Lemma 16.5.57.8B9E394D5AE5 Let $\varphi ,\psi \in \mathcal{E}^{1}(X,\theta ;\phi )$, and set
\[ \varphi _ t\coloneqq (1-t)\varphi +t\psi ,\qquad t\in [0,1]. \]
Then the function
\[ t\mapsto E_{\theta }^{\phi }(\varphi _ t) \]
is differentiable on $[0,1]$, where the derivative at the endpoints is understood as the one-sided derivative. Moreover,
B09DAFFD8EA2
16.5.58
\begin{equation} \frac{\mathrm d}{\mathrm dt}E_{\theta }^{\phi }(\varphi _ t) = \int _ X(\psi -\varphi )\, \theta _{\varphi _ t}^ n. \end{equation}
Proof.
By corollary 16.5.25, $\varphi _ t\in \mathcal E^1(X,\theta ;\phi )$ for all $t\in [0,1]$. For $a=0,\ldots ,n$, put
\[ M_ a=\int _ X(\psi -\varphi )\, \theta _{\psi }^ a\wedge \theta _{\varphi }^{n-a}. \]
Note that each $M_ a$ is finite by corollary 16.5.28. By the multi-linearity of the non-pluripolar product proposition 13.11.4 6, the quantity
\[ \int _ X(\psi -\varphi )\, \theta _{\varphi _ t}^ n \]
is therefore a polynomial in $t$ with coefficients given by the $M_ a$'s. In particular, the right-hand side of (16.5.58) is finite and continuous in $t$.
We prove the formula for the right derivative at a point $t\in [0,1)$. Let $s>0$ be small enough so that $t+s\leq 1$. By the cocycle formula proposition 16.5.30, we have
\[ \begin{aligned} \frac{E_{\theta }^{\phi }(\varphi _{t+s})-E_{\theta }^{\phi }(\varphi _ t)}{s} =& \frac{1}{(n+1)s}\sum _{j=0}^ n \int _ X(\varphi _{t+s}-\varphi _ t)\, \theta _{\varphi _{t+s}}^ j\wedge \theta _{\varphi _ t}^{n-j} \\ =& \frac{1}{n+1}\sum _{j=0}^ n \int _ X(\psi -\varphi )\, \theta _{\varphi _{t+s}}^ j\wedge \theta _{\varphi _ t}^{n-j}. \end{aligned} \]
Again by proposition 13.11.4 6, for each fixed $j$ the last integral is a polynomial in $s$, expressed as a linear combination of the $M_ a$'s with coefficients depending on $t$. Letting $s\to 0+$, it converges to
\[ \int _ X(\psi -\varphi )\, \theta _{\varphi _ t}^ n. \]
Therefore,
\[ \lim _{s\to 0+} \frac{E_{\theta }^{\phi }(\varphi _{t+s})-E_{\theta }^{\phi }(\varphi _ t)}{s} = \int _ X(\psi -\varphi )\, \theta _{\varphi _ t}^ n. \]
The proof for the left derivative is identical, using $s<0$ with $t+s\in [0,1]$. Thus the left and right derivatives agree for $t\in (0,1)$, and the endpoint derivatives are given by the same formula.
$\square$
Pome
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