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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.