Proposition 11.2.10.4793125897B2 For $u\in \mathcal H_\omega $ and $v\in C^\infty (X,\mathbb R)$, put $\Omega _ t=(\omega +t\mathrm{dd}^{\mathrm{c}}u)^{n-1}$ for $0\leq t\leq 1$. Then
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11.2.11
\begin{equation} \begin{split} DE_{\omega ,u}(v)-\int _ Xv\omega _ u^ n =n\int _0^1t\int _ Xv\bigl (& u\mathrm{dd}^{\mathrm{c}}\Omega _ t +\mathrm{d}u\wedge \mathrm d^{\mathrm c}\Omega _ t\\ & -\mathrm d^{\mathrm c}u\wedge \mathrm{d}\Omega _ t\bigr )\, \mathrm{d}t. \end{split} \end{equation}
Proof.
Differentiating the radial integral in (11.2.7) in the direction $v$ gives
\[ DE_{\omega ,u}(v) =\int _0^1\left(\int _ Xv(\omega +t\mathrm{dd}^{\mathrm{c}}u)^ n +nt\int _ Xu\mathrm{dd}^{\mathrm{c}}v\wedge \Omega _ t\right)\mathrm dt. \]
On the other hand,
\[ \begin{aligned} \int _ Xv\omega _ u^ n & =\int _0^1\frac{\mathrm d}{\mathrm dt} \left(t\int _ Xv(\omega +t\mathrm{dd}^{\mathrm{c}}u)^ n\right)\mathrm dt\\ & =\int _0^1\left(\int _ Xv(\omega +t\mathrm{dd}^{\mathrm{c}}u)^ n +nt\int _ Xv\mathrm{dd}^{\mathrm{c}}u\wedge \Omega _ t\right)\mathrm dt. \end{aligned} \]
The terms without a factor $nt$ cancel in the difference, leaving
\[ \begin{aligned} DE_{\omega ,u}(v)-\int _ Xv\omega _ u^ n & =n\int _0^1t\int _ X (u\mathrm{dd}^{\mathrm{c}}v-v\mathrm{dd}^{\mathrm{c}}u)\wedge \Omega _ t\, \mathrm dt\\ & =n\int _0^1t\int _ Xv \bigl (\mathrm{dd}^{\mathrm{c}}(u\Omega _ t)-\mathrm{dd}^{\mathrm{c}}u\wedge \Omega _ t\bigr )\, \mathrm dt. \end{aligned} \]
The second equality is lemma 11.1.2. Expanding the last parenthesis by the product rule in the proof of lemma 11.2.2 gives (11.2.11). When $n=1$, one has $\Omega _ t=1$ and the right-hand side vanishes.
$\square$
Pome
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