## q-open projections of $C(X)$

Let $A$ be a C*-algebra. We call a projection $p$ in the atomic part of $A$ *q-open* if there is an increasing net of positive elements of $A$ converges to $p$ respect to the strong topology. $\newcommand{\Supp}{\operatorname{Supp}}$ **Lemma 1.** If $\{f_i\}$ is an increasing net of positive functions in the closed unit ball of $\ell^\infty(X)\subset B(\ell^2(X))$, then $f_i$ converges to some $f\in\ell^\infty(X)$ strongly is equivalent to $f_i$ converges to $f$ pointwise. *Proof.* \begin{align*} & f_i\xrightarrow{strongly}f\\ \Leftrightarrow & f_i\xrightarrow{weakly}f\\ \Leftrightarrow & \langle f_i h,h\rangle\to \langle fh,h\rangle, \forall h\in \ell^2(X)\\ \Leftrightarrow & \sum_{x\in X}f_i(x)|h(x)|^2\to \sum_{x\in X}f(x)|h(x)|^2, \forall h\in \ell^2(X)\\ \Leftrightarrow & \sum_{x\in X}(f_i(x)-f)g(x)\to 0, \forall g\in \ell^1(X) \end{align*} Suppose $f_i$ converges to $f$ pointwise. For all $g\in \ell^1(X)$ and $\varepsilon >0$, there is a finite set $F$ of $X$ such that $\sum_{x\notin F}|g(x)|<\varepsilon$ and there is a $i_0$ such that $|f_i(x)-f(x)|<\varepsilon$ for all $x\in F$ whenever $i>i_0$. Hence $|\sum_{x\in X}(f_i(x)-f)g(x)|\leq \varepsilon (\|g\|_1+ \#F)$. **Theorem 2.** The mapping $f\to \Supp f$ is a bejection between q-open projections of $C(X)$ and the open set of $X$, where $X$ is a compact Hausdorff space and $\Supp(f):=\{x\in X|f(x)\neq 0\}$ denotes the support of $f\in \ell^\infty(X)$. *Proof.* Suppose $f$ is a q-open projection of $C(X)$, then there is an increasing net $(f_i)$ of positive continuous functions on $X$ converges to $f$ respect to SOT. This is equivalent to $f_i$ increasingly converges to $f$ pointwise by lemma 1. Therefore, $\Supp f=\cup_i\Supp f_i$ is open as the support of each continuous function is open. Suppose $U$ is an open set of $X$, we will prove that the projection with supprot $U$ in the atomic part $\ell^\infty(X)$ of $C(X)$ is open. For every compact set $F$ contained in $U$, by Urysohn's lemma, there is a continuous function $f_F$ such that $f_F({K})=\{1\}$ and ${\Supp f}\subset U$. For each finite collection $\{F_1, F_2, \cdots, F_n\}$ of compact sets contained in $U$, let $f_{\{F_1,\cdots, F_n\}}=\max\{f_{F_1},\cdots, f_{F_n}\}$, then $f_{\{F_1,\cdots, F_n\}}$ forms a increasing net, respect to containment of collections, of positive continuous functions converges pointwise to the projection with support $U$.

## the atomic part of $C(X)$


## the category of open projections and that of hereditary C*-subalgebras


$\CI$: - Obejects: hereditary C*-subalgebras $\CI(A)$ of a C*-algebra $A$. - Arrows: Subset orthomorphisms. By a *subset orthomorphism* between $\CI(A)$ and $\CI(B)$, we mean a mapping $\Theta$ that preserrves orthogonality, in the sense $IJ=0$ implies that $\Theta(I)\Theta(J)=0$ for all $I, J\in \CI(A)$.

