1. ## 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. $\arg \max_{\omega} \{|F(\omega)|\}=0$

2. Originally Posted by Ed209
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

3. 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