Proposition 21.12.39

Proposition 21.12.39.75730375E037 Let $\varphi \in \mathrm{PSH}_{\mathrm{tor}}(X,\omega )$. Then

F02AE1451B98 21.12.40

\begin{equation} E_{\omega }(\varphi )=n!\int _ P \left(G_{0}-G_{\varphi }\right)\, \mathrm{d}\operatorname{vol}. \end{equation}

Proof. For $\psi \in \mathrm{PSH}_{\mathrm{tor}}(X,\omega )$, write

\[ \mathcal C(\psi )\coloneqq n!\int _ P\left(G_0-G_\psi \right)\, \mathrm{d}\operatorname{vol}. \]

Step 1. We assume that $\varphi $ is bounded.

Put $h=F_{\varphi }-F_0$, $\varphi _ t=t\varphi $ and

\[ F_ t\coloneqq F_0+th \]

for $t\in [0,1]$. Then $F_ t=F_{\varphi _ t}$ and $G_ t\coloneqq F_ t^*=G_{\varphi _ t}$. Since $\varphi $ is bounded, $h$ is bounded and continuous; in particular, $F_ t\in \mathcal E^\infty (N_{\mathbb R},P)$ for every $t\in [0,1]$.

We claim that

3654D70C9AC4 21.12.41

\begin{equation} \frac{\mathrm d}{\mathrm dt}\mathcal C(\varphi _ t)=\int _{N_{\mathbb R}}h\, \operatorname{MA}_{\mathbb R}(F_ t). \end{equation}

Let $s>0$ and $t,t+s\in [0,1]$. At a point $m\in \operatorname{Int}P$ where both $G_ t$ and $G_{t+s}$ are differentiable, we have

\[ \begin{aligned} G_{t+s}(m)& \geq G_ t(m)-s h\bigl (\nabla G_ t(m)\bigr ),\\ G_ t(m)& \geq G_{t+s}(m)+s h\bigl (\nabla G_{t+s}(m)\bigr ). \end{aligned} \]

Therefore,

\[ h\bigl (\nabla G_{t+s}(m)\bigr )\leq \frac{G_ t(m)-G_{t+s}(m)}{s}\leq h\bigl (\nabla G_ t(m)\bigr ). \]

Integrating the last inequality and using 21.11.57 gives

\[ \int _{N_{\mathbb R}}h\, \operatorname{MA}_{\mathbb R}(F_{t+s})\leq \frac{\mathcal C(\varphi _{t+s})-\mathcal C(\varphi _ t)}{s}\leq \int _{N_{\mathbb R}}h\, \operatorname{MA}_{\mathbb R}(F_ t). \]

Letting $s\to 0+$ and using theorem 21.11.55 for $F_{t+s}\to F_ t$, we obtain the right derivative in (21.12.41). The left derivative follows in the same way. Hence (21.12.41) holds.

By proposition 21.12.24, the right-hand side of (21.12.41) is

\[ \int _ X\varphi \, \omega _{\varphi _ t}^ n. \]

On the other hand, 16.5.57 gives

\[ \frac{\mathrm d}{\mathrm dt}E_{\omega }(\varphi _ t)=\int _ X\varphi \, \omega _{\varphi _ t}^ n. \]

Thus $E_{\omega }(\varphi _ t)-\mathcal C(\varphi _ t)$ is constant in $t$. It vanishes at $t=0$, hence (21.12.40) follows by taking $t=1$.

Step 2. We handle the general case.

Both sides of (21.12.40) change by the same additive constant when we replace $\varphi $ by $\varphi +A$, namely by

\[ A\int _ X\omega ^ n=A\, n!\operatorname{vol}P. \]

We may therefore assume that $\varphi \leq 0$. For $C>0$, put

\[ \varphi _ C\coloneqq \varphi \lor (-C). \]

Then $\varphi _ C\in \mathrm{PSH}_{\mathrm{tor}}(X,\omega )\cap L^{\infty }(X)$ and $\varphi _ C\searrow \varphi $ as $C\to \infty $. By proposition 21.12.23, $G_{\varphi _ C}\nearrow G_{\varphi }$. Applying Step 1 to $\varphi _ C$, we get

\[ E_{\omega }(\varphi _ C)=n!\int _ P \left(G_0-G_{\varphi _ C}\right)\, \mathrm{d}\operatorname{vol}. \]

Letting $C\to \infty $, the left-hand side converges to $E_{\omega }(\varphi )$ by proposition 16.5.20. On the right-hand side, since $\varphi _ C\leq 0$, we have $G_{\varphi _ C}\geq G_0$, and the monotone convergence theorem applied to $G_{\varphi _ C}-G_0\nearrow G_{\varphi }-G_0$ gives (21.12.40). $\square$

Proposition 21.12.39`75730375E037`

Comments

Proposition 21.12.3975730375E037

Comments

Proposition 21.12.39`75730375E037`