**The Functor between the Two Categories**: We know that the mapping $F:p\mapsto pA''p\cap A$ from $\OP(A)$ onto $\CI(A)$ is a bijection, where $A''$ is the universal enveloping von Neumannn algebra of $A$. The functor from $\OP$ to $\CI$ is defined to be a function $\F$ assigns to each projection orthomorphism $\theta:\OP(A)\to \OP(B)$ a subset orthomorphism $$\F\theta:\CI(A)\to \CI(B), F(p)\mapsto F(\theta(p)).$$ It is a isomorphism between the two categories. We must prove that for each subset isomorphism $\Theta:\CI(A)\to \CI(B)$, $\theta: \CI(A)\to \CI(B), F^{-1}(H)\mapsto F^{-1}(\Theta(H))$ is a projection orthomorphism. Suppose $F(p), F(q)\in \CI(A)$ satisfies that $F(p)F(q)=0$. Then any approximate unit $(u_\lambda)$ of $F(p)$ will strongly converges to the unit $p$ of $pA''p,$ the strong closure of $F(p)$. Since the composition of operators in a bounded set is continuous, $pq=0$. \begin{xy} \xymatrix{ \OP(A)\ar[r]^{F_A}\ar[d]_\theta & \CI(A)\ar[d]^{\F\theta}\\ \OP(B)\ar[r]^{F_B} & \CI(B) } \end{xy}

## the bidual of a C*-algebra is linear isometric onto the universal enveloping von Nemann algebra

The following is the collation of the discussion here.

Theorem 1 (3.7.8,[1]). The enveloping von Nemann algebra $A''$ of a C*-algebra $A$ is isomorphic, as a Banach space, to the second dual $A^{\ast \ast}$ of $A$.

Proof. Let $S$ denote all the states on the C* algebra and $\left( {\pi_{S},H_{S}} \right)$ denotes the universal representation of $A$.

For each state $\tau \in S$, $\tau\left( a \right) = \left\langle {\pi_{\tau}\left( a \right)x_{\tau},x_{\tau}} \right\rangle$, where $\left( {\pi_{\tau},H_{\tau}} \right)$ is the GNS representation associated with $\tau$ and $x_{\tau}$ is the cyclic vector for the representation. Hence the vector state $\left. \tau^{\prime}:u\mapsto\left\langle {ux_{\tau},x_{\tau}} \right\rangle \right.$ in $B\left( {H\tau} \right)$, which is also normal, makes the triangle above in the following diagram communicative. We know that there exists an normal *-homomorphism from $B\left( H_{S} \right)$ onto $B\left( H_{\tau} \right)$, say $i^{\ast} \circ p$, making the triangle below communicative, Therefore, $\widetilde{\tau}: = \tau^{\prime} \circ i^{\ast} \circ p$ is an normal extension of $\tau$ on $B\left( H_{S} \right)$.

\begin{xy} \xymatrix{ & \mathbb{C}\\ A \ar[ur]^{\tau} \ar[r]_{\pi_\tau} \ar[rd]_{\pi_S}& B(H_\tau) \ar[u]_{{\tau'}} \\ & B(H_S) \ar[u]_{i^*\circ p} } \end{xy}

Since by Jordan Decomposition each element of $A^{\ast}$ is a linear combination of elements from $S$ we can therefore define a map $\Phi$ from $A^{\ast}$ into $A_{\ast}'': = \left\{ \tau \in {A{}''}^{\ast} \middle| \tau\text{ is a }\sigma - \text{weakly continous functional on }A'' \right\}$ such that $\Phi\left( \tau \right) = \widetilde{\tau}|_{A''},\forall\tau \in S$. Note that a state on a von Neumann algebra is normal if and only if it is $\sigma$-weakly continuous(3.6.4, [1]).

Since $\pi_{S}$ is none-degenerate, the strong closure of the closed unit ball of $A$ is equal to the closed unit ball of $A''$ by Kaplansky Density Theorem. We also know that the $\sigma$-weak topology coincides with the weak topology on a bounded set, and $WOT$ coincides with $SOT$ on convex sbuset, hence the $\sigma$-weak closure of the closed unit ball of $A$ is equal to the closed unit ball of $A''$. Therefore, $\parallel \widetilde{\tau} \parallel = \parallel \tau \parallel$, i.e., the linear $\Phi$ is an isometry.

