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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

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    $\begingroup$ Before thinking about this question: could you clarify whether you really want this particular statement to be true? If you are happy to allow modified versions, what are they? It is not always easy to hit a moving target $\endgroup$ – Yemon Choi May 10 '16 at 0:07
  • $\begingroup$ Sorry, I will write something about the motivation for my question in future. I no longer need a modification of this statement. $\endgroup$ – user62639 May 10 '16 at 15:03
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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.

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  • $\begingroup$ thank you! There are easier counterexamples then I expected $\endgroup$ – user62639 May 10 '16 at 14:56

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