int _ E\omega _ v^ n\leq (1+Cs)\int _ E\omega _ u^ n$ with $C$ depending only on $(X,\omega )$.

Step 4. Positivity. Let $u\in \mathrm{PSH}(X,\omega )$ be bounded, put $c=\inf _ Xu$, fix $s\in (0,s_0)$, and apply the inequality to the constant function $v=c+s$. Then $\omega _ v=\omega \geq \tfrac 12\omega $, and $E=\{ u<c+s\} $ is nonempty by the definition of the infimum and open because $u$ is upper semicontinuous; moreover $0<v-u\leq s$ on $E$ since $u\geq c$. Hence $0<\int _ E\omega ^ n\leq (1+Cs)\int _ E\omega _ u^ n$, and therefore $\int _ X\omega _ u^ n\geq \int _ E\omega _ u^ n>0$. $\square$