Given $\Psi =\{ u \in H^1_0(\Omega) : \mathrm{div}(u)=0 \in \Omega\}$ equipped with $ H^1(\Omega)$ norm , and $\Phi$ is the closure of $\Psi$ for the $L^2(\Omega)$ norm ,it is equipped with $L^2(\Omega)$ norm.
We have $\xi_m \in \Psi $ for all $m$. Such that it is weakly convergent to $\xi$ in $L^2(\Omega)$
How can I prove that $\mathrm{div}(\xi) =0 $ in $\Omega$
I have proceed as follows: From the density of $\Psi$ in $\Phi$ , I can take a subsequence $\xi_{m,k}$ that converges to $\xi_m$ strongly in $L^2(\Omega)$ but we also can say that $\xi_{m,k}$ is weakly convergent to $\xi $ in $L^2(\Omega)$
So we get $\xi_{m,k}$ is strongly convergent to $\xi$ in $L^2(\Omega)$ this strong convergence leads to convergence in the sense of distributions, and since divergence operator is continuous in distribution sense then we deduce that the $\mathrm{div }(\xi)$ is zero.
Is this correct? Any help please
The first step is to state the goal precisely. The identity $\operatorname{div}\xi=0$ can only be expected to hold in the weak sense, as $$ \int_\Omega \xi\cdot \nabla \phi = 0 \quad \forall \phi\in H^1_0(\Omega) \tag1 $$ Weak convergence $\xi_m\to \xi$ in $L^2(\Omega)$ preserves (1) since $\nabla \phi\in L^2(\Omega)$; this is essentially the definition of weak convergence.
One cannot expect $\operatorname{div}\xi=0$ to hodl in the classical sense; $\xi$ need not be differentiable.