TITLE: Non-negative function with non-negative compactly supported Fourier transform
QUESTION [5 upvotes]: In a paper I'm trying to understand, the author claimed (without proof) the existence of a Schwartz function $\varphi \colon \mathbb{R} \rightarrow \mathbb{R}$ with the following properties:

$0 \leq \varphi(x) \leq 2$ for every $x \in \mathbb{R}$,
$\varphi(x) \geq 1$ for every $x \in [0,1]$,
$\widehat{\varphi} (\xi) \geq 0$ for every $\xi \in \mathbb{R}$,
$\widehat{\varphi}(\xi)$ is compactly supported on (say) $[-10,10]$.

I tried two methods to prove this, but in both cases there's been a small thing I've been unable to show.
Firstly, I took an appropriately scaled Gaussian $G$ that satisfied properties 1 and 2 (with a little leeway) in real space, and considered its Fourier transform $\widehat{G}$, which is another Gaussian. I then added a smooth "remainder" term $\widehat{r}$, defined by
$$\widehat{r}(\xi) = \begin{cases} 0 & |\xi| \leq 9\\ \textrm{smooth in-between} \\ -\widehat{G}(\xi) & |\xi|> 10\end{cases}$$ and defined $\widehat{\varphi} = \widehat{G} + \widehat{r}$, hence $\varphi = G + r$. Since the $L^\infty$ norm of $r$ is bounded by the $L^1$ norm of $\widehat{r}$ (which we know to be very small), we know that $\varphi$ satisfies properties 2, 3, and 4, and that $\varphi\leq 2$. However, I can't seem to show that $\varphi\geq0$ (and I'm not 100% convinced that it's the case anyway).
The second approach I tried was to take a smooth, non-negative, even function in Fourier space, $\widehat{\eta}$, that is supported on $[-5,5]$, so that $\widehat{F} = \widehat{\eta}\ast\widehat{\eta}$ is supported on $[-10,10]$. Then $F = \eta^2 \geq 0$, and by scaling $\widehat{F}$ appropriately, we have $F\leq 2$. Then, assuming that the maximum of $F$ occurs at $x=0$, we can "stretch" $F$ horizontally such that it is greater than 1 on the interval $[0,1]$. However, I'm not sure how to show that the maximum is indeed at $x=0$, or whether that is even the case.
Any ideas on how to plug the hole in either of these proofs, or alternative proofs, would be most welcome.

REPLY [4 votes]: The second construction works beautifully. Since $\eta\ge0$ we have
$$
|F(x)|=\Bigl|\int_{\Bbb R}e^{-\pi ix\xi}\,(\eta\ast\eta)(\xi)\,d\xi\Bigr|\le\int_{\Bbb R}(\eta\ast\eta)(\xi)\,d\xi=\|\eta\|_1^2.
$$
On the other hand
$$
F(0)=\int_{\Bbb R}(\eta\ast\eta)(\xi)\,d\xi=\|\eta\|_1^2,
$$
so that in fact $F$ attains its maximum at $x=0$.