Two densely defined traces on a $C^*$-algebra coinciding on a dense subalgebra are equal

Question

Let $t_1$ and $t_2$ be lower semicontinuous semifinite densely defined traces on a $C^*$-algebra $A$. Let us denote by $\mathcal{R}_1$ and $\mathcal{R}_2$ their ideals of definition, i.e. $\mathcal{R}_i=\left\{x\in A\ \middle|\ t_i(x^*x)<+\infty \right\}$. Next, let $B$ be an involutive dense subalgebra of $A$ such that $B\subset \mathcal{R}_1,\mathcal{R}_2$ and $t_1(b)=t_2(b)$ for any $b\in B^+$. Is it true that then $t_1=t_2$ on the whole algebra $A$?

I have found a similar statement in “C*-algebras” by J.Dixmier 1977: Lemma 6.5.3 on page 139. But he formulates this claim in terms of bitraces, which look artificial to me. Can anyone provide a straightforward and concise argument without bitraces? Is there a better reference for this?

Thank you in advance!

Answer 1

The closest I can get to the conclusion you ask for is Corollary 6.2.5 and 6.2.6 in arXiv:2204.01125. My guess is that it can fail without some additional assumption on B, but I don't have an example to support this. There is also a closely related result in a paper by Kustermans and Vaes. See Proposition 1.13 in Ann. Scient. Éc. Norm. Sup., 4e série, t. 33, 2000, p. 837 à 934.