This post is rehearsal about Proposition 2.47 in *Lectures in Coarse Geometry* by Roe. Any comment is appreciated as well as usage of language.
**PROPOSITION 2.47.** Let $(X, d)$ be a proper metric space. Then the bounded coarse structure on $X$ is the topological coarse structure associated to its Higson compactification.
Poof.$\newcommand{\d}{\operatorname{d}}$
We only need to verify that the bounded coarse structure on $X$ is coarser than the topological coarse structure associated to its Higson compactification.
Suppose $E\subset X\times X$ is continuously controlled by $\operatorname{h} X$ but not boundedly controlled, then for each $n$ there is a pair $(x_n,y_n)\in E$ with $d(x_n,y_n)>2n$. Without loss of generality, suppose $x_n$ tends to infinity of $X$. Then $y_n$ tends to infinity of $Y$ since $E$ is continuously controlled.
Fix a point $O$ in $X$. Because $X$ is a proper metric space (that is, any bounded set is relatively compact), we can choose a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $x_{n_1}=x_1$ and $x_{n_{k+1}}$ is out of the closed ball $B(O, r_k)$ where $r_k=\max\{d(O, x_{n_k}), d(O,y_{n_k})\}+2k+2$. It is easy to see that closed balls $B(x_{n_k}, k)$ are disjoint pairwise.
Define $f(x)=\begin{cases}\frac{k-d(x,x_{n_k})}{k} & \mbox{if}~ d(x,x_{n_k})\sup_{(x,y)\in F}d(x,y)$ and $(x,y)$ is an arbitrary point in $F$.
- $d(x, x_{n_k})<k$ and $d(y, x_{n_k})<k$ for some $k$. $|\d f(x,y)|=\frac{|d(x,x_{n_k)}-d(y,x_{n_k})|}{k}\leq \frac{d_0}{k}.$
- $d(x, x_{n_k})<k$ for some $k$ while $d(y, x_{n_k})\geq k$. $0\leq f(x)=\frac{k-d(x,x_{n_k})}{k}\leq \frac{d(y,x_{n_k})-d(x,x_{n_k})}{k}\leq \frac{d_0}{k}.$ If $d(y, x_{n_j})<j$ for some $j\neq k$, then $|\d f(x,y)|\leq \frac{2d_0}{\min \{j,k\}}.$ If $d(y, x_{n_j})\geq j$ for all $j$, then $|\d f(x,y)|\leq \frac{d_0}{k}.$
- $d(x, x_{n_k})\geq k$ and $d(y, x_{n_k})\geq k$ for all $k$. $f(x)=f(y)=0$, so $\d f(x,y)=0$.
Let $B$ be the closed ball $B(O, r_k)(\supset \cup_{i=1}^k B(x_{n_i},i))$, then $K:=(B\times \overline{F^{-1}[B]})\cup (\overline{F[B]}\times B)$ is a compact set of $X\times X$. If $(x,y)\in F\backslash K$, then $x\notin B $ and $y\notin B$, thus $|\d f(x,y)|\leq \frac{2d_0}{k+1}.$ Therefore, $f\in C_h(X)$.
Hence $\d f(x_n,y_n)$ tends to zero as $(x_n,y_n)$ tends to infinity of $X\times X$. However, $\d f(x_{n_k}, y_{n_k})=-1$ by the definition of $f$, a contradiction.

