Let $$C_{0,\sigma}^\infty(\mathbb{R}^n) := \{f \in C_0^\infty(\mathbb{R}^n)^n \mid \operatorname{div}f = 0\}$$ be the set of all divergence free smooth vector fields and define $$ L_\sigma^2(\mathbb{R}^n) := \overline{C_{0,\sigma}^\infty(\mathbb{R^n})}^{\|\cdot\|_2}. $$
I want to prove the following statement in Sohr - Navier-Stokes:
If $n \geq 2$ then $$ L_\sigma^2(\mathbb{R}^n) = \{f \in L^2(\mathbb{R}^n)^n \mid \operatorname{div} f = 0\}, $$ i.e. $$ \overline{\{f \in C_0^\infty(\mathbb{R}^n)^n \mid \operatorname{div}f = 0\}}^{\|\cdot\|_2} = \{f \in \overline{C_0^\infty(\mathbb{R}^n)^n}^{\|\cdot\|_2} \mid \operatorname{div}f = 0\} $$
The mentioned book provides a proof for both inclusions but the proof for "$\subseteq$" is rather short...
Of course, $$C_{0,\sigma}^\infty(\mathbb{R}^n) \subseteq C_0^\infty(\mathbb{R^n})^n\subseteq L^2(\mathbb{R^n})^n$$ and hence $$C_{0,\sigma}^\infty(\mathbb{R}^n) \subseteq L := \{f \in L^2(\mathbb{R}^n)^n \mid \operatorname{div} f = 0\}.$$
But is $L$ closed? We can see $L = \ker\,\operatorname{div}$ if we interpret the divergence as mapping $$ \operatorname{div} \colon L^2(\mathbb{R}^n)^n \to (C_0^\infty(\mathbb{R^n})^n)' $$ but I am not sure how to prove continuity of this mapping (author never specified a topology on the target space, maybe weak-*, but anyway this seems to complicated...)
Another perspective on the problem: Let $f \in C_{0,\sigma}^\infty(\mathbb{R}^n)$. By definition $\operatorname{div}f = 0$. Here, we interpreted the divergence as a linear form on $C_{0, \sigma}^\infty(\mathbb{R}^n)$. As a constant and hence continous linear mapping the divergence operator can be extended to the closure of its domain of definition i.e. $L_\sigma^2(\mathbb{R^n})$. Of course $$L_\sigma^2(\mathbb{R^n}) \subseteq L_2(\mathbb{R}^n)^n$$ but the interpretation of $$\operatorname{div} f = 0$$ in the definition of $L$ is in the sense of distributions, i.e. $$\operatorname{div} f \in (C_0^\infty(\mathbb{R^n}^n)'$$ which is somewhat confusing.
So I have this question: If $\operatorname{div} f = 0$ holds for all $f \in L_\sigma^2$ does it also hold in the distributional sense, i.e. $(\operatorname{div} f )(\varphi) = 0$ for all $\varphi \in C_0^\infty(\mathbb{R}^n)^n$?