I hope my question is ok for mathoverflow. I first asked on math.stackexchange but received no answer and then delated it.


I want to understand the proof of the theorem (which you can find in the paper ``Completely positive maps of order zero", by Winter and zacharias, Theorem 2.3 ):  



Theorem: Let $A$ and $B$ $C^\*-$algebras and $\phi:A\to B$ be a completely postive map of order zero. Set $C:=C^\*(\Phi(A))\subset B$.  

Then there is a positive element $h\in M(C)\cap C'$ with $\|h\|=\|\phi\|$ and a $\*-$homomorphism
$$\pi\_{\phi}:A\to M(C)\cap \{h\}'$$ such that $$\phi(a)=\pi\_{\phi}(a)h $$ for $a\in A$.


If $A$ is unital, then $\phi(1\_A)=h\in C$.


The beginning of the proof: First we assume that $A$ is unital to apply the result for the unital case, as you can find in Wolff's paper about "disjointness preserving operators on $C^\*$-algebras". Set $\phi(1\_A)=h\in C$. Let $C$ acts nondegenerate on its universal Hilbert space $H$. Wollf's theorem gives us that $h$ lies in the center of $C$. Now it is claimed: (i) h is strictly positive ( this means
$\overline{hCh}=C$) and (2) the support projection of $h$ is $1\_H$. 
These claims I want to understand


ad. (i): Let $c\_1\in C,\epsilon >0$, why there is $c\_2\in C$ such that $\|c\_1-hc\_2h\|<\epsilon$? I know if you consider $g\in hCh$, it is $g=\phi(1\_A)t\phi(1\_A)$ with $t\in C$ and t commutes with $\phi(1\_A)$. But I'm stuck here, $\phi$ isn't multiplicative. 


ad. (ii): What is the exact definition of a support projection here and why is the support projection $1\_H$? Maybe the support projection of h can be understood as $\Phi(\chi\_{\sigma(h)\setminus\{0\}})=\chi\_{\sigma(h)\setminus\{0\}}(h)$, with $\Phi$ is the borel measurable functional calculus of $h$ and $\chi\_{\sigma(h)\setminus\{0\}}$ the characteristic function on $\sigma(h)\setminus \{0\}$ and $\chi\_{\sigma(h)\setminus\{0\}}(h)$ is an orthogonal projection, which is the identity on $\sigma{(h)}\setminus\{0\}$. But what is the definition/ characteristic property of a support projection in detail and why is it $1\_H$? Here it should be important that $C$ acts nondegenerate on $H$.


My third problem is how Stinespring's Theorem is apllied. First they define $$\pi\_{\phi}:A\to C^{\*\*}\subseteq B(H),\; a\mapsto s.o. \lim\limits\_{n\to\infty}(h+\frac{1}{n}1\_H)^{-1}\varphi(a)$$ (s.o.-lim is the strong operator topology limit) which is complitely positve, therefore you can apply Stinespring ( <https://en.wikipedia.org/wiki/Stinespring_factorization_theorem> ). They say that there is a unital $C^\*-$algebra $D$ containing $C^{\*\*}$ and a $\*$-homomorphism 
$$\rho :A\to D$$ such that $$\pi\_\phi(a)=1\_{C\*\*}\rho(a)1\_{C^{\*\*}}$$ for $a\in A$.


My questions: Why does $D$ contain $C^{\*\*}$ (With "$D$ containing $C^{\*\*}$" they mean there is an embedding $i:C^{\*\*}\to D$ which is a nonunital $\*-$homomorphism I think)? I think $D$ corresponds to $B(K)$ as stated in Stinespring. Because $A$ and $\pi\_\phi$ are unital, $H$ can be considered as a sub Hilbert space of $K$ and it is, up to an unitary, $\pi\_{\phi}(a)=P\_H\rho(a)\_{|H}$ for all $a\in A$. Here is $P\_H$ the projection of K onto H. But why is it $\pi\_\phi(a)=1\_{C\*\*}\rho(a)1\_{C^{\*\*}}$ for $a\in A$?


