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 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)&= \int_S\sup_{\psi,\|\psi\|_{L^\infty(\mathbb{R}^d)}=1} |(\psi, \hat{\phi})|^{2k} d\mu(\phi)? \end{aligned}\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.