Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 2G / i / Solution

Because $\overline{\Omega }$ is compact, $\varphi$ attains a maximum. If a maximizer lay in $\Omega$, its Hessian matrix would be negative semidefinite, so its trace would satisfy $\Delta \varphi \le 0$, contradicting $\Delta \varphi >0$. Hence every strictly subharmonic function under the stated hypotheses attains a maximum on $\partial \Omega$.

Statement (i) is true. For $ϵ>0$, the functionhas $\Delta {\varphi }_{ϵ}=4ϵ>0$, soLetting $ϵ\downarrow 0$ givesThis is the maximum principle for harmonic functions.