Weighted two-sided comparison between the rectangular Bregman and Mazur-distance objectives

Let $E,F$ be finite-dimensional complex inner-product spaces and let $p$ be any real number (no hypothesis is placed on $p$ in this statement). For a complex-linear map $T:E\to F$, write $\widehat T$ for its Hermitian dilation, the self-adjoint operator on $E\oplus F$ given by $\widehat T(x,y)=(T^\ast y,\,Tx)$ (hermitianDilation). For a self-adjoint operator $C$ on $E\oplus F$ and $h:\mathbb R\to\mathbb R$, write $h(C)$ for the operator obtained by applying $h$ to the eigenvalues of $C$ in any orthonormal eigenbasis of $C$; this operator itself does not depend on which eigenbasis is chosen, since it acts as $h(\mu)$ on every $\mu$-eigenvector of $C$, not only on the vectors of one particular chosen eigenbasis — which is what lets $\psi_p(\widehat S)\circ\psi_p(\widehat T)$ below be a single well-defined operator even though $\widehat S$ and $\widehat T$ generally have different eigenbases. With $F_p(x)=|x|^p/p$, $G_p(x)=|x|^{p-2}x$, and the scalar Mazur map $\psi_p(x)=\operatorname{sign}(x)\,|x|^{p/2}$ (powerPotential, powerGradient, scalarMazur; all three are totalized real powers, defined for every real $p$ and $x$), set

(dilatedBregmanTrace, the trace-level Bregman combination of $F_p$ between $\widehat S$ and $\widehat T$) and

(dilatedMazurDistanceSq, the squared Hilbert–Schmidt distance $\operatorname{Tr}\big((\psi_p(\widehat S)-\psi_p(\widehat T))^2\big)$ between the Mazur images of the two dilations). Separately, let $\Psi_p(T):E\to F$ be the rectangular Mazur map of $T$ — the unique complex-linear map whose dilation is $\psi_p(\widehat T)$, i.e. $\widehat{\Psi_p(T)}=\psi_p(\widehat T)$ (rectangularMazurMap); it satisfies $D_p(S,T) = 2\,\|\Psi_p(S)-\Psi_p(T)\|_2^2$, where $\|\cdot\|_2$ is the Schatten-$2$ (Hilbert–Schmidt) norm.

For $T:E\to F$ with singular values $\operatorname{sv}_k(T)$, the Schatten $p$-power sphere is $\{T\mid\sum_{k:\,\operatorname{sv}_k(T)\ne0}\operatorname{sv}_k(T)^p=1\}$ (schattenPowerSphere; the sum runs over the nonzero singular values, which for $p>0$ is the same as summing over all of them). Let $\iota$ be a finite index type, let $a:\iota\to\mathbb R$ with $a_i\ge0$ for every $i$, let $(u_i)_{i\in\iota}$ be a family on the Schatten $p$-power sphere, and let $v$ be a further point on that sphere. Define the rectangular Bregman objective and rectangular Mazur-distance objective

(the second equals $\sum_i a_i\,D_p(u_i,v)/2$ as well, since $D_p(S,T)=2\|\Psi_p(S)-\Psi_p(T)\|_2^2$). Assume real numbers $m,M$ satisfy the pointwise two-sided bound

for every pair of complex-linear maps $S,T:E\to F$. Then the theorem states

This lifts a pairwise two-sided bound between $B_p$ and $D_p$, assumed uniformly over rectangular operators, to the weighted finite-family objectives that the variational argument minimizes over the Schatten power sphere, with the same constants $m,M$.

Formalization Note Both named objectives already carry the same $\tfrac12$ normalization relative to $B_p$ and $D_p$ respectively (one directly, the other through the identity $D_p(S,T)=2\|\Psi_p(S)-\Psi_p(T)\|_2^2$), so the displayed two-sided bound needs no extra factor on either side. No convexity of $F_p$ is claimed or needed here, since $p$ ranges over all of $\mathbb R$; $B_p$ and $D_p$ are used purely as totalized algebraic quantities, and the two-sided bound between them is assumed as a hypothesis on this page, not derived from convexity.