I hope that the topic is specific enough for MO. I appreciate your help. Regards


**Edit**: I found a criterion for (i). For positive elements p in a $C^\*$-algebra $A$ it is equivalent:   
-$p$ is strictly positive   

-Let $u\_n=p(\frac{1}{n}+p)^{-1}$, $n\in\mathbb{N}$. It is $(u\_n)$ a approximative unit for $A$


h is clearly positive and maybe with this criterion it is easier to prove (i)? I only obtain $\|hu\_n-h\|\to 0$ and $\|u\_nh-h\|\to 0$, but I need $\|xu\_n-x\|\to 0$ and $\|u\_nx-x\|\to 0$ for all $x\in A$. Here is $u\_n=h(\frac{1}{n}+h)^{-1}$



questions about the proof of the theorem of completely positive order zero maps
@ii:


Step 1: The kernel of $h$ is trivial:


Step 1.1: If $hv=0$, then $0=\langle v,hv\rangle = \langle v\phi(1)v\rangle$ and therefore $0=\langle v,\phi(c1)v\rangle$ for all $c\in\mathbb{C}$. Since $a\mapsto \langle v,\phi(a)v\rangle$ is positive, it follows $0\leq a\leq \|a\|1 \implies 0\leq\langle v,\phi(a)v\rangle\leq \langle v,\phi(\|a\|1)v\rangle=0$ so that $\langle v,\phi(a)v\rangle=0$ for all $a\in A\_+$.


Step 1.2: But then $\phi(a)v = 0$ for all $a\in A\_+$ since $\phi(a)\geq 0$. By linearity $\phi(a)v = 0$ for *all* $a\in A$.


Step 1.3: $\phi(A)v=0$ implies $Cv=0$. Since $C$ operates non-degenerated on $H$, there is no non-trivial subspace that is annihilated by $C$ (this is equivalent to operating non-degenerate). Thus $v=0$. This proves that $\ker(h)=0$.


Step 2: The support projection is $1$. Since $1=\chi\_{\sigma(h)\setminus\{0\}} + \chi\_{\{0\}}$, it suffices to show that $\chi\_{\{0\}}(h) = 0$. But $\chi\_{\{\lambda\}}(h)$ is the projection onto the $\lambda$-eigenspace of $h$ and we have just shown that the kernel of $h$ is trivial so that this projection must indeed be zero.


@i:


Now we can also prove that $\overline{hCh}=C$, because we have just shown that $\frac{h}{h+\frac{1}{n}} \xrightarrow{s.o.t} 1$ (because $\frac{x}{x+\frac{1}{x}} \xrightarrow{pointwise} \chi\_{\sigma\setminus\{0\}}(x)$.)


This implies that $\frac{h}{h+\frac{1}{n}} x \frac{h}{h+\frac{1}{m}} \xrightarrow{n,m\to\infty} x$ for all $x\in C$. Since $C$ is an ideal in $C^+$, $\frac{1}{h+1/n}x\frac{1}{h+1/m} \in C$ for all $x\in C$ and therefore the left hand side expressions are elements of $hCh$ for all $n,m$.


@iii:


In Springstine's theorem you have an operator $V: H\to K$ with $\pi\_\phi(a)=V^\ast\rho(a)V$. Note $\rho$ is unital, since $\pi\_\phi$ is defined on an unital $C^\ast$-algebra. Thus $\pi\_\phi(1\_A)=V^\ast \rho(1\_A) V=V^\ast 1\_{B(K)} V$ so that $V^\ast V = 1\_{B(H)}$ since $\pi\_\phi$ maps the unit to the unit. Thus $V$ is an isometry $H\hookrightarrow K$.


Via $T\mapsto VTV^\ast$ this induces a (non-unital!) injective $\ast$-homomorphism $B(H)\to B(K)$ that is used to establish an identification of operators $H\to H$ with operators $K\to K$.


(If you think of $K$ as $H\oplus H^\perp$ and $V$ as the inclusion $h\mapsto (h,0)$, then this identification maps an operator $T:H\to H$ to the operator $K\to K, (h,h')\mapsto(Th,0)$. In matrix language this is the embedding $T \mapsto \begin{pmatrix} T & 0 \\ 0 & 0 \end{pmatrix}$)


With this identification $\pi\_\phi(a)=V^\ast \rho(a) V$ gets identified with $VV^\ast \rho(a) VV^\ast$ and $VV^\ast$ gets identified with $1\_{B(H)} = 1\_{C^{\ast\ast}}$ so that the equation reads $\pi\_\phi(a) = 1\_{C^{\ast\ast}} \rho(a) 1\_{C^{\ast\ast}}$.


I agree, that it isn't the most clever way to simply ignore this identification.


