Schwarz and trace-norm channel contraction

By [F2], ${\mathrm{\Phi }}^{†}$ is positive. For any Hermitian $H\in M_e(\mathbb{C})$, its finite spectral decomposition [F3] is $H={\sum }_{\lambda }\lambda P_{\lambda }$, where each $\lambda$ is real and each spectral projection $P_{\lambda }$ is positive semidefinite. Hence every ${\mathrm{\Phi }}^{†}(P_{\lambda })$ is Hermitian, so ${\mathrm{\Phi }}^{†}(H)$ is Hermitian. Write an arbitrary $A$ as $A=H+iK$, where $H=(A+A^{\ast })\mathrm{/}2$ and $K=(A-A^{\ast })\mathrm{/}(2i)$ are Hermitian. By complex linearity, ${\mathrm{\Phi }}^{†}(A^{\ast })={\mathrm{\Phi }}^{†}(H)-i{\mathrm{\Phi }}^{†}(K)={\mathrm{\Phi }}^{†}(A{)}^{\ast }$.

For any $n\ge 1$ and $Z\in M_n(\mathbb{C})$, use the standard inner product $⟨u,v⟩={\sum }_k{u}_k\overline{v_k}$, linear in its first argument, and define $\mathrm{\mid }u⟩⟨v\mathrm{\mid }$ as the map $x\mapsto ⟨x,v⟩u$. Since $Z^{\ast }Z$ is positive, with $⟨Z^{\ast }Zv,v⟩=\mathrm{\parallel }Zv{\mathrm{\parallel }}^2$, [F3] diagonalizes it in an orthonormal basis $(v_j)$ with eigenvalues $s_j^2\ge 0$. For each $s_j>0$, put $u_j=Z{v}_j\mathrm{/}{s}_j$. Then $⟨u_j,u_k⟩={\delta }_{jk}$, and $Z={\sum }_{s_j>0}{s}_j\mathrm{\mid }{u}_j⟩⟨v_j\mathrm{\mid }$. The positive square root of $Z^{\ast }Z$ has eigenvalues $s_j$, so $\mathrm{\parallel }Z{\mathrm{\parallel }}_1={\sum }_j{s}_j$.

For $A\in M_e(\mathbb{C})$, the block matrix $\left(\begin{array}{cc}{A}^{\ast }A& A^{\ast }\\ A& I_e\end{array}\right)$ is positive semidefinite because its quadratic form at $(x,y)$ is $\mathrm{\parallel }Ax+y{\mathrm{\parallel }}^2$. Apply ${\mathrm{id}}_2\otimes {\mathrm{\Phi }}^{†}$: by [F2] the trace dual is completely positive, and the definition in [F1] makes its extension by the two-dimensional ancilla positive. By unitality in [F2] and adjoint preservation from step 1.1, the resulting block is $\left(\begin{array}{cc}{\mathrm{\Phi }}^{†}(A^{\ast }A)& {\mathrm{\Phi }}^{†}(A{)}^{\ast }\\ {\mathrm{\Phi }}^{†}(A)& I_d\end{array}\right)\ge 0$.

For any $n\ge 1$, $Z\in M_n(\mathbb{C})$, and $B\in M_n(\mathbb{C})$ with $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$, write $Z_{ab}={\sum }_{s_j>0}{s}_j(u_j{)}_a\overline{(v_j{)}_b}$ using step 1.2. Expanding both diagonal-sum traces gives $\mathrm{Tr}(BZ)={\sum }_{a,b}{B}_{ab}{Z}_{ba}={\sum }_{s_j>0}{s}_j{\sum }_{a,b}{B}_{ab}(u_j{)}_b\overline{(v_j{)}_a}$ and $\mathrm{Tr}(ZB)={\sum }_{a,b}{Z}_{ab}{B}_{ba}={\sum }_{s_j>0}{s}_j{\sum }_{a,b}{B}_{ab}(u_j{)}_b\overline{(v_j{)}_a}$. The inner sum is $⟨B{u}_j,v_j⟩$ for the inner product fixed in step 1.2. For each unit $v_j$, $0\le \mathrm{\parallel }B{u}_j-⟨B{u}_j,v_j⟩v_j{\mathrm{\parallel }}^2=\mathrm{\parallel }B{u}_j{\mathrm{\parallel }}^2-\mathrm{\mid }⟨B{u}_j,v_j⟩{\mathrm{\mid }}^2$, so both trace magnitudes are at most ${\sum }_j{s}_j=\mathrm{\parallel }Z{\mathrm{\parallel }}_1$. Extend the two finite orthonormal lists $(u_j)$, $(v_j)$ to orthonormal bases and choose a unitary $B$ carrying each $u_j$ to $v_j$; then both trace magnitudes equal ${\sum }_j{s}_j$. Thus $\mathrm{\parallel }Z{\mathrm{\parallel }}_1={\sup }_{\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1}\mathrm{\mid }\mathrm{Tr}(BZ)\mathrm{\mid }={\sup }_{\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1}\mathrm{\mid }\mathrm{Tr}(ZB)\mathrm{\mid }$, including $Z=0$.

In the positive block from step 2.1, evaluate the quadratic form at $(x,-{\mathrm{\Phi }}^{†}(A)x)$. This gives $⟨({\mathrm{\Phi }}^{†}(A^{\ast }A)-{\mathrm{\Phi }}^{†}(A{)}^{\ast }{\mathrm{\Phi }}^{†}(A))x,x⟩\ge 0$ for every $x$, so ${\mathrm{\Phi }}^{†}(A{)}^{\ast }{\mathrm{\Phi }}^{†}(A)\le {\mathrm{\Phi }}^{†}(A^{\ast }A)$. Also $A^{\ast }A\le \mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}^2{I}_e$; positivity and unitality in [F2] then give ${\mathrm{\Phi }}^{†}(A{)}^{\ast }{\mathrm{\Phi }}^{†}(A)\le \mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}^2{I}_d$, proving $\mathrm{\parallel }{\mathrm{\Phi }}^{†}(A){\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}$.

Apply the second dual norm formula from step 2.2 in output dimension $e$ to $\mathrm{\Phi }(X)$. For each $A\in M_e(\mathbb{C})$ with $\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$, step 3.1 gives $\mathrm{\parallel }{\mathrm{\Phi }}^{†}(A){\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$, so the same formula in input dimension $d$ bounds $\mathrm{\mid }\mathrm{Tr}(X{\mathrm{\Phi }}^{†}(A))\mathrm{\mid }\le \mathrm{\parallel }X{\mathrm{\parallel }}_1$. The trace-pairing identity [F2] therefore gives $\mathrm{\parallel }\mathrm{\Phi }(X){\mathrm{\parallel }}_1={\sup }_{\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1}\mathrm{\mid }\mathrm{Tr}(\mathrm{\Phi }(X)A)\mathrm{\mid }={\sup }_{\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1}\mathrm{\mid }\mathrm{Tr}(X{\mathrm{\Phi }}^{†}(A))\mathrm{\mid }\le \mathrm{\parallel }X{\mathrm{\parallel }}_1$.