Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 2I (Topics in Analysis)

The closure is closed and bounded in ${\mathbb{R}}^2$, hence compact by the Heine-Borel theorem. It is nonempty. The extreme value theorem therefore gives a point of the closure at which the continuous function attains its maximum. The closure, rather than the open domain, is essential.

At an interior local maximum, the second derivatives in both coordinate directions are nonpositive: restrict the function to each coordinate line and apply the one-variable second-derivative test. Thus ${\nabla }^2\varphi \le 0$ there, contradicting ${\nabla }^2\varphi \ge ϵ>0$. Since a maximum exists on the closure, a maximum lies on the boundary.

For $\eta >0$, use the weak maximum principle for elliptic operators perturbation ${\varphi }_{\eta }(x)=\varphi (x)+\eta |x{|}^2$. Its Laplacian is at least $4\eta$, so the preceding argument puts its maximum on the boundary. If $R^2={\max }_{\bar{\Omega }}|x{|}^2$, then, for any $x\in \bar{\Omega }$,Letting $\eta \downarrow 0$ gives $\varphi (x)\le {\max }_{\partial \Omega }\varphi$. The boundary is compact, so this maximum is attained. This proves the weak weak maximum principle for elliptic operators even when the Laplacian may vanish.

Apply the weak weak maximum principle for elliptic operators to the harmonic function $\varphi$ and to $-\varphi$. Their boundary maxima are both zero, giving $\varphi \le 0$ and $\varphi \ge 0$ on the closure. Consequently $\varphi \equiv 0$.