Problem 02 Goal — Compressed strict convex equality reduces

For every $n\in\mathbb N$, including $n=0$, let $Q=\{x:\{0,1\}\to\mathbb R\mid 0\le x_0,x_1\le1\}$. Let $S$ consist of an $n\times n$ complex matrix $U$ satisfying $U^{*}U=I_n$ and $UU^{*}=I_n$, together with points $\lambda_i\in Q$ for every $i\in\operatorname{Fin}(n)$. Define $A_c=U\operatorname{diag}_i(\lambda_i(c))U^{*}$ for $c\in\{0,1\}$, and, for every total function $f:\mathbb R^{\{0,1\}}\to\mathbb R$, define $\Phi_S(f)=U\operatorname{diag}_i(f(\lambda_i))U^{*}$. Let $P$ be an $n\times n$ complex matrix satisfying $P^{*}=P$ and $P^2=P$. A supplied compression witness $C$ consists of another spectral datum $\widetilde S=(V,(\mu_i)_i)$, where $V^{*}V=I_n$, $VV^{*}=I_n$, and every $\mu_i\in Q$, whose coordinate operators $\widetilde A_c=V\operatorname{diag}_i(\mu_i(c))V^{*}$ satisfy $\widetilde A_0=PA_0P$ and $\widetilde A_1=PA_1P$; the statement does not assert that such a witness exists. If $f$ is continuous on $Q$, is strictly convex there—meaning in particular that for distinct $x,y\in Q$ and positive $a,b\in\mathbb R$ with $a+b=1$, $f(ax+by)<af(x)+bf(y)$—and satisfies $P\Phi_S(f)P=P\Phi_{\widetilde S}(f)P$, where $\Phi_{\widetilde S}(f)=V\operatorname{diag}_i(f(\mu_i))V^{*}$, then $P$ commutes separately with both original coordinate operators: $PA_0=A_0P$ and $PA_1=A_1P$. No positivity condition on $n$, nonzero or proper-rank condition on $P$, or condition on $f$ outside $Q$ is imposed; in particular, $n=0$ is included, with empty spectral families and automatic $0\times0$ matrix equalities, and $P=0$ is included whenever a compression witness is supplied.