I am considering the when the following first order differential operator is a closed operator $$Au=b(x)\dfrac{\partial u}{\partial x_i},$$ on $L^2(\Omega)$ with the domain $D(A)=H^1(\Omega)$. Here I assume $\Omega$ is bounded and has smooth boundary.
Here is what I thought:
If $A$ is closed, then for any $f_k\in D(A)$, $f_k\to f$ and $Af_k\to g$ in $L^2(\Omega)$, we have $Af=g$ and $f\in D(A)$. If I obtain $Af=g$, that is $b(x)\dfrac{\partial u}{\partial x_i}=g\in L^2(\Omega)$, then if $|b(x)|>b_0>0$, we have $f\in D(A)$. So I think maybe we need $|b(x)|>b_0>0$ as an assumption.
I know if $b\in C^{0,1}(\bar{\Omega})$, the following Green's formula holds,
$$-\int_\Omega f_k \partial_i (b\psi)=\int_\Omega (\partial_i f_k) b\psi,\quad \forall \psi\in C_c^\infty.$$ Let $k\to \infty$, we obtain $\int_\Omega f\partial_i (b\psi)= \int_\Omega g\psi$. Then I got stucked since I can't use Green's formula again, since I haven't proved $f\in D(A)$.
Are my assumptions necessary? Can we relax the assumptions for $b$ by changing the domain properly to make $A$ a closed operator? May you point out how to finish the proof? Thanks.
In general you can not obtain $f\in D(A)$ for each $b:\Omega\rightarrow\mathbb{R}$. Consider $f_k(x)=\sum_{l\neq i}h_k(x_l)$, where $h_k$ is a smooth sequence which converges to a non smooth function in $L^2$ in one dimension subspace. This is possible, for instance using mollifier. Then $f_k$ converges to a function in $L^2\setminus H^1$, and you still obtain $Af=g=0$, no matter what your $b$ is.