given there exists an $u \in C^2(\Omega) \cap C(\Omega)$ such that $$0=\int_\Omega \nabla u \nabla v - fv ~dx,~ \forall v \in C^\infty_0(\Omega) $$
show that $u$ also solves pointwise
$$-\Delta u =f,~on~ \Omega$$ with zero boundary conditons.
I have to show that $u$ is twice continuous differentiable and continuous on the closure of $\Omega$. My idea is just to integrate (parital) $$\int_\Omega \nabla u \nabla v~ dx=-\int_\Omega v \Delta u~dx$$ and $$0=-\int_\Omega v (\Delta u-f)~dx~\forall v \in C^\infty_0(\Omega)$$ therefore $$\Delta u-f=0$$ almost everywhere. but now I stuck, can someone help? I think I can't use any sobolev embeddings because I dont know in which space $f$ lies.
My approach to go from weak form to the strong form of equation is to use the distributions notations,
We already have $v\in\mathcal{D(\Omega)}=C_0^\infty(\Omega)$, and we can observe easily that $u$ defines a distribution because it is a $C^2(\Omega)\cap C^1(\Omega)$.
So the integrals can be written as duality crochets as follows, $$ \left< \nabla u,\nabla v\right>_{D'(\Omega),D(\Omega)}=\left< f, v\right>_{D'(\Omega),D(\Omega)}$$ Now, we can use the derivation of the distribution to get the Laplacien operator
$$\left< \nabla u,\nabla v\right>_{D'(\Omega),D(\Omega)}=-\left< \Delta u,v\right>_{D'(\Omega),D(\Omega)} $$ Hence, we obtain $$ -\Delta u=f$$ Given that $f\in L^2(\Omega)$, then we have our result $$-\Delta u=f\quad \mbox{on } \Omega $$