Re: Need help with Abel's convergence theorem

David C. Ullrich wrote:
On 17 Nov 2006 20:16:09 -0800, anon5874@xxxxxxxxx wrote:

(Abel's Theorem) Given that SUM(0,infty)a_k is convergent, a_k in C (or
equivalently that the convergence set of SUM(0,infty)a_kz^k contains
Delta(0;1) U {1}), let r>=1, and let E={|1-z|<=r(1-|z|)}. We have E a
subset of Delta(0;1) U {1}. Show that f is continuous at 1, f being the
sum of SUM(0,infty) a_kz^k on E. Sketch E for r=1, and r>1.

I am trying to work out the significance of the set
E={|1-z|<=r(1-|z|)}? How does this fit into the theorem?

Have you figured out what E looks like? It's a "nontangential
approach region" - a certain subset of the disk which near the
point 1 looks like the region bounded by two rays starting
at 1 and pointing into the disk at a certain angle.

How it fits into the theorem is that it's possible to show that
f _is_ continuous on E; this says that the sum has a limit
as you approach 1 from the interior of the disk in a "non-
tangential" way. If you allow arbitrary approach to 1,
for example along a curve in the open disk which is
_tangent_ to the unit circle at 1, then the theorem
becomes false.

Possibly you meant to ask how the definition of the
region E fits into the _proof_ of the theorem. That
would raise the question have you got any ideas for
how the proof might go...


************************

David C. Ullrich

Thank you very much. That was most helpful. (I couldn't quite picture
what the theorem was actually saying before).

In answer to your question, the author is kind enough to offer a hint:
f(1) - f(z) = SUM(0,infty)a_k(1-z^k) = SUM(1,m)a_k(1-z^k) +
SUM(m+1,infty)a_k(1-z^k)
Then apply the Abel identity to SUM(m+1,m+p)a_k(1-z^k).

I'll give it a go!

Thanks again

-Darren

.

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