Proposition 41.2.6.label Let $A, B$ be $C^{*}$-algebra, $S \subset A$ and $T \subset B$ be operator systems, and $\phi: S \to T$ be a positive map, then
- (1)
For any $x \in S_{sa}$, $\norm{\phi(x)}_{B}\le \norm{\phi(1_A)}_{B} \cdot \norm{x}_{A}$.
- (2)
$\norm{\phi}_{L(S; T)}\le 2\norm{\phi(1_A)}_{B}$.
Proof. (1): Let $x \in T_{+}$, then $0 \le x \le \norm{x}_{A}$, and $0 \le \phi(x) \le \norm{x}_{A} \cdot \norm{\phi(1_A)}_{B}$, so $\norm{\phi(x)}_{B} \le \norm{x}_{A} \cdot \norm{\phi(1_A)}_{B}$.
For any $x, y \in T_{+}$, $x - y \le x$ and $y - x \le y$. As such,
\[\norm{x - y}_{B}\le \norm{x}_{B} \vee \norm{y}_{B}\]
In particular, for any $s \in S_{sa}$,
\begin{align*}\phi(s)&= \frac{1}{2}\braks{\phi(\norm{s}_A + s) - \phi(\norm{s}_A - s)}\\ \norm{\phi(s)}_{B}&\le \frac{1}{2}\braks{\norm{\phi(\norm{s}_A + s)}_B \vee \norm{\phi(\norm{s}_A - s)}_B}\\&\le \norm{s}_{A} \norm{\phi(1_A)}_{B}\end{align*}
(2): For any general $s \in S$,
\begin{align*}\phi(s)&= \phi(\text{Re}(s)) + i\phi(\text{Im}(s)) \\ \norm{\phi(s)}_{B}&\le \norm{\phi(\text{Re}(s))}_{B} + \norm{\phi(\text{Im}(s))}_{B}\\&\le \norm{\phi(1_A)}_{B} \cdot (\norm{\text{Re}(s)}_{A} + \norm{\text{Im}(s)}_{A}) \le 2\norm{\phi(1_A)}_{B}\norm{s}_{A}\end{align*}
$\square$
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