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