TITLE: Fourier transform of non-vanishing $f \in C^\infty([-1,1])$: does it always look like $\frac{\sin x}{x}$?
QUESTION [5 upvotes]: 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 
$$
  \widehat{ 1_{\overline B{}^d}}(\xi) = |\xi|^{-\frac{d}{2}} J_{\frac d 2}{|\xi|} \quad ?
$$

REPLY [2 votes]: 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.