Proof. Let $\mathcal {H}$ be a Hilbert space such that $\mathcal {R}\subseteq {{\rm B}(\mathcal {H})}$. It follows from [Reference Kadison and Ringrose20, 5.5.4] that there is a natural $*$-isomorphism $\iota$ from $\mathcal {R}\mathcal {R}^{\prime }$ (the subalgebra of ${{\rm B}(\mathcal {H})}$ generated by $\mathcal {R}\cup \mathcal {R}^{\prime }$) onto the algebraic tensor product $\mathcal {R}\odot _{\mathcal {Z}}\mathcal {R}^{\prime }$, given by $rr^{\prime }\mapsto r\otimes _{\mathcal {Z}} r^{\prime }$. By [Reference Blanchard6, 2.9], the tensor norm on $\mathcal {R}{\otimes }_{\mathcal {Z}}\mathcal {R}^{\prime }$ restricted to $\mathcal {R}\odot _{\mathcal {Z}}\mathcal {R}^{\prime }$ is minimal among all C$^{*}$-tensor norms on $\mathcal {R}\odot _{\mathcal {Z}}\mathcal {R}^{\prime }$, hence the $*$-homomorphism $\iota$ extends uniquely to the norm closure $\overline {\overline {\mathcal {R}\mathcal {R}^{\prime }}}$. Since $\mathcal {A}$ is injective and commutes with $\mathcal {R}^{\prime }$ the multiplication $\mu _0:\mathcal {A}\otimes \mathcal {R}^{\prime }\to \overline {\overline {\mathcal {A}\mathcal {R}^{\prime }}}\subseteq {{\rm B}(\mathcal {H})}$ is a completely contractive $*$-homomorphism [Reference Brown and Ozawa8, 9.3.3, 3.8.5]. But more is true: by [Reference Giordano and Mingo13, 4.2], the natural map $\mathcal {A}\odot _{\mathcal {Z}}\mathcal {R}^{\prime }\to \mathcal {A}\mathcal {R}^{\prime }$ extends (uniquely) to a $*$-isomorphism $\mathcal {A}\otimes _{\mathcal {Z}}\mathcal {R}^{\prime }\to \overline {\overline {\mathcal {A}\mathcal {R}^{\prime }}}$. It follows that the composition

\[ {{\rm B}(\mathcal{H})}\supseteq\overline{\overline{\mathcal{R}\mathcal{R}^{\prime}}}\to\mathcal{R}{\otimes}_{\mathcal{Z}}\mathcal{R}^{\prime}\stackrel{\psi{\otimes}_{\mathcal{Z}} {\rm id}}{\longrightarrow}\mathcal{A}{\otimes}_{\mathcal{Z}}\mathcal{R}^{\prime}\cong\overline{\overline{\mathcal{A}\mathcal{R}^{\prime}}}\subseteq{{\rm B}(\mathcal{H})} \]

is completely contractive and clearly, it is an $\mathcal {R}^{\prime }$-bimodule map, hence extending to such a map $\phi$ on ${{\rm B}(\mathcal {H})}$ by the Wittstock extension theorem (see [Reference Wittstock30] or [Reference Blecher and Le Merdy7, 3.6.2]). By [Reference Effros and Kishimoto11], $\phi$ can be approximated in the point-weak* topology by a net of elementary complete contractions of the form

(2.1)\begin{equation} x\mapsto \displaystyle\sum_ia_i^{*}(k)xb_i(k)=a(k)^{*}xb(k)\quad (x\in{{\rm B}(\mathcal{H})}) \end{equation}

where $a(k)=(a_1(k),\,\ldots,\,a_n(k))^{T}$ and $b(k)=(b_1(k),\,\ldots,\,b_n(k))^{T}$ are columns with the entries $a_i(k),\,b_i(k)\in \mathcal {R}$ and

(2.2)\begin{equation} a^{*}(k)a(k)=\displaystyle\sum_ia_i^{*}(k)a_i(k)\leq 1,\quad b^{*}(k)b(k)=\sum_ib_i^{*}(k)b_i(k)\leq 1. \end{equation}

Thus, $\psi$ (=$\phi |\mathcal {R}$) can also be approximated by such maps.

Assume now in addition that $\psi$ is unital and consider a point-weak* approximation of $\psi$ of the form (2.1), (2.2). Since

\begin{align*} 0& \leq(b(k)-a(k))^{*}(b(k)-a(k))=b(k)^{*}b(k)+a(k)^{*}a(k)-a(k)^{*}b(k)-b(k)^{*}a(k)\\ & \leq 2-a(k)^{*}b(k)-b(k)^{*}a(k)\to 2-2\psi(1)=0,\\ \end{align*}

it follows that $b(k)-a(k)$ tends to $0$ in the strong operator topology. Hence, $\psi$ can be approximated by maps of the form $x\mapsto a(k)^{*}xa(k)$ in the point-weak* operator topology. To see this, write

\[ \psi(x)-a(k)^{*}xa(k)=(\psi(x)-a(k)^{*}xb(k))+(a(k)^{*}x(b(k)-a(k))) \]

and note that $\|a(k)^{*}x(b(k)-a(k))\xi \|\leq \|x\|\|(b(k)-a(k))\xi \|$ for each vector $\xi \in \mathcal {H}$. Finally, as $a(k)^{*}a(k)$ tends to $\psi (1)=1$ in the strong operator topology, $\psi$ can be approximated by maps of the form

\[ x\mapsto a(k)^{*}xa(k)+\sqrt{1-a(k)^{*}a(k)}x\sqrt{1-a(k)^{*}a(k)}, \]

that is, by unital completely positive elementary maps.

Proof. For each $c\in Z_+$ and $\theta \in (A^{\sharp })_+$, we have that $\|c\theta \|=(c\theta )(1)=\theta (c)$ and it is also well-known that for each hermitian functional $\sigma$ the equality $\|\sigma \|=\sigma _+(1)+\sigma _-(1)=\|\sigma _+\|+\|\sigma _-\|$ holds, hence

\begin{align*} & \rho_+(c)+\rho_-(c)=\|c\rho\|\leq\|c\omega\|=\omega_+(c)+\omega_-(c),\\ & \rho_+(c)-\rho_-(c)=\rho(c)=\omega(c)=\omega_+(c)-\omega_-(c). \end{align*}

Adding and subtracting these two relations, we find that $\rho _+(c)\leq \omega _+(c)$ and $\rho _-(c)\leq \omega _-(c)$ for all $c\in Z_+$. If $Z$, $\omega$, $\rho$, $p^{+}$ and $p^{-}$ are as in the second part of the lemma, we may regard $Z$ as $L^{\infty }(\mu )$ for some positive measure $\mu$ and then the existence of elements $c_+$ and $c_-$ in $Z$ satisfying $0\leq c_+\leq p_+$, $0\leq c_-\leq p^{-}$ and $\rho _+|Z=c_+\omega _+|Z$, $\rho _-|Z=c_-\omega _-|Z$ follows easily, so we will verify here only the last equality in the lemma. The condition $\rho |Z=\omega |Z$ can be written as $(c_+\omega _+-c_-\omega _-)|Z=(\omega _+-\omega _-)|Z$, hence $(1-c_+)\omega _+|Z=(1-c_-)\omega _-|Z$. But $\omega _+=p^{+}\omega _+$ and $\omega _-=p^{-}\omega _-$, since $p^{+}$ and $p^{-}$ are the support projections of $\omega _+|Z$ and $\omega _-|Z$, hence the required equality follows.