For any $\tau \in A_{\ast}''$, $\Phi\left( {\tau|_{A}} \right) = \widetilde{\tau|_{A}}$ coincides with $\tau$ on $A$, so $\Phi\left( {\tau|_{A}} \right) = \tau$. Hence, $A^{\ast} = A_{\ast}''$. □

### References

[1] Gert Kjaergard. Pedersen. C*-algebras and their automorphism groups. Academic Press London ; New York, 1979.

## An closed left ideal $L$ is the intersection of the maximal modular left ideals containing $L$

**Lemma 1.** Let $L_1$ and $L_2$ be closed ideals of a C* algebra $A$. Suppose that $L_1\subset L_2$ and that every pure state of $A$ that vanishes on $L_1$ vanishes on $L_2$. Then $L_1=L_2$. *Proof.* It is a corollary of Corollary 5.1.9 and Theorem 5.3.2 in [1]. **Lemma 2.** Let $\tau$ be a state of a C*-algebra $A$ and $N_\tau:=\{a|\tau(a^*a)=0\}$. Then a closed left ideal $L$ of $A$ is contained in $N_\tau$ if and only if it is contained in $\ker\tau$. *Proof.* For any element $a$ in $A$, $\tau(a^*a)=0$ implies $\tau(a)=0$ since $|\tau(a)|^2\leq \|\tau\|\tau(a^*a)$. Thus $N_\tau\subset \ker \tau$. Suppose $l\in L$, then $l^*l\in L$ since $L$ is a left ideal. Hence if $L\subset\ker \tau$ then $l^*l \in \ker\tau$, i.e. $l\in N_\tau$. So $L\subset N_\tau$. >**Theorem.** Suppose $L$ is a closed left ideal of a C*-algebra $A$. If the set $$\vee L:=\{N_\tau|\tau \mbox{ is a pure state of } A\}$$ is non-empty, then $L=\cap\vee L$. *Proof.* $L\subset \cap\vee L$ is obvious. Let $\tau$ be a pure state of $A$ and suppose $L\subset \ker\tau$. Since \begin{align*} & L\subset \ker \tau & \\ \Leftrightarrow & L\subset N_\tau & (\mbox{ Lemma 2 })\\ \Rightarrow & \cap\vee L\subset N_\tau & (\mbox{ the definition of } \vee L)\\ \Leftrightarrow & \cap \vee L\subset \ker \tau & (\mbox{ Lemma 2 }), \end{align*} $L=\cap\vee L$ by Lemma 1.

### References

[1] Gerald J Murphy. C*-algebras and operator theory. Academic press, 2014.

## Zariski Topology

Suppose $R$ is a ring(not necessarily unital nor abelian) and denote all (proper) prime ideals of $R$ by $\operatorname{Prim}\left( R \right)$. For each subset $S$ of $R$, define $\vee \left( S \right): = \left\{ P \in \operatorname{Prim}\left( R \right) \middle| S \subset P \right\}.$

Theorem 1. $\left\{ \operatorname{Prim}\left( R \right) \backslash \vee \left( S \right) \middle| S \subset R \right\}$ forms a topology of $R$.

Proof. In fact, (a)$\vee \left\{ 0 \right\} = \operatorname{Prim}\left( R \right)$, $\vee \left( R \right) = \varnothing$.

(b)$\cap_{\lambda} \vee \left( S_{\lambda} \right) = \vee \left( {\cup_{\lambda}S_{\lambda}} \right),S_{\lambda} \subset R.$ This can be seen straightforwardly by definition without using any property of an ideal.

(c)$\vee \left( S_{1} \right) \cup \vee \left( S_{2} \right) = \vee \left( {S_{1}S_{2}} \right),S_{1},S_{2} \subset R$.

