# Thread: complex analysis question

1. ## 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.

2. What would this mean? $\displaystyle \sum_{n=0}^{\infty} || a_n z^n ||$ converges? If so then how do we interpert $\displaystyle ||a_n z^n||$?

I can show $\displaystyle \sum_{n=0}^{\infty} |a_nz^n|$ would converge, for $\displaystyle |z|<r$. But I am not sure if this is what you want.

3. 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.

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

It does converge by using the theorem that if $\displaystyle \sum_{n=0}^{\infty} a_n z^n$ is convergent on $\displaystyle |z|<\rho$ then it is absolutely convergent on $\displaystyle |z| \leq r < \rho$.

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

6. 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.

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

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

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