# Non-negative function with non-negative compactly supported Fourier transform

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:

1. $0 \leq \varphi(x) \leq 2$ for every $x \in \mathbb{R}$,
2. $\varphi(x) \geq 1$ for every $x \in [0,1]$,
3. $\widehat{\varphi} (\xi) \geq 0$ for every $\xi \in \mathbb{R}$,
4. $\widehat{\varphi}(\xi)$ 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.

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 $x=0$.