If $P \in \vee \left( S_{1} \right) \cup \vee \left( S_{2} \right)$, then $S_{1} \subset P$ or $S_{2} \subset P$, thus $S_{1}S_{2} \subset P$ since $P$ is an ideal, and hence $\vee \left( S_{1} \right) \cup \vee \left( S_{2} \right) \subset \vee \left( {S_{1}S_{2}} \right)$.

Suppose $P \in \vee \left( {S_{1}S_{2}} \right)$,i.e., $S_{1}S_{2} \subset P$. If $S_{1} \nsubseteq P$ and $S_{2} \nsubseteq P$, then there exists some $s_{1} \in S_{1} \backslash P$ and $s_{2} \in S_{2} \backslash P$, thus $s_{1}s_{2} \notin P$ since the complement of a prime ideal is multiplication closed, a contradiction. Hence $\vee \left( {S_{1}S_{2}} \right) \subset \vee \left( S_{1} \right) \cup \vee \left( S_{2} \right)$. □

We call this topology Zariski topology of $\operatorname{Prim}\left( R \right)$, and call $\operatorname{Prim}\left( R \right)$ with Zariski topology the spectrum of $R$.

We can verify the following properties directly:

$\begin{matrix} & {\cap \vee S \supset S,S \subset R;} & & \\ & {\vee \cap \mathcal{S} \supset \mathcal{S},\mathcal{S} \subset \operatorname{Prim}\left( R \right);} & & \\ & {\vee S_{1} \supset \vee S_{2},S_{1} \subset S_{2};} & & \\ & {\cap \mathcal{S}_{1} \supset \cap \mathcal{S}_{2},\mathcal{S}_{1} \subset \mathcal{S}_{2}.} & & \\ \end{matrix}$

(For the empty set $\varnothing$ of $\operatorname{Prim}\left( R \right)$,define $\cap \varnothing = R$.)

Property 2. For each subset $\mathcal{S}$ of $\operatorname{Prim}\left( A \right)$, $\vee \cap \mathcal{S}$ is the closure of $\mathcal{S}$.

Proof. Suppose $\mathcal{S} \subset \vee S$ for some $S \subset R$, then $\cap \mathcal{S} \supset \cap \vee S$, and thus $\vee \cap \mathcal{S} \subset \vee \cap \vee S = \vee S$(here is the reason for last equation). Therefore, $\vee \cap \mathcal{S}$ is the smallest closed set in $\operatorname{Prim}\left( R \right)$ that contains $\mathcal{S}$. □

For each semiprime ideal $P$ of $R$,

$\cap \vee P = P.$

Recall that an ideal $P$ of a ring $R$ is called semiprime if $P$ is the intersection of some prime ideals of $R$.

If $R$ is a unital abelian ring, $I$ is an ideal of $R$, then

$\cap \vee I = \sqrt{I},$

where $\sqrt{R} = \left\{ r \in R \middle| r^{n} \in I\text{ for some positive interger }n \right\}$.

## Closed ideals of $C(X)$ and closed sets of $X$

Theorem 1 (2.13,[1]). Suppose $V_{1},\cdots\;,V_{n}$ are open subsets of a locally compact Hausdorff space, $K$ is compact, and

$K \subset V_{1} \cup \cdots \cup V_{n}.$

Then there exists functions $h_{i} \prec V_{i}\left( {i = 1,\cdots\;,n} \right)$ such that $h_{1}\left( x \right) + \cdots + h_{n}\left( x \right) = 1\left( {x \in K} \right).$

The collection $\left\{ {h_{1},\cdots\;,h_{n}} \right\}$ is called a partion of unity on $K$, subordinate to the cover $\left\{ {V_{s_{1}},\cdots\;,V_{s_{n}}} \right\}$.

Theorem 2. There is a bijection between closed subsets of $X$ and closed ideals of $C\left( X \right)$, where $X$ is a compact Hausdorff space.