Proof. Since maps in ${{\rm E}(\mathcal {R})}$ are unital and completely positive, they are also completely contractive. Each map in ${{\rm E}(\mathcal {R})}$ is of the form $\psi (x)=\sum \nolimits _{i=1}^{n}a_i^{*}xa_i$, where $a_i\in \mathcal {R}$ and $\sum \nolimits _{i=1}^{n}a_i^{*}a_i=1$, hence weak* continuous and the corresponding map $\psi _{\sharp }$ on the predual $\mathcal {R}_{\sharp }$ of $\mathcal {R}$ is given by $\psi _{\sharp }(\omega )=\sum \nolimits _{i=1}^{n}a_i\omega a_i^{*}$ and is a $\mathcal {Z}$-module map with $\|\psi _{\sharp }\|=\|\psi \|=1$. Hence $\|c\psi _{\sharp }(\omega )\|=\|\psi _{\sharp }(c\omega )\|\leq \|c\omega \|$ for each $c\in \mathcal {Z}$. This means that the inequality $\|(c \omega )\circ \psi \|\leq \|c\omega \|$ holds for all $\psi \in {{\rm E}(\mathcal {R})}$, hence also for all $\psi$ in the point-weak* closure of ${{\rm E}(\mathcal {R})}$ since $c\omega$ is weak* continuous. If $\psi$ is a weak* continuous such map and $\rho =\psi _{\sharp }(\omega )$, then $\|c\rho \|=\|(c\omega )\circ \psi \|\leq \|c\omega \|$. Furthermore, since $\psi |\mathcal {Z}={{\rm id}}$ for each such map $\psi$, it follows that $\rho |\mathcal {Z}=\omega |\mathcal {Z}$ for each $\rho \in \overline {\omega \circ {{\rm E}(\mathcal {R})}}$.

Assume now that the condition (2.4) holds. Decompose each of the functionals $\omega _+,\,\omega _-,\,\rho _+,\,\rho _-$ as described in (2.3), so that

\[ \omega=(\omega_+|\mathcal{Z})\circ\omega_{\mathcal{Z}}^{+}-(\omega_-|\mathcal{Z})\circ\omega_{\mathcal{Z}}^{-}\ \mbox{and}\ \rho=(\rho_+|\mathcal{Z})\circ\rho_{\mathcal{Z}}^{+}-(\rho_-|\mathcal{Z})\circ\rho_{\mathcal{Z}}^{-}, \]

where $\rho _{\mathcal {Z}}^{+},\,\rho _{\mathcal {Z}}^{-},\,\omega _{\mathcal {Z}}^{+},\,\omega _{\mathcal {Z}}^{-}$ are $\mathcal {Z}$-module homomorphisms from $\mathcal {R}$ to $\mathcal {Z}$ such that $p^{+}:=\omega _{\mathcal {Z}}^{+}(1)$ and $p^{-}:=\omega _{\mathcal {Z}}^{-}(1)$ are the support projections of $\omega _+|\mathcal {Z}$ and $\omega _-|\mathcal {Z}$. Let $p_+$ and $p_-$ be the support projections of $\omega _+$ and $\omega _-$. Observe that $p_+\leq p^{+}$ and $p_-\leq p^{-}$. (Namely, $\omega _+ (1-p^{+})=(\omega _+|\mathcal {Z})(1-p^{+})=0$ implies that $1-p^{+}\leq 1-p_+$, hence $p_+\leq p^{+}$.) By Lemma 2.2, there exists $c_+,\, c_-\in \mathcal {Z}$ such that $0\leq c_+\leq p^{+}$, $0\leq c_-\leq p^{-}$,

(2.5)\begin{equation} \rho_+|\mathcal{Z}=c_+\omega_+|\mathcal{Z},\ \rho_-|\mathcal{Z}=c_-\omega_-|\mathcal{Z}\ \mbox{and}\ (p^{+}-c_+)\omega_+|\mathcal{Z}=(p^{-}-c_-)\omega_-|\mathcal{Z}. \end{equation}

When we first tried to find a map $\psi$ satisfying the requirements of the theorem to be of the form $\psi =a\rho _{\mathcal {Z}}^{+}+b\rho _{\mathcal {Z}}^{-}$, where $a,\,b\in \mathcal {R}_+$, we found that it is not always possible to simultaneously satisfy the conditions $\psi (1)=1$ and $\omega \circ \psi =\rho$ by maps of such a form. But after several attempts we arrived to the following map:

(2.6)\begin{equation} \psi=c_+p_+\rho_{\mathcal{Z}}^{+}+(1-c_+p_+)(\rho_{\mathcal{Z}}^{-}+(1-p^{-})\theta). \end{equation}

Here, $\theta$ is any fixed normal positive unital $\mathcal {Z}$-module map from $\mathcal {R}$ to $\mathcal {Z}$. (Such a map exists even on $\mathcal {Z}^{\prime }\supseteq \mathcal {R}$ since $\mathcal {Z}^{\prime }$ is of type $I$, hence isomorphic to a direct sum of matrix algebras of the form ${{\rm M}}_n(\mathcal {Z})$, where $n$ can be infinite.) This map $\psi$ is positive, weak* continuous, $\mathcal {Z}$-module map, with the range contained in the commutative C$^{*}$-algebra generated by $\mathcal {Z}\cup \{p_+\}$, hence completely positive. We can immediately verify that $\psi$ is also unital:

\[ \psi(1)=c_+p_+\rho_{\mathcal{Z}}^{+}(1)+(1-c_+p_+)(\rho_{\mathcal{Z}}^{-}(1)+1-p^{-})=c_+p_+{+}(1-c_+p_+)(p^{-}+1-p^{-})=1. \]

Now, we are going to compute

(2.7)\begin{equation} \psi_{\sharp}(\omega)=\omega\circ\psi=(\omega_+|\mathcal{Z})\circ\omega_{\mathcal{Z}}^{+}\circ\psi-(\omega_-|\mathcal{Z})\circ\omega_{\mathcal{Z}}^{-}\circ\psi. \end{equation}

For this, first observe that if $f,\,g:\mathcal {R}\to \mathcal {Z}$ are $\mathcal {Z}$-module maps and $a\in \mathcal {R}$, then $(f\circ (ag))(x)=f(ag(x))=f(a)g(x)=(f(a)g)(x)$, that is, $f\circ (ag)=f(a)g$. Note also that $p^{+}\rho _{\mathcal {Z}}^{+}=\rho _{\mathcal {Z}}^{+}$ and $p^{-}\rho _{\mathcal {Z}}^{-}=\rho_{\mathcal{Z}}^{-}$ since Lemma 2.2 implies that the support projection of $\rho _+|\mathcal {Z}$ is dominated by the support projection of $\omega _+|\mathcal {Z}$ and similarly for $\rho _-|\mathcal {Z}$ and $\omega _-|\mathcal {Z}$. From the definition (2.6) of $\psi$ and using that $\omega _{\mathcal {Z}}^{+}$ and $\omega _{\mathcal {Z}}^{-}$ are $\mathcal {Z}$-module maps with ranges contained in $\mathcal {Z}$ and mutually orthogonal support projections $p_+$ and $p_-$ (which are just the support projections of $\omega _+$ and $\omega _-$, respectively), we now compute

