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 $\mathcal{B}_+$.
Jordan decomposition for mappings between C*algebras
operator-algebrasself-adjoint-operators
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Here's an answer for your question 1. which has two parts:
a) [An example of a] Banach *-algebra that is not a C*-algebra for which there exists a positive linear functional (it takes $x^*x$ to numbers $ \geq 0$) that is not norm continuous.
As far as I know all “explicit” counterexamples are based on Dixon's construction, and they are quite complicated. In 1982 H.G. Dales proved:
There is a Banach *-algebra with a norm-discontinuous trace, i.e. a positive linear functional $\tau$ such that $\tau(ab) = \tau(ba)$ for all $a,b \in A$.
See Example 4 of H.G. Dales, The Continuity of Traces, in Radical Banach algebras and automatic continuity, Lecture Notes in Mathematics, 975, Springer Verlag, (1982), 451–458. To understand the construction in that article you will need a copy of P.G. Dixon's paper, Non-separable Banach algebras whose squares are pathological, J. Functional Analysis, 26 (1977), 190–200.
Remark. That this example must be rather complicated is not really surprising since it is impossible to “write down” a discontinuous linear map, and it is obviously even harder to write down a discontinuous linear map satisfying further algebraic constraints: a discontinuous linear map necessarily involves the axiom of choice in a strong form, e.g. existence of algebraic bases, since it is consistent with ZF+DC that all linear maps between Banach spaces are continuous. In fact, the (non)existence of discontinuous homomorphisms on $A = C(K)$ is independent of the usual axioms of set theory, see also Andres Caicedo's answer here.
b) Apparently if a Banach *-algebra so much as even has a bounded approximate identity, then all positive linear functionals are continuous. Does anybody have a proof of this?
Yes. More generally, Theorem 5.5.10 on page 698 of H.G. Dales, Banach Algebras and Automatic Continuity states the automatic continuity of a positive linear functional on a Banach algebra with a not necessarily continuous involution and a bounded approximate left identity. Here's an outline of the argument in the simpler case of an isometric involution:
Suppose that $A$ is a Banach *-algebra with a bounded approximate identity. Arguably the most important single result in automatic continuity theory is the Cohen-Hewitt factorization theorem. In one of its most basic forms it can be phrased as follows:
Let $A$ be a Banach algebra with a bounded approximate left identity $(u_i)_{i \in I}$ and let $M$ be a left Banach $A$-module. For $m \in M$ the following are equivalent:
There are $a \in A$ and $n \in M$ such that $m = an$
$\lim_{i} \lVert u_i m - m\rVert = 0$.
A short proof (very close to Hewitt's argument) can be found in Cigler–Losert–Michor, Banach modules and functors on categories of Banach spaces, Theorem III.1.15 on page 108 (they are assuming that the approximate identity is in the unit ball of $A$ but the proof is easy to adapt to the more general statement given here).
Over time the factorization theorem has been sharpened in many ways and entire books were written about it: R.S. Doran, J. Wichmann Approximate Identities and Factorization in Banach Modules. See also Chapter 2 of Dales's comprehensive Banach Algebras and Automatic Continuity.
Some further remarks:
If all $m \in M$ satisfy the conditions of the factorization theorem then $M$ is called an essential $A$-module.
Since the definition of an approximate identity is property 2. for all $a \in A$ when $A$ is considered as a left $A$-module, one immediate consequence of the theorem is that every $a \in A$ factors as $a = bc$. For $A = L^1(G)$ this is Cohen's original form of the factorization theorem.
An easy lemma on essential modules $M$ is that the space $c_0(M) = c_0(\mathbb{N},M)$ of null sequences in $M$ with the norm $\lVert (m_n)_{n \in \mathbb{N}}\rVert_{c_0(M)} = \sup_{n \in \mathbb{N}} \lVert m_n \rVert_M$ is an essential module (check that property 2. is satisfied for $c_0(M)$). This implies that every null sequence $m_n \to 0$ in $M$ can be factored as $m_n = a x_n$ with $a \in A$ and $x_n \to 0$ in $M$.
[Incidentally, this can be used to show that a right $A$-linear map $\varphi\colon A \to N$ into a right Banach $A$-module is automatically continuous by observing that $a_n \to 0$ implies $a_n = ab_n$ with $b_n \to 0$ so that $\varphi(a_n) = \varphi(a)b_n \to 0$.]
The proof of continuity of positive linear functionals on a Banach *-algebra with a bounded approximate identity is now relatively easy:
Recall the Cauchy-Schwarz inequality for positive linear functionals $f$: $$ \lvert f(ab a^\ast)\rvert \leq f(aa^\ast) \cdot \sqrt{\rho(bb^\ast)}, $$ for all $a, b \in A$, where $\rho(x) = \lim_{n\to\infty} \lVert x^n\rVert^{1/n}$ is the spectral radius of $x \in A$.
It follows from this that $x \mapsto f(a x a^\ast)$ is continuous for all $a \in A$ and thus $x \mapsto f(ax b^\ast)$ is continuous for all $a,b \in A$.
Now let $x_n \to 0$. We want to show that $f(x_n) \to 0$. By the factorization theorem (remark 3. above), we can write $x_n = a c_n$ with $c_n \to 0$. Now $c_n^\ast \to 0$, so we can write $c_n^{\ast} = b d_n$ with $d_n \to 0$. Thus, $x_n = a d_n^\ast b^\ast$ with $d_n \to 0$, so $f(x_n) = f(a d_n^\ast b^\ast) \to 0$, as desired.
If you scroll to the bottom of the PDF (e.g. the one in https://www.jstor.org/stable/24714007 ), it says:
Added in proofs. E. Kirchberg has recently shown by a non-constructive proof that there exists a counter-example to Woronowicz's Conjecture. Hence, Conjecture 3.4 is also false by virtue of Proposition 3
So, the conjecture is false.
Best Answer
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 $\phi: E \rightarrow A$.