Upgraded Kadison-Schwarz.

src/op/system/completely.tex

@@ -111,38 +111,43 @@
 
 \begin{proposition}[Kadison-Schwarz Inequality]
 \label{proposition:cauchy-schwarz-2-positive}
-    Let $A, B$ be unital $C^*$-algebras, $\phi: A \to B$ be a unital $2$-positive map, then for each $x \in A$,
+    Let $A, B$ be unital $C^*$-algebras, $\phi: A \to B$ be a $2$-positive map, then for each $x \in A$,
     \[
-    \phi(x)^*\phi(x) \le \phi(x^*x)
+    \phi(x)^*\phi(x) \le \norm{\phi(1_A)}_B \phi(x^*x)
     \]
+
+    xxx
 \end{proposition}
 \begin{proof}
     Since
     \[
     \begin{bmatrix}
-            1 &x \\
+            1_A &x \\
             x^* &x^*x
         \end{bmatrix} = \begin{bmatrix}
-            1 &x \\
+            1_A &x \\
             0 &0
         \end{bmatrix}^*\begin{bmatrix}
-            1 &x \\
+            1_A &x \\
             0 &0
         \end{bmatrix} \ge 0
     \]
 
-    and $\phi$ is unital and $2$-positive, 
+    and $\phi$ is $2$-positive, 
     \[
     \begin{bmatrix}
-            1 &\phi(x) \\
+            1_B\norm{\phi(1_A)}_B &\phi(x) \\
+            \phi(x)^* &\phi(x^*x)
+        \end{bmatrix} \ge \begin{bmatrix}
+            \phi(1_A) &\phi(x) \\
             \phi(x)^* &\phi(x^*x)
         \end{bmatrix} = \phi_2\begin{bmatrix}
-            1 &x \\
+            1_A &x \\
             x^* &x^*x
         \end{bmatrix} \ge 0
     \]
 
-    By \autoref{lemma:norm-matrix-test}, $\phi(x)^*\phi(x) \le \phi(x^*x)$. 
+    By \autoref{lemma:norm-matrix-test}, $\phi(x)^*\phi(x) \le \norm{\phi(1_A)}_B\phi(x^*x)$. 
 \end{proof}