Proposition 41.3.10.label Let $A, B$ be $C^{*}$-algebras with $A$ being unital, $S \subset A$ be an operator system, and $\phi: S \to B$ be a completely positive map, then
\[\norm{\phi}_{\text{cb}}= \norm{\phi}_{L(S; B)}= \norm{\phi(1_A)}_{B}\]
Proof. After taking a unitisation, assume without loss of generality that $B$ is unital.
It is sufficient to show that $\norm{\phi}_{\text{cb}}\le \norm{\phi(1_A)}_{B}$. To this end, let $T \in M_{n}(S)$ with $\norm{T}_{M_n(S)}\le 1$, then
\[\begin{bmatrix}1_{M_n(A)}&T \\ T^{*}&1_{M_n(A)}\end{bmatrix} \ge 0\]
in $M_{2}(M_{n}(S)) = M_{2n}(S)$ by Lemma 41.3.6. Since $\phi$ is completely positive,
\[\phi_{2n}\begin{bmatrix}1_{M_n(A)}&T \\ T^{*}&1_{M_n(A)}\end{bmatrix} = \begin{bmatrix}\phi_{n}(1_{M_n(A)})&\phi_{n}(T) \\ \phi_{n}(T)^{*}&\phi_{n}(1_{M_n(A)})\end{bmatrix} \ge 0\]
First suppose that $\phi_{n}(1_{M_n(A)})$ is invertible, then
\begin{align*}\phi_{n}(1_{M_n(A)})&\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}\cdot 1_{M_n(B)}\\ 1_{M_n(B)}&\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}\cdot \phi_{n}(1_{M_n(A)})^{-1}\\ \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1}&\le \phi_{n}(1_{M_n(A)})^{-1}\end{align*}
By the Schur complement condition,
\begin{align*}\phi_{n}(1_{M_n(A)})&\ge \phi_{n}(T)\phi_{n}(1_{M_n(A)})^{-1}\phi_{n}(T)^{*} \\&\ge \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1}\phi_{n}(T)\phi_{n}(T)^{*} \\ \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{2}&\ge \normn{\phi_n(T)\phi_n(T)^*}_{M_n(B)}= \normn{\phi_n(T)}_{M_n(B)}^{2}\end{align*}
and $\normn{\phi_n(T)}_{M_n(B)}\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}$. Therefore $\norm{\phi}_{\text{cb}}\le \normn{\phi(1_A)}_{B}$.
Now suppose that $\phi$ is an arbitrary completely positive map. Let $\psi \in S(A)$ be a state and $\eps > 0$, then $\phi + \eps 1_{B}\psi$ is also a completely positive map. In which case,
\[\norm{\phi}_{\text{cb}}\le \inf_{\eps > 0}\norm{\phi + \eps1_B\psi}_{\text{cb}}\le \inf_{\eps > 0}\norm{\phi(1_A) + \eps 1_B}_{B} = \norm{\phi(1_A)}_{B}\]
$\square$
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