# complex analysis question

• May 11th 2008, 07:25 PM
squarerootof2
complex analysis question
p is the radius of convergence of Σa_n*z^n (n≥0) and let r be a real number less than p. prove that the series converges normally on the disc abs (z)<r and diverges for abs (z)>p. *converging normally means that the sup-norm converges.
• May 11th 2008, 07:43 PM
ThePerfectHacker
What would this mean? $\sum_{n=0}^{\infty} || a_n z^n ||$ converges? If so then how do we interpert $||a_n z^n||$?

I can show $\sum_{n=0}^{\infty} |a_nz^n|$ would converge, for $|z|. But I am not sure if this is what you want.
• May 11th 2008, 07:50 PM
squarerootof2
i'm sorry for not being clear, but this would mean that llull=sup abs (u(z)) where z is in E(subset of complex C) converges. here llull would be called the sup-norm of u.
• May 11th 2008, 08:00 PM
ThePerfectHacker
Quote:

Originally Posted by squarerootof2
i'm sorry for not being clear, but this would mean that llull=sup abs (u(z)) where z is in E(subset of complex C) converges. here llull would be called the sup-norm of u.

The sup-norm, $||a_nz^n||$, over $E = \{ z\in \mathbb{C}: |z| < r\}$ would be $\sup\{ |a_n z^n| : z\in E \} = \sup \{ |a_n||z^n| : z\in E\} = |a_n|r^n$.
Thus, we need to show $\sum_{n=0}^{\infty} |a_n| r^n$ converges.

It does converge by using the theorem that if $\sum_{n=0}^{\infty} a_n z^n$ is convergent on $|z|<\rho$ then it is absolutely convergent on $|z| \leq r < \rho$.
• May 11th 2008, 08:04 PM
squarerootof2
is this theorem we're supposed to use the abel's lemma?
• May 11th 2008, 08:09 PM
ThePerfectHacker
Quote:

Originally Posted by squarerootof2
is this theorem we're supposed to use the abel's lemma?

I am not sure what Abel's lemma you are using here.
If you never seen that before, how about you try proving this.
It happens to be not so difficult to show.
• May 11th 2008, 08:09 PM
squarerootof2
oh and how would we deal with divergence of the series? i was actually getting more stuck on that part... thanks.
• May 11th 2008, 08:13 PM
ThePerfectHacker
Quote:

Originally Posted by squarerootof2
oh and how would we deal with divergence of the series? i was actually getting more stuck on that part... thanks.

Same idea. If $E = \{ z\in \mathbb{C} : |z| < r\}$ then $||a_n z^n||_E = \sup \{ |a_n z^n|: z\in E \} = |a_n|r^n$.

Suppose that $\sum_{n=0}^{\infty} |a_n|r^n$ would converge then by the comparision test it would mean $\sum_{n=0}^{\infty}a_n r^n$ would converge. But that is clearly impossible since $r>\rho$, so it diverges.