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I was reading this paper by Evans and Hanche-Olsen. In theorem 2, there are six equivalent statements given. I write just two of them, which I want to use.

Let $L$ be a bounded self-adjoint linear map on a unital $C^*$ algebra $\mathcal{A}$. The following conditions are equivalent

(iii) If $y\in\mathcal{A}_+$ (set of positive semi-definite objects), $a\in\mathcal{A}$ satisfy $ya=0$, then $a^*L(y)a\geq0$.

(iv) For some full invariant set of states $S$ on $\mathcal{A}$, that $y\in\mathcal{A}_+$, $f\in S$ with $f(y)=0$ imply $fL(y)\geq0$.

My question is on condition (iii). If $L$ is a positive map, the condition holds (by definition). However if it is not, what extra condition(s) a general linear bounded map $L$ must satisfy such that condition (iii) holds.

If we replace $\mathcal{A}$ by $\mathcal{B(H)}$ for some Hilbert space $\mathcal{H}$ (finite or infinite dimensions) can we find such conditions. If some works are already in literature, please refer. Advanced thanks for any help, suggestions etc.

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    $\begingroup$ Absolutely baffling question. You have just cited a paper that gives six equivalent versions of this condition. $\endgroup$
    – Nik Weaver
    Commented Nov 22, 2012 at 20:39
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    $\begingroup$ Perhaps OP actually wants a class of examples of conditionally positive maps which are not positive? $\endgroup$
    – Ollie
    Commented Nov 22, 2012 at 21:30
  • $\begingroup$ @Nik Weaver and @Ollie Margetts The conditions are all equivalent. If $L$ were a conditionally completely positive, from condition (i) $e^{tL}$ becomes completely positive. I wanted to relax the conditionally completely positive criteria, such that (iii) still holds, and I can get, perhaps not a completely positive but a general positive map. $\endgroup$
    – RSG
    Commented Nov 23, 2012 at 3:27
  • $\begingroup$ 1. Yes, if $L$ is CCP it generates a CP semigroup. 2. Yes you can relax this condition on $L$ to (iii) and, as Evans shows, this is equivalent to the semigroup being positive. 3. I'm afraid I still don't know what you're asking for? Possibly you can edit the question to make this clearer $\endgroup$
    – Ollie
    Commented Nov 24, 2012 at 21:54

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