Contractive local expectations and their commutator bound

On ${\mathbb{C}}^q$ in the standard basis $e_j$, $0\le j<q$, define $S{e}_j=e_{j+1\mathrm{m}\mathrm{o}\mathrm{d}q}$ and $D{e}_j=e^{2\pi ij\mathrm{/}q}{e}_j$. The $q^2$ unitaries $S^a{D}^b$, $0\le a,b<q$, satisfy $q^{-2}{\sum }_{a,b=0}^{q-1}(S^a{D}^b{)}^{\ast }R(S^a{D}^b)=q^{-1}\mathrm{Tr}(R)I_q(R\in M_q(\mathbb{C}))\mathrm{.}$ Indeed, averaging over $b$ cancels each off-diagonal matrix unit because its phase is a nontrivial $q$-th root sum, and averaging the surviving diagonal units over $a$ gives their arithmetic mean times $I_q$; for $q=1$ the formula is the identity. Applying this one-site twirl on each factor in $Y\setminus X$ yields the normalized partial trace of [F1]. Each twirl is an average of unitary conjugations, so it is unital, positive, and contractive. The formula is unchanged when an extra identity factor is added, since its normalized trace is $1$; hence $E_X$ is consistent on ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$. Contractivity extends it uniquely to the completion, and its range remains in the closed finite-dimensional subalgebra ${\mathcal{A}}_X$.

Enumerate $Y\setminus X=\{z_1,\dots ,z_k\}$ and write $E_{z_j}$ for the one-site twirl. The commuting twirls satisfy $B-E_X(B)={\sum }_{j=1}^k{E}_{z_1}\cdots E_{z_{j-1}}(B-E_{z_j}(B))\mathrm{.}$ For any $C\in {\mathcal{A}}_Y$, the average formula gives $\mathrm{\parallel }C-E_{z_j}(C)\mathrm{\parallel }\le {\sup }_{V_z}\mathrm{\parallel }C-V_z^{\ast }C{V}_z\mathrm{\parallel }={\sup }_{V_z}\mathrm{\parallel }[C,V_z]\mathrm{\parallel }$, where $V_z$ ranges over the one-site unitaries. Each preceding twirl is contractive and commutes with conjugation by $V_z$, so its term in the displayed sum is at most ${\sup }_{V_z}\mathrm{\parallel }[B,V_z]\mathrm{\parallel }$. The triangle inequality and the finite sum over sites give the stated bound.