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Math Help - integrability

  1. #1
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    integrability

    let f and g be integrable functions on [a,b]. Show that

    |\int_{a}^{b} fg| \leq {({\int_{a}^{b}} f^{2})^{1/2} {({\int_{a}^{b}} g^{2})^{1/2}
    Last edited by rondo09; August 7th 2010 at 11:08 PM.
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  2. #2
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    Quote Originally Posted by rondo09 View Post
    let f and g be integrable functions on [a,b]. Show that

    |\int_{a}^{b} fg| \leq {({\int_{a}^{b}} f^{1/2})^2 {({\int_{a}^{b}} g^{1/2})^2

    First, there's a (rather huge and important) mistake: it must be |\int_{a}^{b} fg| \leq {({\int_{a}^{b}} f^2)^{1/2} {({\int_{a}^{b}} g^2)^{1/2}.

    Second, you can google "Cauchy-Schwartz inequality" , or: let \lambda\in\mathbb{R} , and then

    0\leq\int\left(f-\lambda g\right)^2=\int\left(f^2-2fg\lambda+g^2\lambda^2\right) , and now take \lambda:=\frac{\int f^2}{\int fg} , so substituting above:

    0\leq \int \!\!f^2 -\frac{\int f^2}{\int fg}\,2\int \!\!fg+\frac{\left(\int f^2\right)^2}{\left(\int fg\right)^2}\,\int \!\!g^2 =-\int \!\!f^2+\frac{\left(\int f^2\right)^2}{\left(\int fg\right)^2}\,\int \!\!g^2\iff

    \iff \left(\int\!fg\right)^2\int\!f^2\leq \left(\int\!f^2\right)^2\int\!g^2

    And from here your inequality follows at once ( note that we must assume something to justify the above proof: what is this and what happens if the assumption isn't fulfilled?)

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    First, there's a (rather huge and important) mistake: it must be |\int_{a}^{b} fg| \leq {({\int_{a}^{b}} f^2)^{1/2} {({\int_{a}^{b}} g^2)^{1/2}.

    Second, you can google "Cauchy-Schwartz inequality" , or: let \lambda\in\mathbb{R} , and then

    0\leq\int\left(f-\lambda g\right)^2=\int\left(f^2-2fg\lambda+g^2\lambda^2\right) , and now take \lambda:=\frac{\int f^2}{\int fg} , so substituting above:

    0\leq \int \!\!f^2 -\frac{\int f^2}{\int fg}\,2\int \!\!fg+\frac{\left(\int f^2\right)^2}{\left(\int fg\right)^2}\,\int \!\!g^2 =-\int \!\!f^2+\frac{\left(\int f^2\right)^2}{\left(\int fg\right)^2}\,\int \!\!g^2\iff

    \iff \left(\int\!fg\right)^2\int\!f^2\leq \left(\int\!f^2\right)^2\int\!g^2


    And from here your inequality follows at once ( note that we must assume something to justify the above proof: what is this and what happens if the assumption isn't fulfilled?)

    Tonio
    Yeah.. I have mistyped it.. Sorry... It should be
    |\int_{a}^{b} fg| \leq {({\int_{a}^{b}} f^2)^{1/2} {({\int_{a}^{b}} g^2)^{1/2}
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  4. #4
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    You can prove it also similarly to what tonio did and using using ax^2+bx+c>=0 iff b^2-4ac<0
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