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 $t\_1(b^\*b)=t\_2(b^\*b)$ 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?



Two densely defined traces on a $C^*$-algebra coinciding on a dense subalgebra are equal
Yes, this is true.


Let $a\in A\_+$. By lower semicontinuity it suffices to show that $t\_1((a-\delta)\_+) = t\_2((a-\delta)\_+)$ for all $\delta>0$ (where $(a-\delta)\_+$ is the positive part of $a-\delta 1$ in the unitisation of $A$ (which is an element of $A$ and not actually in the unitisation)). Fix $\delta>0$.


The key trick for getting the approximations to work, is to use that $t\_i$ extends canonically to a positive linear functional on $\mathrm{span} R\_i^\ast R\_i$. Hence if $x\in R\_i$ and $b,c\in A\_+$ then
\begin{equation}
- \| b-c\| t\_i(x^\ast x) \leq t\_i(x^\ast(b-c)x) \leq \| b-c\| t\_i(x^\ast x)
\end{equation}
and thus
\begin{equation}
|t\_i(x^\ast b x) - t\_i(x^\ast c x) | \leq \| b-c\| t\_i(x^\ast x).
\end{equation}


This will be used a couple of times (in particular, using that $(a-\delta)\_+^{1/2}\in R\_i$, since $(a-\delta)\_+$ is in the Pedersen ideal; the smallest dense two-sided ideal of $A$).


Let $\epsilon>0$. Pick $e\in A\_+$ be such that $(a-\delta)\_+ e = (a-\delta)\_+$, and let $x\in B$ be close enough to $e^{1/2}$ that $\| e - x^\ast x\| \max\{ t\_1((a-\delta)\_+) , t\_2((a-\delta)\_+)\} < \epsilon$. By the trick mentioned above we have
\begin{equation}
t\_i((a-\delta)\_+) = t\_i((a-\delta)\_+^{1/2} e (a-\delta)\_+^{1/2}) \approx\_\epsilon t\_i((a-\delta)\_+^{1/2} x^\ast x (a-\delta)\_+^{1/2}).
\end{equation}
Now pick $y\in B$ close enough to $(a-\delta)\_+^{1/2}$ so that
\begin{equation}
\| y y^\ast - (a-\delta)\_+ \| \max\{ t\_1(x^\ast x) , t\_2(x^\ast x)\} < \epsilon.
\end{equation}
Then by the above trick
\begin{equation}
t\_i((a-\delta)\_+^{1/2} x^\ast x (a-\delta)\_+^{1/2}) = t\_i(x(a-\delta)\_+ x^\ast) \approx\_\epsilon t\_i (xyy^\ast x^\ast).
\end{equation}
As $xy \in B$ we get $t\_1(xyy^\ast x^\ast) = t\_2(xyy^\ast x^\ast)$ by assumption, and therefore
\begin{equation}
| t\_1((a-\delta)\_+) - t\_2((a-\delta)\_+) | < 4 \epsilon.
\end{equation}
As $\epsilon>0$ was arbitrary we get $t\_1((a-\delta)\_+) = t\_2((a-\delta)\_+)$, and thus $t\_1(a) = t\_2(a)$ by lower semicontinuity.


