States in a $C^*$-algebra bounded?

Question

A functional $\phi$ on a $C^*$-algebra $A$ with unit element, i.e. $\phi: A \rightarrow \mathbb{C}$ is called a state if $\phi(T^*T) \ge 0$ for all $T \in A$ and $\phi( \operatorname{id}) = 1.$ Now, I started wondering whether such a functional $\phi$ is automatically bounded or not?

Answer 1

Yes, this is standard. Let $T\in A$ selfadjoint, i.e. with $T=T^*$. Note first that $\phi(T)$ is real: since $T+\|T\|\,\text{id}$ is positive, we have that $$ \phi(T)+\|T||\in\mathbb R, $$ so $\phi(T)\in \mathbb R$. Now, as $-T+\|T\|\,\text{id}\geq0$, we get $-\phi(T)+\|T\|\geq0$, so $$\phi(T)\leq\|T\|.$$ Since $-T$ is also selfadjoint, we can also get $$-\phi(T)\leq\|T\|.$$ Combining both inequalities, we get $$|\phi(T)|\leq\|T\|.$$

Now, for arbitrary $T$ we write $T=T_1+iT_2$ with $T_1,T_2$ selfadjoint. Then $$ |\phi(T)|\leq|\phi(T_1)|+|\phi(T_2)|\leq2\|T\|. $$ So $\phi$ is bounded.

With a little more subtlety, one can actually show that $\|\phi\|=1$. Or, more generally if the map positive but not unital, $\|\phi\|=\phi(\text{id}).$