(2.8)\begin{align} \omega_{\mathcal{Z}}^{+}\circ\psi& =\omega_{\mathcal{Z}}^{+}(c_+p_+)\rho_{\mathcal{Z}}^{+}+\omega_{\mathcal{Z}}^{+}(1-c_+p_+)(\rho_{\mathcal{Z}}^{-}+(1-p^{-})\theta)\\ & =c_+\rho_{\mathcal{Z}}^{+}+(p^{+}-c_+)(\rho_{\mathcal{Z}}^{-}+(1-p^{-})\theta) \nonumber \end{align}

and similarly

(2.9)\begin{equation} \omega_{\mathcal{Z}}^{-}\circ\psi=\omega_{\mathcal{Z}}^{-}(1-c_+p_+)\rho_{\mathcal{Z}}^{-}=p^{-}\rho_{\mathcal{Z}}^{-}=\rho_{\mathcal{Z}}^{-}. \end{equation}

From (2.7), (2.8) and (2.9) we have, using also (2.5) and (2.6),

\begin{align*} \omega\circ\psi& =(\omega_+|\mathcal{Z})\circ[c_+\rho_{\mathcal{Z}}^{+}+(p^{+}-c_+)(\rho_{\mathcal{Z}}^{-}+(1-p^{-})\theta)]-(\omega_-|\mathcal{Z})\circ\rho_{\mathcal{Z}}^{-}\\ & =(c_+\omega_+|\mathcal{Z})\circ\rho_{\mathcal{Z}}^{+}+[(p^{+}-c_+)\omega_+|\mathcal{Z}]\circ\rho_{\mathcal{Z}}^{-}+[(p^{+}-c_+)(1-p^{-})\omega_+|\mathcal{Z}]\circ\theta\\ & \quad-(\omega_-|\mathcal{Z})\circ\rho_{\mathcal{Z}}^{-}\\ & =(\rho_+|\mathcal{Z})\circ\rho_{\mathcal{Z}}^{+}+[(p^{-}-c_-)\omega_-|\mathcal{Z}]\circ\rho_{\mathcal{Z}}^{-}+[(1-p^{-})(p^{-}-c_-)p^{-}\omega_-|\mathcal{Z}]\circ\theta\\ & \quad -(p^{-}\omega_-|\mathcal{Z})\circ\rho_{\mathcal{Z}}^{-}\\ & =\rho_+{-}(c_-\omega_-|\mathcal{Z})\circ\rho_{\mathcal{Z}}^{-}=\rho_+{-}(\rho_-|\mathcal{Z})\circ\rho_{\mathcal{Z}}^{-}=\rho_+{-}\rho_-{=}\rho. \end{align*}

It follows from Theorem 2.1 that $\psi$ is in the point-weak* closure of ${{\rm E}(\mathcal {R})}$. Thus, $\rho =\omega \circ \psi$ is in the weak closure of the convex set $\omega \circ {{\rm E}(\mathcal {R})}$, which is the same as the norm closure by the Hahn–Banach theorem and the fact that $\mathcal {R}$ is the dual of $\mathcal {R}_{\sharp }$.

Proof. By Theorem 2.3, we only need to verify that the condition $\rho |\mathcal {Z}=\omega |\mathcal {Z}$ implies that $\|c\rho \|\leq \|c\omega \|$ for all $c\in \mathcal {Z}_+$ and conversely. Since $\omega$ and $\rho$ are positive, we have $\|c\rho \|=(c\rho )(1)=\rho (c)$ and $\omega (c)=\|c\omega \|$ for all $c\in \mathcal {Z}_+$. If $\|c\rho \|\leq \|c\omega \|$ for all $c\in \mathcal {Z}_+$, then $\rho (c)\leq \omega (c)$. Applying this to $1-c$ instead of $c$, where $0\leq c\leq 1$, it follows that $\rho (c)=\omega (c)$ for all such $c$. But such elements span $\mathcal {Z}$, hence it follows that $\rho |\mathcal {Z}=\omega |\mathcal {Z}$ if and only if $\|c\rho \|\leq \|c\omega \|$ for all $c\in \mathcal {Z}_+$.

Proof. (i)$\Rightarrow$(ii) If $\rho =\omega \circ \phi$, where $\phi$ is of the form (2.10), then let $\tilde {\omega }$ be any state on ${{\rm B}(\mathcal {H}_{\pi })}$ extending $\omega \circ \pi ^{-1}$, let $\tilde {\phi }$ be the map on ${{\rm B}(\mathcal {H}_{\pi })}$ defined by $\tilde {\phi }(x)=\sum \nolimits _{j\in \mathbb {J}}\pi (a_j^{*})x\pi (a_j)$ and set $\tilde {\rho }=\tilde {\omega }\circ \tilde \phi$. Then, $\tilde {\phi }(x)=x$ for each $x\in \pi (\mathcal {R})^{\prime }$, hence $\tilde {\rho }|\pi (\mathcal {R})^{\prime }=\tilde {\omega }|\pi (\mathcal {R})^{\prime }$. Moreover, $\tilde {\rho }$ extends $\rho \circ \pi ^{-1}$.

(ii)$\Rightarrow$(iii) Take for $\pi$ a faithful normal representation on a Hilbert space $\mathcal {H}$ such that $\omega$ is the restriction of a vector state $\tilde {\omega }$ on ${{\rm B}(\mathcal {H})}$. (For example, $\mathcal {R}$ may be in the standard form [Reference Takesaki28, Chapter IX] so that all normal states on $\mathcal {R}$ and $\mathcal {R}^{\prime }$ are vector states.) For simplicity of notation, we may assume that $\mathcal {R}\subseteq {{\rm B}(\mathcal {H})}$, that is, $\pi ={{\rm id}}$. Then, with $\tilde {\rho }$ as in (ii), we have $\tilde {\rho }|\mathcal {R}=\rho$ and $\tilde {\rho }|\mathcal {R}^{\prime }=\tilde {\omega }|\mathcal {R}^{\prime }$.

