Let a functionbe analytic on a bounded region
and continuous on
Then 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 so
for some M >0.
Alsosince all points outside
are exterior to R and so
Thusis a compact set. By the Extreme Value Theorem, as
is continuous on
there exists
such that
for
Ifthen the proof is complete. Otherwise
so
must be constant on
by the Local Maximum Principle. It follows by the continuity of
on
that
is constant on
also. In fact if
is any point of
then there exists a sequence
in
such that
and hence
by the continuity of
at
Thus if
say, for
then
for
so that
also.
Henceis constant on
so that (1) holds for any point of
If the functionis non – constant and analytic on
then (1) can be strengthened to
for
In fact, iffor some
then
must have a local maximum at
which is not possible by the Local Maximum Theorem.