$\newcommand{\zhang}[1]{{\Large{\text{#1}}}}
\def \A{\mathscr A} \def \F{\mathbb{F}} \def \N{\mathbb{N}} \def \R{\mathbb{R}} \def \C{\mathbb{C}}\def \H{\mathscr{H}}\def \d{\mathrm{d}} \def \Inv{\mathrm{Inv}}
\def \ni{^{-1}} \def \jx{\lim_{n\to\infty}} \def \yb{\frac{1}{2}}
\newcommand{\x}{\times} \newcommand{\wt}{\textbf{Question.}} \newcommand{\bh}[1]{\textbf{#1.}} \newcommand{\xl}[2]{#1_1, #1_2, \cdots, #1_{#2}} \newcommand{\xl}[2]{#1_1, #1_2, \cdots, #1_{#2}} \newcommand{\tm}[1]{\textbf{ #1.}} \renewcommand{\l}[1]{\lVert #1\rVert}\newcommand{\zm}{Proof.} \newcommand{\zb}{}$
$\bh{3.5}$An element a of $A_+$ is strictly positive if the hereditary $C^*$-subalgebra of $A$ generated by $a$ is $A$ itself, that is, if $(aAa)^- = A$.
(a) Show that if $A$ is unital, then $a \in A_+ $ is strictly positive if and only if $a$ is invertible.
(b) If $H$ is a Hilbert space, show that a positive compac operator on $ H$ is strictly positive in $K(H)$ if and only if it has dense range.
(c) Show that if $a$ is strictly positive in $A$, then $\tau( a) > 0$ for all non-zero positive linear functionals $\tau$ on $A$.
$\zm$ (a) If $a$ is invertible, then $aAa=A$, thus $a$ is strictly positive.
Conversely, if $a$ is strictly positive, then $(aAa)^-=A$.
Since the set $G$ of all invertible elements is open, $aAa\cap G\neq\emptyset$. Hence there exists some $b\in A$ such that $aba$ is invertible, thus $a$ is invertible.
(b)If $u$ is strictly positive in $K(H)$, that is $(uK(H)u)^-=K(H)$, then $\forall x\in H,$ there exists a sequence $\{v_n\} $ in $B(H)$ such that $$ x\otimes x=\lim_{n\to \infty} uv_nu.$$
Hence $$\l{x}^2x=x\otimes x(x)=\lim_{n\to\infty}u(v_nu(x))\in (uH)^-.$$
Therefore, $x\in (uH)^-$ and $ H=(uH)^-$.
Conversely, if $(uH)^-H$, then for all $x\in H,$ then there exists a sequence $\{h_n\}$ in $H$ such that $uh_n\to x$, hence $$ u(h_n\otimes h_n )u= (uh_n)\otimes(uh_n)\to x\otimes x.$$
Therefore, every rank-one projection belongs to $(uK(H)u)^-$.
Since all the rank-one projections is the total subset of $K(H)$, $K(H)=(uK(H)u)^-$.
(c)Suppose $\tau $ is a positive functional such that $ \tau(a)=0$,
then $$\tau((a^{\frac{1}{2}})^*a^\yb)=0,$$ so $$\tau(a^* a)=\tau((a^\frac{1}{2})^3 a^\yb)=0.$$
Hence for all $b\in A$, $$\tau(aba)=0.$$ Therefore, $\tau=0$ for $A=(aAa)^-$.
$\bh{3.3}$If $\varphi:A\to B$ is a positive linear map between $C^*$ algebra, show that $\varphi$ is bounded.
$\zm$ For all positive linear functional $\tau$ on $B$, $\tau\circ\varphi $ is a positive linear functional on $A$ and thus bounded, i.e.,$$\lVert\tau\circ\varphi(x)\rVert\leq M_\tau ~~ (\forall x\in A \text{ with } \l{x}\leq 1) $$ for some positive constant $M_\tau$.
Every bounded linear functional on $B$ is a linear combination of 4 positive linear functional on $B$ by Jordan Decomposition, thus for all $b^*\in B^*,$ $$\l{b^*(\varphi(x))}\leq M_{b^*} ~~( \forall x\in A \text{ with } \l{x}\leq 1)$$ for some positive constant $M_{b^*}$. Hence $\l{\varphi(x)}\leq M$ for some positive constant $M$ by the Principle of Uniform Boundedness, i.e. $\varphi$ is bounded.

