17.3 Relative Monge–Ampère energyEC60D9ED3C42

Proof. Let $w\in \mathcal E^\infty (X,\theta ,\phi )$. For $0\leq r\leq 1$ the potential $\phi +r(w-\phi )=(1-r)\phi +rw$ again belongs to $\mathcal E^\infty (X,\theta ,\phi )$, and multilinearity gives

Since $\binom nj\int _0^1r^ j(1-r)^{n-j}\, \mathrm dr=1/(n+1)$, as one sees by integrating by parts $j$ times, we obtain

Now put $h=v-u$ and $q_{r,t}=\phi +r(u_ t-\phi )$. Since $\theta _{q_{r,t}}=(1-r)\theta _\phi +r(1-t)\theta _ u+rt\theta _ v$, the measures $\mu _{q_{r,t}}$ are polynomials in $(r,t)$ whose coefficients are mixed measures of total mass $m$, and the weights $u_ t-\phi =(u-\phi )+th$ are bounded. All expressions below are therefore polynomials in $(r,t)$ with finite coefficients, and the derivatives are derivatives of polynomials. Identities between such polynomials of measures follow from the corresponding identities between polynomials in commuting variables $x,y,z$, by substituting the mixed measure $\langle \theta _\phi ^ a\wedge \theta _ u^ b\wedge \theta _ v^ c\rangle $ for the monomial $x^ ay^ bz^ c$; multilinearity shows that this substitution is compatible with the products occurring here. For instance, the identities $\partial _ t\ell ^ n=nr(z-y)\ell ^{n-1}$ and $\partial _ r\ell ^ n=n\bigl ((1-t)y+tz-x\bigr )\ell ^{n-1}$ for $\ell =(1-r)x+r(1-t)y+rtz$ give

in the sense of (17.2.1). Differentiating (17.3.7) with $w=u_ t$ and using the first of these identities, we obtain

By theorem 17.2.5, applied to the bounded quasi-psh differences $u_ t-\phi $ and $h$, the second integral equals $\int _ Xh\, \mathrm{dd}^{\mathrm{c}}(u_ t-\phi )\wedge \langle \theta _{q_{r,t}}^{n-1}\rangle $, which is $\frac1n\partial _ r\int _ Xh\, \mu _{q_{r,t}}$ by the second identity. The integrand in $r$ is therefore $\partial _ r\bigl [r\int _ Xh\, \mu _{q_{r,t}}\bigr ]$, and since $q_{1,t}=u_ t$,

This is (17.3.5). Integrating it from $t=0$ to $t=1$, and expanding $\mu _{u_ t}=\sum _ j\binom nj t^ j(1-t)^{n-j}\langle \theta _ v^ j\wedge \theta _ u^{n-j}\rangle $ with the same beta integrals, gives (17.3.6). If $u\leq v$, every term of the cocycle formula is nonnegative, so the energy is increasing. Every mixed measure has total mass $m$, so taking $v=u+C$ in the cocycle formula gives the last assertion. $\square$

Proof. Let $u_ t=(1-t)u+tv$ and $h=v-u$. Then $\partial _ t\mu _{u_ t}=n\, \mathrm{dd}^{\mathrm{c}}h\wedge \langle \theta _{u_ t}^{n-1}\rangle $. Indeed, by multilinearity and (17.2.1), both sides expand into combinations of the mixed measures $\langle \theta _ u^ a\wedge \theta _ v^{n-a}\rangle $, obtained by substituting $\langle \theta _ u^ a\wedge \theta _ v^{n-a}\rangle $ for $y^ az^{n-a}$ in the two sides of the identity $\partial _ t((1-t)y+tz)^ n=n(z-y)((1-t)y+tz)^{n-1}$ in commuting variables $y,z$. Differentiating (17.3.5) once more, again as a polynomial in $t$, and using this identity, we obtain

by lemma 17.2.7. Thus $t\mapsto E_\theta ^\phi (u_ t)$ is concave, and its derivative decreases from $\int _ Xh\, \mu _ u$ at $t=0$ to $\int _ Xh\, \mu _ v$ at $t=1$. Integrating the derivative over $[0,1]$ gives the tangent bounds (17.3.10). Since $\int _ X\langle \theta _{u_ t}^ n\rangle =m$, the first variation formula also gives $|E_\theta ^\phi (v)-E_\theta ^\phi (u)|\leq m\sup |v-u|$. Finally, the cocycle formula (17.3.6) applied to the pair $\psi ,u$ reads $E_\theta ^\phi (u)-E_\theta ^\phi (\psi )=E_\theta ^\psi (u)$. $\square$

