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

  • $\begingroup$ 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! $\endgroup$ – math112358 Oct 17 '18 at 2:53
  • $\begingroup$ If $\tau$ is not faithful,does there must exist $a$ such that the equality holds? $\endgroup$ – math112358 Oct 17 '18 at 16:15
  • $\begingroup$ Yes. If $\tau$ is not faithful, there exists $b\geq0$ with $\tau(b)=0$. Take $a=b^{1/2}$, and then $\tau(a^*a)=0$. $\endgroup$ – Martin Argerami Oct 17 '18 at 16:30
  • $\begingroup$ Can $\tau(a^*a)=|\tau(a)|^2$= nonzero positive number if $\tau $ is not faithful. $\endgroup$ – math112358 Oct 17 '18 at 16:49
  • $\begingroup$ Yes. There is an example in the answer. $\endgroup$ – 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.

  • $\begingroup$ 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? $\endgroup$ – math112358 Oct 17 '18 at 16:23
  • $\begingroup$ Any state is completely positive (the GNS construction basically gives you a factorization $V^\ast \pi(a) V$) $\endgroup$ – Adrián González-Pérez Oct 17 '18 at 17:01

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