Typical-subspace compression with a failure flag ($F_e({\pi }^{\otimes n},{\mathcal{N}}_n)\ge [\mathrm{Tr}({\pi }^{\otimes n}\Pi ){]}^2$)

Measure the quantum typical subspace projector $\Pi$. Encode its successful sector isometrically and reserve one orthogonal compressed flag for failure; decode the flag to a fixed state. The composite channel is ${\mathcal{N}}_n(\tau )=\Pi \tau \Pi +\mathrm{Tr}[(I-\Pi )\tau ]|{\phi }_0⟩⟨{\phi }_0|$. It is trace preserving. The Kraus formula for entanglement fidelity gives the displayed lower bound, while convexity of the square gives the same bound for average pure-source fidelity. One flag adds only a vanishing asymptotic rate overhead.

For a quantum state ensemble of pure vectors and an orthogonal projection $P$, measure $\{P,I-P\}$, preserve the successful subspace, and decode failure to a fixed normalized vector $\phi$. The resulting quantum channel is $\Lambda (X)=PXP+\mathrm{Tr}[(I-P)X]|\phi ⟩⟨\phi |$. For ${\alpha }_k=\Vert P{\psi }_k\Vert$, its squared quantum fidelity on the $k$th source vector is ${\alpha }_k^4+(1-{\alpha }_k^2)|⟨{\psi }_k|\phi ⟩{|}^2$. The inequality $a^2\ge 2a-1$ on $[0,1]$ proves the displayed average bound. It depends only on the source density operator's weight in $P$ and does not require the emitted vectors to be orthogonal.