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Math Help - complex analysis continuity

  1. #1
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    complex analysis continuity

    Suppose that f: A \rightarrow \mathbb{C} is continuous, where A is a region in \mathbb{C}. Let \gamma be a simple closed curve in A. Show that for each n, f_n(z)=\int_\gamma \frac{f(\zeta)}{(\zeta-z)^n}d\zeta is continuous on A-\gamma.

    Let \epsilon>0. Suppose that there is \delta>0 such that |z-z_0|<\delta. I need to show |f_n(z)-f_n(z_0)|<\epsilon.
    |f_n(z)-f_n(z_0)|=|\int_\gamma \frac{f(\zeta)}{(\zeta-z)^n}d\zeta-\int_\gamma \frac{f(\zeta)}{(\zeta-z_0)^n}d\zeta|=\int_\gamma f(\zeta)(\frac{1}{(\zeta-z)^n}-\frac{1}{(\zeta-z_0)^n})d\zeta.

    I got stuck here... can I get some help please?
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  2. #2
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    Quote Originally Posted by dori1123 View Post
    Suppose that f: A \rightarrow \mathbb{C} is continuous, where A is a region in \mathbb{C}. Let \gamma be a simple closed curve in A. Show that for each n, f_n(z)=\int_\gamma \frac{f(\zeta)}{(\zeta-z)^n}d\zeta is continuous on A-\gamma.

    Let \epsilon>0. Suppose that there is \delta>0 such that |z-z_0|<\delta. I need to show |f_n(z)-f_n(z_0)|<\epsilon.
    |f_n(z)-f_n(z_0)|=|\int_\gamma \frac{f(\zeta)}{(\zeta-z)^n}d\zeta-\int_\gamma \frac{f(\zeta)}{(\zeta-z_0)^n}d\zeta|=\int_\gamma f(\zeta)(\frac{1}{(\zeta-z)^n}-\frac{1}{(\zeta-z_0)^n})d\zeta.

    I got stuck here... can I get some help please?
    There's also a more general proof using uniform continuity; but let's do it your way.

    Remember a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+a b^{n-2} + b^{n-1}). We have

    \frac{1}{(\zeta-z)^n}-\frac{1}{(\zeta-z_0)^n}=\frac{(\zeta-z_0)^n-(\zeta-z)^n}{(\zeta-z)^n(\zeta-z_0)^n}

    hence, using the above formula,

    \left|\frac{1}{(\zeta-z)^n}-\frac{1}{(\zeta-z_0)^n}\right|\leq  |z-z_0|\frac{\sum_{k=0}^{n-1}|\zeta-z|^k |\zeta-z_0|^{n-1-k}}{|\zeta-z|^n|\zeta-z_0|^n}.

    Since the curve \gamma is a bounded closed set (thus a compact set), you can choose \delta>0 such that the closed ball \bar{B}(z_0,\delta) does not meet \gamma. Then we can find m,M>0 such that, for any z in this ball, m<|\zeta-z|<M for all \zeta\in \gamma. Using these bounds in the previous inequality enables to bound the right-hand side by C|z-z_0|\leq C\delta for some constant C, and to conclude quickly from there.
    Last edited by Laurent; April 29th 2010 at 03:28 AM. Reason: No power n on |z-z_0|
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  3. #3
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    Thank you very much.
    Last edited by dori1123; April 29th 2010 at 05:16 PM.
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