3
$\begingroup$

Let $\mu$ be a Borel probability measure on $L^2(\mathbb{R}^d)$ for $d\ge 1$ which is moreover supported on the unit sphere $$S=\{\phi\in L^2(\mathbb{R}^d): \| \phi\|_{ L^2(\mathbb{R}^d)}=1\}.$$ Let us denote $$(f,g) = \int_{\mathbb{R}^d} fg\,dx$$ and $\hat{f}$ is the Fourier transform on $\mathbb{R}^d$. Under which conditions on $\mu$, if any, can we guarantee that for any $k\ge 1$. \begin{aligned} A_k\equiv \sup_{\psi,\|\psi\|_{L^\infty(\mathbb{R}^d)}=1} \int_S |(\psi, \hat{\phi})|^{2k}d\mu(\phi)&\le C_k \int_S\sup_{\psi,\|\psi\|_{L^\infty(\mathbb{R}^d)}=1} |(\psi, \hat{\phi})|^{2k} d\mu(\phi)? \end{aligned} Note that the RHS can be written as $$\int_S \left(\sup_{\psi,\|\psi\|_{L^\infty(\mathbb{R}^d)}=1} |(\psi, \hat{\phi})|\right)^{2k}d\mu(\phi)=\int_S \| \hat{\phi}\|_{L^1(\mathbb{R}^d)}^{2k}d\mu(\phi)$$ by duality.

Assume that $A_k$ is well-defined for each $k\ge 1$ and in fact $A_k\le C \varepsilon^{2k}$ for some $C, \varepsilon>0$ and all $k\ge 1$ if that is helpful.

Note: this question came up in my study of the Gross-Pitaevski hierarchy for infinitely many bosons.

$\endgroup$
4
  • 3
    $\begingroup$ Why the downvote? I am definitely open to editing the question given suggestions to improve it. $\endgroup$
    – Dispersion
    Commented Apr 22 at 19:14
  • $\begingroup$ The Fourier transform is isometry on $L^2(\Bbb R^d)$. Should we better replace $\hat\phi$ by $\phi$? $\endgroup$
    – Liding Yao
    Commented Apr 23 at 1:17
  • $\begingroup$ I actually want $\hat{\phi}$ since I am interested for various reason in studying the Gross-Pitaevksi hierarchy with initial data lying in some tensorized version of the space $\mathcal{F}L^1(\mathbb{R}^d)$, one of them being that the space is an algebra. $\endgroup$
    – Dispersion
    Commented Apr 23 at 1:29
  • 1
    $\begingroup$ The reason we consider a measure supported on $L^2$ comes from a quantum de Finetti theorem which tells us that for a natural initial datum, it may be written as the convex combination of tensorized states with respect to exactly a Borel measure supported on $S$. $\endgroup$
    – Dispersion
    Commented Apr 23 at 1:37

0

You must log in to answer this question.

Browse other questions tagged .