Let a functionbe analytic on a bounded regionand continuous onThen there exists(the boundary of) such that
for(1) (the interior together with the boundary of).
This is the Maximum Principle
Proof
Note that( the interior together with the boundary of R) is closed. R is also bounded and sofor some M >0.
Alsosince all points outsideare exterior to R and so
Thusis a compact set. By the Extreme Value Theorem, asis continuous onthere existssuch thatfor
Ifthen the proof is complete. Otherwisesomust be constant onby the Local Maximum Principle. It follows by the continuity ofonthatis constant on also. In fact ifis any point ofthen there exists a sequenceinsuch thatand henceby the continuity ofatThus ifsay, forthenforso thatalso.
Henceis constant onso that (1) holds for any point of
If the functionis non – constant and analytic onthen (1) can be strengthened tofor
In fact, iffor somethenmust have a local maximum atwhich is not possible by the Local Maximum Theorem.