Proof.
We may assume that $\varphi $ is $\mathcal{I}$-model by replacing $\varphi $ by $P_{\theta }[\varphi ]_{\mathcal{I}}$.
Note that it suffices to prove (18.28.28) at $t=0$, which is equivalent to
94F4F91642C8
18.28.29
\begin{equation} \lim _{t\to 0}\frac{E_{\theta }^{\varphi }\left(P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v+tf)\right)-E_{\theta }^{\varphi }\left(P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)\right)}{t}=\int _ K f\, \theta ^ n_{P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)}. \end{equation}
It suffices to prove the right derivative at $0$ for an arbitrary $f$; applying the same argument to $-f$ then gives the left derivative. In other words, we need to prove that
8353FF09478E
18.28.30
\begin{equation} \lim _{t\to 0+}\frac{E_{\theta }^{\varphi }\left(P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v+tf)\right)-E_{\theta }^{\varphi }\left(P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)\right)}{t}=\int _ K f\, \theta ^ n_{P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)}. \end{equation}
We first reduce the proof of (18.28.30) to the case $f\leq 0$. Assume that this case is known. Choose $A\in \mathbb R$ such that $f-A\leq 0$ on $K$. Since for any $t>0$,
\[ P_{\theta ,K}[\varphi ]_{\mathcal I}\Bigl ( v+t(f-A) \Bigr )= P_{\theta ,K}[\varphi ]_{\mathcal I}(v+tf)-tA, \]
then we have
\[ \int _ K f\, \theta ^ n_{P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)}=\int _ K (f-A)\, \theta ^ n_{P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)}+A\int _ X \theta ^ n_{P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)}. \]
By (16.5.13), for any $t>0$,
\[ E_{\theta }^{\varphi }\Bigl (P_{\theta ,K}[\varphi ]_{\mathcal{I}}\left(v+t(f-A)\right)\Bigr )=E_{\theta }^{\varphi }\left(P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v+tf)\right)-tA\int _ X\theta _{\varphi }^ n. \]
So the limit in (18.28.30) exists and is equal to the left-hand side of (18.28.30) with $f-A$ in place of $f$ plus $A\int _ X\theta _{\varphi }^ n$. So (18.28.30) for $f-A$ implies the desired formula for $f$ in view of 18.28.15 and the monotonicity theorem theorem 13.13.11. Hence we may assume that $f\leq 0$.
Fix $t>0$. Set
\[ \eta _ t\coloneqq P_{\theta ,K}[\varphi ]_{\mathcal I}(v+tf),\quad \eta _0\coloneqq P_{\theta ,K}[\varphi ]_{\mathcal I}(v). \]
By the comparison principle (16.5.6) and proposition 16.5.30, we find
\[ \begin{aligned} E_{\theta }^{\varphi }(\eta _ t)-E_{\theta }^{\varphi }(\eta _0) ={}& \frac{1}{n+1}\sum _{i=0}^ n\int _ X(\eta _ t-\eta _0)\, \theta _{\eta _ t}^ i\wedge \theta _{\eta _0}^{n-i}\\ \leq {}& \int _ X(\eta _ t-\eta _0)\, \theta _{\eta _0}^ n. \end{aligned} \]
By 18.28.21 and corollary 18.28.14, the measure $\theta _{\eta _0}^ n$ is carried by $K\cap \{ \eta _0=v\} $, and the measure $\theta _{\eta _ t}^ n$ is carried by $K\cap \{ \eta _ t=v+tf\} $. On $K\cap \{ \eta _0=v\} $, we have $\eta _ t\leq v+tf=\eta _0+tf$, hence
\[ \eta _ t-\eta _0\leq tf. \]
On $K\cap \{ \eta _ t=v+tf\} $, we have $\eta _0\leq v$, hence
\[ \eta _ t-\eta _0\geq tf. \]
It follows that
\[ \int _ X(\eta _ t-\eta _0)\, \theta _{\eta _0}^ n \leq t\int _ K f\, \theta _{\eta _0}^ n \]
and
\[ \int _ X(\eta _ t-\eta _0)\, \theta _{\eta _ t}^ n \geq t\int _ K f\, \theta _{\eta _ t}^ n. \]
Thus, letting $t\to 0+$, we get the inequality
\[ \varlimsup _{t\to 0+}\frac{E_{\theta }^{\varphi }\left(P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v+tf)\right)-E_{\theta }^{\varphi }\left(P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)\right)}{t}\leq \int _ K f\, \theta ^ n_{P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)}. \]
Similarly, for $t>0$, we have
\[ \begin{aligned} & E_{\theta }^{\varphi }\left(P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v+tf)\right)-E_{\theta }^{\varphi }\left(P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)\right)\\ \geq & \int _ X \left(P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v+tf)-P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)\right)\, \theta _{P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v+tf)}^ n\\ \geq & t\int _ K f\, \theta _{P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v+tf)}^ n. \end{aligned} \]
Letting $t\to 0+$, we find
\[ \varliminf _{t\to 0+}\frac{E_{\theta }^{\varphi }\left(P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v+tf)\right)-E_{\theta }^{\varphi }\left(P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)\right)}{t}\geq \int _ K f\, \theta ^ n_{P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)}. \]
Here we used the following elementary continuity consequence:
\[ P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)-t\| f\| _{C^0(K)} \leq P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v+tf) \leq P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v) \]
for $t>0$, and hence $P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v+tf) \nearrow P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)$ almost everywhere when $t\to 0+$. Next we claim that
\[ \int _ K f\, \theta ^ n_{P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)}\leq \lim _{t\to 0+} \int _ K f\, \theta _{P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v+tf)}^ n. \]
Choose a continuous extension $\tilde f\in C^0(X)$ of $f$. Since 18.28.21, corollary 18.28.14 show that all the measures appearing here are carried by $K$, the integrals over $K$ against $f$ are the same as the integrals over $X$ against $\tilde f$. Thus corollary 13.13.12 applies.
We conclude (18.28.29).
$\square$
Pome
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