(iii)$\Rightarrow$(i) Assume that $\mathcal {R}$ is represented faithfully on a Hilbert space $\mathcal {H}$ such that $\omega$ is the restriction of a vector state $\tilde {\omega }$ on ${{\rm B}(\mathcal {H})}$ and that $\tilde {\rho }$ is a normal state on ${{\rm B}(\mathcal {H})}$ such that $\tilde {\rho }|\mathcal {R}=\rho$ and $\tilde {\rho }|\mathcal {R}^{\prime }=\tilde {\omega }|\mathcal {R}^{\prime }$. Let $\xi \in \mathcal {H}$ be such that $\tilde {\omega }(x)=\langle x\xi,\,\xi \rangle$ ($x\in {{\rm B}(\mathcal {H})}$). As a normal state, $\tilde {\rho }$ is of the form

\[ \tilde{\rho}(x)=\langle x^{(\infty)}\eta,\eta\rangle\quad (x\in{{\rm B}(\mathcal{H})}), \]

where $x^{(\infty )}$ denotes the direct sum of countably many copies of $x$ acting on the direct sum $\mathcal {H}^{\infty }$ of countably many copies of $\mathcal {H}$ and $\eta \in \mathcal {H}^{\infty }$. Now, from $\tilde {\omega }(x)=\tilde {\rho }(x)$ for all $x\in \mathcal {R}^{\prime }$, we have

\[ \langle x\xi,\xi\rangle=\langle x^{(\infty)}\eta,\eta\rangle\quad (x\in(\mathcal{R}^{\prime})). \]

Replacing $x$ by $x^{*}x$, it follows that there exists an isometry $u:[\mathcal {R}^{\prime }\xi ]\to [(\mathcal {R}^{\prime })^{(\infty )}\eta ]$ such that $u\xi =\eta$ and $uy=y^{(\infty )}u$ for all $y\in \mathcal {R}^{\prime }$. This $u$ can be extended to a partial isometry from $\mathcal {H}$ into $\mathcal {H}^{\infty }$, denoted again by $u$, by declaring it to be $0$ on the orthogonal complement of $[\mathcal {R}^{\prime }\xi ]$ in $\mathcal {H}$. Then, $u$ intertwines the identity representation ${{\rm id}}$ of $\mathcal {R}^{\prime }$ and the representation ${{\rm id}}^{\infty }$ and, is therefore, a column $(u_j)$, where $u_j\in \mathcal {R}$. For $r\in \mathcal {R}$, we have

\[ \rho(r)=\tilde{\rho}(r)=\langle r^{(\infty)}\eta,\eta\rangle=\langle r^{(\infty)}u\xi,u\xi\rangle=\langle u^{*}r^{(\infty)}u\xi,\xi\rangle=\omega(u^{*}r^{(\infty)}u). \]

Thus, $\rho =\omega \circ \psi$, where $\psi$ is a map on $\mathcal {R}$, defined by $\psi (r)=u^{*}r^{(\infty )}u=\sum \nolimits _ju_j^{*}ru_j$. This map $\psi$ is not necessarily unital, but from

\[ \omega(1)=1=\rho(1)=\omega(\psi(1))=\omega\bigg(\sum_ju_j^{*}u_j\bigg)=\omega(u^{*}u)\text{ and } u^{*}u\leq 1 \]

we infer that $1-u^{*}u\leq 1-p$, where $p$ is the support projection of $\omega$. Hence, $p\leq u^{*}u$ and we may replace $\psi$ by the unital map $\phi$ defined by $\phi (r)=p\psi (r)p+(1-p)r(1-p)$, which satisfies $\omega \circ \phi =\omega \circ \psi =\rho$ and has the required form:

\[ \psi(r)=\sum_jpu_j^{*}ru_jp+p^{{\perp}}rp^{{\perp}},\quad\text{where }\sum_jpu_j^{*}u_jp+p^{{\perp}}p^{{\perp}}=pu^{*}up+p^{{\perp}}=1. \]

The equivalence (i)$\Leftrightarrow$(iv) is proved by similar arguments and we will omit the details, just note that $\mathcal {R}\cong \pi _{\omega }(\mathcal {R})\oplus \ker \pi _{\omega }$.

Proof. To prove the non-trivial direction of the theorem, suppose that the condition (A) is satisfied, but that $\rho \notin \overline {\omega \circ {{\rm E}(A)}}$. Then, by the Hahn–Banach theorem, there exist $h\in A_h$ and $\alpha,\,\delta \in \mathbb {R}$, $\delta >0$, such that

(3.1)\begin{equation} \omega(\psi(h))\leq\alpha\ \forall\psi\in{{\rm E}(A)}\text{ and } \rho(h)\geq\alpha+\delta. \end{equation}

Since $\rho |Z=\omega |Z$, in particular $\rho (1)=\omega (1)$, we may replace $h$ by $h+\gamma 1$ (and $\alpha$ with $\alpha +\gamma \omega (1)$) for a sufficiently large $\gamma \in \mathbb {R}$ and thus assume that $h$ is positive in invertible. Given $\varepsilon >0$, let $a\in A_h$ be such that

\[{-}1\leq a\leq 1\text{ and } \omega_+(a)-\omega_-(a)=\omega(a)>\|\omega\|-\varepsilon=\omega_+(1)+\omega_-(1)-\varepsilon. \]

By a well-known argument, which we now recall, this implies the relations (3.2). Namely, from the above, we have $\omega _+(1-a)<-\omega _-(1+a)+\varepsilon$ and (since $1-a\geq 0$ and $1+a\geq 0$) this implies that $\omega _+(1-a)<\varepsilon$ and $\omega _-(1+a)\leq \varepsilon$. Thus $\omega _+(a_+)\geq \omega _+(a)>\omega _+(1)-\varepsilon$ and $\omega _-(a_+)=\omega _-(1+a)-\omega _-(1-a_-)\leq \omega _-(1+a)\leq \varepsilon$. In conclusion,

(3.2)\begin{equation} \omega_+(1-a_+)<\varepsilon\text{ and } \omega_-(a_+)\leq\varepsilon. \end{equation}

For each $t\in \check {A}$ let $m(t)$ and $M(t)$ be the smallest and the largest point in the spectrum $\sigma (h(t))$ of $h(t)\in A/t$. Since $\check {A}$ is Hausdorff by assumption, the two functions $M$ and $m$ (given by $M(t)=\|h(t)\|$ and $m(t)=\|h(t)^{-1}\|^{-1}$) are continuous [Reference Pedersen26, 4.4.5] and therefore define elements of the centre $Z$ of $A$ by the Dauns–Hoffman theorem. Set

\[ b=Ma_+{+}m(1-a_+). \]

For each $t\in \check {A}$, the spectrum of $b(t)$ is $\sigma (m(t)1+(M(t)-m(t))a_+(t))$ and is contained in $m(t)+(M(t)-m(t))[0,\,1]\subseteq [m(t),\,M(t)]$ since $\sigma (a_+(t))\subseteq \sigma (a_+)\subseteq [0,\,1]$. Thus, the numerical range $W(b(t))$ of $b(t)$ (which for normal elements coincides with the convex hull of the spectrum) is contained in $W(h(t))=[m(t),\,M(t)]$. Therefore, by [Reference Magajna23, 4.1], $b$ is in the norm closure of the set $\{\psi (h):\, \psi \in {{\rm E}(A)}\}$, hence

(3.3)\begin{equation} \omega(b)\leq\alpha \end{equation}

by the first relation in (3.1). On the other hand, we can estimate $\omega (b)$ as

\begin{align*} \omega(b)& =\omega_+(Ma_+)-\omega_-(Ma_+)+\omega_+(m(1-a_+))-\omega_-(m(1-a_+))\\ & =\omega_+(M)-\omega_-(m)-\omega_+((M-m)(1-a_+))-\omega_-((M-m)a_+)\\ & \geq\omega_+(M)-\omega_-(m)-\|M-m\|(\omega_+(1-a_+)+\omega_-(a_+)) \end{align*}

(since $0\leq (M-m)(1-a_+)\leq \|M-m\|(1-a_+)$ and $0\leq (M-m)a_+\leq \|M-m\|a_+$)

\[{\geq}\omega_+(M)-\omega_-(m)-2\|M-m\|\varepsilon\ \ \mbox{(by (3.2))}.\]

