Lemma 17.5.9B4AA6C4C1C94

Proof. Let $f\in C(X,\mathbb R)$ and $u_ t=P_\theta (u+tf)$ for $t\in \mathbb R$. By proposition 17.5.4, $u_ t\in \mathcal E^1(X,\theta ,\phi )$, $u_0=u$ and $u_ t\leq u+tf$ everywhere. Hence $\mathrm e^{\lambda u_ t}\leq \mathrm e^{\lambda (u+tf)}$, and $F_\lambda (u_ t)\leq F_\lambda (u)$ gives

The left side vanishes at $t=0$. Its energy term is differentiable, with derivative $\int _ Xf\, \mu _{u_ t}$, by proposition 17.5.4. For $|t|\leq 1$ the $t$-derivative $\lambda f\mathrm e^{\lambda (u+tf)}$ of the integrand is bounded by $\lambda \sup _ X|f|\, \mathrm e^{\lambda \sup _ Xu+\lambda \sup _ X|f|}$, which is integrable for the finite measure $\nu $; by the mean value theorem and dominated convergence, the integral is differentiable as well, with derivative $\lambda \int _ Xf\mathrm e^{\lambda (u+tf)}\, \mathrm d\nu $. Since the left side is maximal at $t=0$, its derivative vanishes there: $\int _ Xf\, \mu _ u=\int _ Xf\mathrm e^{\lambda u}\, \mathrm d\nu $. Since $f$ is arbitrary and $\mathrm e^{\lambda u}\nu $ is a finite measure, lemma 17.5.3(1) gives $\mu _ u=\mathrm e^{\lambda u}\nu $. $\square$