composition and strong limits of completely positive maps is completely positive

Question

I have two claims about completely positive maps. Let $X$, $Y$, $Z$ be $C^\ast$-algebras.

1) Let $f:X\to Y$ and $g:Y\to Z$ be completely positive maps. I want to know, why $g\circ f$ is completely positve? First of all, it is clear, that the composition is linear. Next, for all $n\in\mathbb{N}$, the map $g^{(n)} \circ f^{(n)} :M_n(X)\to M_n(Z),\; (x_{ij})\mapsto (g^{(n)} \circ f^{(n)})(x_{ij})=(g\circ f (x_{ij}))$ has to be positive. But if $(x_{ij})\ge 0$, it is $f^{(n)}(x_{ij})\ge 0$ for all $n$ and therefore $(g^{(n)} \circ f^{(n)})(x_{ij})=g^{(n)}(f(x_{ij})\ge 0$ for all $n\in\mathbb{N}$, because g is completely positive. Is that all?

2) Now I want to know, why the strong limit of completely positive maps is completely positive. I'll try to formulate my problem. Let $H$ be a complex Hilbert space, and let $(f_n)$ be a sequence such that $f_n:H\to H$ is completely positive for all $n\in\mathbb{N}$, and such that the strong limit $f_n(x)\to f(x)$ for all $x\in H$ exists. Why is f completely positive too? It's clear that f is linear. Now, for every $n\in\mathbb{N}$, it is to show that $\langle f^{(n)}(h_{ij})x,x\rangle =\langle (f(h_{ij}))x,x\rangle \ge 0$ für every $x\in \mathbb{C}^n$. But how to do this? If $f_n\to f$ strong, is it $f_n^{(k)}\to f^{(k)}$ for every $k\in\mathbb{N}$ and $n\to\infty$? Regards

Regards

Answer 1

1) Yes.

2) CP on a Hilbert space does not make sense. You need $X\subset B (H) $ a C$^*$-algebra, and $Y\subset B (K) $ on the codomain. There, you have to convince yourself that entry wise sot convergence in $M_n (Y) $ implies sot convergence in $M_n (Y) $.

Note that $\langle T_jx,x\rangle \to\langle Tx,x\rangle$ is wot convergence, not sot.

Let $A\subset B(H)$, $B\subset B(K)$ be C$^*$-algebras. Let $\varphi_j:A\to B$ be completely positive maps. Usually, when one says that $\varphi_j\to\varphi$ strongly it means "pointwise strong": for every $a\in A$, $\varphi_j(a)\to\varphi(a)$ strongly, which in turn means that $\varphi_j(a)x\to\varphi(a)x$ for all $x\in H$. Now fix $n\in \mathbb N$, let $\hat a\in M_n(A)$ be positive. Fix $\hat x\in H^n$. Then $$ \langle \varphi^{(n)}(\hat a)\hat x,\hat x\rangle=\lim_j\langle \varphi_j^{(n)}(\hat a)\hat x,\hat x\rangle\geq0, $$ and so $\varphi$ is completely positive. The facts used for the equality above are that sot-convergence in $M_n(A)$ is equivalent to sot-entrywise convergence; and that sot-convergence implies wot-convergence.