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Math Help - Equality holding in Cauchy's Inequality

  1. #1
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    Equality holding in Cauchy's Inequality

    Hello, this was a problem on an exam:

    From Cauchy's Integral Formula, we get Cauchy's Inequality |f^{(n)}(z_0)| \leq \frac{n!M}{R^n}.
    Prove that equality holds if and only if f(z)= \frac{aMz^n}{R^n} for some a \in \mathbb{C} with |a| = 1.

    Proving (\Leftarrow) is easy, just take the modulus of the nth derivative of f.
    How do we prove the other direction?

    Thanks!
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  2. #2
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    Re: Equality holding in Cauchy's Inequality

    What are M and R?
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    Re: Equality holding in Cauchy's Inequality

    Ah, From Cauchy's Integral Formula and Inequality.
    f is analytic on a ball of radius R
    |f(z)| < M for all z in the ball
    Last edited by Bingk; September 21st 2012 at 04:07 AM. Reason: correction
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    Re: Equality holding in Cauchy's Inequality

    So the statement is:

    Let \Omega = \{z \in \mathbb{C} \ | \ \lVert z - z_0 \rVert < R \}. Let f \in H(\bar{\Omega}) satisfy \lVert f(z) \rVert \le M \ \forall z \in \Omega.

    Then \lVert f^{(n)}(z_0) \rVert = \frac{n!M}{R^n} \Leftrightarrow \exists a \in \mathbb{C}, \lVert a \rVert = 1, such that f(z) = \frac{aMz^n}{R^n} \ \forall z \in \Omega.

    Correct?

    Are you sure it isn't supposed to be f(z) = \frac{aM(z-z_0)^n}{R^n} ?
    Last edited by johnsomeone; September 21st 2012 at 04:38 AM.
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  5. #5
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    Re: Equality holding in Cauchy's Inequality

    What I wrote is what was on the exam, but I think that was justified by assuming that z0 = 0.

    And yes, what you have is correct, as far as I understood the problem
    Last edited by Bingk; September 21st 2012 at 04:54 AM. Reason: Added
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