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.
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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 $|w|<\sqrt6$.
Julián Aguirre
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Thank you for your answer! – TriSSSe Mar 18 '13 at 21:39
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Good answer, but the region of negativity is the reverse of what is stated; It's when $|\omega| > \sqrt{6}$. A positive function should have $\hat f(\omega)>0$ in at least a small region around $\omega=0$ because the integral of a positive function is positive. – qgp07 Apr 04 '17 at 00:30
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@qgp07 Thanks for noticing. I have edited the answer. – Julián Aguirre Apr 04 '17 at 10:02
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As the answer by Julián Aguirre shows, the result that you are planning on proving is not true. What is true is that the Fourier transform of a real-valued even function is a real-valued even function; but one of the functions being nonnegative does not imply that its transform is also nonnegative. See this answer on stats.SE for details of what additional conditions you might need to make the result be true.
More to the point, since you are interested in probability density functions that are even functions, here is a result that might help you.
If $X$ is a continuous random variable that can be expressed as $Y-Z$ where $Y$ and $Z$ are independent identically distributed random variables, then the density of $X$ is related to the densities of $Y$ and $Z$ as $$f_X(x) = \int_{-\infty}^\infty f_Y(y)f_Z(y-x)\,\mathrm dy = \int_{-\infty}^\infty f_Y(y)f_Y(y-x)\,\mathrm dy$$ which is the autocorrelation function of the common density of $Y$ and $Z$, and so is a real-valued even nonnegative function. Its Fourier transform is thus a power spectral density, and so is also a real-valued even nonnegative function as you desire. Note that it is not necessary for $f_Y$ to be an even function (though of course it is nonnegative).
Thus, it suffices to find random variables that can be decomposed into the difference of two independent identically distributed random variables to get instances of real-valued nonnegative even functions whose Fourier transforms are also real-valued nonnegative even functions.
Dilip Sarwate
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