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Let $M$ be a von Neumann algebra and $T: M\rightarrow \mathbb{C}$ be a finite normal faithful tracial map, s.t., if $\phi : M \rightarrow A \cap A^{*}$ is a conditional expectation, (A being a weak star closed subalgebra of $M$), $T(\phi(x))=T(x) \quad \forall x \in M$.

Does this imply that ,
$T(\phi(x))=T(x) \quad \forall x \in L^{1}(M)$ ?
where $L^{1}(M)$ is the completion of $M$ in $||.||_{1}$.

I know that we can extend $\phi$ continuously to $L^{1}(M)$. How can I use this fact to answer my question?

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  • $\begingroup$ Any help or tip will be appreciated. Thanks in advance. $\endgroup$ – prince Aug 21 '18 at 11:21
  • $\begingroup$ I think the answer is yes. As you say, you can extend $\phi$ continuously to $L^1(M)$. The same goes for $T$. Therefore $F(x) = T(\phi(x)) - T(x)$ is a continuous function in $L^1(M)$ that is zero over $M \subset L^1(M)$, but that subset is dense and therefore $F$ is identically zero. $\endgroup$ – Adrián González-Pérez Aug 21 '18 at 11:22
  • $\begingroup$ I was thinking the same but wasn't that confident. Thank a lot. $\endgroup$ – prince Aug 21 '18 at 11:24

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