Positivity of Fourier transform of a perturbed Gaussian Consider the function $f = \mathbb{R} \rightarrow \mathbb{R}$, with  $f(x) = e^{- c x^2}$, where $c >0$ is a constant. It is well known that the Fourier transform of this function is a  positive function.
Now consider a new function $g(x)$ which is obtained from $f(x)$ by introducing a very small deformation. For example, $g(x)$
is such that 
$g(x) = f(x)$ when $|x|> 1$ and 
$g(x) = 1 - c \, x^2 + \delta$ when $|x|< 1$, where $\delta = \delta(c)$ is such that 
$1 - c + \delta = e^{-c}$. This ensures that $g(x)$ is continuous.
Question: Is there a $c$ small enough such that also $g(x)$ has a positive Fourier transform?
Comment: the deformation becomes smaller and smaller as $c$ is smaller.
 A: No, there is no such $c$. The transform of the Gaussian decays super-exponentially at infinity (being another Gaussian), while the perturbation $h=g-f$ has Fourier transform of variable sign, with magnitude about $|\xi|^{-2}$. So the sign of the sum is determined by the sign of the latter. 
Explanation: $h$ is a function with compact support, which is continuous but is not continuously differentiable; $h'$ jumps at $\pm 1$. Therefore, $\widehat{h'}$ is not in $L^1(\mathbb{R})$. Also, $h'(0)=0$ and the Fourier inversion formula imply that 
$$
\lim_{\epsilon\to 0}\int_\mathbb{R} \widehat {h'}(\xi)e^{-\epsilon \xi^2}\,d\xi \to 0
$$
Thus, $\widehat{h'}$ is "zero on average" but its tail is  about the size of $1/|\xi|$, because the discontinuities at $\pm 1$ add a multiple of $\operatorname{cos}(\xi)/\xi$ to the transform, which dominates the contribution of the smooth part of $h'$. (Recall the Fourier transform of $\chi_{[-1,1]}$ is $\sin (\xi)/\xi$; this one is more like the transform of $x\chi_{[-1,1]}(x)$, since $h'$ is odd, so the main part of the transform is $\cos(\xi)/\xi$.) Returning to $\hat h$, i.e., dividing by $i\xi $ or so, we see that $\hat h$ behaves like $\cos (\xi)/\xi^2$ at infinity. 
