 Solution

Use the classical regularity setting $u\in C^2(D)\cap C(\overline{D})$, with $D$ bounded. Continuity on the compact set $\overline{D}$ guarantees existence of a minimum. The inequality $\Delta u=f\le 0$ means that $u$ is superharmonic function.

A strict inequality $\Delta u<0$ would immediately exclude an interior minimum, because its Hessian matrix there would be positive semidefinite and have nonnegative trace. To handle equality, introduce the quadratic perturbation proof of the minimum principle:Its minimum must lie on $\partial D$. Choose $R$ so that $x^2+y^2\le R^2$ on $\overline{D}$. For any $p\in D$,Letting $\epsilon \downarrow 0$ gives $u(p)\ge {\min }_{\partial D}u$. ThereforeThis asserts that a minimum is attained on the boundary, not that it cannot also be attained inside; a constant function attains it everywhere.