Unnumbered result: tightness for absent sender/receiver side information

Let $A,C,B,R$ be nonempty finite types, let $\psi\in\mathbb C^{R\times(A\times(C\times B))}$ satisfy $\|\psi\|=1$, and assume $|A|=1$ or $|B|=1$, allowing both equalities. Form $\rho_{uv}=\psi_u\overline{\psi_v}$, obtain $\rho_U$ for any register collection $U$ by summing over equal complementary labels, and write $S(U)=-\operatorname{Re}\operatorname{tr}(f(\rho_U))$, where $f(t)=t\log_2t$ is applied by continuous functional calculus and $0\log_20=0$. For every $Q,E\in\mathbb R$, the two inequalities

hold if and only if $Q\ge0$ and the following operational condition holds: for every $\varepsilon,\delta>0$, some natural $N$ works for every $n\ge N$ with $n\ge1$, giving natural numbers $q,k,l$, an encoder channel from $A^nC^n$ and $k$ qubits to $A^n$, $l$ qubits, and a $q$-qubit message, and a decoder channel from that message, $B^n$, and $k$ qubits to $C^nB^n$ and $l$ qubits, such that $q\le n(Q+\delta)$, $k-l\le n(E+\delta)$, and their output on $\psi^{\otimes n}\otimes\Phi_k$, leaving $R^n$ unchanged, differs from the projector onto $\psi^{\otimes n}\otimes\Phi_l$ by trace norm at most $\varepsilon$. Here $\Phi_h=2^{-h/2}\sum_{z\in\{0,1\}^h}|z,z\rangle$, the target places $A^n$ at the encoder and $C^nB^n$ at the decoder, each channel has a finite Kraus representation with $\sum_jK_j^\dagger K_j=I$, and $\|M\|_1=\operatorname{Re}\operatorname{tr}\sqrt{M^\dagger M}$. Zero values of $q,k,l$ and arbitrary real $E$ are included.