If I have proved that $\operatorname{div}(f)=0$ in the weak sense as follows $$\int_{\Omega}{\operatorname{div}(f)\phi }=0 , \forall \phi \in H^1_0(\Omega) $$
And if moreover I proved that $\operatorname{div}(f) \in L^2(\Omega) $
What additional information can I get? Can I obtain that $\operatorname{div}(f)=0 $ in the classical sense?
If $f \in H^1$ then $\operatorname{div}(f) \in L^2$. So in $L^2(\Omega)$ you have for all $\phi \in H^1_0(\Omega)$: $$\langle \operatorname{div}(f),\phi \rangle=0$$ but $H^1_0(\Omega)$ is a dense subset of $L^2(\Omega)$ so by density for any $\varphi \in L^2(\Omega)$: $$\langle \operatorname{div}(f),\varphi \rangle=0$$ i.e: $$\operatorname{div}(f)=0 \text{ in } L^2(\Omega)$$