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
\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}
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.