Thus, by (3.1), (3.3) and since $m\leq h\leq M$ implies that $\rho _+(h)\leq \rho _+(M)$ and $\rho _-(h)\geq \rho _-(m)$, we have now

\[ \omega_+(M)-\omega_-(m)-2\varepsilon\|M-m\|+\delta\leq\rho(h)=\rho_+(h)-\rho_-(h)\leq\rho_+(M)-\rho_-(m). \]

This can be rewritten as

(3.4)\begin{equation} (\omega_+{-}\rho_+)(M)\leq(\omega_-{-}\rho_-)(m)+2\varepsilon\|M-m\|-\delta \end{equation}

or (since $\omega |Z=\rho |Z$ implies that $(\omega _--\rho _-)|Z=(\omega _+-\rho _+)|Z$ and since $M,\,m\in Z$)

\[ (\omega_+{-}\rho_+)(M-m)\leq 2\varepsilon\|M-m\|-\delta. \]

Since this holds for all $\varepsilon >0$, by choosing small enough $\varepsilon$, it follows that $(\omega _+-\rho _+) (M-m)<0$. But, $Z\ni M-m\geq 0$ and $\omega _+|Z\geq \rho _+|Z$ by Lemma 2.2, hence $(\omega _+-\rho _+) (M-m)\geq 0$, which is a contradiction.

Proof. By definition, the central carrier $z$ of $h$ is the infimum of all $c$ in $\mathcal {Z}$ such that $h\leq c$. If $\Delta$ is the maximal ideal space of $\mathcal {Z}$, then $z$ corresponds (via the Gelfand isomorphism) to the function $\Delta \ni t\mapsto \|h(t)\|$, where $h(t)$ is the coset of $h$ in $\mathcal {R}/t\mathcal {R}$. (This function is continuous by [Reference Glimm14].) Thus, we will regard $z$ as a function on $\Delta$. Let $[m,\,M]$ be an interval containing the range of $z$, where $m\geq 0$ and $M=\|h\|=\|z\|$. Given $a\in A_+$, the set $U=\{t\in \delta :\ a(t)\ne 0\}$ is open since the function $\Delta \ni t\mapsto \|a(t)\|\in \mathbb {R}$ is continuous. The weak* closure of the ideal generated by $a$ in $\mathcal {R}$ is of the form $p\mathcal {R}$ for a unique projection $p\in \mathcal {Z}$ and $p$ is open by definition. Since the quotient algebras $\mathcal {R}/t\mathcal {R}$ have only scalars in their centres, $p(t)=1$ for each $t\in U$, hence also for each $t\in \overline {U}$ by continuity, so $p\geq q$, where $q\in \mathcal {Z}$ is the projection that corresponds to the characteristic function of $\overline {U}$. But from the definition of $U$, we see that $qa=a$ and this implies that $qb=b$ for each $b$ in the ideal generated by $a$. Hence, $qp=p$ and it follows that $q=p$. In particular, for each $r\in \mathbb {R}_+$ the projection that corresponds to the closure of the set $U_r=\{t\in \Delta :\, z(t)>r\}$ is open since $U_r$ is just the set $\{t\in \Delta :\, a(t)\ne 0\}$, where $a=(h-r)_+$. (This has been observed already by Halpern in [Reference Halpern18, proof of Lemma 6].) Given $\varepsilon >0$, for each $k\in \mathbb {N}$ let $p_k$ be the projection corresponding to the closure of the set $U_k=\{t\in \Delta :\, z(t)>M-k\varepsilon \}$. Then, $0=p_0\leq p_1\leq p_2\leq \ldots \leq p_n=1$, where $n\in \mathbb {N}$ is such that $M-n\varepsilon < m$ and $M-(n-1)\varepsilon \geq m$. Now, from $1=(p_1-p_0)+(p_2-p_1)+\ldots +(p_n-p_{n-1})$, we have that $F_k:=\overline {U}_{k}\setminus \overline {U}_{k-1}$ are disjoint closed and open sets that cover $\Delta$ and for $t\in F_k$, we have that $M-k\varepsilon \leq z(t)\leq M-(k-1)\varepsilon$. Thus, if we choose in each interval $[M-k\varepsilon,\,M-(k-1)\varepsilon ]$ a point $\lambda _k\geq 0$ and set $c:=\sum \nolimits _{k=1}^{n}\lambda _k(p_k-p_{k-1})$, it follows that $\|z-c\|\leq \varepsilon$. Finally, observe that

\[ c=(\lambda_1-\lambda_2)p_1+(\lambda_2-\lambda_3)p_2+\ldots+(\lambda_{n-1}-\lambda_n)p_{n-1}+\lambda_np_n \]

is a linear combination with positive coefficients of open projections.

Proof. Evidently, $\rho \in \overline {\omega \circ {{\rm E}(A)}}$ implies that $\|\rho |J\|\leq \|\omega |J\|$ for each ideal $J$ in $A$ since maps in ${{\rm E}(A)}$ are contractive and preserve ideals. For the converse, suppose that $\rho \notin \overline {\omega \circ {{\rm E}(A)}}$. Then, by the Hahn–Banach theorem there exist $h\in A_h$ and $\alpha \in \mathbb {R}$ such that (3.1) holds, that is $\omega (\psi (h))\leq \alpha$ for all $\psi \in {{\rm E}(A)}$, while $\rho (h)>\alpha.$ Replacing $h$ by $h+\beta 1$ for a sufficiently large $\beta \in \mathbb {R}$ (and consequently $\alpha$ by $\alpha +\beta$), we may assume that $h$ is positive.

Let $\mathcal {R}$ be the universal von Neumann envelope of $A$ and denote the unique weak* continuous extensions of $\omega$ and $\rho$ to $\mathcal {R}$ by the same two letters. We will use the same notation as in the proof of Lemma 3.3. Thus, $z$ is the infimum of all $c$ in $\mathcal {Z}$ such that $h\leq c$. Since $W(z(t)1)=\{z(t)\}\subseteq W(h(t))$ for each $t\in \Delta$, it follows by [Reference Magajna23, 3.3] that $z\in \overline {{{\rm co}}_{\mathcal {R}}}(h)$ (= the weak* closure of the $\mathcal {R}$-convex hull of $h$), hence by the first relation in (3.1)

(3.5)\begin{equation} \omega(z)\leq\alpha, \end{equation}

since each map $\psi$ of the form $x\mapsto \sum \nolimits _ib_i^{*}xb_i$ ($b_i\in \mathcal {R}$, $\sum \nolimits _ib_i^{*}b_i=1$) can be approximated by maps of the form $x\mapsto \sum \nolimits _ia_i^{*}xa_i$ ($a_i\in A$, $\sum \nolimits _ia_i^{*}a_i=1$). (This follows by using the Kaplansky density theorem in ${{\rm M}}_n(\mathcal {R})$ to approximate the column $b=(b_1,\,\ldots,\,b_n)^{T}$ by $(a_1,\,\ldots,\,a_n)^{T}$.) Since $\omega$ and $\rho$ are states, the hypothesis $\|\rho |J\|\leq \|\omega |J\|$ for each ideal $J$ in $A$ means that $\rho (p)\leq \omega (p)$ for each open projection $p\in \mathcal {Z}$. Then, it follows by Lemma 3.3 that $\rho (z)\leq \omega (z)$. But from $h\leq z$ and using (3.5), we have now that $\rho (h)\leq \rho (z)\leq \omega (z)\leq \alpha$, which is in contradiction with the previously established relation $\rho (h)>\alpha$.

