Self-adjointness of generator and semigroup of an SDE

Question

$ \newcommand{\bR}{\mathbb{R}} \newcommand{\bE}{\mathbb{E}} \newcommand{\bT}{\mathbb{T}} \newcommand{\bP}{\mathbb{P}} \newcommand{\bF}{\mathbb{F}} \newcommand{\cF}{\mathcal{F}} \newcommand{\eps}{\varepsilon} \newcommand{\diff}{\mathop{}\!\mathrm{d}} \newcommand{\qtext}[1]{\quad\text{#1}} \newcommand{\qtextq}[1]{\quad\text{#1}\quad} \DeclareMathOperator*{\trace}{tr} $ We fix $T \in (0, \infty)$ and let $\bT$ be the interval $[0, T]$. We consider measurable functions \begin{align} b &: \bT \times \bR^d \to \bR^d, \\ \sigma &: \bT \times \bR^d \to \bR^d \otimes \bR^m. \end{align}

We assume that $b$ $\sigma$ are regular enough. Let $(B_t, t \ge 0)$ be a $m$-dimensional Brownian motion and $\bF := (\cF_t, t \ge 0)$ an admissible filtration on a probability space $(\Omega, \mathcal A, \bP)$. For $(s, x) \in [0, T) \times \bR^d$, we consider the SDE $$ \begin{cases} \diff X^x_{s,t} & = b (t, X^x_{s, t}) \diff t + \sigma (t, X^x_{s, t}) \diff B_t \qtextq{for} t \in [s, T] , \\ X^x_{s,s} & = x . \end{cases} $$

Let $a:= \sigma \sigma^\top$. The differential operator $(L_t)_{t \in \bT}$ is defined for $f \in C^2 (\bR^d)$ and $x \in \bR^d$ by \begin{align} L_t f (x) := \langle b (t, x), \nabla f (x) \rangle + \frac{1}{2} \trace ( a (t, x) \nabla^2 f (x) ) . \end{align}

The semigroup $(P_{s, t})_{0 \le s < t \le T}$ is defined for $x \in {\bR}^d$ and $f \in C_b ({\bR}^d)$ by $$ P_{s, t} f (x) := \bE [f(X^x_{s, t})] . $$

We consider two statements:

1. It holds for $f, g \in C^2_c (\bR^d)$ that $$ \int_{\bR^d} (L_t f) g \diff x = \int_{\bR^d} f (L_t g) \diff x . $$

2. It holds for $f, g \in C_c (\bR^d)$ that $$ \int_{\bR^d} (P_{s, t} f) g \diff x = \int_{\bR^d} f (P_{s, t} g) \diff x . $$

The backward Kolmogorov equation holds, i.e., $\partial_s P_{s, t} f + L_s P_{s, t} f =0$ for $f \in C^2_b (\bR^d)$.

Is it true that (1) implies (2)?

Thank you for your elaboration.

Answer 1

Let's at least elaborate on why $P_{s,t} $ in general is not self-adjoint, even if $L_t $ is. $P_{s,t} $ can be composed from infinitesimal time evolutions, $$ P_{s,t} = \lim_{N\rightarrow \infty } \left( 1-\epsilon L_{t+N\epsilon } \right) \left( 1-\epsilon L_{t+(N-1)\epsilon } \right) \left( 1-\epsilon L_{t+(N-2)\epsilon } \right) \ldots \left( 1-\epsilon L_{t+\epsilon } \right) \ , $$ where $\epsilon =(s-t)/N $ has been introduced. This satisfies $\partial_{s} P_{s,t} + L_s P_{s,t} =0$, as one sees by evaluating $$ \partial_{s} P_{s,t} = \frac{1}{\epsilon } \left( P_{s,t} - P_{s-\epsilon ,t} \right) \ . $$ Now, consider the adjoint, \begin{eqnarray} P^{\dagger }_{s,t} &=& \lim_{N\rightarrow \infty } \left( 1-\epsilon L^{\dagger }_{t+\epsilon } \right) \ldots \left( 1-\epsilon L^{\dagger}_{t+(N-2)\epsilon } \right) \left( 1-\epsilon L^{\dagger}_{t+(N-1)\epsilon } \right) \left( 1-\epsilon L^{\dagger}_{t+N\epsilon } \right) \\ &=& \lim_{N\rightarrow \infty } \left( 1-\epsilon L_{t+\epsilon } \right) \ldots \left( 1-\epsilon L_{t+(N-2)\epsilon } \right) \left( 1-\epsilon L_{t+(N-1)\epsilon } \right) \left( 1-\epsilon L_{t+N\epsilon } \right) \end{eqnarray} using the self-adjointness of $L_t $. Thus, the ordering of the factors in $P^{\dagger }_{s,t} $ is reversed compared to $P_{s,t} $, and therefore, in general, $P^{\dagger }_{s,t} \neq P_{s,t}$. If the $L_t $ at different times commute, $[L_t ,L_{t^{\prime } }]=0$, then one can reorder the factors in $P^{\dagger }_{s,t} $, obtaining $P^{\dagger }_{s,t} = P_{s,t}$.