Lemma 41.3.5.label Let $A, B$ be $C^{*}$-algebras with $A$ being unital, $S \subset A$ be an operator system, $\phi: S \to B$ be a linear map, and $n \in \natp$.
- (1)
If $\phi$ is $n$-positive, then $\phi$ is $k$-positive for all $1 \le k \le n$.
- (2)
For each $1 \le k \le n$, $\norm{\phi_k}_{L(M_k(S); M_k(B))}\le \norm{\phi_n}_{L(M_n(S); M_n(B))}$.
Proof. For any $n \in \natp$ and $1 \le k < n$, there is a natural *-homomorphism
such that the following diagram commutes:
(1): Since $e \otimes \text{Id}_{S}$ is the restriction of a *-homomorphism, it is positive. On the other hand, for any $x \in M_{k}(B)$, $x \ge 0$ if and only if $(e \otimes \text{Id}_{B})(x) \ge 0$. Given that $\phi_{n}$ is positive, $\phi_{k}$ must also be positive.
(2): The maps $e \otimes \text{Id}_{S}$ and $e \otimes \text{Id}_{B}$ are both isometric. As such, the norm of $\phi_{k}$ is bounded above by the norm of $\phi_{n}$.$\square$
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