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immanuel78
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 About Maximum Modulus Theorem(2nd version)   « on: Jun 3rd, 2007, 9:23am » Quote Modify

Let G be a bounded open set in C and suppose f is a continuous function on the closure of G which is analytic in G.
Then max{|f(z)| : z in the closure of G} = max{|f(z)|:z in the boundary of G}

If G is connceted, the proof is not hard to me.
But if G is not connected, I don't know the proof.
I want to know the proof when G is not connected.
 « Last Edit: Jun 3rd, 2007, 9:25am by immanuel78 » IP Logged
Obob
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 Re: About Maximum Modulus Theorem(2nd version)   « Reply #1 on: Jun 3rd, 2007, 10:25am » Quote Modify

Write G as the disjoint union of its connected components.  Now we know by compactness that |f| has a maximum somewhere in the closure of G.  If it takes this maximum on the interior of one of its components but not also somewhere on the boundary of that component, this violates the maximum modulus principle applied to that component.  So the maximum of |f| occurs either on the boundary of some component, or on some point of the closure of G that is not in the boundary of any component (of course it could also occur in some component, but in that case we showed it also occurs on that component's boundary).  Since boundary points of components are evidently boundary points of G, this says that the maximum of |f| occurs on the boundary of G.

Notice that it might possibly happen that G has boundary points which are not boundary points of any of the connected components.  By playing around with a sequence of disjoint "wedges" approaching zero, you should be able to construct a
set G whose boundary contains zero, even though zero is not a boundary point of any component.  Moreover, it is possible to come up with an analytic function on this domain whose maximum modulus occurs at zero, but nowhere else.
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