Does the weak divergence exist for each $\mathcal L^2(\Omega;\mathbb R^d)$-function?

Question

Let $\Omega\subseteq\mathbb R^d$ be open.

$v:\Omega\to\mathbb R$ is called weak divergence of $u:\Omega\to\mathbb R^d$ $:\Leftrightarrow$ $$\int_\Omega v\varphi\;{\rm d}\lambda=-\int_\Omega\langle u,\nabla\varphi\rangle\;{\rm d}\lambda\;\;\;\text{for all }\varphi\in C_c^\infty(\Omega)\;.\tag 1$$ In that case, we write $v=\operatorname{div}u$.

Since $$\left\|v\varphi\right\|_{L^1(\Omega)}\le\left\|v\right\|_{L^1(\Omega)}\left\|\varphi\right\|_{L^\infty(\Omega)}$$ by Hölder's inequality and $\varphi\in\mathcal L^\infty(\Omega)$, the integral on the left-hand side of $(1)$ is well-defined, if $v\in\mathcal L^1(\Omega)$. Moreover, since $$\langle u,\nabla\varphi\rangle_{L^2(\Omega;\mathbb R^d)}\le\left\|u\right\|_{L^2(\Omega;\mathbb R^d))}\left\|\nabla\varphi\right\|_{L^2(\Omega;\mathbb R^d))}$$ by the Cauchy–Schwarz inequality and $\nabla\varphi\in\mathcal L^\infty(\Omega;\mathbb R^d)$, the integral on the right-hand side of $(1)$ is well-defined, if $u\in\mathcal L^2(\Omega)$.

Question:

1. Is $v=\operatorname{div}u$ possible for more general $(u,v)$?
2. Does $\operatorname{div}u$ exist for each $u\in\mathcal L^2(\Omega;\mathbb R^d)$? If that's the case: How can we prove it (or where can I find a rigorous proof of that) and does the same hold true for more general $u$?
3. Does $\operatorname{div}u$, if each component of $u$ is weakly differentiable?

Answer 1

1. In the sense of (1), it can be defined for $u_i,v\in L^1_{\text{loc}}(\Omega)$. More generally, it can be defined for arbitrary distributions.
2. There are $u \in L^2(\Omega; \mathbb{R}^d)$, such that there is no $v \in L^1_{\text{loc}}(\Omega)$ with $v = \operatorname{div} u$. But $u$ has always a distributional divergence.
3. If each component of $u$ is weakly differentiable with weak derivative $\partial_i u \in L^1_{\text{loc}}(\Omega)$, then $\operatorname{div} u = \sum_{i=1}^d \partial_i u$.