In the setting below, I want to compute $$\int_0^t\operatorname E_\nu\left[f(X_s)g(X_t)\right]\:{\rm d}s\tag0.$$
Let
- $E$ be a $\mathbb R$-Banach space;
- $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $E$;
- $A$ denote the generator$^1$ of $(\kappa_t)_{t\ge0}$;
- $(\Omega,\mathcal A)$ be a measurable space;
- $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$;
- $(X_t)_{t\ge0}$ be an $E$-valued right-continuous process on $(\Omega,\mathcal A)$;
- $\operatorname P_x$ be a probability measure on $(\Omega,\mathcal A)$ with $$\operatorname P_x\left[X_0=x\right]=1\tag1$$ and $$\operatorname E_x\left[f(X_{s+t}\mid\mathcal F_s\right]=(\kappa_tf)(X_s)\;\;\;\text{for all }f\in\mathcal E_b\text{ and }s,t\ge0\tag2$$ for $x\in E$;
- $\mu,\nu$ be probability measures on $E$.
Let $f,g\in\mathcal E_b$. If I'm not totally mistaken, we should be able to write$^2$ $$\operatorname E_\nu\left[f(X_s)g(X_t)\right]=\int\left(\kappa_s\left(f\kappa_{t-s}g\right)\right)\:{\rm d}\mu\tag3$$ for all $t>s\ge0$. Now, for my particular choice of $\mu,\nu$, I was able to show the identity $$\operatorname E_\nu\left[f(X_s)g(X_t)\right]=-\frac1c\int A\kappa_s\left(f\kappa_{t-s}g\right)\:{\rm d}\mu\tag4$$ for all $t>s\ge0$. Now, since $$\frac{\rm d}{{\rm d}r}\kappa_rh=A\kappa_rh\tag5,$$ the right-hand side of $(4)$ should actually be $$-\frac1c\frac{\rm d}{{\rm d}r}\left.\int\kappa_r\left(f\kappa_{t-s}g\right)\:{\rm d}\mu\right|_{r=s}\tag6.$$ Now, my problem is that when I insert $(6)$ into $(0)$, there is still another $s$ in the function and so I cannot apply the fundamental theorem of calculus to get rid of the differentiation. So, what can I do here?
$^1$ We csonder $(\kappa_t)_{t\ge0}$ as a contraction semigroup on $$\mathcal E_b:=\left\{f:E\to\mathbb R\mid f\text{ is bounded and Borel measurable}\right\}$$ equipped with the supremum norm.
$^2$ As usual, $$\operatorname E_\mu:=\int\mu({\rm d}x)\operatorname E_x.$$