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.

  • $\begingroup$ The answer to the first question is yes; this follows from the bi-polar theorem. $\endgroup$ – Jochen Glueck Aug 20 '20 at 20:45
  • $\begingroup$ Could you explain, what is the bounded-weak topology? $\endgroup$ – Mateusz Wasilewski Aug 21 '20 at 11:57
  • $\begingroup$ 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"). $\endgroup$ – Manish Kumar Aug 21 '20 at 12:21
  • $\begingroup$ @JochenGlueck Can you please elaborate? $\endgroup$ – Manish Kumar Aug 21 '20 at 14:04
  • $\begingroup$ @ManishKumar: I added the details in an answer. $\endgroup$ – Jochen Glueck Aug 21 '20 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.

  • $\begingroup$ Thank you for this nice answer. $\endgroup$ – Manish Kumar Aug 22 '20 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.

  • 1
    $\begingroup$ In the proof of the theorem, why is the polar of $X_+$ equal to the dual wedge $X_+'$? $\endgroup$ – Manish Kumar Aug 21 '20 at 17:38
  • $\begingroup$ @ManishKumar: Oops, sorry - I actually meant minus the dual wedge. Corrected. $\endgroup$ – Jochen Glueck Aug 21 '20 at 18:38
  • $\begingroup$ Thanks for this nice explanation. $\endgroup$ – Manish Kumar Aug 21 '20 at 19:03

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