Each pure state on $M_n=B(\mathbb{C}^n)$ is of form $$w_x:=\langle \cdot ~ x, x\rangle,$$ thus each left maximal modular ideal is of form $$\{u|w_x(u^*u)=0\}=\{u|\|ux\|=0\}.$$
Particularly, corresponding to $e_1=(1,0)$ in $\mathbb{C}^2$, the left maximal modular ideal is is of form
$$\left( \begin{array} & 0 & a\\
0 & b \end{array}\right). $$

$\newcommand{\P}{\operatorname{Proj}}$$\newcommand{\qCP}{\operatorname{CP}}$
Suppose $\theta$ is a projection orthoisomorphism between the set $\qCP(A)$ and $\qCP(B)$ of all q-closed projections of $A=C(X,\mathcal M_m)$ and $B=C(Y,\mathcal M_n)$.
We know that [it preserves 0, order and rank-one projections.][1]
**Theorem 1.** $\theta$ can be extended to a projection orthoisomorphism between
$\ell^\infty(X,\P(A))$ and $\ell^\infty(Y,\P(B))$.
*Proof*.
For each projection $p\in \ell^\infty(X,\P(\mathcal M_m))$, denote by $p_1$ the set of all the rank-one projections majorised by $p$, and define $\tilde\theta(p)\in \ell^\infty(Y, \P(\mathcal M_n))$ by $\tilde\theta(p)=\sup_{m\in p_1}\theta(m)$. This extends $\theta$.
Since any projection in $\mathcal M_m$ is the supremum of rank-one projctions majorised by it,
any projection in $\ell^\infty(X,\P(\mathcal M_m))$ is the supremum of rank-one projctions majorised by it.
From the definition we konw that $\tilde\theta$ preserves order, hence $\{\theta(m)|m\in p_1\}$ consists of all the rank-one projections majorised by $\tilde \theta(p)$. Therefore, $\tilde\theta$ is a bijection.
Since $\tilde\theta$ preserves rank-one projections and order, and a projection $p$ is orthogonal to another projection $q$ in $\ell^\infty(X,\mathcal M_m)$ iff $rs=0$ for all rank-one projection $r$ majorised by $p$ and $s$ majorised by $q$, we have $\tilde\theta$ preserves orthogonality.
By Dye's theorem, we have
**Corollary.** $\theta$ is implemented by a Jordan *-isomorphism between $\ell^\infty(X,\mathcal M_m)$ and $\ell^\infty(Y,\mathcal M_n)$.
[1]: https://math.liveadvances.com/c-ding/properties-of-a-projection-orthoisomorphism-on-the-set-of-closed-projections/

By a *projection orthoisomorphism* we mean a one-one mapping $\theta$ that sends the set of projections of a C*-algebra to that of another C*-algebra which preserves orthogonality, in the sense that $pq=0$ if and only if $\theta(p)\theta(q)=0$.
$\newcommand{\P}{\operatorname{Proj}}$
Let $X$ is a compact Hausdorff space and $A=C(X,\mathcal M_m)$.
The atomic part of $A$ is of form $\ell^\infty(X, \mathcal M_m)$, so the q-closed projections of $A$ is of form $\ell^\infty(X, \P(\mathcal M_m))$, where $\P(\mathcal M_m)$ denotes the set of all projections of $\mathcal M_m$.
A rank-one projection $m$ of $\ell^\infty(X,M_m)$ is such a projection that $m(x_0)$ is a rank-one projection in $\mathcal M_m$ for some point $x_0\in X$ and $m(x)=0$ for others. It is q-closed.
$\newcommand{\qCP}{\operatorname{CP}}$
**Proposition 1.** Suppose $A=C(X, \mathcal M_m)$, $B=C(Y, \mathcal M_n)$, and $\theta$ is a projection orthoisomorphism between the set $\qCP(A)$ and $\qCP(B)$ of q-closed projections of $A$ and $B$ respectively. Some properties are as follows:
(a) $\theta(0)=0$;

(b) $\theta$ preserves order in the sense that $p\leq q$ implies that $\theta(p)\leq \theta(q)$;

