Jordan decomposition of trace functional consists of tracial functionals

Question

Let $\mathfrak{A}$ be a C*-algebra, and let $\tau \in \mathfrak{A}'$ be a self-adjoint bounded tracial linear functional on $\mathfrak{A}$. Let $\tau = \tau^+ - \tau^-$ be the Jordan decomposition of $\tau$. Does it follow that $\tau^+, \tau^-$ are also traces?