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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.

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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 $x=0$.

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  • $\begingroup$ Wow, that's actually really simple. And there I was trying to find the zeroes of the derivative... Many thanks $\endgroup$ – Capybara Mar 30 '17 at 12:50

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