Proof. If $u\leq u'$, every majorant of $u'$ in the definition is a majorant of $u$, so $E_\theta ^\phi (u)\leq E_\theta ^\phi (u')$. Translating majorants by a constant and using the shift formula on $\mathcal E^\infty (X,\theta ,\phi )$ gives the shift formula. The potentials $u_ k=\max \{ u,\phi -k\} $ belong to $\mathcal E^\infty (X,\theta ,\phi )$, decrease with $k$ and dominate $u$; hence $E_\theta ^\phi (u_ k)$ decreases and is bounded below by $E_\theta ^\phi (u)$. Conversely, let $w$ be any majorant in the definition, and choose $C$ with $w\geq \phi -C$. For $k\geq C$ both $u$ and $\phi -k$ lie below $w$, so $u_ k\leq w$ and $E_\theta ^\phi (u_ k)\leq E_\theta ^\phi (w)$. Letting $k\to \infty $ and taking the infimum over $w$ gives $\lim _ kE_\theta ^\phi (u_ k)\leq E_\theta ^\phi (u)$. $\square$

Proof. First let the net be a sequence $(u_ j)$. By monotonicity $E_\theta ^\phi (u_ j)$ decreases to a limit $L\geq E_\theta ^\phi (u)$. Let $w\in \mathcal E^\infty (X,\theta ,\phi )$ with $w\geq u$, and put $v_ j=\max \{ u_ j,w\} $. Then $v_ j$ decreases pointwise to $\max \{ u,w\} =w$. Since $u_ j\leq u_1\leq \phi +C$ and $w\geq \phi -C$ for some constant $C$, the differences $v_ j-w$ are uniformly bounded. The tangent bound (17.3.10) gives

by bounded convergence. Since $E_\theta ^\phi (u_ j)\leq E_\theta ^\phi (v_ j)$, we get $L\leq E_\theta ^\phi (w)$, and taking the infimum over $w$ gives $L\leq E_\theta ^\phi (u)$.

For a general net, corollary 10.3.6 shows that $u_ i\to u$ in $L^1(X)$; its hypothesis holds because $\sup _ Xu_ i\geq \sup _ Xu>-\infty $. Choose indices $i_1\leq i_2\leq \cdots $ with $\| u_{i_ k}-u\| _{L^1}<1/k$. The decreasing pointwise limit $\tilde u$ of $(u_{i_ k})$ is a $\theta $-psh function by theorem 5.10.1, applied in charts after adding smooth local potentials of $\theta $, since it dominates $u$ and so is not identically $-\infty $. By monotone convergence, $\int _ X(\tilde u-u)\, \omega ^ n=\lim _ k\int _ X(u_{i_ k}-u)\, \omega ^ n=0$, so $\tilde u=u$ almost everywhere, hence everywhere. The sequence case gives $E_\theta ^\phi (u_{i_ k})\to E_\theta ^\phi (u)$. The net $E_\theta ^\phi (u_ i)$ is decreasing, so its limit is its infimum, which is at most $\lim _ kE_\theta ^\phi (u_{i_ k})=E_\theta ^\phi (u)$; the reverse inequality is monotonicity. (The selected indices need not be cofinal in $I$; this is why we compare infima.)

For concavity let $u,v\preceq \phi $ and $0<t<1$; we may assume that $E_\theta ^\phi (u)$ and $E_\theta ^\phi (v)$ are finite, since otherwise there is nothing to prove. The potentials $(1-t)\max \{ u,\phi -k\} +t\max \{ v,\phi -k\} $ belong to $\mathcal E^\infty (X,\theta ,\phi )$ and decrease to $(1-t)u+tv$ as $k\to \infty $. Concavity on $\mathcal E^\infty (X,\theta ,\phi )$ and decreasing continuity give

