Proof that $\| \varphi \| = \varphi(1)$ for positive linear functionals on operator systems

Question

Let $M \subset \mathcal{B}(H)$ be an operator system, i.e., $M$ is a self-adjoint unital subspace, and let $\varphi: M \rightarrow \mathbb{C}$ be a positive linear functional. How does one prove that $\varphi$ is bounded with $\| \varphi \| = \varphi(1)$?

Answer 1

Here is an argument. All the time, $\varphi:M\to\mathbb C$ is linear and positive, and $M\subset B(H)$ is an operator system.

Lemma. If $a\in M$ is selfadjoint, then $\varphi(a)\in\mathbb R$. Also, $\varphi(x^*)=\overline{\varphi(x)}$, for all $x\in M$.

Proof. Let $a\in M$ selfadjoint; then, as $1\geq0$ and $a+\|a\|\,1\geq0$, $$\varphi(a)=\varphi(a+\|a\|\,1)-\|a\|\varphi(1)\in \mathbb R.$$ Now, for arbitrary $x\in M$, write $x=a+ib$ with $a,b\in M$ selfadjoint. Then $$ \varphi(x^*)=\varphi(a-ib)=\varphi(a)-i\varphi(b)=\overline{\varphi(a)+i\varphi(b)}=\overline{\varphi(a+ib)}=\overline{\varphi(x)}.\ \ \ \ \ \ \Box$$

Now fix $a\in M$. Choose $t\in\mathbb R$ such that $e^{it}\varphi(a)=|\varphi(a)|$. Then $|\varphi(a)|=\varphi(e^{it}a)$. By the Lemma, $$\text{Re}\,\varphi(x)=\varphi(\text{Re}\,x),\ \ \ \ \ x\in M. $$ Then, using the inequality $\text{Re}\,x\leq\|x\|\,1$, $$ |\varphi(a)|=\text{Re}\,|\varphi(a)|=\text{Re}\,\varphi(e^{it}a)=\varphi(\text{Re}\,e^{it}a)\leq\varphi(\|e^{it}a\|\,1)=\|a\|\,\varphi(1). $$ So $\|\varphi\|\leq\varphi(1)\leq\|\varphi\|$.