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Recall that a map $\phi: M\rightarrow N$ of von Neumann algetras is normal if $$\phi(sup x_{i})=sup\phi(x_{i})$$ for all norm bounded, monotone increasing nets of self-adjoint elements $\{x_{i}\}\subset M$.

Let $N$ be a von Neumann algebras and $M_{n}{(\mathbb{C})}$ be a compelx matrix, if $$\phi: M_{n}{(\mathbb{C})} \rightarrow N$$ is a contractive completely positive map, then $\phi $ is a normal?

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Yes. Because $M_n(\mathbb C)$ is a finite-dimensional von Neumann algebra, so if $x_i\nearrow x$ for $x_i$ selfadjoint for all $i$, then $x_i\to x$ in norm. So $$ \phi(x)=\phi(\lim x_i)=\lim \phi(x_i)\ \ \ \text{ (in norm)}. $$ As $\phi$ preserves positivity, we have $\phi(x_i)\leq\phi(x)$, and with the limit being in norm, it is also a strong limit; so $\phi(x_i)\nearrow\phi(x)$.

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  • $\begingroup$ Maybe, in $M_{n}(\mathbb{C})$ the norm topology and the ultraweak topology are equivalent, so a c.c.p map is also continuous in ultraweak topology. Hence, it is normal. $\endgroup$ – Yan kai Apr 1 '14 at 7:20

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