# state on a non-unital $C^*$ algebra

Suppose $$\tau$$ is a state on a non-unital $$C^*$$ algebra $$A$$.There is a well-known inequality: $$\tag{*}|\tau(a)|^2\leq\tau(a^*a),\ \text{ for all } a\in A.$$

1. Does there exist some nonzero element $$a_0$$ such that $$\tau(a_0^*a_0)-|\tau(a_0)|^2$$ is small enough?

2. Does there exist a nonzero elements $$a_0$$ such that equality holds in $$(*)$$?

For any state and any approximate unit $$\{e_j\}$$ of $$A$$, you have $$\lim_j\tau(e_j)=\|\tau\|$$ (cfr. Davidson's C$$^*$$-Algebras by Example, Lemma I.9.5). So here $$\tau(e_j)\to1$$. Since one can take $$e_j\geq0$$ and $$\|e_j\|\leq1$$ for all $$j$$, you have $$\tau(e_j^*e_j)=\tau(e_j^2)\leq\|e_j\|\,\tau(e_j)\leq1.$$ Thus $$1=\lim_j|\tau(e_j)|^2\leq\limsup_j\tau(e_j^*a_j)\leq1,$$ giving us equality. This gives the existence of your $$a_0$$.

The answer to your second question is no when $$\tau$$ is faithful. Given your $$\tau$$, you can always extend it to the unitization $$\tilde A$$ of $$A$$, and the extension is still faithful. So if $$\|a\|\leq1$$ and $$|\tau(a)|^2=\tau(a^*a)$$, this is equality in the Cauchy-Schwarz inequality, so you get that $$a=\lambda\,1$$.

When $$\tau$$ is not faithful, equality can occur. For instance take $$A=M_2(\mathbb C)\oplus K(\ell^2(\mathbb N))$$, and $$\tau(a\oplus k)=a_{22}$$. Then the matrix unit $$E_{22}\oplus 0$$ satisfies the equality

• For any $\epsilon > 0$,there exists $e_0$(depends on $\epsilon$) such that $\tau(e_0^*e_0)-|\tau(e_0)|^2＜\epsilon$.For different $\epsilon$s ,we have different $e_0$s,But I want to find a fixed constant ,the answer is no if $\tau$ is faithful.Thanks,Pro Argerami! – math112358 Oct 17 '18 at 2:53
• If $\tau$ is not faithful,does there must exist $a$ such that the equality holds? – math112358 Oct 17 '18 at 16:15
• Yes. If $\tau$ is not faithful, there exists $b\geq0$ with $\tau(b)=0$. Take $a=b^{1/2}$, and then $\tau(a^*a)=0$. – Martin Argerami Oct 17 '18 at 16:30
• Can $\tau(a^*a)=|\tau(a)|^2$= nonzero positive number if $\tau$ is not faithful. – math112358 Oct 17 '18 at 16:49
• Yes. There is an example in the answer. – Martin Argerami Oct 17 '18 at 17:13

Regarding your question about the $$a$$'s for which $$\tau(a^\ast a) = |\tau(a)|^2$$: It seems that it can be solved using Choi's Theorem on multiplicative domains of cp maps, see [BO; Proposition 1.5.7]. In particular $$a$$ satisfies the identity above iff for every $$b \in A$$, $$\tau(b a) = \tau(b) \tau(a) = \tau(a) \tau(b) = \tau(a b)$$.

[BO] Brown, Nathanial P.; Ozawa, Narutaka, $$C^*$$-algebras and finite-dimensional approximations, Graduate Studies in Mathematics 88. Providence, RI: American Mathematical Society (AMS). xv, 509 p. (2008). ZBL1160.46001.

• Would you pleasae tell me the details, the proposition is true for c.c.p. maps,but here $\tau$ is a state.How to apply Choi's thorem? – math112358 Oct 17 '18 at 16:23
• Any state is completely positive (the GNS construction basically gives you a factorization $V^\ast \pi(a) V$) – Adrián González-Pérez Oct 17 '18 at 17:01