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.

Any helpful comments or partial answers are appreciated.


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.

  • $\begingroup$ I have edited my answer; I hope it helps. $\endgroup$ – kimchi lover Jun 2 at 21:05
  • 1
    $\begingroup$ $f \ge 0$ 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 $\endgroup$ – reuns Jun 2 at 21:33
  • $\begingroup$ @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 $f(x)dx$. $\endgroup$ – kimchi lover Jun 2 at 21:45
  • 1
    $\begingroup$ For $f \in L^1$ it works too by looking at $\int_{-\infty}^\infty f(t) |\psi(t)|^2 t^2 dt$ with $\hat{\psi} \in C^\infty_c([-1/10,1/10])$ $\endgroup$ – reuns Jun 2 at 21:47
  • $\begingroup$ @reuns Post it as another answer. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.