Jordan decomposition for mappings between C*algebras The term "Jordan decomposition" typically refers to decomposition of a real measure or a self-adjoint element of some $C^*$-algebra into a difference of two positive elements. Such a decomposition holds also for self-adjoint functionals on $C^*$-algebras. My question concenrs bounded operators acting between $C^*$-algebras which are self-adjoint, i.e. $\phi\colon\mathcal{A}\to\mathcal{B}$ is $\textit{self-adjoint}$ if $\phi(\mathcal{A}_{sa})\subset\mathcal{B}_{sa}$. Are there any results showing that such mapping are differences of two positive ones? A $\textit{positive}$ map means that it maps $\mathcal{A}_+$ into $\mathcal{B}_+$. 
 A: Cool question.  You should check out this paper by Tsui.  I haven't read over the paper carefully, but it seems that his section 1.3 gives a few counter-examples.  In particular, it looks like there are self-adjoint maps from $B(H)$ to itself (and also from $C(\mathbb{N}\cup \{\infty\})$ to itself, where $\mathbb{N}\cup \{\infty\}$ is the one-point compactification) which don't admit such a positive decomposition.  So it seems we can't expect this decomposition to exist even for Type I factors or commutative C$^*$-algebras.  
The main result of part 1 of that paper gives a condition on the range von Neumann algebra $B$ that guarantees that all self-adjoint $\phi: A \rightarrow B$ (with $A$ a C$^*$-algebra) admit a positive decomposition.  It seems like this happens if and only if $B$ is Type I and strictly finite, i.e. $B = \oplus_i (M_{n_i} \otimes A_i)$ where $A_i$ are commutative and $\sup_i n_i < \infty$.
I wasn't previously aware of those results, so it's possible there's now a fuller answer to your question.  The only other potentially relevant result I spotted in my quick search was this paper by Smith and Williams on what they call the decomposition property for a C$^*$-algebra.  They study when a C$^*$-algebra $A$ admits completely positive decompositions for all self-adjoint completely bounded maps $\phi: E \rightarrow A$.
