Feasibility and range of transpose entanglement fidelity

The logical transpose decoder is CPTP, and its recovered channel has the Kraus operators $K_{ab}$. The trace-fidelity formula gives the asserted sum, while the fidelity definition puts it in $[0,1]$. The reference-extended physical recovered outputs are independent of completion; applying the fixed logical compression on the output factor preserves their equality, hence preserves their overlap defining $f$.

For every $\mathcal{R}\in \mathfrak{D}$, the map $\mathcal{R}\mathcal{C}$ is complex-linear and satisfies $\mathrm{tr}(\mathcal{R}(\mathcal{C}(X)))=\mathrm{tr}(\mathcal{C}(X))=\mathrm{tr}X$ for every logical operator $X$. For each integer $r\ge 1$, its ancilla amplification is ${\mathcal{R}}^{(r)}{\mathcal{C}}^{(r)}$: both sides apply $\mathcal{R}\mathcal{C}$ to each ancilla-indexed block. This is a composition of positive maps, so it preserves nonnegative operators. Thus $\mathcal{R}\mathcal{N}{\mathrm{Ad}}_V$ is a channel on ${\mathbb{C}}^d$, and its entanglement fidelity is defined and belongs to $[0,1]$.

By [F3], $\mathcal{D}\in \mathfrak{D}$, so $f\in \mathcal{F}$ and $\mathcal{F}$ is nonempty. Step 1.2 gives $\mathcal{F}\subseteq [0,1]$, so $1$ is a real upper bound. By [F6], $g=\sup \mathcal{F}$ exists in $\mathbb{R}$. Since $g$ is an upper bound and $f\in \mathcal{F}$, $f\le g$; since $1$ is an upper bound and $g$ is the least upper bound, $g\le 1$. Together with step 1.1 this proves $0\le f\le g\le 1$.