Let $\mathfrak{A}$ be a C\*-algebra, and let $\phi \in \mathfrak{A}^\*$ be a self-adjoint bounded linear functional on $\mathfrak{A}$. Then there exists a unique pair $\phi^+, \phi^-$ of positive bounded linear functionals on $\mathfrak{A}$ such that $\phi = \phi^+ - \phi^-$ and $\| \phi \| = \left\| \phi^+ \right\| - \left\| \phi^- \right\|$.


My question is: If $\phi$ is tracial (i.e. $\phi(xy)=\phi(yx)$ for all $x, y \in \mathfrak{A}$), then are $\phi^+, \phi^-$ necessarily tracial?



Jordan decomposition of tracial functionals on a C*-algebra
Assuming $\mathfrak{A}$ is unital, every algebra element is a linear combination of unitaries, so a linear functional being tracial is equivalent to $\phi = \phi \circ \operatorname{Ad} u$ for every unitary $u$. In this case, for every unitary $u$ we have


$$
\phi = \phi \circ \operatorname{Ad} u = \phi^+ \circ \operatorname{Ad} u - \phi^- \circ \operatorname{Ad} u.
$$


By the uniqueness of the decomposition you have $\phi^\pm = \phi^\pm \circ \operatorname{Ad} u$ and so $\phi^{\pm}$ are tracial.


