is a linear map on an operator system into a $C^*$-algebra (+ extra conditions) positive?

Question

First of all, sorry for my bad english. I tried to find out whether the following statement is true or not:

Let $X$ be a operator system, $B$ a $C^*$-algebra and $f:X\to B$ linear such that $f(1)\ge 0$ and $\|f(1)\|_B=\|f\|_{op}$. Then $f$ is positive.

I already know that the statement is true if $B=\mathbb{C}$ (see for example corollary 2.6 in Paul Skoufranis notes about completely positive maps) but it is not possible to transfer the proof directly to the more general situation, if $B$ is arbitrary. But maybe it is possible with small modifications...

An other try is to apply the statement for functionals, i.e. I considered $\phi\circ f:X\to \mathbb{C}$, where $\phi:B\to \mathbb{C}$ is a state. But then, in general, the condition $\phi (f(1))=\|\phi\circ f\|_{op}$ fails, or am I wrong?

If the statement should be wrong, I need a counterexample. I also tried to modify the standard example of a positive operator defined on an operator system: $$f:S\to M_2(\mathbb{C})$$ $$f(az+b+c\overline{z})=\begin{pmatrix} b & 2a \\ 2c & b \end{pmatrix},$$ where $S:=\{p\in C(\mathbb{T}): p(z)=az+b+c\overline{z},\; a,b,c\in \mathbb{C}\}$. But I was not successful (until now).

It would be very helpful for me to know if I need a counterexample for the statement or if it is possible to proof it. Do you have an idea or a reference?

Best

Answer 1

It's false. Take $X = B = \mathbb{C}^2 \cong l^\infty(\{0,1\})$, and define $f: X \to B$ by $f(a,b) = (a,.5(a-b))$. Then $f(1,1) = (1,0) \geq (0,0)$ and $\|f\| = \|f(1,1)\|_\infty = 1$, but $f(0,1) = (0, -.5) \not\geq 0$.

Edit: but maybe it is worth pointing out that if $f(1_X) = 1_B$ and $\|f\| = 1$ then $f$ must be positive. You can reduce this to the scalar case by composing $f$ with arbitrary states on $B$, as dr. mop suggested in the question.