# Are (completely) positive maps approximated by normal (completely) positive maps?

Let $$\mathcal{H}$$ denote a Hilbert space and $$B(\mathcal{H})$$ denote the algebra of all bounded operators on $$\mathcal{H}$$. By recognizing the (Banach) dual of $$B(\mathcal{H})$$ with the double dual of trace-class operators, one can show using standard result of Banach space theory that, any bounded linear functional $$\phi$$ on $$B(\mathcal{H})$$ can be approximated in weak$$^*$$ topology by (bounded) trace-class operators. In other words, $$\phi$$ is approximated by normal linear functionals on $$B(\mathcal{H})$$. My question is the following:

If the linear functional $$\phi$$ is positive, can $$\phi$$ be approximated by positive normal linear functionals in weak$$^*$$ topology?

Moreover, can this be generalized to completely positive maps? The topology here in consideration is bounded-weak topology. More specifically, if $$M$$ is a von Neumann algebra, then can every completely positive map $$\Phi:M\to B(\mathcal{H})$$ be approximated by normal completely positive maps in bounded-weak topology?

Some reference would be appreciated on these topics as I am new to them. Thank you.

• The answer to the first question is yes; this follows from the bi-polar theorem. Aug 20, 2020 at 20:45
• Could you explain, what is the bounded-weak topology? Aug 21, 2020 at 11:57
• Bounded-weak topology is given by the following convergence: a net $\Phi_i$ converges to $\Phi$ if $\Phi_i(a)\to \Phi(a)$ in weak operator topology for all $a\in M$. This is also a weak$^*$ topology (see Paulsen's book "Completely bounded maps and operator algebra"). Aug 21, 2020 at 12:21
• @JochenGlueck Can you please elaborate? Aug 21, 2020 at 14:04
• @ManishKumar: I added the details in an answer. Aug 21, 2020 at 15:17

Also the answer to the second question is yes, and the approximation may be chosen to converge in the point-ultrastrong$$^*$$ topology.

First, by choosing a net of finite rank orthogonal projections $$p_i \in B(\mathcal{H})$$ such that $$p_i \rightarrow 1$$ strongly, the completely positive maps $$\Phi_i : M \rightarrow B(p_i H) : \Phi_i(a) = p_i \Phi(a) p_i$$ converge to $$\Phi$$ in the point-ultrastrong$$^*$$ topology. So it suffices to deal with completely positive maps $$\Phi : M \rightarrow M_n(\mathbb{C})$$. This can be found in [BO, Corollary 1.6.3]. By [BO, Proposition 1.5.14], $$\omega : M_n(\mathbb{C}) \otimes M \rightarrow \mathbb{C} : \omega(A) = \sum_{i,j} \Phi(A_{ij})_{ij}$$ is a positive functional. Choose a net $$\omega_k$$ of normal positive functionals on $$M_n(\mathbb{C}) \otimes M$$ that converge pointwise to $$\omega$$. Again by [BO, Proposition 1.5.14], there is a corresponding net of completely positive maps $$\Phi_k : M \rightarrow M_n(\mathbb{C}) : (\Phi_k(a))_{ij} = \omega_k(e_{ij} \otimes a) \; .$$ By construction, the maps $$\Phi_k$$ are normal and they converge to $$\Phi$$ in the point-norm topology.

[BO] N.P. Brown and N. Ozawa, C$$^*$$-algebras and finite-dimensional approximations. Graduate Studies in Mathematics 88. American Mathematical Society, Providence, 2008.

• Thank you for this nice answer. Aug 22, 2020 at 18:27

The answer to the first question is yes. This follows from the following more general result.

Terminology I: Ordered Banach spaces. By a pre-ordered Banach space I mean a pair $$(X,X_+)$$ where $$X$$ is a real Banach space and $$X_+$$ is a non-empty closed subset of $$X$$ such that $$X_+ + X_+ \subseteq X_+$$ and $$\alpha X_+ \subseteq X_+$$ for each scalar $$\alpha \ge 0$$ (in other words: $$X_+$$ is a so-called wedge in $$X$$.)

