Defining the set
Let $n \geq 2$, and let $\Omega$ be a domain in $\mathbb{R}^{n}$.
Define the set $ C^{\infty}_{0,\sigma}(\Omega) := \{ f \in C^{\infty}_{0}(\Omega)^{n} \ | \ \text{div}(f) = 0 \}$. ie the set of smooth $n$-vector functions, with compact support and $0$ divergence.
We use this to define $L^{2}_{\sigma}(\Omega) := \overline{C^{\infty}_{0,\sigma}(\Omega)}^{||\cdot||_{L^{2}}}$.
Questions
(1.) It seems clear that $L^{2}_{\sigma}(\Omega) \subsetneq L^{2}(\Omega)$, but could someone give me a hint as to how I can show this?
(2.) Obviously, not every function in $L^{2}_{\sigma}(\Omega)$ will be differentiable in the strong sense, but is it possible we have div$(f) = 0$, for all $f \in L^{2}_{\sigma}(\Omega)$ when we take the derivative of $f$ in the weak sense?
2) Yes, every $f\in L^2_\sigma(\Omega)$ satisfies $\mathrm{div} f=0$ in the weak (distributional) sense. By definition, there is a sequence $(f_n)\subset C_{0,\sigma}^\infty(\Omega)$ such that $f_n\rightarrow f$ in $L^2(\Omega)$. For any test function $\phi\in C_0^\infty(\Omega)$, you obtain $$ \langle\mathrm{div} f,\phi\rangle = -\int_\Omega f\cdot\nabla \phi\, dx = -\lim_{n\rightarrow \infty}\int_\Omega f_n\cdot\nabla \phi\, dx = \lim_{n\rightarrow \infty}\int_\Omega \mathrm{div} f_n\, \phi\, dx = 0. $$
1) By 2), you need only take any $f\in L^2(\Omega)$ with $\mathrm{div}f \neq 0$.