MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $C$ be a convex subset of $\mathbb R^{2n}$ and $\mathbf 1_C$ be the characteristic function of $C$. Is it true that $$\forall u\in\mathscr S(\mathbb R^n),\quad \langle\mathbf 1_C^{Weyl}u,u\rangle\le \Vert u\Vert_{L^2(\mathbb R^n)}^2,\tag F $$ or more explicitly, $ \iint_{\mathbb R^n\times\mathbb R^n}\mathbf 1_C(x,\xi)\mathscr H(u,u)(x,\xi) dx d\xi\le \Vert u\Vert_{L^2(\mathbb R^n)}^2, $ where $\mathscr H(u,u)$ is the Wigner function given by $$ \mathscr H(u,u)(x,\xi)=\int_{\mathbb R^n}e^{-2i\pi z\cdot \xi} u(x+\frac{z}{2})\bar u(x-\frac{z}{2}) dz. \tag W$$ Since $(W)$ is real-valued, the lhs of $(F)$ is also real. The property is easy for $C$ equal to a half-space, since, due to Segal identity on symplectic transformations of Weyl symbols, the operator with Weyl symbol $\mathbf 1_C$ is, in that case, unitarily equivalent to the multiplication by $H(x_1)$.

Inequality (F) seems to have received the name of Flandrin's conjecture.

share|cite|improve this question
+1, but... What is intuition behind ? Can you write it in terms of Weyl-Moyl product ? May be it can generalized to arbitrary star-product on symplectic manifold, if yes, then how ? What are references ? – Alexander Chervov Sep 13 '12 at 17:36
Intuition: think about the probability measure with density $\mathbf 1_C/\vert C\vert$. Then the expectation of the Wigner function should be smaller than 1 for a normalized $u$. Note that the Wigner function is real-valued but not positive in general and with integral 1. References: it seems to me that Philippe Jaming spoke about these problems, somewhat linked to his work on the uncertainty principle. I think that Flandrin is an engineer, specialized in signal business. – Bazin Sep 13 '12 at 19:02
Here is a reference : – jjcale Sep 13 '12 at 19:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.