The dual wedge of $$X_+$$ is the wedge $$X'_+ := \{x' \in X': \, \langle x',x \rangle \ge 0 \text{ for each } x \in X_+\}.$$ Note that $$(X', X'_+)$$ is a pre-ordered Banach space, too. Moreover, for each $$x \in X$$ it follows from the Hahn-Banach theorem that $$x \in X_+$$ if and only if $$\langle x', x\rangle \ge 0$$ for each $$x' \in X'_+$$.

By iterating this procedure, one can also define the bi-dual wedge $$X''_+$$ of $$X_+$$ in $$X''$$.

Terminology II: Polars Let $$\langle X,Y\rangle$$ be a dual pair of two real vector spaces; in other words, $$\langle \cdot, \cdot \rangle: X \times Y \to \mathbb{R}$$ is a bi-linear map such that $$X$$ separates $$Y$$ and $$Y$$ separates $$X$$ via this map.

For every subset $$A \subseteq X$$ the subset $$A^\circ := \{y \in Y: \, \langle x, y \rangle \le 1 \text{ for all } x \in A \}$$ of $$Y$$ is called the polar of $$A$$ in $$Y$$. Similarly, for each set $$B \subseteq Y$$ the subset $${}^\circ B := \{x \in X: \, \langle x, y\rangle \le 1 \text{ for all } y \in B \}$$ of $$X$$ is called the polar of $$B$$ in $$X$$.

Now, the bi-polar theorem (see for instance the theorem on page 126 in H. H. Schaefer's book "Topological vector spaces" (1971)) says the following:

Theorem. The so-called bi-polar $$\left({}^\circ B \right)^\circ$$ of a subset $$B \subseteq Y$$ is the closure of the convex hull of $$B \cup \{0\}$$ with respect to the topology on $$Y$$ induced by $$X$$ via the duality mapping $$\langle \cdot, \cdot \rangle$$.

Now we can apply this result to pre-ordered Banach spaces:

Density of wedges in their bi-dual wedges Let $$(X,X_+)$$ be a pre-ordered Banach space, and identify $$X_+$$ with a subset of $$X''_+$$ by means of evaluation.

Theorem. The wedge $$X_+$$ is weak$${}^*$$-dense in the bi-dual wedge $$X''_+$$.

Proof. We consider the dual pair $$\langle X', X'' \rangle$$ with respect to the usual duality. Then it is easily checked that the polar of $$X_+ \subseteq X''$$ in $$X'$$ equals the negative dual wedge $$-X'_+$$. Similarly, it is easy to see that the polar of $$-X'_+$$ in $$X''$$ equals the bi-dual wedge $$X''_+$$. Hence, the bi-polar theorem implies that $$X''_+$$ is the weak$${}^*$$-closure of $$X_+$$ in $$X''$$.

Remark. I believe that the same still works if we intersect the wedge with the unit ball, i.e. the intersection of $$X_+$$ with the unit ball is weak$${}^*$$-dense in the intersection of $$X''_+$$ with the unit ball. I have not checked the details, though.

Application to the first question of the OP. The space $$B(\mathcal{H})$$ is the complexification of the space of self-adjoint operators on $$\mathcal{H}$$. So to apply the general result above, one can choose $$X$$ to be the set of all those trace class operators that yield real values when applied to self-adjoint operators; then $$X'$$ is simply the self-adjoint part of $$B(\mathcal{H})$$, and $$X''$$ is the set of all bounded linear functionals on $$B(\mathcal{H})$$ that map self-adjoint operators to real values. The wedges $$X_+$$, $$X'_+$$ and $$X''_+$$ are the standard cones in these spaces. Since we have seen above that $$X_+$$ is weak$${}^*$$-dense in $$X''_+$$, this yields that desired result.

• In the proof of the theorem, why is the polar of $X_+$ equal to the dual wedge $X_+'$? Aug 21, 2020 at 17:38
• @ManishKumar: Oops, sorry - I actually meant minus the dual wedge. Corrected. Aug 21, 2020 at 18:38
• Thanks for this nice explanation. Aug 21, 2020 at 19:03