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Math Help - A very annoying Inequality

  1. #1
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    A very annoying Inequality

    Can anyone help me with this question:

    Show that if f is continuous on the closed interval [0,1] then,

    [IMG]file:///C:/DOCUME%7E1/OWNER%7E1.UNC/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/DOCUME%7E1/OWNER%7E1.UNC/LOCALS%7E1/Temp/moz-screenshot-1.jpg[/IMG]
    |f| < sqrt ( f^2)

    the lower/upper limits of both integrals are 0 and 1 respectively.
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  2. #2
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    Quote Originally Posted by davidmccormick View Post
    Can anyone help me with this question:

    Show that if f is continuous on the closed interval [0,1] then,

    i guess this is the inequality you want to prove: \int_0^1 |f(x)| \ dx \leq \sqrt{\int_0^1|f(x)|^2 \ dx}.
    this is a special case of Holder's inequality. however, it can be easily proved as an inequality on its own: clearly for any real number \lambda we have: \int_0^1(\lambda - |f(x)|)^2 \ dx \geq 0, which gives us:

    p(\lambda)=\lambda^2 - \left(2\int_0^1 |f(x)| \ dx \right) \lambda + \int_0^1 |f(x)|^2 \ dx \geq 0, \ \ \forall \lambda \in \mathbb{R}. thus the discriminant of the quadratic function p(\lambda) must be \leq 0, that is: \left(\int_0^1 |f(x)| \ dx \right)^2 \leq \int_0^1 |f(x)|^2 \ dx. \ \ \Box
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    this is a special case of Holder's inequality. however, it can be easily proved as an inequality on its own: clearly for any real number \lambda we have: \int_0^1(\lambda - |f(x)|)^2 \ dx \geq 0, which gives us:

    p(\lambda)=\lambda^2 - \left(2\int_0^1 |f(x)| \ dx \right) \lambda + \int_0^1 |f(x)|^2 \ dx \geq 0, \ \ \forall \lambda \in \mathbb{R}. thus the discriminant of the quadratic function p(\lambda) must be \leq 0, that is: \left(\int_0^1 |f(x)| \ dx \right)^2 \leq \int_0^1 |f(x)|^2 \ dx. \ \ \Box
    I'm more impressed that you gleaned what the question was than your work.
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