Proof. Suppose that $\rho \in \overline {\omega \circ {{\rm E}(A)}}$ and let $(\psi _k)$ be a net in ${{\rm E}(A)}$ such that $\rho (a)=\lim _k\omega (\psi _k(a))$ for all $a\in A$. Extend $\omega$, $\rho$ and each $\psi _k$ weak* continuously to the universal von Neumann envelope $\mathcal {R}$ of $A$ and denote the extensions by the same symbols. Let $\psi$ be a weak* limit point of the net $(\psi _k)$ and note that $\psi$ is a unital completely positive (hence contractive) module map over the centre $\mathcal {Z}$ of $\mathcal {R}$. Set $\rho _1=\omega _+\circ \psi |A$ and $\rho _2=\omega _-\circ \psi |A$. Then $\rho _1\in \overline {\omega _+\circ {{\rm E}(A)}}$, $\rho _2\in \overline {\omega _-\circ {{\rm E}(A)}}$ and $\rho =\omega \circ \psi |A=\rho _1-\rho _2$. This proves the inclusion $\overline {\omega \circ {{\rm E}(A)}}\subseteq \overline {\omega _+\circ {{\rm E}(A)}}-\overline {\omega _-\circ {{\rm E}(A)}}$.

To prove the reverse inclusion, suppose that $\rho _1\in \overline {\omega _+\circ {{\rm E}(A)}}$ and $\rho _2\in \overline {\omega _-\circ {{\rm E}(A)}}$. Then, there exist nets of maps $\phi _k$ and $\psi _k$ in ${{\rm E}(A)}$ such that $\rho _1=\lim _k\omega _+\circ \phi _k$ and $\rho _2=\lim _k\omega _-\circ \psi _k$. Let $p$ and $q$ be the support projections in $\mathcal {R}$ of $\omega _+$ and $\omega _-$ (where $\omega _+$ and $\omega _-$ have been weak* continuously extended to $\mathcal {R}$). Let $(a_n)$ be a net of positive contractions in $A$ strongly converging to $p$ in $\mathcal {R}$, set $b_n=\sqrt {1-a_n^{2}}$ and define maps $\phi _{k,n}$ and $\psi _{k,n}$ on $A$ by

\[ \phi_{k,n}(x)=a_n\phi_k(x)a_n\text{ and }\psi_{k,n}(x)=b_n\psi_k(x)b_n. \]

The nets $(\omega _+(b_n^{2}))=(\omega _+(1-a_n^{2}))$ and $(\omega _-(a_n^{2}))= (\omega _-(1-b_n^{2}))$ all converge to $0$. From this, we will verify in the following by using the Cauchy–Schwarz inequality for positive functionals that $\lim _{k,n}\omega _+\circ \phi _{k,n}=\rho _1$, $\lim _{k,n}\omega _-\circ \psi _{k,n}=\rho _2$, $\lim _{k,n}\omega _+\circ \psi _{k,n}=0$ and $\lim _{k,n}\omega _-\circ \phi _{k,n}=0$ pointwise on $A$, hence

\[ \rho:=\rho_1-\rho_2=\lim_{k,n}[(\omega_+{-}\omega_-)\circ(\phi_{k,n}+\psi_{k,n})]=\lim_{k,n}\omega\circ\theta_{k,n}, \]

where $\theta _{k,n}:=\phi _{k,n}+\psi _{k,n}$. Evidently each $\theta _{k,n}$ is elementary completely positive map and also unital since $\theta _{k,n}(1)=a_n\phi _k(1)a_n+b_n\psi _k(1)b_n=a_n^{2}+b_n^{2}=1$. Thus, $\rho \in \overline {\omega \circ {{\rm E}(A)}}$, verifying the inclusion $\overline {\omega \circ {{\rm E}(A)}}\supseteq \overline {\omega _+\circ {{\rm E}(A)}}-\overline {\omega _-\circ {{\rm E}(A)}}$. Now, we will verify that $\lim _{k,n}\omega _+\phi _{k,n}=\rho _1$, the verification of the other three limits that we have used is similar. For each $x\in A$, we estimate

\begin{align*} |\rho_1(x)-\omega_+(\phi_{k,n}(x))|& =|\rho_1(x)-\omega_+(a_n\phi_k(x)a_n)|\\ & \leq|\rho_1(x)-\omega_+(\phi_k(x))|+|\omega_+((1-a_n)\phi_k(x))|\\& \quad +|\omega_+(a_n\phi_k(x)(1-a_n))|\\ & \leq|\rho_1(x)-\omega_+(\phi_k(x))|+\omega_+((1-a_n)^{2})^{1/2}\omega_+(\phi_k(x)^{*}\phi_k(x))^{1/2}\\ & \quad+\omega_+(\phi_k(x)^{*}a_n^{2}\phi_k(x))^{1/2}\omega_+((1-a_n)^{2})^{1/2}\\ & \leq|\rho_1(x)-\omega_+(\phi_k(x))|+2\omega_+((1-a_n)^{2})^{1/2}\|\omega_+\|^{1/2}\|x\|. \end{align*}

Both terms in the last line of the above expression converge to $0$.

Proof. Suppose that $\rho \in \overline {\omega \circ {{\rm E}(A)}}$. Using the notation introduced in the first part of the proof of Lemma 3.6, we have observed that the map $\psi$ on $\mathcal {R}$ introduced in that proof is a contractive unital $\mathcal {Z}$-bimodule map. Thus, for any ideal $J$ in $A$, if $p\in \mathcal {Z}$ is the projection satisfying $\overline {J}=p\mathcal {R}$, then $\psi (\overline {J})=\psi (p\mathcal {R})=p\psi (\mathcal {R})\subseteq \overline {J}$. Since $\omega$ (and hence also $\omega _+$ and $\omega _-$) are weak* continuous on $\mathcal {R}$, we have $\|\omega _+|\overline {J}\|=\|\omega _+|J\|$ and $\|\omega _-|\overline {J}\|=\|\omega _-|J\|$. With $\rho _1=\omega _+\circ \psi |A$ and $\rho _2=\omega _-\circ \psi |A$ (as in the proof of Lemma 3.6), we have $\rho =\rho _1-\rho _2$, $\rho _1(1)=\omega _+(1)$, $\rho _2(1)=\omega _-(1)$,

\[ \|\rho_1|J\|\leq\|\omega_+{\circ}\psi|\overline{J}\|\leq\|\omega_+|\overline{J}\|=\|\omega_+|J\| \]

and similarly $\|\rho _2|J\|\leq \|\omega _-|J\|$. Therefore, also

\[ \|\rho|J\|=\|\rho_1|J-\rho_2|J\|\leq\|\rho_1|J\|+\|\rho_2|J\|\leq\|\omega_+|J\|+\|\omega_-|J\|=\|\omega|J\|. \]

Conversely, assume the existence of positive functionals $\rho _1$ and $\rho _2$ on $A$ satisfying the norm inequalities in condition (B). Then, by Theorem 3.4 $\rho _1\in \overline {\omega _+\circ {{\rm E}(A)}}$ and $\rho _2\in \overline {\omega _-\circ {{\rm E}(A)}}$, hence by Lemma 3.6 $\rho \in \overline {\omega \circ {{\rm E}(A)}}$.

