Consider a separable Hilbert space $\mathcal{H}$, we can first define $T(\mathcal{H})$ the trace class of $\mathcal{H}$, then $D(\mathcal{H})$ to denote the set of positive operator with trace less than 1. 


The set of positive maps consists of all elements that maps positive operator to positive operator. The set of completely positive maps contains all elements $a$ that $id\_k\otimes a$ is still positive for any integer $k$.


We are interested in $C$, a subset of the completely positive maps contains all $c$ such that for all $b\in D(\mathcal{H})$,
$$\mathrm{tr}[c(b)]\leq \mathrm{tr}[b].$$


In finite dimensional case, we always have the so-called Kraus decomposition that any $c$ can be written as 
$$c(b)=\sum E\_i bE\_i^{\*},$$
with $E\_i^{\*}E\_i\leq I$.


Moreover, we have $c^\*$, the dual map of $c$, satisfied
$$c^\*(b)=\sum E\_i^{\*} bE\_i.$$


The dual map is defined as 
$$\langle d,c(b)\rangle = \langle c^\*(d),b \rangle$$ 
for all $b,d$.


In the separable Hilbert space, do we have similar version of Kruas decomposition, maybe with infinite sum? Do we have similar version of dual map?



Completely Positive Maps and their dual in Separable Hilbert Space
This is true; though I don't know a canonical reference (perhaps someone else can provide this).


But let me sketch a proof. In infinite dimensions, we need to be a little careful about topologies. Let $\newcommand{\mc}{\mathcal}\mc T(H)$ be the trace-class operators on $H$ with norm $\newcommand{\tr}{\operatorname{tr}}$$\|x\|\_1 = \tr(|x|)$, and $\mc B(H)$ the algebra of all bounded operators with norm $\|\cdot\|$. As usual, $\mc T(H)$ is the predual of $\mc B(H)$ with duality $\langle x, t \rangle = \tr(xt)$.


Let $c:\mc T(H) \rightarrow \mc T(H)$ be a linear map which is completely positive. But what exactly do we mean by "completely positive" here? For $M\_n(\mc T(H))$ is not a C$^\*$-algebra, and so doesn't immediately have a notion of positivity (but see note 2 below). For now, we proceed as follows.
As $c$ is positive, it is bounded (see note 1). Thus we can form $c^\*$, a map on $\mc B(H)$, and demand that this is completely bounded.


We now apply the usual Stinespring construction to obtain an auxiliary Hilbert space $H'$, a unital $\*$-homomorphism $\pi:\mc B(H)\rightarrow\mc B(H')$, and a bounded linear map $V:H\rightarrow H'$ with $c^\*(x) = V^\*\pi(x)V$. We now make some observations:


* You define $\mc D(H)$ as $\{x\in\mc T(H) : x\geq 0, \tr(x)\leq 1\}$ which is just the closed unit ball of $\mc T(H)$ intersected with $\mc T(H)\_+$. Notice now that the requirement that $\tr(c(x)) \leq \tr(x)$ for all $x\in \mc D(H)$ can be multiplied by any positive number, and so is equivalent to $\tr(c(x)) \leq \tr(x)$ for all $x\in\mc T(H)\_+$. This is equivalent to $\langle c^\*(1), x \rangle \leq \langle 1, x \rangle$ for all $x\in\mc T(H)\_+$. This is equivalent to just that $c^\*(1) \leq 1$ in $\mc B(H)\_+$.
* Thus $c^\*(1) = V^\*\pi(1)V = V^\*V \leq 1$.
* Careful examination of the Stinespring construction shows that $\pi$ is *normal* because $c^\*$ is.
* A normal unital $\*$-homomorphism $\mc B(H)\rightarrow\mc B(H')$ has a special form. $H' = H \otimes K$ for some Hilbert space $K$ and $\pi(x) = x\otimes 1$.
* In fact, in this case, we can streamline the Stinespring construction (see for example page 9 of [arXiv:1107.5244 math.OA](https://arxiv.org/abs/1107.5244)).


So we have $V:H\rightarrow H\otimes K$ with $V^\*V\leq 1$ and $c^\*(x) = V^\*(x\otimes 1)V$. Let $(e\_i)$ be an orthonormal basis of $K$, and check that there are $E\_i\in\mc B(H)$ with
$$ V(\xi) = \sum\_i E\_i(\xi)\otimes e\_i \qquad (\xi\in H). $$
That $V^\*V\leq 1$ means exactly that $\sum\_i E\_i^\*E\_i\leq 1$. Then
$$ c^\*(x) = \sum\_i E\_i^\* x E\_i. $$
It is now easy to check that $c(t) = \sum\_i E\_itE\_i^\*$ to $t\in\mc T(H)$.
Finally, if $H$ is separable, then $K$ will be separable and so the family $(E\_i)$ will be a sequence (or finite).


**Note 1:** If $c$ is positive but not bounded, then firstly observe that every $t\in\mc T(H)$ can be written as the linear combination of 4 positive trace-class operators, so $c$ is not bounded on the positives. Thus we can find $(x\_n)\subseteq \mc T(H)\_+$ with $\tr(x\_n)\leq n^{-2}$ but $\tr(c(x\_n))\geq n$. Then $x=\sum\_n x\_n$ converges absolutely in $\mc T(H)\_+$, and for any $N$, $\sum\_{n\leq N}x\_n \leq x$ so $\sum\_{n\leq N} c(x\_n) \leq c(x)$ so $\tr(c(x)) \geq \sum\_{n\geq N} \tr(c(x\_n)) \geq N$, a contradiction.


**Note 2:** We identify $M\_n(\mc B(H))$ with $\mc B(H\otimes \mathbb C^n)$ which gives the canonical $C^\*$-algebra structure on $M\_n(\mc B(H))$ and hence the positive cone. The predual is $\mc T(H\otimes \mathbb C^n)$ which is of course just $M\_n(\mc T(H))$. Furthermore, duality behaves as you expect, namely if $x=(x\_{i,j})\in M\_n(\mc B(H))$ and $t=(t\_{i,j}) \in M\_n(\mc T(H))$ then
$$ \tr(xt) = \sum\_{i,j} \tr(x\_{i,j} t\_{j,i}). $$
Then $c^\*$ being completely positive is equivalent to $c$ being completely positive, where $M\_n(\mc T(H))$ is given this positive cone, i.e. as operators on $H\otimes\mathbb C^n$. (Be aware that it is *also* common to use $\langle x,t\rangle = \sum\_{i,j} \langle x\_{i,j} , t\_{i,j} \rangle = \sum\_{i,j} \tr(x\_{i,j}t\_{i,j})$ which is of course different.)