In particular the finite locus of the energy is convex. $\square$

Proof. Both conditions are invariant under adding a constant to $u$, so we may assume $u\leq \phi $. Put $u_ k=\max \{ u,\phi -k\} $ and $A_ k=\{ u>\phi -k\} $. The potentials $u_ k$ belong to $\mathcal E^\infty (X,\theta ,\phi )$ and have mass $m$. Since $u_ k\leq \phi $, all terms in (17.3.3) are nonpositive, and keeping only the term $j=n$ gives

Suppose that $E_\theta ^\phi (u)>-\infty $. Since $E_\theta ^\phi (u)\leq E_\theta ^\phi (u_ k)\leq E_\theta ^\phi (\phi )=0$, the right side is bounded by $C=-(n+1)E_\theta ^\phi (u)$. On $X\setminus A_ k$ the weight $\phi -u_ k$ equals $k$ outside a pluripolar set; on the plurifine open set $A_ k$ we have $u_ k=u$, so $\mu _{u_ k}=\mu _ u$ there by locality. Hence

The sets $A_ k$ increase to the set where $u$ is finite, which carries $\mu _ u$, so $\int _ X\langle \theta _ u^ n\rangle \geq m$; since $u\leq \phi $, mass monotonicity (theorem 13.8.7) gives the reverse inequality, and therefore $u\in \mathcal E(X,\theta ,\phi )$. Locality also gives $\int _{A_ k}(\phi -u)\mu _ u\leq C$, and monotone convergence yields $\int _ X(\phi -u)\mu _ u\leq C$. This proves that $u\in \mathcal E^1(X,\theta ,\phi )$ and the first inequality in (17.3.15).

Conversely, suppose $u\in \mathcal E^1(X,\theta ,\phi )$, and let $I=\int _ X(\phi -u)\mu _ u<\infty $. The measures $\mu _{u_ k}$ and $\mu _ u$ have the same total mass $m$ and agree on $A_ k$, so $\mu _{u_ k}(X\setminus A_ k)=\mu _ u(X\setminus A_ k)$. Thus

The lower tangent bound in (17.3.10), applied to $u_ k$ and $\phi $, gives $E_\theta ^\phi (u_ k)\geq -I$. Letting $k\to \infty $ and using proposition 17.3.12, we obtain $E_\theta ^\phi (u)\geq -I>-\infty $, which is the second inequality in (17.3.15). The first inequality was proved above. $\square$

Proof. Convex combinations of potentials with the same singularity type as $\phi $ have that singularity type. For $\mathcal E^1(X,\theta ,\phi )$, convexity follows from the concavity of the extended energy (proposition 17.3.13) and theorem 17.3.14. If $u,v\in \mathcal E(X,\theta ,\phi )$ and $0<t<1$, then $(1-t)u+tv\preceq \phi $, and by multilinearity $\mu _{(1-t)u+tv}$ is a combination of the mixed measures $\langle \theta _ u^ j\wedge \theta _ v^{n-j}\rangle $ with coefficients $\binom nj(1-t)^ jt^{n-j}$. Each of these measures has total mass $m$, as noted at the beginning of this section, so $\int _ X\langle \theta _{(1-t)u+tv}^ n\rangle =m$. The last assertion follows from the monotonicity of the extended energy and theorem 17.3.14. $\square$

Proof. For $t\geq 0$ we have

because $\gamma \leq v+t$ and $v\leq u+t$ imply $\gamma \leq u+2t$. The weights $\gamma -u$ and $\gamma -v$ are Borel functions with values in $[0,\infty ]$ on the complements of the pole sets of $u$ and $v$, respectively, and these pole sets carry no mass; the level sets above are taken on these complements. Since $u\in \mathcal E(X,\theta ,\phi )$ and $v-t\preceq \phi $, the comparison principle of proposition 16.25.4, applied with $p=n$ to $u$ and $v-t$, gives

Integrating over level sets, we obtain

where in the last step we used $(v-u)_+\leq \gamma -u$, which holds because $\gamma \geq \max \{ u,v\} $. All functions involved are nonnegative, so Tonelli's theorem applies without any integrability assumption.

