Channel Tensor Error Bound

Put $a=\mathrm{\parallel }{\mathrm{\Phi }}_1{\mathrm{\parallel }}_{\diamond }$ and $b=\mathrm{\parallel }{\mathrm{\Phi }}_2{\mathrm{\parallel }}_{\diamond }$. For any finite ancilla $R$, let $S_{\mathrm{m}\mathrm{i}\mathrm{d}}:K_1\otimes H_2\otimes R\to H_2\otimes K_1\otimes R$ and $S_{\mathrm{o}\mathrm{u}\mathrm{t}}:K_2\otimes K_1\otimes R\to K_1\otimes K_2\otimes R$ be the wire swaps from [L3]. For each product matrix unit $X=X_1\otimes X_2\otimes X_R$ on $H_1\otimes H_2\otimes R$, factorwise tensor-map composition [L5] shows that first applying ${\mathrm{\Phi }}_1\otimes {\mathrm{id}}_{H_2\otimes R}$, conjugating by $S_{\mathrm{m}\mathrm{i}\mathrm{d}}$, applying ${\mathrm{\Phi }}_2\otimes {\mathrm{id}}_{K_1\otimes R}$, and conjugating by $S_{\mathrm{o}\mathrm{u}\mathrm{t}}$ gives ${\mathrm{\Phi }}_1(X_1)\otimes {\mathrm{\Phi }}_2(X_2)\otimes X_R$. This is exactly $({\mathrm{\Phi }}_1\otimes {\mathrm{\Phi }}_2\otimes {\mathrm{id}}_R)(X)$. By the matrix-unit basis [L7], equality holds for every operator $X$. The two amplified maps have induced trace norms at most $a$ and $b$; conjugating by the wire swaps preserves trace norm by [L3]. Hence the composite output has trace norm at most $ab\mathrm{\parallel }X{\mathrm{\parallel }}_1$, and taking the suprema proves $\mathrm{\parallel }{\mathrm{\Phi }}_1\otimes {\mathrm{\Phi }}_2{\mathrm{\parallel }}_{\diamond }\le ab$.

By [L1], choose unit rank-one operators $X_i=R_{x_i,y_i}$ on an ancilla of dimension $\dim H_i$ attaining $\mathrm{\parallel }{\mathrm{\Phi }}_i{\mathrm{\parallel }}_{\diamond }$. The tensor input $X_1\otimes X_2$ has trace norm $\mathrm{\parallel }{X}_1{\mathrm{\parallel }}_1\mathrm{\parallel }{X}_2{\mathrm{\parallel }}_1=1$ by [L2]. After the factor reordering, its output is $({\mathrm{\Phi }}_1\otimes {\mathrm{id}}_{{\mathbb{C}}^{\dim H_1}})(X_1)\otimes ({\mathrm{\Phi }}_2\otimes {\mathrm{id}}_{{\mathbb{C}}^{\dim H_2}})(X_2)$ by [L5] and [L7]; its trace norm is the product of the two output trace norms by [L2]. It is therefore an admissible test showing $\mathrm{\parallel }{\mathrm{\Phi }}_1\otimes {\mathrm{\Phi }}_2{\mathrm{\parallel }}_{\diamond }\ge ab$. Together with step 1.1 this proves multiplicativity.

Use the exact identity ${\mathrm{\Phi }}_1\otimes {\mathrm{\Phi }}_2-{\mathrm{\Psi }}_1\otimes {\mathrm{\Psi }}_2=({\mathrm{\Phi }}_1-{\mathrm{\Psi }}_1)\otimes {\mathrm{\Phi }}_2+{\mathrm{\Psi }}_1\otimes ({\mathrm{\Phi }}_2-{\mathrm{\Psi }}_2)\mathrm{.}$ For each ancilla, [F1] gives the triangle inequality for the amplified trace norm; taking its supremum gives the diamond-norm triangle inequality. By [F2] the four factor maps are channels, so step 2.1 bounds the two tensor terms by $\mathrm{\parallel }{\mathrm{\Phi }}_1-{\mathrm{\Psi }}_1{\mathrm{\parallel }}_{\diamond }\mathrm{\parallel }{\mathrm{\Phi }}_2{\mathrm{\parallel }}_{\diamond }$ and $\mathrm{\parallel }{\mathrm{\Psi }}_1{\mathrm{\parallel }}_{\diamond }\mathrm{\parallel }{\mathrm{\Phi }}_2-{\mathrm{\Psi }}_2{\mathrm{\parallel }}_{\diamond }$. Both channel norms equal one by [L6], giving the stated sum.