Existence of a certain Schwartz function with compact Fourier support My question is if there exists a Schwartz function $f$ on the real line such that the following hold:


*

*$f\ge 0$

*$\widehat{f}$ is equal to $1$ on $[-1,1]$ 

*$\widehat{f}$ is compactly supported


(I'm not requiring $\widehat{f}$ to be non-negative, or even real-valued.)
I know how to construct functions satisfying 1 and 3 by taking $\widehat{f}=\phi*\phi$ for an appropriate $\phi$, but I don't see how to make 2 work, so I suspect that the answer is that there is no such function.
Any helpful comments or partial answers are appreciated.
 A: All probabilists know there can be no such $f$, not even in $L^1$, or for measures $\mu(dx)$ more generally than $f(x)dx$.  This can be seen many different ways. The usual argument starts by showing that if $\hat f(x)=1$ on a neighborhood of $0$ then $\int_{\mathbb R} (1-\cos(tx))f(x)dx = 0$ for all $t\in[-1,1]$.  Since the integrand is non-negative, $f$ must vanish for all $x$ for which $tx$ is not an integer multiple of $\pi/2$, for all $|t|\le 1$, and so for all $x\ne 0$.
But I forgot this fact, and came up with this, instead:
The function $\hat f$ must satisfy the Bochner condition, that for each finite set $\{t_i\}$ of reals, the   matrices $A=(a_{ij})$  where $a_{ij} = \hat f(t_i-t_j)$ are positive semidefinite. This, and the assumption that $\hat f=1$ on $[-1,1]$  is enough to show that $\hat f(t)=1$ for all $t$, as follows.
If $t_2-t_1\in [-1,1]$ and $t_3-t_2\in[-1,1]$ but $t_3-t_1\notin[-1,1]$, one can learn what $a = \hat f(t_3-t_1)$ is by noting that $A=\pmatrix{1&1&a\\1&1&1\\\bar a&1&1}$ is psd, and hence $q=(1,-2,1)A(1,-2,1)^T\ge0$.  But $q=2(\Re a-1)$ so $\Re a\ge1$ and so $|a|\ge1$.  But also the submatrix $\pmatrix{1&a\\\bar a&1}$ is psd, so its determinant $1-|a|^2$ cannot be negative, so  $|a|\le1$.  Hence so $A$  psd implies $a=1$.  So now we know $\hat f(t)=1$ for all $t\in[-2,2]$.  The same argument, applied inductively, shows $\hat f = 1$ on each set $[-2^k,2^k]$, and thus on $\mathbb R$. 
The Bochner theorem is a basic fact in harmonic analysis.  The Fourier transform of a non-negative measure, or function, is "positive definite", in the sense that the matrices formed from the FT as above, must be positive semidefinite.  The Wikipedia article is perhaps too high-brow for beginners, the Wolfram MathWorld one is too terse, but there are expositions out there that should match your level.
The basic idea is that $\hat f(t-s) = \int_{\mathbb R} \exp(-2\pi i (t-s)x) f(x)dx$ (give or take some norming constants), so a sum like $$\sum_k\sum_l \bar x_k x_l \hat f ( t_k-t_l) = \int_{\mathbb R}\left| \sum_k x_k e^{-2\pi t_k x}\right|^2 f(x)dx\ge 0.$$
This is the easy part of the theorem.  The hard part, that the collection of all such inequalities hold, and the requirement that $\hat f$ be continuous, is enough to imply that $f\ge0$ everywhere, is not needed in your case.
A: As a partial answer, by Uryshonn's lemma you can find a continuous function $\phi$ such that $\phi(x)=1$ for $|x| \leq 1$ and $\phi(x)=0$ for $|x| \geq 2$ and $0 \leq \phi \leq 1$.
Then by inversion formula take $f(x)=\int_{\Bbb{R}}\phi(\xi)e^{2 \pi i \xi x}d \xi$
This function has properties $2,3$ in your post.