Proof. If $\rho \in \overline {\omega \circ {{\rm E}(A)}}$, then by Theorem 3.4 $\|\rho |J\|\leq \|\omega |J\|$ for each ideal $J$ in $A$. Denoting by $p$ the projection in $\mathcal {R}:=A^{\sharp \sharp }$ such that $\overline {J}=p\mathcal {R}$, this means that $\rho (p)\leq \omega (p)$ for each open central projection $p$, where $\omega$ and $\rho$ have been weak* continuously extended to $\mathcal {R}$. Since $A$ is liminal, such projections are strongly dense in the set of all central projections by [Reference Digernes and Halpern10], hence it follows that $\rho (p^{\perp })\leq \omega (p^{\perp })$. Since $\rho (p)+\rho (p^{\perp }=\rho (1)=1=\omega (1)=\omega (p)+\omega (p^{\perp })$, we conclude that $\rho (p)=\omega (p)$, that is $\|\rho |J\|=\|\omega |J\|$. By Theorem 3.4, this implies that $\omega \in \overline {\rho \circ {{\rm E}(A)}}$.

Proof. Suppose that $\omega |M=0$ and let $\rho \in \overline {\omega \circ {{\rm E}(A)}}$. Then, $\rho |M\!=\!0$, hence also $\rho (J)\!=\!0$ for each proper ideal $J$ of $A$ since $J\subseteq M$. Thus, $\|\omega |J\|=\|\rho |J\|$ for each ideal $J$ of $A$, so $\omega \in \overline {\rho \circ {{\rm E}(A)}}$ by Theorem 3.4.

Suppose now that $\omega |M\ne 0$. Let $\rho$ be any state on $A$ such that $\rho |M=0$. Then $\|\rho |J\|\leq \|\omega |J\|$ for all ideals $J$, hence $\rho \in \overline {\omega \circ {{\rm E}(A)}}$ by Theorem 3.4. But $\omega \notin \overline {\rho \circ {{\rm E}(A)}}$ since $\rho |M=0$ and $\omega |M\ne 0$, thus $\omega$ is not maximally mixed.

Proof.

1. (i) Let $D=S(A/J_A)$ and $\omega$ a maximally mixed state on $A$. Suppose that $\omega \notin D$. Then, $\overline {\omega \circ {{\rm E}(A)}}\cap D=\emptyset$, otherwise this intersection would be a weak* closed proper ${{\rm E}(A)}$-invariant subset of $\overline {\omega \circ {{\rm E}(A)}}$, which would contradict the fact that $\omega$ is maximally mixed. Thus, by the Hahn–Banach theorem, there exist $\alpha,\,\beta \in \mathbb {R}$ and $h\in A_h$ such that

   (4.1)\begin{equation} \rho(h)\leq\alpha\ \forall\rho\in D\text{ and } \omega(\psi(h))\geq\beta>\alpha\ \forall\psi\in{{\rm E}(A)}. \end{equation}
   Replacing $h$ by $h+\gamma 1$ for a sufficiently large $\gamma \in \mathbb {R}_+$ (and modifying $\alpha,\, \beta$), we may assume that $h$ is positive. Then, the first relation in (4.1) means that $\|\dot {h}\|\leq \alpha$, where $\dot {h}$ denotes the coset of $h$ in $A/J_A$. The (algebraic) numerical range $W_{A/J_A}(h)$ of $\dot {h}$ is an interval, say $[c,\,d]$, contained in the numerical range $W_A(h)$ of $h$, which is an interval, say $[a,\,b]$; note that $a\leq c\leq d=\|\dot {h}\|\leq b=\|h\|$. Let $f:[a,\,b]\to [c,\,d]$ be the function, which act as the identity on $[c,\,d]$, and maps $[a,\,c]$ into $\{c\}$ and $[d,\,b]$ into $\{d\}$. For every proper ideal $K$ in $A$ the quotient $A/(K+J_A)$ is non-zero, for $K$ is contained in a maximal ideal $M$ and hence $K+J_A\subseteq M+J_A=M\ne A$. Since $W_{A/(K+J_A)}(h)\subseteq W_{A/K}(h)\cap W_{A/J_A}(h)$, this intersection is not empty, hence the interval $W_{A/K}(h)$ intersects $[c,\,d]$ and is therefore mapped by $f$ into itself. The numerical range $W_{A/K}(f(h))$ of the coset of $f(h)$ in $A/K$ is just the convex hull of the spectrum $\sigma _{A/K}(f(h))=f(\sigma _{A/K}(h)$, hence $W_{A/K}(f(h))\subseteq f(W_{A/K}(h))\subseteq W_{A/K}(h)$. This inclusion implies that $f(h)\in \overline {{{\rm E}(A)}(h)}$ by [Reference Magajna23], hence $\omega (f(h))>\alpha$ by the second relation in (4.1). Since $\omega$ is a state, it follows that $W_A(f(h))$ intersects $(\alpha,\,\infty )$. But this is a contradiction since $W_A(f(h))$ is the convex hull of the spectrum $\sigma _A(f(h))=f(\sigma _A(h))\subseteq [c,\,d]=[c,\,\|\dot {h}\|]\subseteq [c,\,\alpha ]$. Thus, $\omega \in D$.

2. (ii) By the Chinese remainder theorem [Reference Grove15, 6.3], there is a natural isomorphism $A/\cap _{j=1}^{n}M_j\cong \oplus _{j=1}^{n}A/M_j$, thus we may regard $\omega$ as a state on $\oplus _{j=1}^{n}A/M_j$. Since the algebras $A/M_j$ are simple, all states on them are maximally mixed by Example 4.2. The same then holds for their direct sum, so all states on $A/\cap _{j=1}^{n}M_j$ are maximally mixed and (ii) follows by Remark 4.3.

The set of all states that annihilate some finite intersection of maximal ideals of $A$ is convex and norming for $A/J_A$ (since the natural map $A/J_A\to \oplus _MA/M$, where the sum is over all maximal ideals in $A$, is a monomorphism, thus isometric), hence weak* dense in $S(A/J_A)$ [Reference Kadison and Ringrose20, 4.3.9].

Proof. By Remark 4.3 a state $\omega$ on $A/J_A$ is maximally mixed if and only if it is maximally mixed on $A$. By [Reference Archbold and Gogić3, 3.10], the quotients of weakly central C$^{*}$-algebras are weakly central, so in particular $A/J_A$ is weakly central. In this way, we reduce the proof to the algebra $A/J_A$ (instead of $A$), which has strong radical $0$. Thus, we may assume that $J_A=0$.