Proof. Suppose $I$ is a closed ideal of $C\left( X \right)$. Let $I^{\top}: = \left\{ x \in X \middle| f\left( x \right) = 0\forall f \in I \right\}$, and $\varepsilon > 0$. If $s \notin I^{\top}$, then there is a $f_{s}^{\prime} \in I$ such that $f_{s}^{\prime}\left( s \right) \neq 0$ and thus there is a $f \in I$ such that $f_{s}\left( s \right) = 1$. Let $V_{s} = \left\{ x \middle| \middle| f_{s}\left( x \right) - 1 \middle| < \varepsilon \right\}$, then $V_{s},s \in K$ is a open cover of $K$. Hence we have a finite open cover $V_{s_{1}},\cdots\;,V_{s_{n}}$ of $K$.

Let $\left\{ {h_{1},\cdots\;,h_{n}} \right\}$ is called a partion of unity on $K$ subordinate to the cover $\left\{ {V_{s_{1}},\cdots\;,V_{s_{n}}} \right\}$, and let

$f_{K}: = \sum\limits_{i = 1}^{n}f_{s_{i}}h_{i}.$

Then $f_{K} \in I$, and $\left| f_{K}\left( x \right) - 1 \middle| = \middle| \sum_{i = 1}^{n}f_{s_{i}}\left( x \right)h_{i}\left( x \right) - \sum_{i = 1}^{n}h_{i}\left( x \right) \middle| \leq \sum_{i = 1}^{n} \middle| f_{s_{i}}\left( x \right) - 1 \middle| h_{i}\left( x \right) \leq \varepsilon\left( {x \in K} \right) \right.$.

For any $g \in I^{\top\bot}: = \left\{ f \in C\left( X \right) \middle| f\left( x \right) = 0\forall x \in I^{\top} \right\}$, set $K = \left\{ x \in X \middle| \middle| g\left( x \right) \middle| \geq 0 \right\}$. Since

$\begin{array}{rlrl} \left| g\left( x \right) - g\left( x \right)f_{K}\left( x \right) \right| & \left. \leq \parallel g \parallel \middle| 1 - f_{K}\left( x \right) \middle| < \parallel g \parallel \varepsilon\left( {x \in K} \right),\qquad \right. & & \qquad \\ \left| g\left( x \right) - g\left( x \right)f_{K}\left( x \right) \right| & \left. \leq \middle| g\left( x \right) \middle| \left( 1 + \middle| f_{s_{i}}\left( x \right) \middle| h_{i}\left( x \right) \right)\qquad \right. & & \qquad \\ & {\leq \varepsilon\left( \left. 1 + \middle| f_{s_{i}}\left( x \right) \right| \right) \leq \varepsilon\left( {2 + \varepsilon} \right)\left( {x \in \cup_{i = 1}^{n}V_{s_{i}} \backslash K} \right),\qquad} & & \qquad \\ \left| g\left( x \right) - g\left( x \right)f_{K}\left( x \right) \right| & \left. = \middle| g\left( x \right) \middle| < \varepsilon\left( {x \notin \cup_{i = 1}^{n}V_{s_{i}}} \right),\qquad \right. & & \qquad \\ \end{array}$

$\left| g\left( x \right) - g\left( x \right)f_{K}\left( x \right) \middle| \leq \left( {\parallel g \parallel + 2 + \varepsilon} \right)\varepsilon \right.$. Therefore, $g \in I$ and $I^{\top\bot} = I$.

Suppose $B$ is a closed set in $X$. If there exists some $x_{0} \in B^{\bot\top} \backslash B$, then by Uryshon’s lemma, there exists some $f \in C\left( X \right)$ such that $f\left( x \right) = 0\forall x \in B$ and $f\left( x_{0} \right) \neq 0$, i.e., $f \in B^{\bot}$ but $x_{0} \notin B^{\bot\top},$ a contradiction. Hence $B^{\bot\top} = B$. □

### References

[1]   Walter Rudin. Real and complex analysis. Tata McGraw-Hill Education, 1987.

