We follow the exposition in [BB10, §4.3] very closely. Arguing as in Corollary 7.3 we may assume that $f,g\in\mathcal{D}(X)$. Set $\mu:=\MA(P_{\omega}(f))$. We need to prove that

(7.3)

$$\frac{d}{dt}\bigg|_{t=0+}E_{\omega}\circ P_{\omega}(f+tg)=\int g\,\mu.$$

As a first step, we linearize the problem and prove that

(7.4)

$$\frac{d}{dt}\bigg|_{t=0+}E_{\omega}\circ P_{\omega}(f+tg)=\frac{d}{dt}\bigg|_{t=0+}\int P_{\omega}(f+tg)\,\mu.$$

Denote the left and right hand sides of (7.4) by $a$ and $b$, respectively. Note that the one-sided derivatives exist since both $E_{\omega}$ and $P_{\omega}$ are concave.

Since $E_{\omega}$ is concave on the space of bounded $\omega$-psh functions, the function

$$[0,1]\ni s\mapsto h(s):=E_{\omega}\left(sP_{\omega}(f+tg)+(1-s)P_{\omega}(f)\right)$$

is also concave, hence

$$E_{\omega}(sP_{\omega}(f+tg)+(1-s)P_{\omega}(f))\leq\\ E_{\omega}\circ P_{\omega}(f)+\frac{d}{dt}\bigg|_{t=0}h=E_{\omega}\circ P_{\omega}(f)+s\left(\int(P_{\omega}(f+tg)-P_{\omega}(f))\mu\right)$$

by Proposition 6.3. Taking $s=1$ and letting $t\to 0$ yields $a\leq b$.

To prove the reverse inequality, fix $\varepsilon>0$. Then there exists $\delta>0$ such that

$$D:=\int_{X}P_{\omega}(f+\delta g)\mu-\int_{X}P_{\omega}(f)\mu\geq\delta(b-\varepsilon).$$

Since $\mu$ is the differential of $E_{\omega}$, there exists $\gamma>0$ such that

$$E\left((1-t)P_{\omega}(f)+tP_{\omega}(f+\delta g)\right)\geq E_{\omega}(P_{\omega}(f))+t(D-\delta\varepsilon)\geq E_{\omega}(P_{\omega}(f))+t\delta(b-2\varepsilon)$$

for $0\leq t\leq\gamma$. The concavity of $P$ yields $P_{\omega}(f+t\delta g)\geq(1-t)P_{\omega}(f)+tP_{\omega}(f+\delta g)$. Since $E_{\omega}$ is non-decreasing we get

$$E_{\omega}\circ P_{\omega}(f+t\delta g)\geq E\left((1-t)P_{\omega}(f)+tP_{\omega}(f+\delta g)\right)\geq E(P_{\omega}(f))+t\delta(b-2\varepsilon)$$

for $0\leq t\leq\gamma$. Letting $t\to 0$ and $\varepsilon\to 0$ we conclude $a\geq b$. This shows that (7.4) holds.

In view of (7.4) it remains to show that

(7.5)

$$\int_{X}(P_{\omega}(f+tg)-P_{\omega}(f))\,\mu=t\int_{X}g\mu+o(t)$$

as $t\to 0+$.

Since $P_{\omega}(f)\leq f$, the orthogonality property implies $P_{\omega}(f)=f$ for $\mu$-a.e. point. We thus have $P_{\omega}(f+tg)\leq f+tg=P_{\omega}(f)+tg$ $\mu$-a.e. We claim that $\mu(\Omega_{t})=O(t)$ with

$$\Omega_{t}:=\{P_{\omega}(f+tg)<P_{\omega}(f)+tg\}.$$

Observe that $|P_{\omega}(f+tg)-P_{\omega}(f)|\leq t\sup|g|$ so that the claim implies

$$\int_{X}(P_{\omega}(f+tg)-P_{\omega}(f)-tg)\mu\\ =\int_{\Omega_{t}}(P_{\omega}(f+tg)-P_{\omega}(f))\mu+\int_{X\setminus\Omega_{t}}(P_{\omega}(f+tg)-P_{\omega}(f))\mu\\ =\int_{\Omega_{t}}(P_{\omega}(f+tg)-P_{\omega}(f)-tg)+\int_{X}tg\,\mu\\ \leq t\!\int_{X}\!g\,\mu+\mu(\Omega_{t})\sup_{X}|P_{\omega}(f+tg)-P_{\omega}(f)-tg|=t\!\int_{X}\!g\,\mu+O(t^{2})$$

which proves (7.5).

The estimate of $\mu(\Omega_{t})$ is based on the comparison principle. Since $g$ is a model function, there exists $C\gg 1$, $\psi\in\mathcal{D}(X)$ such that $\psi$ and $\psi+g$ are $C\omega$-psh by Proposition 2.6. Note that $\Omega_{t}=\{P_{\omega}(f+tg)+t\psi<P_{\omega}(f)+t(\psi+g)\}$, and both functions $P_{\omega}(f+tg)+t\psi$ and $P_{\omega}(f)+t(\psi+g)$ are $(1+Ct)\omega$-psh. The comparison principle then yields

$$\int_{\Omega_{t}}\left((1+Ct)\omega+dd^{c}\left(P_{\omega}(f)+t(\psi+g)\right)\right)^{n}\leq\int_{\Omega_{t}}((1+Ct)\omega+dd^{c}\left(P_{\omega}(f+tg)+t\psi\right))^{n}$$

By expanding as polynomials in $t$, we get

$$((1+Ct)\omega+dd^{c}\left(P_{\omega}(f)+t(\psi+g)\right))^{n}=(\omega+dd^{c}P_{\omega}(f))^{n}+O(t)$$

and

$$((1+Ct)\omega+dd^{c}\left(P_{\omega}(f+tg)+t\psi\right))^{n}=(\omega+dd^{c}P_{\omega}(f+tg))^{n}+O(t).$$

From these three estimates we conclude

$$\mu(\Omega_{t})=\int_{\Omega_{t}}\MA(P_{\omega}(f))\leq\int_{\Omega_{t}}\MA(P_{\omega}(f+tg))+O(t).$$

But $\Omega_{t}\subseteq\{P_{\omega}(f+tg)<f+tg\}$, so the orthogonality property implies that the last integral vanishes. This concludes the proof. ∎