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 $B(H)$)


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.