Let
- $(E,\mathcal E)$ be a measurable space
- $\mathcal E_b$ denote the set of all bounded $\mathcal E$-measurable $f:E\to\mathbb R$
- $\lambda$ be a $\sigma$-finite measure on $(E,\mathcal E)$.
If $p,q\in[1,\infty]$ with $\frac1p+\frac1q=1$, then $$\langle f,g\rangle:=\int fg\:{\rm d}\lambda$$ for $f\in\mathcal L^p(\lambda)$ and $g\in\mathcal L^q(\lambda)$.
Now let $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$ and, as usual, $$(\kappa_tf)(x):=\int\kappa_t(x,{\rm d}y)f(y)\;\;\;\text{for }x\in E,f\in\mathcal E_b\text{ and }t\ge0.$$ We can define a linear operator $(\mathcal D(A),A)$ via $$\mathcal D(A):=\left\{f\in\mathcal E_b\mid\exists g\in\mathcal E_b:\forall x\in E:\frac{(\kappa_tf)(x)-f(x)}t\xrightarrow{t\to0+}g(x)\right\}$$ and $$(Af)(x):=\frac{(\kappa_tf)(x)-f(x)}t\xrightarrow{t\to0+}g(x)\;\;\;\text{for }x\in E\text{ and }f\in\mathcal E_b.$$
Is it possible to show that $A$ (considered as an unbounded operator from $\mathcal E_b$ to $L^1(\lambda)$) has an adjoint operator $A^\ast$ with respect to $\langle\;\cdot\;,\;\cdot\;\rangle$, i.e. $$\langle Af,g\rangle=\langle f,A^\ast g\rangle\tag2$$ for all $f\in\mathcal D(A)$ and $g\in\mathcal D(A^\ast)$?
$\mathcal D(A^\ast)$ should clearly be a subspace of $\mathcal L^1(\lambda)$. But what is $\mathcal D(A^\ast)$ precisely and how can we guarantee exisgtence of $A^\ast$?
EDIT: My guess is $$\mathcal D(A^\ast)=\{g\in\mathcal L^1(\lambda)\mid\exists f^\ast\in\mathcal L^1(\lambda):\forall f\in\mathcal D(A):\langle Af,g\rangle=\langle f,f^\ast\rangle\}\tag3.$$