# Existence of a certain Schwartz function with compact Fourier support

My question is if there exists a Schwartz function $$f$$ on the real line such that the following hold:

1. $$f\ge 0$$
2. $$\widehat{f}$$ is equal to $$1$$ on $$[-1,1]$$
3. $$\widehat{f}$$ is compactly supported

(I'm not requiring $$\widehat{f}$$ to be non-negative, or even real-valued.)

I know how to construct functions satisfying 1 and 3 by taking $$\widehat{f}=\phi*\phi$$ for an appropriate $$\phi$$, but I don't see how to make 2 work, so I suspect that the answer is that there is no such function.

All probabilists know there can be no such $$f$$, not even in $$L^1$$, or for measures $$\mu(dx)$$ more generally than $$f(x)dx$$. This can be seen many different ways. The usual argument starts by showing that if $$\hat f(x)=1$$ on a neighborhood of $$0$$ then $$\int_{\mathbb R} (1-\cos(tx))f(x)dx = 0$$ for all $$t\in[-1,1]$$. Since the integrand is non-negative, $$f$$ must vanish for all $$x$$ for which $$tx$$ is not an integer multiple of $$\pi/2$$, for all $$|t|\le 1$$, and so for all $$x\ne 0$$.

But I forgot this fact, and came up with this, instead:

The function $$\hat f$$ must satisfy the Bochner condition, that for each finite set $$\{t_i\}$$ of reals, the matrices $$A=(a_{ij})$$ where $$a_{ij} = \hat f(t_i-t_j)$$ are positive semidefinite. This, and the assumption that $$\hat f=1$$ on $$[-1,1]$$ is enough to show that $$\hat f(t)=1$$ for all $$t$$, as follows.

If $$t_2-t_1\in [-1,1]$$ and $$t_3-t_2\in[-1,1]$$ but $$t_3-t_1\notin[-1,1]$$, one can learn what $$a = \hat f(t_3-t_1)$$ is by noting that $$A=\pmatrix{1&1&a\\1&1&1\\\bar a&1&1}$$ is psd, and hence $$q=(1,-2,1)A(1,-2,1)^T\ge0$$. But $$q=2(\Re a-1)$$ so $$\Re a\ge1$$ and so $$|a|\ge1$$. But also the submatrix $$\pmatrix{1&a\\\bar a&1}$$ is psd, so its determinant $$1-|a|^2$$ cannot be negative, so $$|a|\le1$$. Hence so $$A$$ psd implies $$a=1$$. So now we know $$\hat f(t)=1$$ for all $$t\in[-2,2]$$. The same argument, applied inductively, shows $$\hat f = 1$$ on each set $$[-2^k,2^k]$$, and thus on $$\mathbb R$$.

The Bochner theorem is a basic fact in harmonic analysis. The Fourier transform of a non-negative measure, or function, is "positive definite", in the sense that the matrices formed from the FT as above, must be positive semidefinite. The Wikipedia article is perhaps too high-brow for beginners, the Wolfram MathWorld one is too terse, but there are expositions out there that should match your level.

The basic idea is that $$\hat f(t-s) = \int_{\mathbb R} \exp(-2\pi i (t-s)x) f(x)dx$$ (give or take some norming constants), so a sum like $$\sum_k\sum_l \bar x_k x_l \hat f ( t_k-t_l) = \int_{\mathbb R}\left| \sum_k x_k e^{-2\pi t_k x}\right|^2 f(x)dx\ge 0.$$ This is the easy part of the theorem. The hard part, that the collection of all such inequalities hold, and the requirement that $$\hat f$$ be continuous, is enough to imply that $$f\ge0$$ everywhere, is not needed in your case.

• I have edited my answer; I hope it helps. – kimchi lover Jun 2 at 21:05
• $f \ge 0$ implies $0<\int_{-\infty}^\infty f(t) t^2 dt = \frac{-1}{4\pi^2}\hat{f}''(0)$ so $\hat{f}$ isn't constant around $0$ @arrian.salmaan – reuns Jun 2 at 21:33
• @reuns For Schwartz functions the existence of the integral $\int f (t)t^2dt$ is guaranteed, and your simpler argument works. I suppose I was distracted by the "harder" problem where $f\in L^1$ or where one has an arbitrary measure $\mu(dx)$ instead of one of form $f(x)dx$. – kimchi lover Jun 2 at 21:45
• For $f \in L^1$ it works too by looking at $\int_{-\infty}^\infty f(t) |\psi(t)|^2 t^2 dt$ with $\hat{\psi} \in C^\infty_c([-1/10,1/10])$ – reuns Jun 2 at 21:47
• @reuns Post it as another answer. – kimchi lover Jun 2 at 22:16

As a partial answer, by Uryshonn's lemma you can find a continuous function $$\phi$$ such that $$\phi(x)=1$$ for $$|x| \leq 1$$ and $$\phi(x)=0$$ for $$|x| \geq 2$$ and $$0 \leq \phi \leq 1$$.

Then by inversion formula take $$f(x)=\int_{\Bbb{R}}\phi(\xi)e^{2 \pi i \xi x}d \xi$$

This function has properties $$2,3$$ in your post.