Consider the Poisson equation with zero Dirichlet boundary condition: Find $u \in C^2 (\Omega)\cap C(\overline{\Omega})$ s.t. $$\begin{cases} -\Delta u=f&in\ \Omega {,}\\ u\mid _{\partial \Omega }=0,& \end{cases} $$ where $\Omega \subset \mathbb{R}^2$ and $f \in L^2 (\Omega)$. To obtain a weak formulation, one typically multiplies the equation by a test function $\varphi \in C_0^{\infty}(\Omega)$, integrates over $\Omega$ by parts and obtains $$ \int_{\Omega} \nabla u \cdot \nabla \varphi dx = \int_{\Omega} f \varphi dx.$$ This makes sense for $u \in H_0^1 (\Omega)$, thus the problem is: Find $u \in H_0^1 (\Omega)$ s.t. $$ \int_{\Omega} \nabla u \cdot \nabla \varphi dx = \int_{\Omega} f \varphi dx$$ for every test function $\varphi \in C_0^{\infty}(\Omega)$. As the test functions are dense in $H_0^1 (\Omega)$, one may equivalently write the weak problem as: Find $u \in H_0^1 (\Omega)$ s.t. $$ \int_{\Omega} \nabla u \cdot \nabla v dx = \int_{\Omega} f v dx$$ for every $v \in H_0^1 (\Omega)$.
Is there a reason for using smooth functions in the derivation instead of $H_0^1 (\Omega)$? As far as I understand, the derivation makes sense for $H_0^1 (\Omega)$ functions just the same.