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First, some notation: let $X = (X_{t})_{t\geq 0}$ be some strong Markov process in $E = \mathbb R^n$ with cadlag paths. Let us denote by $P_t$ the transition semigroup of $X$ and by $\mathbb B$ the Banach space of all bounded measurable real-valued functions $f$ with a norm $\|f\| = \sup\limits_{x\in \mathbb E}|f(x)|$. Put: $$ A f(x) = \lim\limits_{t\downarrow 0}\frac{P_tf(x) - f(x)}{t}\tag{IG} $$ to be an infinitesimal generator of $P_t$, defined on $\mathscr D(A)\subset \mathbb B$ being the space of function for which the limit in $(\mathrm{IG})$ exists in norm. It is well-known that for any $f\in \mathscr D(A)$ and $x\in E$ the process $$ C^{f}_t = f(X_t) - f(x)-\int\limits_0^t(Af)(X_s)ds $$ is a martingale.

For $f\in \mathbb B$ we further say that $f\in\mathscr D(\mathscr A)$ if there is $h\in \mathbb B$ such that for any $x\in E$ the process $$ C^{f,h}_t = f(X_t) - f(x)-\int\limits_0^t h(X_s)ds $$ is a local martingale. On $\mathscr D(\mathscr A)$ we define the extended generator as $$ \mathscr A:f\to h.\tag{EG} $$ Note that $\mathscr A$ is defined up to some sort of uniqueness: i.e. if $h_1,h_2$ are candidates for $\mathscr A f$ then $$ \int\limits_0^\infty 1\{s:h_1(X_s) \neq h_2(X_s)\}\;ds = 0 \text{ a.s.} $$

Conversely, let $(L,\mathscr D(L))$ be some linear operator with $\mathscr D(L)\subset \mathbb B$. The Markov process $Y$ with cadlag paths is said to satisfy the martingale problem with $L$ if $$ f(Y_t) - f(x)-\int\limits_0^t(Lf)(Y_s)ds \tag{LMP} $$ is a local martingale for any $x\in E$ and $f\in \mathscr D(L)$. The local martingale problem is said to be well-posed if it has a solution and any two solutions have the same family of finite-dimensional distributions.


My question is the following. Suppose we start with some given process $X$ and $\mathscr D(\mathscr A)$ precisely. Clearly, there is at least one solution of the local martingale problem for $\mathscr D(\mathscr A)$ which is given by the process $X$. I wonder if it is unique, or what are the sufficient conditions for the uniqueness.

My thoughts are: since we know $\mathscr D(\mathscr A)$ - we know also $\mathscr D( A)$ and the operator $A$ since $A = \mathscr A$ on the domain of definition of the former operator. Then we can switch to the martingale problem for the infinitesimal generator $A$ rather than dealing with the extended generator and the local martingale problem. I think I've seen somewhere that it such problem then have the unique solution - but I am not sure where have I've seen it and whether it's true.

Any help or hints are appreciated.

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