complex analysis question

**squarerootof2** · May 11th 2008, 06:25 PM

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.

**ThePerfectHacker** · May 11th 2008, 06:43 PM

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 $|z|<r$ . But I am not sure if this is what you want.

**squarerootof2** · May 11th 2008, 06:50 PM

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.

**ThePerfectHacker** · May 11th 2008, 07:00 PM

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 $|z| \leq r < \rho$ .

**squarerootof2** · May 11th 2008, 07:04 PM

is this theorem we're supposed to use the abel's lemma?

**ThePerfectHacker** · May 11th 2008, 07:09 PM

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.

**squarerootof2** · May 11th 2008, 07:09 PM

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

**ThePerfectHacker** · May 11th 2008, 07:13 PM

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.

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