(c) $\theta$ sends a rank-one projection to a rank-one projection.
*Proof*. (a)
$0$ can be characterized as the only q-closed projection orthogonal to itself;
(b) If $p\leq q$ but $\theta(p)\nleq\theta(q)$, then there is a rank-one projection $n$ in $\ell^\infty(Y,\mathcal M_n)$ such that $n$ is not orthogonal to $\theta(p)$ but $n\perp \theta(q)$, thus $\theta^{-1}(n)$ is not orthogonal to $p$ but $\theta^{-1}(n)$ is orthogonal to $q$, a contradiction.
(c) A rank-one projection can be characterized as follows: itself is the only one non-zero q-closed projection less than it. By (a) and (b), $(c)$ is obtained.

Let \(A\) be a C*-algebra, \(X\) be a compact subset of \(\mathbb{C}\), and \(c\) be a hermitian element in \(A\) such that \(\sigma\left( c \right) \subset X\). The restriction of a function on \(X\) to \(\sigma\left( c \right)\) is a *-homomorphism from \(C\left( X \right)\) to \(C\left( {\sigma\left( c \right)} \right)\), the Gelfand representation of \(C\left( {\sigma\left( c \right)} \right)\) is a *-isomorphism between \(C\left( {\sigma\left( c \right)} \right)\) and the unital C*-algebra, say \(C^{\ast}\left( c \right)\) generated by \(a\), the inclusion of \(C^{\ast}\left( c \right)\) into \(A\) and the atomic representation \(\left( {\pi_{a},H_{a}} \right)\) of \(A\) are both *-homomorphism. Hence the composition of them defines a *-homomorphism \(H\) from \(C\left( X \right)\) to \(B\left( H_{a} \right)\). Let \(\widetilde{H}\) be the extension of \(H\) to \(C\left( X \right)^{\ast \ast}\), then \(\widetilde{H}\left( f \right)\) coincides with \(f|_{\sigma{(c)}}\left( c \right)\) for each continuous function on \(X\). Since \(f_{i} \in C\left( X \right)\) \(\sigma\left( {C\left( X \right)^{\ast \ast},C\left( X \right)^{\ast}} \right)\) converges to a Borel measurable bounded function \(f\) implies that \(f_{i}|_{\sigma{(c)}}\) \(\sigma\left( {C\left( {\sigma\left( c \right)} \right)^{\ast \ast},C\left( {\sigma\left( c \right)} \right)^{\ast}} \right)\) converges to \(f|_{\sigma{(c)}}\), \(\widetilde{H}\left( f \right)\) coincides with \(f|_{\sigma{(c)}}\left( c \right)\) for each Borel measurable bounded function on \(X\). Hence there is no ambiguity to write \(f\left( c \right)\) for \(\widetilde{H}\left( f \right)\).

Let \(a\) and \(b\) whose spectrums are both contained in \(X\) be two hermitian elements such that \(ab = 0\). If the spectrum of \(a + b\) is also contained in \(X\), then for each bounded Borel measurable function \(f\) on \(X\),

\[f\left( {a + b} \right) = f\left( a \right) + f\left( b \right).\]
This can be verified for polynomials, continuous functions and bounded Borel measurable functions on \(X\) in turn.

