Lemma 16.5.55.88E2DBCC3C35 Let $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$. Define
\[ \eta _ t\coloneqq \Bigl ( (1-t)\varphi +t\psi \Bigr )\land \psi ,\quad t\in [0,1]. \]
Then $E_{\theta }^{\phi }(\eta _ t)$ is differentiable for $t\in [0,1]$ and
F8B858957CF5
16.5.56
\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t}E_{\theta }^{\phi }(\eta _ t)=\int _{X} \Bigl ( \psi -\min \{ \varphi ,\psi \} \Bigr )\, \theta _{\eta _ t}^ n. \end{equation}
Proof.
Let us prove (16.5.56) with right-derivative instead of derivative, and $t\in [0,1)$. The left-derivative case is completely parallel.
For each $t\in [0,1]$, we let
\[ f_ t=\min \{ (1-t)\varphi +t\psi , \psi \} . \]
Fix $t\in [0,1)$ and $s>0$ small enough so that $t+s<1$. Thanks to proposition 16.5.36, we have
\[ \begin{aligned} E_{\theta }^{\phi }(\eta _{t+s})-E_{\theta }^{\phi }(\eta _{t})\leq & \int _ X \left(\eta _{t+s}-\eta _ t\right)\, \theta _{\eta _ t}^ n \\ =& \int _ X \left(\eta _{t+s}-f_ t\right)\, \theta _{\eta _ t}^ n & \textup{by \href{../../tag/4CF7246E6E5A/}{16.3.3}}\\ \leq & \int _ X \left(f_{t+s}-f_ t \right)\, \theta _{\eta _ t}^ n\\ =& s\int _ X \Bigl (\psi -\min \{ \varphi ,\psi \} \Bigr )\, \theta _{\eta _ t}^ n. \end{aligned} \]
Similarly, we get
\[ E_{\theta }^{\phi }(\eta _{t+s})-E_{\theta }^{\phi }(\eta _{t})\geq s\int _ X \Bigl (\psi -\min \{ \varphi ,\psi \} \Bigr )\, \theta _{\eta _{t+s}}^ n. \]
Observe that $\eta _{t+s}$ converges in capacity (in fact, even uniformly) to $\eta _ t$ as $s\to 0+$. The desired result (16.5.56) is then just a consequence of theorem 13.12.8.
$\square$
Pome
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