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Math Help - 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 0\leq t\leq 1 such that

    f(x)=f(a)+\sum_{|\alpha|=1}D^{\alpha}f(a)(x-a)^\alpha+<br />
\sum_{|\alpha|=2}\frac{D^{\alpha}f(a+t(x-a))}{\alpha!}(x-a)^\alpha

    Here \alpha are multiinedex. For |\alpha|=1 all the terms are 0 by the hypothesis. For |\alpha|=2 use the hypothesis that S is convex and compact for finding a bound M for each |D^{\alpha}f(a+t(x-a))| for |\alpha|=2, and clearly each |(x-a)^\alpha|\leq |x-a|^2 for a multiindex \alpha such that |\alpha|=2.
    Last edited by Enrique2; October 30th 2009 at 04:41 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 <br /> <br />
|D^{\alpha}f(a+t(x-a))|<br />
?
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  4. #4
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    a+t(x-a) is in S because S is convex. Since S is also compact and f is C^2 all the derivatives of order 2 are uniformly bounded on 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 S.Take M as the biggest bound for the modulus of each one of all these derivatives!
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