For the second assertion, add constants to $u$ and $v$ so that $u,v\leq \phi $; this changes $|u-v|$ by at most a constant, and the mixed measure has finite mass $m$. Put $w=\frac1n\sum _ jw_ j$, which lies in $\mathcal E^1(X,\theta ,\phi )$ by corollary 17.3.17. By multilinearity, $\langle \theta _{w_1}\wedge \cdots \wedge \theta _{w_ n}\rangle \leq n^ n\mu _ w$. Choose $C$ with $w\leq \phi +C$ and apply the first assertion to $u$, $w$ and $\gamma =\phi +C$: its right side is finite because $u,w\in \mathcal E^1(X,\theta ,\phi )$. Hence $\phi -u$ is integrable for $\mu _ w$, and so for the mixed measure. The same holds for $\phi -v$, and $|u-v|\leq (\phi -u)+(\phi -v)$. $\square$

Proof. All mixed measures involved have total mass $m$, so both assertions are unchanged if constants are added to $u$ and $v$; we may therefore assume $u,v\leq \phi $. Put $u_ k=\max \{ u,\phi -k\} $ and $v_ k=\max \{ v,\phi -k\} $. Fix $0\leq j\leq n$ and write $\lambda _ k=\langle \theta _{u_ k}^ j\wedge \theta _{v_ k}^{n-j}\rangle $ and $\lambda =\langle \theta _ u^ j\wedge \theta _ v^{n-j}\rangle $. We show that

On the plurifine open set $A_ k=\{ u>\phi -k\} \cap \{ v>\phi -k\} $ the measures $\lambda _ k$ and $\lambda $ agree by locality, and $u_ k-v_ k=u-v$. By lemma 17.3.18, $|u-v|$ is $\lambda $-integrable. The sets $X\setminus A_ k$ decrease to the union of the pole sets of $u$, $v$ and $\phi $, which is pluripolar and therefore $\lambda $-null, so dominated convergence gives $\int _{X\setminus A_ k}|u-v|\lambda \to 0$. It remains to show that the integral of $|u_ k-v_ k|$ against $\lambda _ k$ over $X\setminus A_ k$ tends to zero.

For $z\in \{ u,v\} $ put $D_ k(z)=\{ z\leq \phi -k\} $, so that $X\setminus A_ k=D_ k(u)\cup D_ k(v)$. We claim that

Consider $z=u$ and $j<n$, and put $q=(v_ k+\phi -k)/2$. Since $\phi -k\leq v_ k\leq \phi $, we have $q\leq \phi -k/2$ on $X$, and $u_ k=\phi -k\leq q$ on $D_ k(u)$. Moreover $\theta _{v_ k}\leq 2\theta _ q$, since $2\theta _ q-\theta _{v_ k}=\theta _\phi \geq 0$, so multilinearity gives $\lambda _ k\leq 2^{n-j}\langle \theta _{u_ k}^ j\wedge \theta _ q^{n-j}\rangle $. The potentials $q$ and $u_ k$ belong to $\mathcal E(X,\theta ,\phi )$, and the non-strict comparison principle of proposition 16.25.4, with $n-j$ factors $\theta _ q$ and the $j$ fixed factors $\theta _{u_ k}$, gives

For $j=n$ the same bound holds directly, since then $\lambda _ k=\mu _{u_ k}$ and $D_ k(u)\subset \{ u\leq \phi -k/2\} $. The measures $\mu _{u_ k}$ and $\mu _ u$ have the same total mass and agree on $\{ u>\phi -k/2\} $ by locality, so the last mass equals $\mu _ u(\{ u\leq \phi -k/2\} )$. Since exchanging $u$ and $v$ and replacing $j$ by $n-j$ does not change $\lambda _ k$, the case $z=v$ follows from the case $z=u$. Finally,

because $\phi -z$ is $\mu _ z$-integrable. This proves (17.3.21).

Since $u_ k-\phi $ and $v_ k-\phi $ take values in $[-k,0]$, we have $|u_ k-v_ k|\leq k$, and (17.3.21) shows that the integral of $|u_ k-v_ k|$ against $\lambda _ k$ over $X\setminus A_ k$ tends to zero. This proves (17.3.20). Now pass to the limit in the cocycle formula for $u_ k,v_ k$: its left side converges to $E_\theta ^\phi (v)-E_\theta ^\phi (u)$ by proposition 17.3.12, and each term on the right converges by (17.3.20). The tangent bounds pass to the limit in the same way, using the terms $j=n$ and $j=0$. Taking $v=\phi $ in the cocycle formula gives the last assertion. $\square$