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Thread: [SOLVED] Minimum of a complex function, proof

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Minimum of a complex function, proof

    I've no idea what theorem can be useful to prove the following : Let $\displaystyle f$ be a continuous function in a closed and bounded region $\displaystyle R$, assume also that $\displaystyle f$ is analytic and not constant in the interior of $\displaystyle R$.
    Assuming that $\displaystyle f(z)\neq 0$ in all $\displaystyle R$, prove that $\displaystyle |f(z)|$ has a minimum and that it is reached in the frontier of $\displaystyle R$ and never in the interior of $\displaystyle R$.
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  2. #2
    MHF Contributor Matt Westwood's Avatar
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    I'm not sure, but I suspect this may be an application of Liouville's theorem:

    http://www.proofwiki.org/wiki/Liouvi...mplex_Analysis)

    ... check it out anyway, there may be techniques and insights in there which will give you some inspiration.
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  3. #3
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    If $\displaystyle f(R)\ne 0$, then what can you say about the (analytic) function $\displaystyle g(R)=\frac{1}{f(R)}$ using the Maximum Modulus Principle?
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  4. #4
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by shawsend View Post
    If $\displaystyle f(R)\ne 0$, then what can you say about the (analytic) function $\displaystyle g(R)=\frac{1}{f(R)}$ using the Maximum Modulus Principle?
    That $\displaystyle g$ reaches a minimum in the frontier of $\displaystyle R$.
    Not sure I'm right. By the way I never heard this principle before. That's interesting, it reminds me of Extreme value theorem - Wikipedia, the free encyclopedia. Thanks a lot!
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  5. #5
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    By the Maximum Modulus Principle, $\displaystyle g(z)$ reaches a maximum on the boundary which means that $\displaystyle f(z)$ must reach a minimum on the boundary as well.
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