## Jordan Decomposition of bounded linear functionals on a C*-algebra

Let’s first review Theorem 3.3.6,[1].

Theorem 1. If $a$ is a normal element of a non C*-algebra $A$, then there exists a state $\tau$ of $A$ such that $\left| \tau\left( a \right) \middle| = \parallel a \parallel \right.$.

Corollary 2. Suppose $A$ is a C*-algebra and $a \in A$. If $\tau\left( a \right) = 0$ for all state $\tau$ of $A$, then $a = 0$.

Proof. Suppose $a = b + ic$ where $b$ and $c$ are hermitian elements of $A$. For any state $\tau$ on $A$, we have $\tau\left( b \right),\tau\left( c \right) \in {\mathbb{R}}$, thus if $\tau\left( a \right) = \tau\left( b \right) + i\tau\left( c \right) = 0$, then $\tau\left( b \right) = 0,\tau\left( c \right) = 0$. By the arbitrariness of $\tau$ and Theorem 1, $b = c = 0$. □

By an involutive vector space, we mean a complex vector space $V$ with an involution, that is, a conjugate-linear operator $\left. \ast :V\rightarrow V \right.$ such that $\ast \circ \ast = \operatorname{Id}$. An element $v \in V$ is called self-adjoint if $v^{\ast} = v$.

Lemma 3. Suppose $\left. T:V\rightarrow W \right.$ is a homomorphism between two involutive vector spaces $V,W$. If for a self-adjoint element $w \in W$, there exists some $v \in V$ such that $T\left( v \right) = w$, then there is a self-adjoint element $v^{\prime} \in V$ such that $T\left( v^{\prime} \right) = w$.

Proof. Let $v^{\prime}: = \frac{v + v^{\ast}}{2}$ and $v^{''} = \frac{v - v^{\ast}}{2i}$ be self-adjoint elements in $V$, then $v = v^{\prime} + iv^{''}$. Since $w = T\left( v \right) = T\left( v^{\prime} \right) + iT\left( v^{''} \right)$ is self-adjoint, $T\left( v^{''} \right) = 0$, thus $w = T\left( v \right) = T\left( v^{\prime} \right)$. □

Theorem 4 (Jordan decomposition). Let $\tau$ be a self-adjoint bounded linear functional on a C*-algebra, then there exist some positive linear functionals $\tau_{+},\tau_{-}$ such that $\tau = \tau_{+} - \tau_{-}$ and $\parallel \tau \parallel = \parallel \tau_{+} \parallel + \parallel \tau_{-} \parallel$.

Proof. Let $S$ be the set of all the states on $A$, and

$\begin{matrix} {\Phi:} & \left. A\rightarrow C\left( S \right) \right. & & \\ & \left. a\mapsto\widehat{a}: \right. & \left. S\rightarrow{\mathbb{C}} \right. & \\ & & \left. \tau\mapsto\tau\left( a \right). \right. & \\ \end{matrix}$

then $\Phi$ has following properties:

(a) $\Phi$ is an injection. $\left. \Phi\left( a \right) = \Phi\left( b \right)\Rightarrow\tau\left( a \right) = \tau\left( b \right) \right.$ for all states. By Corollary 2, a=b.

(b) $\Phi$ preserves involution: $\tau\left( a^{\ast} \right) = \overline{\tau\left( a \right)}$ for all states and $\widehat{a^{\ast}}\left( \tau \right) = \tau\left( a^{\ast} \right)$, ${\widehat{a}}^{\ast}\left( \tau \right) = \overline{\widehat{a}\left( \tau \right)} = \overline{\tau\left( a \right)}.$

(c) $\Phi$ is linear and bounded.

