Maximum modulus principle with boundary and infinity control

Fix $\epsilon >0$ and form the superlevel set $K_{\epsilon }:=\{z\in \mathrm{\Omega }:\mathrm{\mid }f(z)\mathrm{\mid }\ge M+\epsilon \}$.

By [L5] and the reverse triangle inequality, $\mathrm{\mid }f\mathrm{\mid }$ is continuous, so $K_{\epsilon }$ is relatively closed in $\mathrm{\Omega }$: at a point where $\mathrm{\mid }f\mathrm{\mid }<M+\epsilon$, that strict inequality persists on a neighbourhood. Boundary control excludes every point of $\mathrm{\partial }\mathrm{\Omega }$ from the closure of $K_{\epsilon }$, so $K_{\epsilon }$ is closed in the plane. It is bounded because $\mathrm{\Omega }$ is bounded or, in the unbounded case, because infinity control excludes all points with sufficiently large modulus. Thus [L2] makes $K_{\epsilon }$ compact and it lies entirely inside $\mathrm{\Omega }$.

If $K_{\epsilon }$ were nonempty, [L3] would give a maximizer of $\mathrm{\mid }f\mathrm{\mid }$ on it. Outside $K_{\epsilon }$ the modulus is smaller than $M+\epsilon$, so this is also an interior global maximum on $\mathrm{\Omega }$; [L1] makes $f$ constant. When $\mathrm{\Omega }$ is bounded, its boundary is nonempty because otherwise it would be a nonempty clopen subset of the connected plane [L4] and hence the whole unbounded plane, so the constant contradicts finite-boundary control. When $\mathrm{\Omega }$ is unbounded, it contradicts infinity control. Hence $K_{\epsilon }$ is empty.

The set $K_{\epsilon }$ is empty for every $\epsilon >0$. If some $z$ had $\mathrm{\mid }f(z)\mathrm{\mid }>M$, taking $\epsilon =(\mathrm{\mid }f(z)\mathrm{\mid }-M)\mathrm{/}2$ would put $z$ in $K_{\epsilon }$, a contradiction; therefore $\mathrm{\mid }f\mathrm{\mid }\le M$ on $\mathrm{\Omega }$.