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Let $A$ be a C*-algebra , $B(H)$ be the bounded linear operator on Hilbert space $H$ and $P_{i}\in B(H)$ be an increasing net of finite-rank projections which converge to the identity in the strong operator topologgy.

If $\phi_{i}: A\rightarrow P_{i}B(H)P_{i}$, $\phi_{i}(a)=P_{i}\phi(e)P_{i}$ is contractive completely positive map, and $\phi_{i}$ converge to a $\Phi$ in the point-ultraweak topology (i.e. $\xi(\phi_{i}(a))\rightarrow \xi(\Phi(a))$, $\forall a\in A, \forall\xi \in B(H)_{*}$). Then, how to prove $\Phi$ is also a contractive completely positive map? (Here, $B(H)_{*}$ denotes all the normal linear functionals on $B(H)$)

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Since the net $\{\phi_i(a)\}$ is bounded ($\|\phi_i(a)\|=\|P_i\phi(a)P_i\|\leq\|\phi(a)\|\leq\|a\|$), it is enough to test convergence on functionals of the form $h\mapsto\langle hx,y\rangle$ (the ultraweak topology agrees with the weak operator topology on bounded sets). Then, given $a\in M_n(A)$, $X=(x_1,\ldots,x_n)\in H^n$, $$ \langle\Phi^{(n)}(a)X,X\rangle=\sum_{j=1}^n\sum_{k=1}^n\langle\Phi(a_{jk})x_k,x_k\rangle =\lim_i\sum_{j=1}^n\sum_{k=1}^n\langle\phi_i(a_{jk})x_k,x_k\rangle=\lim_i\langle\phi_i^{(n)}(a)X,X\rangle\geq0. $$ The same computation shows that $$ \langle\Phi^{(n)}(a)X,X\rangle=\lim_i\langle\phi_i^{(n)}(a)X,X\rangle\leq\|a\|\,\|X\|^2, $$ showing that $\phi$ is contractive.

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  • $\begingroup$ Not sure what you mean. $\endgroup$ – Martin Argerami Mar 19 '14 at 4:58

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