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Thread: A variation of the Mean Value Theorem

  1. #1
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    A variation of the Mean Value Theorem

    Hi, I need help proving this:

    Let U be an open set in R^n, f:U--->R be of class C^2 and Gradient of f(a) = 0, a is an element of S contained in U where S is compact and convex.

    Prove that there exists a constant M such that:
    |f(x)-f(a)| <= M* |x-a|^2 for all x in S.

    Thanks a lot.
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  2. #2
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    Use Taylor theorem for approaching with a degree 1 polynomial. There exists $\displaystyle 0\leq t\leq 1$ such that

    $\displaystyle f(x)=f(a)+\sum_{|\alpha|=1}D^{\alpha}f(a)(x-a)^\alpha+
    \sum_{|\alpha|=2}\frac{D^{\alpha}f(a+t(x-a))}{\alpha!}(x-a)^\alpha$

    Here $\displaystyle \alpha$ are multiinedex. For $\displaystyle |\alpha|=1$ all the terms are 0 by the hypothesis. For $\displaystyle |\alpha|=2$ use the hypothesis that $\displaystyle S$ is convex and compact for finding a bound $\displaystyle M$ for each $\displaystyle |D^{\alpha}f(a+t(x-a))|$ for $\displaystyle |\alpha|=2$, and clearly each $\displaystyle |(x-a)^\alpha|\leq |x-a|^2$ for a multiindex $\displaystyle \alpha$ such that $\displaystyle |\alpha|=2$.
    Last edited by Enrique2; Oct 30th 2009 at 03:41 PM.
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  3. #3
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    How do I use the fact that S is convex and compact to find an M such that$\displaystyle

    |D^{\alpha}f(a+t(x-a))|
    $?
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  4. #4
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    $\displaystyle a+t(x-a)$ is in $\displaystyle S$ because $\displaystyle S$ is convex. Since $\displaystyle S$ is also compact and $\displaystyle f$ is $\displaystyle C^2$ all the derivatives of order 2 are uniformly bounded on $\displaystyle S$.
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  5. #5
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    Yes, but what exactly is the bound for the constant M?
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  6. #6
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    OK, there are a finite number of order 2 derivatives, all of them are continuous on the compact set $\displaystyle S$.Take $\displaystyle M$ as the biggest bound for the modulus of each one of all these derivatives!
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