Suppose now that $S_m(A)=S(A)$. Then, $S_m(A/P)=S(A/P)$ for each primitive ideal $P$ of $A$ by Remark 4.3. If $M$ is a maximal ideal of $A$ containing $P$, then $A/M$ is a quotient of $A/P$, hence each state $\rho \in S(A/M)$ can be regarded as a state on $A/P$ and therefore can be weak* approximated by convex combinations of vector states on $A/P$, where $A/P$ has been faithfully represented on a Hilbert space. Since $A/P$ is primitive, as a consequence of the Kadison transitivity theorem, each vector state is of the form $x\mapsto \theta (u^{*}xu)$ for a fixed state $\theta$ on $A/P$ with $\theta (M/P)\ne 0$, where $u\in A/P$ is unitary [Reference Kadison and Ringrose20, 5.4.5]. Thus, $\rho \in \overline {\theta \circ {{\rm E}}(A/P)}$. But $\rho (M/P)=0$, while $\theta (M/P)\ne 0$ if $M\ne P$, hence $\theta \notin \overline {\rho \circ {{\rm E}}(A/P)}$ if $M\ne P$. Thus, $\rho$ can not be maximally mixed (on $A/P$ and hence also on $A$) if $P$ is not maximal. This argument, which we have found in [Reference Archbold, Robert and Tikuisis4, proof of 3.15], shows that in general the equality $S_m(A)=S(A)$ can hold only if all primitive ideals containing $J_A$ are maximal. If $A$ is weakly central and by our reduction above $J_A=0$, then the assumption that all primitive ideals are maximal implies that the primitive spectrum $\check {A}$ of $A$ is homeomorphic to the maximal ideal space $\Delta$ of $Z$ (via the map $\check {A}\ni M\mapsto M\cap Z\in \Delta$). Thus, $\check {A}$ is Hausdorff and in this case, Corollary 3.2 shows that all states on $A$ are maximally mixed.

Proof. Let $J$ be an ideal in $\mathcal {R}$ and $K=J\cap \mathcal {Z}$. As an ideal in $\mathcal {Z}$, $K$ can be identified with the set of all continuous functions on $\Delta$ than vanish on some closed subset $\Delta _K$ of $\Delta$, hence $K$ is the intersection of a family $\{t: t\in \Delta _K\}$ of maximal ideals of $\mathcal {Z}$. By [Reference Kadison and Ringrose20, 8.7.15], there exists the largest ideal $J(K)$ in $\mathcal {R}$ such that $J(K)\cap \mathcal {Z}=K$, and it follows from [Reference Kadison and Ringrose20, 8.7.16] that $J(K)=\cap _{t\in \Delta _K}M_t$. Now $J\cap \mathcal {Z}=K$ implies that $J\subseteq J(K)$. Thus, if $\phi$ has the property that $\phi (M_t)\subseteq t$ for all $t\in \Delta$, then $\phi (J)\subseteq \phi (J(K))\subseteq \cap _{t\in \Delta _K}\phi (M_t)\subseteq \cap _{t\in \Delta _K}t=K\subseteq J$.

If $\mathcal {R}$ is properly infinite, then $M_t=\mathcal {R} t+J_{\mathcal {R}}$ for each $t\in \Delta$ by [Reference Kadison and Ringrose20, 8.7.21 (1)]. Thus, if $\phi (J_{\mathcal {R}})=0$, then we have $\phi (M_t)=\phi (\mathcal {R})t\subseteq t$ for all $t\in \Delta$. Conversely, if $\phi (M_t)\subseteq t$ for all $t$, then $\phi (J_{\mathcal {R}})=\phi (\cap _{t\in \Delta }M_t)\subseteq \cap _{t\in \Delta }t=0$.

Proof. By [Reference Magajna24, 2.2] and [Reference Magajna25, 2.1] each completely contractive map $\phi :\mathcal {R}\to \mathcal {Z}\subseteq \mathcal {R}$ which preserves all ideals of $\mathcal {R}$ is in the point-norm closure of maps of the form $x\mapsto a^{*}xb=\sum \nolimits _{j=1}^{n}a_j^{*}xb_j$, where $n\in \mathbb {N}$, $a_j,\,b_j\in \mathcal {R}$, $a:=(a_1,\,\ldots,\,a_n)^{T}$, $b:=(b_1,\,\ldots,\,b_n)$, $\|a\|\leq 1$ and $\|b\|\leq 1$. If $\phi$ is unital, then we can modify such maps to unital maps in the same way as in the proof of Theorem 2.1, which shows that $\phi \in \overline {{{\rm E}(\mathcal {R})}}^{{\rm p.n.}}$.

Proof. Suppose that $\rho \in \overline {\omega \circ {{\rm E}(\mathcal {R})}}$. Then, $\rho |\mathcal {Z}=\omega |\mathcal {Z}=\mu$, hence

\[ \omega=\mu\circ\phi=(\rho|\mathcal{Z})\circ\phi=\rho\circ\phi. \]

By Corollary 4.9 $\phi$ can be approximated in the point-norm topology by a net of maps $\phi _k\in \overline {{{\rm E}(\mathcal {R})}}^{{\rm p.n.}}$. Then, $\omega (x)=\lim _k(\rho (\phi _k(x)))$ for all $x\in \mathcal {R}$. This shows that $\overline {\omega \in \rho \circ {{\rm E}(\mathcal {R})}}$, so $\omega$ is maximally mixed.

Any tracial state $\omega$ annihilates the properly infinite part of $\mathcal {R}$, hence we assume that $\mathcal {R}$ is finite. Then, $\omega =(\omega |\mathcal {Z})\circ \tau$, where $\tau$ is the central trace on $\mathcal {R}$ [Reference Kadison and Ringrose20, 8.3.10]. Since $M_t$ is of the form $M_t=\{a\in \mathcal {R}:\, \tau (a^{*}a)\in t\}$ by [Reference Kadison and Ringrose20, 8.7.17], for $a\in M_t$, we have by the Schwarz inequality $\tau (a)^{*}\tau (a)\leq \tau (a^{*}a)\in t$. This implies that $\tau (a)\in t$. Thus, $\tau (M_t)\subseteq t$, hence $\omega$ is maximally mixed by the first part of the corollary.

Proof. Let $\Phi$ be the universal representation of $\mathcal {R}$, so that $\mathcal {R}^{\sharp \sharp }$ is the weak* closure of $\Phi (\mathcal {R})$. Then, the $*$-homomorphism $\Phi ^{-1}:\Phi (\mathcal {R})\to \mathcal {R}$ can be weak* continuously extended to a $*$-homomorphism $\Psi :\mathcal {R}^{\sharp \sharp }\to \mathcal {R}$; set $\tilde {\omega }=\omega \circ \Psi$ [Reference Kadison and Ringrose20, 10.1.1, 10.1.12]. Let $\tilde {\mathcal {Z}}$ be the centre of $\mathcal {R}^{\sharp \sharp }$. Since $\tilde {\omega }$ is weak* continuous, by [Reference Halpern17] or [Reference Strătilă and Zsidó27, 1.4], there exists a unique $\tilde {\mathcal {Z}}$-module homomorphism $\psi :\mathcal {R}^{\sharp \sharp }\to \tilde {\mathcal {Z}}$ such that $\tilde {\omega }=(\tilde {\omega }|\tilde {\mathcal {Z}})\circ \psi$ and $\psi (1)$ is the support projection $q$ of $\tilde {\omega }|\tilde {\mathcal {Z}}$. It is not hard to verify that $\phi :=(\Psi |\tilde {\mathcal {Z}})\circ \psi \circ \Phi$ has the properties stated in the lemma.