Let $\Delta:D(\Delta)\subset L^2\to L^2$ where $D(\Delta)=\left\{u\in L^2:\Delta u\in L^2\right\}$. Let $f\in L^2$.
Let $u\in L^2$ a weak solution of $\Delta u=f$, then $$(u,\Delta^*v)=(f,v),\quad \forall v\in D(\Delta^*)$$
Because $\Delta=\Delta^{*}$ ( self adjoint on $D(\Delta)$ maximal domain) then \begin{align} (f,v)=(u,\Delta^*v)\\ =(u,\Delta v)\\ =(\Delta u,v) \end{align} Therefore, $(f,v)=(\Delta u,v) \forall v\in D(\Delta)$ implies $\Delta u=f$?