Let $A$ be a unital $C^{\*}$-algebra and $\phi:A \rightarrow A$ be a completely positive map, i.e. $\phi^{(n)}:M\_{n}(A) \rightarrow M\_{n}(A)$ preserves positivity for any natural number $n$, where $\phi^{(n)}((A\_{ij})\_{ij})=(\phi(A\_{ij}))\_{ij}$. It is well-known that the norm $\|\phi\|=sup\_{\|x\|\leq 1}\{\|\phi(x)\|\}=\|\phi(I)\|$.


Also recall that, given an element $x$ in a $C^{\*}$-algebra $A$, the element $\|x\|-x$ is always a positive element.


It is often mighty tempting to want to treat a unital c.p. map like it is an element of a $C^{\*}$-algebra and say that $\|\phi\|id-\phi$ is a positive map. I've heard that there are many circumstances where this is doable. 



> 
> **Question** What are some natural circumstances where $\|\phi\|id-\phi$ is a positive map, with $\phi$ completely positive as above?
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For example, if $A$ is a finite von Neumann algebra, then sometimes the map $\phi$ naturally gives rise to a bounded operator on $L^{2}(A)$. If this operator happens to be a positive operator then we get what we want. (I'd be interested in knowing natural circumstances where we know that precisely this happens.)



Completely positive maps as "positive operators"
This answer deals with the case that $\phi$ is non-unital. In this case, $\phi$
must be of the form $\phi(a)=ha$, where $h$ is a positive element in the center 
of $A$.


Unfortunately, the solution I've got is somewhat long (hopefully correct): Let us assume that $\phi$ is contractive (otherwise one can rescale). Since $\phi(a)\leq a$ for any positive element $a$, $\phi$ preserves orthogonality of positive elements. There's a nice structural result for such maps proven in ``Completely positive maps of order zero", by Winter and zacharias. By Theorem 2.3 of that paper $\phi$ has the form $\phi(a)=h\pi(a)$, where $h=\phi(1)$,
$$\pi:A\to M(C^\*(\phi(A)))$$


is a unital homomorphism into the multiplier algebra of $C^\*(\phi(A))$ and $h$ commutes with the image of $\pi$. 


From $\phi(a)\leq a$ one gets $h\pi(a)\leq a$ for any positive $a$. 
Suppose that $\pi(b)=0$, with $b\in A\_+$ positive contraction. Then $h=h\pi(1-b)\leq 1-b$.
Similarly, $h\leq 1-b^{1/n}$ for all $n$. This implies that $hb=0$. So $h$ is orthogonal to $\ker \pi$. In particular, $\pi$ is injective on $\overline{hAh}$. 


Let us show that $\pi$ is the identity on $\overline{hAh}$. Taking $n$-root in $h\pi(h^n)\leq h^n$ we get $h^{1/n}\pi(h)\leq h$. Since $h$ is a strictly positive element of $C^\*(\phi(A))$ and $\pi(h)$ is a multiplier for that algebra, we have that $h^{1/n}\pi(h)\to \pi(h)$ in the strict topology. So, $\pi(h)\leq h$, and so $\pi$ maps $\overline{hAh}$ into itself. Passing to the limit in $h^{1/n}\pi(a)\leq a$, with $a\in \overline{hAh}$, one gets $\pi(a)\leq a$ for all such $a$. But if $\alpha$ and $\beta$ are homomorphisms such that $\alpha\leq \beta$ on all positive elements, then $\alpha+\tilde\alpha=\beta$, where $\tilde\alpha$ is a homomorphism with orthogonal range to $\alpha$. (There's probably a reference for this; I'll skip the argument to keep this answer short.) In our case, since $\pi$ is injective, we get $\pi(a)=a$ for $a\in \overline{hAh}$. In particular, $\pi(h)=h$.


We have that
$$
\phi(a)=h\pi(a)=\pi(h^{1/2} a h^{1/2})=h^{1/2} a h^{1/2}
$$for all $a\in A$. Finally, let us show that $h$ is in the center of $A$. 
We have $\pi(ah-ha)=\pi(a)\pi(h)-\pi(h)\pi(a)=0$. So 
$ha-ah\in \ker \pi$. It was argued above that this implies
$h(ha-ah)=0$. So $h^2a=ah^2$. So $h$ is in the center.


