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Math Help - Cauchy-Schwarz inequality for Integrals

  1. #1
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    Cauchy-Schwarz inequality for Integrals

    Suppose that the functions f,g,f^2,g^2, fg are integrable on the closed, bounded interval [a,b]. Prove that:

     \int ^b _a fg \leq \sqrt { \int ^b_a f^2 } \sqrt { \int ^b_a g^2 } .

    Proof.

    Now,  0 \leq \sqrt { \int _a ^b (f- \lambda g)(f- \lambda g) } ^2 = \int ^b_a (f^2- \lambda fg - \lambda fg + \lambda ^2 g^2 )
    = \int ^b_a f^2 - \lambda \int ^b _a fg - \lambda \int ^b _a fg + \lambda ^2 \int ^b _a g^2

    Choose  \lambda = \int _a ^b fg [ \int ^b _a g^ {2}]^{-1}

    then the previous equation becomes  \int ^b _a f^2 - \int ^b _a (fg)^2 [ \int ^b _a  g^2 ]^{-1} - \int ^b _a (fg)^2 [ \int ^b _a  g^2 ] ^{-1} + \int ^b _a (fg)^2
    which implies  0 \leq \int ^b _a f^2 - \int ^b _a (fg)^2 (2 [ \int ^b _a g^2 ]^{-1})
     \int ^b _a (fg)^2 (2 [ \int ^b _a g^2 ] ^{-1} \leq \int ^b _a f^2
     \int ^b _a (fg)^2 \leq \frac { \int ^b _a f^2  \int ^b _a g^2 } {2}

    How do I get rid of the 2 at the bottom?

    Thanks.
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  2. #2
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    The Cauchy-Schwarz inequality is a general inequality for inner product spaces.
    Thus, what you wrote is a consequence of this general inequality found here*.



    *)I am just a little worried because the space of all integral functions is not an inner product space with norm \left< f,g\right> = \smallint_a^b fg.
    However, the proof on that page seems to work.
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  3. #3
    Moo
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    Hello,

    A more general view on this inequality is Hölder's inequality.
    Cauchy-Schwarz is then a specific situation, with p=q=2.

    Hölder's inequality - Wikipedia, the free encyclopedia
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