Analysis II

**slevvio** · May 1st 2010, 11:03 AM

Let f be analytic on an open disc $D(z_0, R)$ . If $r<R$ , then f is bounded on the circle | $z-z_0| =r$ . 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 $D(z_0,R)$ , so has a power series expansion $f(z) = \sum_{n=0}^{\infty} a_n (z- z_0) ^n$ with radius of convergence $R' \ge R$ (by Taylor).

So on the circle:

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

**awkward** · May 1st 2010, 04:08 PM

Originally Posted by slevvio

Let f be analytic on an open disc $D(z_0, R)$ . If $r<R$ , then f is bounded on the circle | $z-z_0| =r$ . 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 $D(z_0,R)$ , so has a power series expansion $f(z) = \sum_{n=0}^{\infty} a_n (z- z_0) ^n$ with radius of convergence $R' \ge R$ (by Taylor).

So on the circle:

$|f(z)| = |\sum_{n=0}^{\infty} a_n (z- z_0) ^n| \le \sum_{n=0}^{\infty} |a_n (z- z_0) ^n| = \sum_{n=0}^{\infty} |a_n| |(z- z_0) ^n| = \sum_{n=0}^{\infty} |a_n| |r|^n$ , 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 | $z-z_0| =r$ is a compact set, and a continuous function defined on a compact set is bounded?

**slevvio** · May 3rd 2010, 05:12 AM

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!

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