# Nuclear maps and almost conditional expectations

This is an exercise from Brown-Ozawa book:

2.1.8. If $\theta: A\to B$ is nuclear and $C\subseteq B$ is a $C^*$-subalgebra with the properties that:

(a). $\theta(A)\subseteq C$ and

(b). There exist a sequence of c.c.p. maps $\varphi_n: B\to C$ such that $\varphi_n|_{C} \to id_C$ in the point-norm topology,

then $\theta:A\to C$ is also nuclear.

My first question is whether the conditions guarantee us that there exists a condinitional expectation $E: B\to C$ $?$
Is yes, then we're done. I thought some argument with $limsup$ could help.
Second, we hace $\varphi_n \circ \theta$ are nuclear and converges (pointwise in norm) to $\theta$. Is that true that a limit of nuclear maps is again nuclear? If yes, we're done.

Thanks.

Well, my answer does not answer your questions regarding the existence of conditional expectation in this case or if the point-norm limit of nuclear maps is nuclear. These are interesting questions also for me, I hope someone else would answer them.

However, here is a solution to the question:

Recall the equivalent definition of nuclear map, that can be found here:
An exercise about the definition of nuclear maps

So, let $F\subseteq A$ be finite subset and $\epsilon>0$ be given.
As $\varphi_n|_C \to id|_C$ pointwise, and the range of $\theta$ is contained in $C$, there exists $N_0\in \Bbb{N}$ s.t. $||\varphi_{N_0}(\theta(a))-\theta(a)||<\epsilon/2$ for all $a\in F$.
Using the equivalent definition for nuclearity of $\theta$ we also have $n$ and pair of c.c.p. maps $\phi:A\to M_n$ and $\psi:M_n\to B$ s.t. $||\psi\circ \phi(a)-\theta(a)||<\epsilon/2$ for all $a\in F$.
Now, $\phi: A\to M_n$ and $\varphi_{N_0}\circ \psi: M_n\to C$ are c.c.p. maps that satisfy for all $a\in F$:

$||\varphi_{N_0}\circ \psi \circ \phi (a)-\theta(a)||\leq ||\varphi_{N_0}(\psi(\phi(a)))-\varphi_{N_0}(\theta(a))||+||\varphi_{N_0}(\theta(a))-\theta(a)||<\epsilon$.

As required.

I hope this helps .

• BTW I think similar arguments would show that limit of nuclear maps is nuclear. – Shirly Geffen Nov 1 '16 at 13:40