Hybrid bound for a composed channel circuit

For any two maps $A,B$ with matching domains, the triangle inequality for the induced trace norm holds at each finite reference space by [F5]. Taking the supremum over reference spaces in [F1] gives $\mathrm{\parallel }A+B{\mathrm{\parallel }}_{\diamond }\le \mathrm{\parallel }A{\mathrm{\parallel }}_{\diamond }+\mathrm{\parallel }B{\mathrm{\parallel }}_{\diamond }$.

For $j=1,\dots ,L$, let $P_j={\mathrm{\Phi }}_{j-1}\cdots {\mathrm{\Phi }}_1$ and $Q_j={\tilde{\mathrm{\Phi }}}_L\cdots {\tilde{\mathrm{\Phi }}}_{j+1}$. If either product is empty, it is the identity on the corresponding nonzero finite-dimensional space and is a channel by [F10]; otherwise its factors compose to a channel by [F3]. Apply [F2] first to the identical pair $P_j,P_j$ followed by ${\tilde{\mathrm{\Phi }}}_j,{\mathrm{\Phi }}_j$, and then to the pair ${\tilde{\mathrm{\Phi }}}_j\circ P_j,{\mathrm{\Phi }}_j\circ P_j$ followed by the identical pair $Q_j,Q_j$. After dividing by $2$, the two applications give $d_{\diamond }(Q_j{\tilde{\mathrm{\Phi }}}_j{P}_j,Q_j{\mathrm{\Phi }}_j{P}_j)\le d_{\diamond }({\tilde{\mathrm{\Phi }}}_j,{\mathrm{\Phi }}_j)\mathrm{.}$ The finite telescoping identity ${\tilde{\mathrm{\Phi }}}_L\cdots {\tilde{\mathrm{\Phi }}}_1-{\mathrm{\Phi }}_L\cdots {\mathrm{\Phi }}_1={\sum }_{j=1}^L(Q_j{\tilde{\mathrm{\Phi }}}_j{P}_j-Q_j{\mathrm{\Phi }}_j{P}_j)$ follows by changing stages one at a time.

Apply the diamond-norm triangle inequality from step 1.1 to the finite sum in step 1.2. Each summand has normalized distance at most its local value, so the final normalized diamond distance is at most ${\sum }_j{d}_{\diamond }({\tilde{\mathrm{\Phi }}}_j,{\mathrm{\Phi }}_j)$.

The identity ${\mathrm{id}}_R$ is a channel by [F10], so tensor-channel closure [F3] makes $({\tilde{\mathrm{\Phi }}}_L\cdots {\tilde{\mathrm{\Phi }}}_1)\otimes {\mathrm{id}}_R$ and $({\mathrm{\Phi }}_L\cdots {\mathrm{\Phi }}_1)\otimes {\mathrm{id}}_R$ channels. Their complete positivity at ancilla dimension one sends the nonnegative $\rho$ to nonnegative operators, and trace preservation keeps trace one, so the outputs ${\tilde{\rho }}_L,{\rho }_L$ are density operators by [F6] and the channel definition. By the supremum in [F1], [F7]–[F9] give $D({\tilde{\rho }}_L,{\rho }_L)\le d_{\diamond }({\tilde{\mathrm{\Phi }}}_L\cdots {\tilde{\mathrm{\Phi }}}_1,{\mathrm{\Phi }}_L\cdots {\mathrm{\Phi }}_1)$ because $\mathrm{\parallel }\rho {\mathrm{\parallel }}_1=1$.

Apply [F4] to ${\tilde{\rho }}_L,{\rho }_L$ and the declared final POVM. Its outcome-law total variation is at most $D({\tilde{\rho }}_L,{\rho }_L)$, hence at most the sum from step 2.1.