When do completely positive maps have a closed image?

Question

Let $\mathcal{A}, \mathcal{B}$ be C*-algebras. A map $\phi \colon \mathcal{A} \rightarrow \mathcal{B}$ is completely positive (cp) if it's linear, * preserving and all of its' coordinatewise extensions to matrices $\phi^{(n)} \colon M_n(\mathcal{A}) \rightarrow M_n(\mathcal{B})$ map positive matrices to positive matrices. For instance, by Stinesprings' Representation, completely positive maps are just of the form $V \pi V^*$, where $\pi$ is a *-homomorphism.

My question is under which conditions do these cp maps have a closed image (see here for a proof that homomorphisms do have a closed image).

Of course this is true if the map is isometric, but I can't seem to prove that, for instance, injective cp maps of norm 1 are isometric. The usual proof for homomorphisms uses that the spectrum remains unchanged under a homomorphism, but does this also hold for cp maps? It does seem false to me.

Thank you for any help you may give. I must be overlooking something.

Answer 1

This is not a complete answer to the original question, since the original question has a rather open-ended phrasing; but I think it addresses Diego's main points, and shows that even quite well-behaved cp maps won't have the desired properties.

First of all: if we restrict attention to cases where either $A$ or $B$ is commutative, then every positive linear map $A\to B$ is automaticaly completely positive: see Lemma 5.1.4 and Theorem 5.1.5 in the book of Effros--Ruan, for instance. To keep things simple I will take both $A$ and $B$ to be commutative ${\rm C^*}$-algebras; I'm sure one can create more fancy examples by modifying the constructions below.

Q1: "Are injective norm-one cp maps isometric?"

Counterexample: take $A=B = \ell_\infty^2$ (viewed as $C(\{0,1\})$) and take the linear map $T:A\to B$ defined by $T(x,y) = (x, y/3)$.

Q2: "Do ucp maps have closed range?"

This counterexample requires a bit more work than I originally thought, so it seems to deserve writing down as an answer rather than merely a comment. We work with $A=B=C[0,1]$ and consider positive, unit-preserving operators $T:C[0,1]\to C[0,1]$ of the form $$ (Tf)(x) = \int_0^1 K(x,t) f(t)\,dt $$ for a suitably chosen positive kernel function $K$. The condition $T1 = 1$ translates into requiring $\int_0^1 K(x,t)\,dt=1$ for all $x$, so that each "row" of $K$ should have total mass 1, and then on the analysts' general principle that "to make things go wrong, get things to pile up more and more" it is natural to try an Ansatz like $$ K(x,t) = \begin{cases} x^{-1} & \hbox{ if $0\leq t\leq x$} \\ 0 & \hbox{if $x< t \leq 1$} \end{cases} $$ With this choice, our operator $T$ becomes $$ (Tf)(x) = \frac{1}{x} \int_0^x f(t)\,dt $$ (sometimes known as the Hardy operator, if I recall correctly) and then we can see directly that $T$ is positive and unit-preserving.

On the other hand, because $T$ is averaging the behaviour of a given $f$ over the initial parts of the interval $[0,1]$, one intuitively feels that $T$ has a good chance of shrinking the norms of functions which are small near $0$ and big near $1$, which will mean that $T$ can take elements of norm $1$ to elements with small norm. Therefore we consider $Tf_n$, where $f_n(t)=t^n$, and calculate that $$ (Tf_n)(x) = \frac{1}{x} \int_0^x t^n\,dt = \frac{1}{n+1} x^n $$ so that each $f_n$ is in fact an eigenvector of $T$ with eigenvalue $(n+1)^{-1}$. In particular, $T$ is not bounded below.

To conclude that $T$ does not have closed range, it only remains to note that $T$ is injective. This is immediately plausible as we are basically integrating and then dividing by a weight, and taking the indefinite integral of a non-zero function should give a non-zero function. More precisely: suppose $f\in C[0,1]$ and $Tf=0$. Then $\int_0^x f(t)\,dt=0$. So by the fundamental theorem of calculus $f=0$.d