I'm looking for a non-trivial $L^1(\mathbb{R})$ function $f\ge 0$ with compact support and such that $$\hat f(\xi) = \frac{1}{(2\pi)^{\frac{d}{2}}}\int_{\mathbb{R}}f(x)e^{-ix\xi}dx\ge0$$ The last condition is giving me trouble. As $\xi\in\mathbb{R}$, the exponential can take any value on the complex unit circle, in particular negative values. Also, as $\xi$ is unknown, I don't see how I can work around this with $f$.
How could such a function $f$ be derived?
On
Same as kimchi lover ... Triangular function:
$$
f(x) = \begin{cases}1-|x|,\quad & -1\le x \le 1
\\ 0,\quad &\text{otherwise}\end{cases}
$$
Fourier transform
$$
\widehat{f}(\xi) = \frac{1-\cos\xi}{\pi \xi^2}
$$
On
$$\text{let } f(x)= e^{-ax^2} \text{ where $a>0$}$$
Fourier transform of $f$
$$\hat{f}(k)=\int_{-\infty}^{\infty}e^{-ax^2}e^{-i2\pi kx}dx$$
$$\hat{f}(k)=\int_{-\infty}^{\infty}e^{-ax^2}cos(2\pi kx)dx-i\int_{-\infty}^{\infty}e^{-ax^2}sin(2\pi kx)dx$$
The second integrand is odd, so integration over a symmetrical range gives 0. The value of the first integral is given by (Abramowitz, M. and Stegun, I. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 302, 1972. )
$$\hat{f}(k)=\sqrt{\frac{\pi}{a}}e^{-\frac{\pi^2 k^2}{a}}$$
Let $g=\chi_{[-1,1]}$, let $f=g*g$.