Let $f(x)$ be a real function, symmetric around $x=0$. From the properties of the Fourier transform, we know that $\hat{f}(\omega)$ (the spectrum of $f(x)$) is also a real function, symmetric around $\omega = 0$. However, I need to prove that $\hat{f}(\omega)$ is positive for certain functions $f(x)$. In particular, I am dealing with what I would call "well-behaved" (Gaussian-like) probability density functions: I can assume that $f(x)$ has its (positive) maximum at $x=0$ and strictly decreases smoothly to zero for $x \to \pm \infty$. Can it then be shown that, under the above conditions on $f(x)$, the spectrum $\hat{f}(\omega)$ is positive? Thank you for your help.
[Math] Proof that the Fourier transform of a positive, real, symmetric function is positive, real, symmetric
fourier analysis
Best Answer
No. Consider for example $$ f(x)=(1+x^2)e^{-x^2}. $$ Then $f(x)>0$ and $f(x)=f(-x)$ for all $x\in\mathbb{R}$. Moreover $$ f'(x)=-2\,x^3e^{-x^2}, $$ so that it is strictly decreasing in $(0,\infty)$ and achieves its maximum at $x=0$. Finally $$ \lim_{x\to\infty}f(x)=0. $$ But $$ \hat f(\omega)=-\frac{e^{-\frac{\omega ^2}{4}} \left(\omega ^2-6\right)}{4 \sqrt{2}}, $$ which is negative if $|w|>\sqrt6$ and positive if $|w|<\sqrt6$.