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Thread: Norm of indicator function in L^2-space

  1. #1
    MHF Contributor Swlabr's Avatar
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    Norm of indicator function in L^2-space

    I am reading a book by one John Conway, and at one point he seems to imply that $\displaystyle \|\chi_{\Delta}\|_2 = \left(\int |\chi_{\Delta}|^2 d \mu\right)^{1/2} = (\mu({\Delta}))^{1/2}$ (where $\displaystyle \chi_{\Delta}$ is the indicator function).

    I can't quite understand this. I feel it should just be $\displaystyle \mu({\Delta})$...

    So, does anyone know why?

    Thanks in advance!
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  2. #2
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    Because $\displaystyle {\chi_\Delta}^2=\chi_{\Delta\bigcap\Delta}=\chi_\D elta, and \int \chi_\Delta d\mu=\mu(\Delta)$
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by karkusha View Post
    Because $\displaystyle {\chi_\Delta}^2=\chi_{\Delta\bigcap\Delta}=\chi_\D elta$
    Yes, of course! Thanks!
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by karkusha View Post
    Because $\displaystyle {\chi_\Delta}^2=\chi_{\Delta\bigcap\Delta}=\chi_\D elta, and \int \chi_\Delta d\mu=\mu(\Delta)$
    Wait a tick - the square is outside the absolute value, not inside...
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  5. #5
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    $\displaystyle |\chi_\Delta|^2=|{\chi_\Delta}^2|=|\chi_\Delta|=\c hi_\Delta$
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