Fourier Theory Problem

Jul 2010
2
0
Is anyone able to prove, or provide a counter example to the following:

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

CaptainBlack

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

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

Now:

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

So:

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

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

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

etc

CB
 
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Reactions: Ed209
Jul 2010
2
0
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