Solution
= Solution
Statement (iii) is false. On the unit disc, take
$$
\phi(x,y)=1-x^2-y^2,
\qquad
f=\Delta\phi=-4.
$$
The function $f$ is continuous on the closed disc, but $\phi$ has its strict maximum $1$ at the interior point $(0,0)$ and equals zero on the boundary.