(d) $\Phi$ is bounded below. Suppose $a \in A$ and $a = b + ic$, where $b = \frac{a + a^{\ast}}{2}$ and $c = \frac{a - a^{\ast}}{2i}$ are hermitian element of $A$. Then

$\begin{array}{rlrl} & {\left. \parallel \widehat{a} \parallel = \sup\limits_{\tau \in S} \middle| \tau\left( a \right) \middle| = \sup\limits_{\tau \in S} \middle| \tau\left( {b + ic} \right) \right|\qquad} & & \qquad \\ = & {\sup\limits_{\tau \in S}\tau\left( b \right) + i\tau\left( c \right) \geq \sup\limits_{\tau \in S}\max\left\{ \left| \tau\left( b \right) \middle| , \middle| \tau\left( c \right) \right| \right\}\qquad} & & \qquad \\ = & {\max\left\{ {\parallel b \parallel , \parallel c \parallel} \right\} \geq \frac{1}{2} \parallel b + ic \parallel \geq \frac{1}{2} \parallel a \parallel .\qquad} & & \qquad \\ \end{array}$

Let

$\begin{array}{rlrl} {\Phi^{\ast}:} & \left. C\left( S \right)^{\ast}\rightarrow A^{\ast}\qquad \right. & & \qquad \\ & \left. \mu\mapsto\mu \circ \Phi\qquad \right. & & \qquad \\ \end{array}$

then $\Phi^{\ast}$ has the following properties: (a) $\Phi^{\ast}$ is linear.

(b) $\Phi^{\ast}$ is preserves involution. $\Phi^{\ast}\left( \mu^{\ast} \right)\left( a \right) = \mu^{\ast}\left( {\Phi\left( a \right)} \right) = \overline{\mu\left( \left( {\Phi\left( a \right)} \right)^{\ast} \right)} = \overline{\mu\left( {\Phi\left( a^{\ast} \right)} \right)},$ and $\left( {\Phi^{\ast}\left( \mu \right)} \right)^{\ast}\left( a \right) = \overline{\Phi^{\ast}\left( \mu \right)\left( a^{\ast} \right)} = \overline{\mu\left( {\Phi\left( a^{\ast} \right)} \right)}$$\left( {\mu \in C\left( X \right)^{\ast},a \in A} \right).$

(c) $\Phi^{\ast}$ is a surjection. Since $\Phi$ is s bounded below, $\left. \Phi^{- 1}:\Phi\left( A \right)\rightarrow A \right.$ is bouded. Suppose $\tau \in A^{\ast}$, then $\tau \circ \Phi^{- 1} \in \Phi\left( A \right)^{\ast}$. Let $\mu \in A^{\ast}$ be the extension of $\tau \circ \Phi^{- 1}$, then $\Phi^{\ast}\left( \mu \right) = \mu \circ \Phi = \tau \circ \Phi^{- 1} \circ \Phi = \tau$.

For a self-adjoint bounded linear functional $\tau \in A^{\ast}$, by lemma 3, there exists some self-ajoint linear functional $\mu \in C\left( S \right)^{\ast}$ such that $\Phi^{\ast}\left( \mu \right) = \tau$. It corresponds to a real regular measure on $S$. □

### References

[1]   Gerald J Murphy. C*-algebras and operator theory. Academic press, 2014.

## If $\sigma$ is a reverse mapping such that $\sigma^2\geq \operatorname{id}$, then $\sigma^3=\sigma$.

Let $S$, $T$ be two partial order sets, and suppose $\sigma:S\to T$ and $\tau:T\to S$ are two maps such that \begin{eqnarray*} \tau\circ\sigma(s)\geq s, \forall s\in S\\ \sigma\circ\tau(t)\geq t, \forall t\in T. \end{eqnarray*} If $\sigma(s_1)\leq \sigma(s_2)\forall s_1\geq s_2$, then $\sigma\circ\tau\circ\sigma=\sigma$.