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Thread: a < oo; b<oo => a+b < oo

  1. #1
    Senior Member
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    a < oo; b<oo => a+b < oo

    Hello!

    Im not in a hurry, but I dont know where to post this (easy) question.

    I know this:

    $\displaystyle |\hat{f}(\xi)|^2 := |\int_{-\infty}^{\infty} f(x)\ e^{- 2\pi i x \xi}\,dx, |^2 < \infty$

    $\displaystyle |\hat{g}(\xi)|^2 := |\int_{-\infty}^{\infty} g(x)\ e^{- 2\pi i x \xi}\,dx, |^2 < \infty$

    (Fourier transform)

    Now I want to show that

    $\displaystyle |\hat{f}(\xi)|^2 + |\hat{g}(\xi)|^2 < \infty$

    I guess this is true, because I think it is obvious, but isn't there a professional proof on this?


    What do you think?

    Any comments would be much appreciated.

    Best regards,
    Rapha
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  2. #2
    Junior Member Ziaris's Avatar
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    The sum of two finite numbers is finite. Both $\displaystyle |\hat{f}(\xi)|^2$ and $\displaystyle |\hat{g}(\xi)|^2$ are finite, so $\displaystyle |\hat{f}(\xi)|^2+|\hat{g}(\xi)|^2$ is finite. Perhaps there is a more 'professional' way of doing this, but wording it like this seems sufficient.
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  3. #3
    MHF Contributor

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    That's as 'professional' as you need to be! The sum of two finite numbers is defined to be a finite number.
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  4. #4
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    Ok
    thank you, I wasn't sure about this.
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