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Math Help - Analysis II

  1. #1
    Senior Member slevvio's Avatar
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    Analysis II

    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
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  2. #2
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    Quote Originally Posted by slevvio View Post
    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?
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  3. #3
    Senior Member slevvio's Avatar
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    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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