# Fourier Theory Problem

#### Ed209

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

MHF Hall of Fame
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

Ed209

#### Ed209

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