Normalized States

Question

A linear functional is normalized iff it preserves identity: $$\|\omega\|=1 \iff \omega(\mathrm{id})=1$$ Can somebody help me proving it?
(I just remember it was kind of an easy thing.)

Answer 1

The way it is stated, the property is not true. For example, consider the linear functional on $M_2(\mathbb C)$, $$ \omega:\begin{bmatrix}a&b\\ c&d\end{bmatrix}\mapsto a-d. $$ Then $\omega(\text{id})=0$, so a normalized version of $\omega$ will fail the assertion.

The result is true, however, when $\omega$ is required to be positive.

For any $x$, we have $x^*x\leq\|x^*x\|\,\text{id}$. Using Kadison's Schwarz inequality ($\omega(x)^*\omega(x)\leq\omega(x^*x)$), $$ |\omega(x)|^2\leq\omega(x^*x)\leq\|x^*x\|\,\omega(\text{id})=\|x\|^2\,\omega(\text{id}). $$ This implies that $\|\omega\|\leq\omega(\text{id})\leq\|\omega\|$. So $\|\omega\|=\omega(\text{id})$.

A less trivial result is the fact that if a linear functional satisfies $\|\omega\|=\omega(\text{id})$, then $\omega$ is positive.