Re: Abel's Convergence Theorem exercise

In article
<1164298989.165130.94840@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
anon5874@xxxxxxxxx wrote:

The World Wide Wade wrote:
In article
<1164241110.855578.204360@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
anon5874@xxxxxxxxx wrote:

I am trying to answer the exercise in my book, but I get an incorrect
result. (I'm probably carrying over some asumptions from real analysis
that I need to shake out). Can any one suggest where I went wrong?

Question:
Given that SUM(0,inf)a_k is convergent, a_k in C. Let r>=1, and let
E={|1-z|<=r(1-|z|)}. We have E subset Delta(0;1)U{1}.
Show that f is continuous at 1, f being SUM(0,inf)a_kz^k on E.

Hint:
f(1) - f(z) = SUM(1,inf)a_k(1-z^k) = SUM(1,m)a_k(1-z^k) +
SUM(m+1,inf)a_k(1-z^k)
Apply the Abel identity to SUM(m+1,m+p)a_k(1-z^k) with the (1-z^k)
being the a's of the identity.

Assume WLOG sum_(n=0,oo) a_n = 0. Then S_n = sum_(n=0,n) a_k -> 0
as n -> oo. Use Abel's identity and take a limit to see
sum(0,inf) a_nz^n = sum(0,inf) S_nz^n(1-z). We want the last sum
-> 0 as z -> 1 within E. Because of the 1-z term, we're free to
consider sum(N,inf) S_nz^n(1-z) for any large N. Estimate that by
|1-z|sum(N,inf) |S_n||z^n| and use the inequality defining E.

(Attempted) Answer:
ok, so I started with:
SUM(m+1,m+p)a_k(1-z^k)

Applying Abel's identity (this was a little tricky), I got:

SUM(m+1,m+p)a_k(1-z^k) = SUM(m+1,m+p-1){(1-z^k) - (1-z^(k+1))}S_k +
(1-z^(m+p))S_n

where S_k = SUM(m+1,k)a_k.

Simplifying, I get:
SUM(m+1,m+p-1){z^(k+1)-z^k}S_k + (1-z^(m+p))S_n

= (z-1)SUM(m+1,m+p-1)z^ks_k + (1-z^(m+p))S_n

Wasn't exactly sure what to do next, so I took the absolute value:

|(z-1)SUM(m+1,m+p-1){z^kS_k} + (1-z^(m+p))S_n|

I note that |s_k|<ep (epsilon) for big enough m, and moved it to the
outside (is that allowed?):

< ep.|(1-z)SUM(m+1,m+p-1)z^k + (1-z^(m+p))| =
ep.|(1-z).z^(m+1).(1-z^p)/(1-z) + (1-z^(m+p))|

= ep.|z^(m+1).(1-z^p) + (1-z^(m+p))| <= 4ep for |z|<=1.

But this would yield a set E={|z|<=1, z!=1}

Which doesn't seem correct? Help!

Isn't that what I did?
I think I am having trouble using the inequality that defines E.
My working out seems to suggest that the sum will be continuous for any
sequence with |z|<=1.

Do you know what I did wrong?

Your proof seemed a little long, so I didn't read it. To see that
|1-z|sum(N,inf) |S_n||z^n| is small for z close to 1, replace
|S_n| by eps. What would your estimate then be? You should get
the inequality defining E staring in your face.
.

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