Suppose $\theta$ is a orthoisomorpihism between closed projection of $A=C(X, \mathcal M_m)=\mathcal M_m(C(X))$ and $B=C(Y,\mathcal M_n)=\mathcal M_n(C(Y))$.$\newcommand{\P}{\operatorname{Proj}}$
We have known that [$\theta$ can be extended to a projection orthoisomorphism $\tilde\theta$ between
$\ell^\infty(X,\P(A))$ and $\ell^\infty(Y,\P(B))$][1]. By Dye's theorem, $\tilde\theta$ is implemented by a Jordan *-isomorphism between $\ell^\infty(X,\mathcal M_m)$ and $\ell^\infty(Y,\mathcal M_n)$.
Hence $\tilde\theta$ also preserves q-open projections in $\ell^\infty(X, \mathcal M_m)$.
Since a projection $p$ is q-open projection and q-closed iff $p$ is a projection in $C(X,\mathcal M_m)$,
by restricting $\tilde\theta$ to the $C(X,\P(\mathcal M_m))$, we get a projection orthoisomorphism between $C(X,\P(\mathcal M_m))$ and $C(Y,\P(\mathcal M_n))$.
Denote the characteristic matrix in $A=\mathcal M_m(C(X))$ of $a\in C(X)$ by $$P_{ij}(a)=\left(\begin{matrix}(1+a^*a)^{-1} & (1+a^*a)^{-1}a^*\\ a(1+a^*a)^{-1} & a(1+a^*a)^{-1}a^*\end{matrix}\right)$$ and write $P_i=P_{ij}(0)$ for simplicity. The following can be varified:
- $\tilde\theta$ preserves the projection of form $P_i$;
- the lattice complement of $P_{j}$ in $P_i+P_j$ is of form $P_{ij}(a)$ if $X$ is connected;
Therefore, if $X$ and $Y$ is connected, projections of form $P_{ij}(a)$ correspond under $\tilde\theta$ as $\tilde\theta$ preserves lattice order and addition.
Hence the restriction of $\tilde\theta$ to $C(X,\P(\mathcal M_m))$ can be implemented by a Jordan *-isomorphism betwwen $A$ and $B$($\color{red}{unique?}$), by [[lemma 6, 1](#dyelemma6)], when $m=n\geq 3$.
\[1\] Dye, H. (1955). On the Geometry of Projections in Certain Operator Algebras. Annals of Mathematics, 61(1), second series, 73-89. doi:10.2307/1969620
[1]: https://math.liveadvances.com/c-ding/a-projection-orthoisomorphism-of-q-closed-projections-can-be-extended-to-a-projection-orthoisomorphism-of-all-projections/

Suppose $A=C(X,\mathcal M_m)$ and $B=C(Y, \mathcal M_n)$ are two unital C*-algebras, $A_a''$ and $B_a''$ are the atomic parts of $A''$ and $B''$ respectively, $z_A$ is a central projection of $A_a''$ and $z_B$ is a central projection of $B_a''$.
Let $J_1:z_AA_a''\to z_B B_a''$ be a *-isomorphism, $J_2: (1-z_A)A_a'' \to (1-z_B)B_a''$ be a *-antiisomorphism, and $J=J_1\oplus J_2:A_a''\to B_a''$ be a Jordan isomorphism such that $\newcommand{\qOP}{\operatorname{OP}}$ $J(\qOP(A))=\qOP(B) $, where $\qOP(A)$ denotes the set of all q-open projections of $A$.Then $J(A)=B$.
Since there is a bijection between the set of q-open spectral projections and the set of open spectral projections, by [1, Theorem 2.2], a hermitian element in $A_a''$ belongs to $A$ if and only if each open set in its spectrum corresponds to a q-open spectral projection.
Suppose $a$ is a hermitian element in $A$. For each open set $U$ in $\mathbb{R}$, $J_1(\chi_U(z_Aa))=\chi_U(J_1(z_Aa))$, and $TJ_2(\chi_U((1-z_A)a))=\chi_U(TJ_2((1-z_A)a))=T(\chi_U(J_2((1-z_A)a)))$, where $Tf(y)$ is the transpose of $f(y)$ $( f\in C(Y,\mathcal M_n), y\in Y)$.
Hence, \begin{align*}J(\chi_U(a)) = &J(\chi_U(z_Aa+ (1-z_A)(a)))\\
= & J(\chi_U(z_Aa)+ \chi_U((1-z_A)a))\\
= &J_1(\chi_U(z_Aa))+ J_2(\chi_U((1-z_A)a)\\
= & \chi_U (J_1(z_Aa))+\chi_U(J_2(1-z_A)a)\\
= & \chi_U(J_1(z_Aa)+J_2((1-z_A)a)=\chi_U(J(a)).\end{align*}Therefore, $\chi_U(J(a))$ is a q-open spectral projection of $B$ and thus $J(a)\in B$.
[1] Akemann, C. A., Pedersen, G. K., & Tomiyama, J. (1973). Multipliers of C∗-algebras. Journal of Functional Analysis, 13(3), 277-301.

