I am reading the following note: https://arxiv.org/abs/1309.3112 (you can just read 37-38, with some notations on p.23)
Simply speaking:
- Define the linear operator $\mathcal{L}: \mathcal{b}^{1}([0,T]\times X) \rightarrow \mathcal{b}([0,T]\times X)$ by $$v \mapsto \mathcal{L}v = \frac{\partial v}{\partial t} + \sum_{i=1}^n \frac{\partial v}{\partial x_i}f_i=\frac{\partial v}{\partial t} + (\nabla v)^Tf$$
- Adjoint $\mathcal{L}':\mathcal{b}([0,T]\times X)' \rightarrow \mathcal{b}^1([0,T]\times X)'$:
$$\langle v,\mathcal{L}'\mu\rangle= \langle \mathcal{L}v,\mu\rangle=\int_0^T\int_X\mathcal{L}v(t,x,u)\mu(dt,dx)$$ This operator can be expressed as:
$$\mu \mapsto\mathcal{L}'\mu=-\frac{\partial \mu}{\partial t} - \mbox{div} (f\mu)$$ where div means divergence.
My question is how to show the above?
I try to do the following:
$$\langle v,\mathcal{L}'\mu\rangle= \langle \mathcal{L}v,\mu\rangle=\int_0^T\int_X\big[\frac{\partial v}{\partial t} + (\nabla v)^Tf\big]\mu(dt,dx)$$ and then
$$=\int_0^T\int_X\frac{\partial v}{\partial t} \mu(dt,dx)+ \int_0^T\int_X(\nabla v)^Tf\mu(dt,dx)$$
It seems I have to use integration by part for both terms; however, I have no idea how to use integration by part to obtain the correct form (integration technique). I try to use the discussion link provided below about the transpose relation between gradient operator and divergence operator; I cannot obtain the desired form.
