Fourier transform of non-vanishing $f \in C^\infty([-1,1])$: does it always look like $\frac{\sin x}{x}$? Given a smooth non-vanishing real even function $f \in C^\infty([-1,1])$ we can compute the Fourier transform (of its product with the indicator of the unit interval):
$$
  \widehat f(\xi)=\int_{-1}^1 e^{i\xi x}f(x) \, dx.
$$
I'm plotting some examples, and it turns out that the qualitative behavior of $\widehat f$ is similar to that of $\frac{\sin x}{x}$ (which corresponds to $f(x) = 1$):
$\hspace{11em}$

Is it always the case or it is possible to find such a function $f$ (smooth, $f(x) \neq 0$ for any $x$, even) for which the Fourier transform $\widehat f$ will be qualitatively different? In particular, is it possible to find such $f$ that $\widehat f$ does not have any zeros on the real line $\mathbb R$?
More generally, if $f \in C^\infty(\overline B{}^d)$, where $\overline B{}^d$ denotes the unit ball in $\mathbb R^d$, $d \geq 2$, we can put
$$
  \widehat f(\xi)=\int_{\overline B{}^d} e^{i\xi x}f(x) \, dx.
$$
If $f$ is spherically symmetric (i.e. $f(x)=f_0(|x|)$ for some scalar $f_0$) and $f(x)>0$ on $\overline B{}^d$, can we conclude that qualitatively $\widehat f$ behaves like 
$$
  \widehat{ 1_{\overline B{}^d}}(\xi) = |\xi|^{-\frac{d}{2}} J_{\frac d 2}{|\xi|} \quad ?
$$
 A: If $f \in C^\infty[-1,1]$ is even non-vanishing, we continue $f$ by zero to $\mathbb R$ so that 
$$
  f = f(1)1_{[-1,1]}+g, \quad f(1) \neq 0,
$$
where $g$ is Lipschitz supported in $[-1,1]$. Hence,
$$
  \widehat f(\xi) = f(1) \tfrac{\sin(\xi)}{\xi}+\widehat g(\xi) = \xi^{-1}(f(1)\sin(\xi)+o(1)), \quad |\xi|\to+\infty.
$$
It follows that qualitatively $\widehat f$ is always similar to $\frac{\sin(\xi)}{\xi}$. In particular, there is always an infinite number of real zeros and one can indicate their asymptotic positions.
On the other hand, I don't see a straightforward generalization to dimension $d \geq 2$. As in the $d=1$ case, we can write
$$
  f = f(x_0) 1_{\overline B{}^d} + g,
$$
where $x_0 \in \partial \overline B{}^d$, and $g(x) = g_0(|x|)$ for some even Lipschitz scalar function $g_0$ supported in $[-1,1]$. I don't see whether the Fourier transform of $g(x)$ decays faster than $|\xi|^{-\frac{d+1}{2}}$ or not.
A: Suppose $f$ is even and $C^1$ on $[-1,1],$ with $f(x) = 0$ for $|x|>1.$ Then integration by parts shows 
$$\hat f (x) = \int_{-1}^1 f(t)\cos (xt)\, dt = f(t)\frac{\sin (xt)}{x}\big |_{-1}^1- \frac{1}{x}\int_{-1}^1 f'(t)\sin (xt)\, dt$$ $$ = 2f(1)\frac{\sin x}{x} + o\left (\frac{1}{x}\right ).$$
We have used the Riemann-Lebesgue lemma to get the $o(1/x)$ term.
A: The Paley-Wiener theorem characterizes such Fourier transforms : if $f \in L^2([-1,1])$ is a square integrable function supported on $[-1,1]$, then its Fourier transform $\hat{f}$ extends to an holomorphic function on $\mathbb{C}$ of exponential type, which means there's a constant $C>0$ such that
$$\hat{f}(z) \le C e^{|z|}$$
And for all $y \in \mathbb{R}$, the function $x \mapsto \hat{f}(x+i y)$ is square integrable.
Conversely, any function $g$ which satisfy those conditions (holomorphic continuation, exponential growth, square integrability) is the Fourier transform of some square integrable function $f$.
A: I don't know the answer for the zeros.
The Fourier transform of $f(x ) =(1-|x|) 1_{x \in [-1,1]}$ is $2\frac{1-\cos(\omega)}{\omega^2}$ which is non-negative and has some zeros at $\pi/2+k\pi$, 
which means the Fourier tranform of $g(x) = f(x)+f(\pi x)$ is positive. 
$g$ is only continuous, not smooth.

$h(x) = e^{-1/(1-x^2)}$ is a typical example of a compactly supported smooth function $C^\infty_c$, which means it is in the Schwartz space and so is its Fourier transform  (Schwartz means smooth and rapidly decreasing)
