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Math Help - Fourier Theory Problem

  1. #1
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    Fourier Theory Problem

    Is anyone able to prove, or provide a counter example to the following:

    let f(t) be a non-negative function, then the Fourier transform of  f(t), F(\omega) say, takes its maximum absolute value at \omega = 0. i.e. \arg \max_{\omega} \{|F(\omega)|\}=0
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  2. #2
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    Quote Originally Posted by Ed209 View Post
    Is anyone able to prove, or provide a counter example to the following:

    let f(t) be a non-negative function, then the Fourier transform of  f(t), F(\omega) say, takes its maximum absolute value at \omega = 0. i.e. \arg \max_{\omega} \{|F(\omega)|\}=0
    \displaystyle F(\omega)=\int_{-\infty}^{\infty}f(x)e^{-i \omega x}\;dx

    Now:

    |F(\omega)|= \left|\int_{-\infty}^{\infty}f(x)e^{-i \omega x}\;dx\right|\le \int_{-\infty}^{\infty}\left|f(x)e^{-i \omega x}\right|\;dx

    So:

    \displaystyle |F(\omega)|\le \int_{-\infty}^{\infty}\left|f(x)\right|\;dx

    but as $$ f(x) is non-negative this is:

    \displaystyle |F(\omega)|\le \int_{-\infty}^{\infty}\left|f(x)\right|\;dx=\int_{-\infty}^{\infty}f(x)\;dx=F(0)

    etc

    CB
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  3. #3
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    Nice!
    I was just about to post the same proof as I figured it out yesterday, but thank you.
    Its quite a nice results actually, and imo not particularly intuitive
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