# Thread: Analysis II

1. ## Analysis II

Let f be analytic on an open disc $\displaystyle D(z_0, R)$. If $\displaystyle r<R$, then f is bounded on the circle |$\displaystyle z-z_0| =r$. My notes make use of this fact but it doesn't prove it so I have tried to do it myself, is this correct?

f is analytic on $\displaystyle D(z_0,R)$, so has a power series expansion $\displaystyle f(z) = \sum_{n=0}^{\infty} a_n (z- z_0) ^n$ with radius of convergence$\displaystyle R' \ge R$ (by Taylor).

So on the circle:

$\displaystyle |f(z)| = |\sum_{n=0}^{\infty} a_n (z- z_0) ^n| \le \sum_{n=0}^{\infty} |a_n (z- z_0) ^n| = \sum_{n=0}^{\infty} |a_n| |(z- z_0) ^n| = \sum_{n=0}^{\infty} |a_n| |r|^n$, which converges since r = |r| < R'. Hence f is bounded on the circle - is this ok? Thanks for any help and advice

2. Originally Posted by slevvio
Let f be analytic on an open disc $\displaystyle D(z_0, R)$. If $\displaystyle r<R$, then f is bounded on the circle |$\displaystyle z-z_0| =r$. My notes make use of this fact but it doesn't prove it so I have tried to do it myself, is this correct?

f is analytic on $\displaystyle D(z_0,R)$, so has a power series expansion $\displaystyle f(z) = \sum_{n=0}^{\infty} a_n (z- z_0) ^n$ with radius of convergence$\displaystyle R' \ge R$ (by Taylor).

So on the circle:

$\displaystyle |f(z)| = |\sum_{n=0}^{\infty} a_n (z- z_0) ^n| \le \sum_{n=0}^{\infty} |a_n (z- z_0) ^n| = \sum_{n=0}^{\infty} |a_n| |(z- z_0) ^n| = \sum_{n=0}^{\infty} |a_n| |r|^n$, which converges since r = |r| < R'. Hence f is bounded on the circle - is this ok? Thanks for any help and advice
Your computation looks correct to me, but is it really necessary?

Why not just use the fact that |$\displaystyle z-z_0| =r$ is a compact set, and a continuous function defined on a compact set is bounded?

3. ahh - we did not cover compactness in lectures, but we were given a sheet about it - i was just confused since i was never told to use compactness to prove this, but i see this now , thanks!