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Math Help - Integral inequality

  1. #1
    Junior Member
    Joined
    Jan 2010
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    Integral inequality

    Hi, I need to show that \|f\|_1=\int_{0}^{1}|f(x)|dx\leq(\int_{0}^{1}|f(x)  |^2dx)^\frac{1}{2}=\|f\|_2

    I think I need to use the Cauchy Schwarz inequality:

    |<x,y>|\leq\|x\|_2\|y\|_2

    Any help would be great. Thanks
    Last edited by ejgmath; April 14th 2010 at 10:56 AM.
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  2. #2
    Super Member
    Joined
    Jan 2009
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    \left(\int_0^1 |f(x)g(x)|~dx \right )^2  \leq \left(\int_0^1 [f(x)^2] ~dx \right )\left(\int_0^1 [g(x)^2] ~dx\right  )

    Let  g(x) = 1 so we have
     <br /> <br />
\left(\int_0^1 |f(x))|~dx\right  )^2 \leq \left(\int_0^1 [f(x)^2] ~dx\right  )\left(\int_0^1 dx\right  ) = \int_0^1 [f(x)^2] ~dx  <br />
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