Let
- $b\in C^1(\mathbb R)$ be Lipschitz continuous
- $\sigma\in C^2(\mathbb R)$ be Lipschitz continuous with $\sigma(\mathbb R)\subseteq\mathbb R\setminus\left\{0\right\}$ and $\sigma''$ being bounded
- $Lf:=bf'+\frac12\sigma^2f''$ for $f\in C^2(\mathbb R)$ and $L^\ast g:=\frac12(\sigma^2g)''-(bg)'$ for $g\in C^2(\mathbb R)$
- $(\kappa_t)_{t\ge0}$ be a semigroup of Markov kernels on $(\mathbb R,\mathcal B(\mathbb R))$ and $$\kappa_t f:=\int\kappa_t(\;\cdot\;,{\rm d}y)f(y)\tag1$$ for bounded Borel measurable $f:\mathbb R\to\mathbb R$.
Assume $(\kappa_t)_{t\ge0}$ is a strongly continuous semigroup (via $(1)$) on $C_0(\mathbb R)$ (continuous functions on $\mathbb R$ vanishing at infinity equipped with the supremum norm) with infinitesimal generator $(\mathcal D(A),A)$ given by the restriction $A$ of $L$ to $$\mathcal D(A):=\left\{f\in C_0(\mathbb R)\cap C^2(\mathbb R):Lf\in C_0(\mathbb R)\right\}.$$ Moreover, assume there is a Borel measurable $p_t:\mathbb R\times\mathbb R\to[0,\infty)$ with $$\kappa_t(x,B)=\int_Bp_t(x,y)\:\lambda({\rm d}y)\;\;\;\text{for all }(x,B)\in\mathbb R\times\mathcal B(\mathbb R)\tag2,$$ where $\lambda$ denotes the Lebesgue measure, for all $t\ge0$ with
- $[0,\infty)\ni t\mapsto p_t(x,y)$ is differentiable for all $x,y\in\mathbb R$ and $$[0,\infty)\times\mathbb R\ni(t,y)\mapsto\frac{\partial p}{\partial t}(t,x,y)\tag3$$ is locally bounded for all $x\in\mathbb R$
- $\mathbb R\ni y\mapsto p_t(x,y)$ is twice continuously differentiable for all $(t,x)\in[0,\infty)\times\mathbb R$.
We can observe the following: Fix $f\in C_c^2(\mathbb R)$. Then, $$\frac{\kappa_{t+h}f-\kappa_tf}h\xrightarrow{h\to0}A(\kappa_tf)=\kappa_t(Af)\tag4,$$ where the convergence is with respect to the supremum norm, for all $t\ge0$. Fix $(t,x)\in[0,\infty)\times\mathbb R$. On the one hand, $$\frac{(\kappa_{t+h}f)(x)-(\kappa_tf)(x)}h\xrightarrow{h\to0}\int\frac{\partial p}{\partial t}(t,x,y)f(y)\:{\rm d}y\tag5$$ by 1. and the dominated convergence theorem. On the other hand, $$\left(\kappa_t\left(Af\right)\right)(x)=\int f(y)\left(L^\ast\left(p_t(x,\;\cdot\;)\right)\right)(y)\:{\rm d}y\tag6$$ by 2. and integration by parts. Thus, $$\int\frac{\partial p}{\partial t}(t,x,y)f(y)\:{\rm d}y=\int f(y)\left(L^\ast\left(p_t(x,\;\cdot\;)\right)\right)(y)\:{\rm d}y\tag7.$$ Are we able to conclude $$\frac{\partial p}{\partial t}(t,x,y)=\left(L^\ast\left(p_t(x,\;\cdot\;)\right)\right)(y)\tag8$$ (since $f$ was arbitrary)?