Housekeeping.

src/op/system/complete-upgrade.tex

@@ -34,60 +34,6 @@ The following is a slight generalisation of one form of the above example, but s
     Let $A$ be a $C^*$-algebra, $S, T \subset A$ be operator systems, and $\phi: S \to T$. If there exists $x \in A$ such that $\phi(s) = x\phi(s)x^*$ for all $s \in S$, then $\phi$ is completely positive. 
 \end{proposition}
 
-
-
-\begin{proposition}
-\label{proposition:completely-positive-norm}
-    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
-    \[
-    \norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; B)} = \norm{\phi(1_A)}_B
-    \]
-\end{proposition}
-\begin{proof}
-    After taking a \hyperref[unitisation]{proposition:c-star-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 \autoref{lemma:norm-matrix-test}. 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 \hyperref[Schur complement condition]{proposition:schur-complement-positive}, 
-    \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, 
-    \[
-    \norm{\phi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi + \eps1_B\psi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi(1_A) + \eps 1_B}_B = \norm{\phi(1_A)}_B
-    \]
-\end{proof}
-
 \begin{proposition}
 \label{proposition:commutative-mn-facts}
     Let $X$ be a compact Hausdorff space, then:

src/op/system/completely.tex

@@ -180,7 +180,58 @@
 \end{proof}
 
 
-% Omitted
+\begin{proposition}
+\label{proposition:completely-positive-norm}
+    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
+    \[
+    \norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; B)} = \norm{\phi(1_A)}_B
+    \]
+\end{proposition}
+\begin{proof}
+    After taking a \hyperref[unitisation]{proposition:c-star-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 \autoref{lemma:norm-matrix-test}. 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 \hyperref[Schur complement condition]{proposition:schur-complement-positive}, 
+    \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, 
+    \[
+    \norm{\phi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi + \eps1_B\psi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi(1_A) + \eps 1_B}_B = \norm{\phi(1_A)}_B
+    \]
+\end{proof}
+
 
 \begin{proposition}
 \label{proposition:unital-2-positive-contractive}