From: Jeff Rubin <jeffrubin@my-deja.com>
Subject: Question about uniform convergence of polynomials
Date: Sun, 17 Sep 2000 18:51:31 GMT
Newsgroups: sci.math

I'm trying to work through a proof in Dunford & Schwartz
Linear Operators.  In this proof, K is a compact subset of
the complex plane and it is assumed that its complement in
the plane is not connected.  G is assumed to be a bounded
component of the complement of K.  Furthermore, there is
shown to be a certain sequence of polynomials, {P_n}, such that
{P_n(z)} converges uniformly for z in K.

Then there appears the sentence "It follows from the theory of
functions of a complex variable that" {P_n(z)} also converges
uniformly for z in G.

What result in complex analysis does this depend upon and how
is it applied in this situation?

Thanks,

Jeff Rubin


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==============================================================================

From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Question about uniform convergence of polynomials
Date: 17 Sep 2000 15:33:53 -0400
Newsgroups: sci.math
Summary: [missing]

In article <8q33rc$3sh$1@nnrp1.deja.com>,
Jeff Rubin  <jeffrubin@my-deja.com> wrote:
[quote of previous message deleted --djr]

Maximum Modulus Principle: here if B is an open bounded component of the
complement of K then every function holomorphic on B and continuous on the
closure of B (the boundary of B is a subset of K) attains the maximum of
its modulus on the boundary (and nowhere else if it is non-constant).
Apply it to the polynomials P_m(z)-P_n(z) for Cauchy condition.

Hope it helps, ZVK(Slavek).
==============================================================================

From: "Fabrice P. Laussy" <laussy@nat.fr>
Subject: Re: Question about uniform convergence of polynomials
Date: Sun, 17 Sep 2000 21:35:29 +0200
Newsgroups: sci.math

Jeff Rubin wrote:

> I'm trying to work through a proof in Dunford & Schwartz
> Linear Operators.  In this proof, K is a compact subset of
> the complex plane and it is assumed that its complement in
> the plane is not connected.  G is assumed to be a bounded
> component of the complement of K.  Furthermore, there is
> shown to be a certain sequence of polynomials, {P_n}, such that
> {P_n(z)} converges uniformly for z in K.

You mean P_n restricted to K converges uniformly, don't you? Uniform
convergence doesn't apply to sequences of numbers (which {P_n(z)} is).

> Then there appears the sentence "It follows from the theory of
> functions of a complex variable that" {P_n(z)} also converges
> uniformly for z in G.
>
> What result in complex analysis does this depend upon

Analytic continuation

> and how
> is it applied in this situation?

Its principle is its own application. It means you can, for an
analytic function (like a polynomial) defined on a subset like K (no
need for it to be compact) extend it over a greater subset so that
it's still analytic (and thus benefit from properties like uniform
convergence, and others). You proceed open subset (which intersect the
one where the analytic function is defined) by open subset until you
cover the whole area you aim.

Those are the basis, for the details, check a textbook of complex
analysis.

F.P.L.
==============================================================================

From: wramey@home.removethis.com (Wade Ramey)
Subject: Re: Question about uniform convergence of polynomials
Date: Sun, 17 Sep 2000 20:33:55 GMT
Newsgroups: sci.math

Jeff, when you state a problem so clearly, it's a pleasure to help.

The result follows from the maximum principle for holomorphic functions:
Let U be a bounded open subset of the complex plane, and suppose f is
continuous on the closure of U, holomorphic on U. Then |f| attains its
maximum value on the boundary of U.

Corollary: If you have two such functions f and g, and |f - g| < c on the
boundary of U, then |f - g| < c  on U.

In your specific case, note that polynomials are holomorphic everywhere,
and that the boundary of G is contained in K.

Wade 

[quote of original message deleted --djr]