 Solution

The maximum principle for subharmonic functions here states that if $u\in C^2(D)\cap C(\overline{D})$ and $\Delta u=f\ge 0$ in $D$, thenIf $f\le 0$, apply this to $-u$ to obtain the corresponding minimum principle. In particular a harmonic function has both its maximum and minimum on the boundary. Without a sign condition on $f$, this conclusion need not hold.

First suppose $\Delta u>0$. By compactness and continuity, $u$ attains a maximum on $\overline{D}$. At an interior maximum, the first derivatives vanish and the Hessian matrix is negative semidefinite. Its trace is then nonpositive, contradicting $\Delta u>0$. Thus a maximum lies on the boundary.

For the non-strict case put $u_{ϵ}=u+ϵ(x^2+y^2)$, with $ϵ>0$. Then $\Delta u_{ϵ}=f+4ϵ>0$, so the strict case yields, for $z\in \overline{D}$,Let $ϵ\downarrow 0$. This gives $u(z)\le {\max }_{\partial D}u$, proving the theorem. The argument proves the weak maximum principle; it does not claim that every maximizer must be on the boundary when $u$ is constant.