Suppose \(X\) and \(Y\) are vector spaces, and \(\left. \sigma:X \times Y\rightarrow{\mathbb{C}} \right.\) is a bilinear function. Denote by \(\sigma\left( {X,Y} \right)\) the smallest topology on \(X\) such that every linear function \(\widehat{y}\left( {y \in Y} \right)\) on \(X\) defined by \(y\left( x \right): = \sigma\left( {x,y} \right)\) is continuous. By saying \(Y\) separates \(X\) (through \(\sigma\)), we mean for any nonzero element \(x\) in \(X\), there is some \(y\) in \(Y\) such that \(\sigma\left( {x,y} \right) \neq 0\). Suppose \(A\) is a subset of \(X\), one can easily check that \(A^{\bot}: = \left\{ y \in Y \middle| \sigma\left( {x,y} \right) = 0\forall x \in A \right\}\) is a \(\sigma\left( {Y,X} \right)\) closed subspace of \(Y\).

Theorem 1. If \(Y\) separates \(X\) and \(X\) also separates \(Y\), then

(a)\(Y\) is linear isomorphic to \(\left( {X,\sigma\left( {X,Y} \right)} \right)^{\ast}\);

(b)For any subspace \(A\) of \(X\), \(A^{\bot\bot}: = \left( A^{\bot} \right)^{\bot}\) is the \(\sigma\left( {X,Y} \right)\)-closure \(A\) of \(A\).

Proof. (a)Let

\[\begin{array}{rlrl}
{\Phi:} & \left. y\rightarrow\left( {X,\sigma\left( {X,Y} \right)} \right)^{\ast}\qquad \right. & & \qquad \\
& \left. y\mapsto\widehat{y}.\qquad \right. & & \qquad \\
\end{array}\]
Obviously \(\Phi\) is linear, and is injective since \(X\) separates \(Y\). We must prove \(\Phi\) is surjective.

The \(\sigma\left( {X,Y} \right)\) topology is deduced by seminorms: \(\left. X\rightarrow{\mathbb{C}},x\mapsto \middle| \widehat{y}\left( x \right) \right|\), so it is locally convex and Hausdorff(since \(Y\) seprates \(X\)). Suppose \(F \in \left( {X,\sigma\left( {X,Y} \right)} \right)^{\ast},\) then there exists some nonezero elements \(y_{1},y_{2},\cdots\;,y_{n} \in Y\) and \(M > 0\) such that

\[\left| F\left( x \right) \middle| \leq M\max\limits_{1 \leq i \leq n} \middle| \widehat{y_{i}}\left( x \right) \middle| . \right.\]
Hence

\[\ker F \supset \bigcap\limits_{i = 1}^{n}\ker\widehat{y_{i}}\]
and thus \(F = \sum_{i = 1}^{n}k_{i}\widehat{y_{i}} = \widehat{\sum_{i = 1}^{n}k_{i}y_{i}} = \Phi\left( {\sum_{i = 1}^{n}k_{i}\widehat{y_{i}}} \right)\) for some scalars \(k_{1},\cdots\;,k_{n}\).

(b)Since \(A^{\bot\bot}\) is a \(\sigma\left( {X,Y} \right)\)-closed subspace of \(X\) containing \(A\), \(A \subset A^{\bot\bot}\). We will verify the opposite direction next.

If there exists some \(x_{0} \in A^{\bot\bot} \backslash A\), the Hahn-Banach theorem follows that there exists a \(\sigma\left( {X,Y} \right)\)-continous functional \(f\) on \(X\) such that \(f\left( x_{0} \right) = 1\) and \(f\left( x \right) = 0\forall x \in A\). By (a), we konw that \(f = y\) for some \(y \in Y\). Thus \(\sigma\left( {x,y} \right) = 0\forall x \in A\), i.e. \(y \in A^{\bot}\). Therefore \(x_{0} \notin A^{\bot\bot}\). Note that \(A^{\bot\bot} = \left\{ x \in X \middle| \sigma\left( {x,y} \right) = 0\forall y \in A^{\bot} \right\} = \bigcap_{y \in A^{\bot}}\ker \widehat{y}\). □

Example 2. Suppose \(H\) is a Hilbert space with the inner product \(\left\langle {\cdot , \cdot} \right\rangle\).

(a)Let \(\sigma\left( {x,y} \right) = \left\langle {x,y} \right\rangle\), then \(\sigma\left( {H,H} \right)\) is precisely the weak* topology on \(H\) by Riesz Representation.

(b)Let \(\sigma\left( {u,v} \right) = \operatorname{Tr}\left( {uv} \right)\left( {u \in B\left( H \right),v \in F\left( H \right)} \right)\). Recall that \(F\left( H \right)\) denotes all the operators in \(B\left( H \right)\) with finite rank, each of which is a linear combination of rank-one projections. For any non-zero element \(u \in B\left( H \right)\), there is a \(h \in H\) such that \(uh \neq 0\). Let \(v = h \otimes uh\), then \(\sigma\left( {u,v} \right) = \operatorname{Tr}\left( {uh \otimes uh} \right) = \parallel uh \parallel^{2} \neq 0\). Hence \(F\left( H \right)\) seperates \(B\left( H \right) \supset A\). Because \(\sigma\left( {u,x \otimes y} \right) = \left\langle {ux,y} \right\rangle\left( {x,y \in H} \right)\), the topology \(\sigma\left( {B\left( H \right),F\left( H \right)} \right)\) is precisely the weak operator topology.

(c)Suppose \(H\) is a Hilbert space, \(A\) is a subspace of \(B\left( H \right)\), \(L^{1}\left( H \right)\) is the space of trace class operators, \(\sigma\left( {u,v} \right) = \operatorname{Tr}\left( {uv} \right)\left( {u \in B\left( H \right),v \in L^{1}\left( H \right)} \right)\), \(A^{\bot} = \left\{ v \in L^{1}\left( H \right) \middle| \sigma\left( {u,v} \right) = 0\forall u \in A \right\}\). Then \(\left. L^{1}\left( H \right)/ A^{\bot} \right.\) is the space of all \(\sigma\)-weakly continuous linear functionals of \(A\). Let \(\widetilde{\sigma}\left( {u,\left\lbrack v \right\rbrack} \right) = \sigma\left( {u,v} \right)\left( u \in A,\left\lbrack v \right\rbrack \in L^{1}\left( H \right)/ A^{\bot} \right)\). It is well-defined and \(A\) and \(\left. L^{1}\left( H \right)/ A^{\bot} \right.\) seperates each other through \(\widetilde{\sigma}\). Therefore, \(\left. \left( {A,\sigma\left( {A,L^{1}\left( H \right)} \right)} \right)^{\ast} = \left( {A,\widetilde{\sigma}\left( A,L^{1}\left( H \right)/ A^{\bot} \right)} \right)^{\ast} = L^{1}\left( H \right)/ A^{\bot} \right.\). We konw that \(\sigma\left( {A,L^{1}\left( H \right)} \right) = \sigma\left( {B\left( H \right),L^{1}\left( H \right)} \right)_{A}\), so \(\left. L^{1}\left( H \right)/ A^{\bot} \right.\) is isomorphic to the space of all \(\sigma\)-weakly continuous linear functionals of \(A\).

$\newcommand{\M}{\operatorname{Max}}\newcommand{\C}{\mathbb{C}}$**Theorem**. If $X$ is a compact Hausdorff space, and $\M=\{\text{non-zero algebra homomorphism of }C(X) \text{to } \C\}$ is the maximal ideal space of $C(X)$, then there is a homeomorphism of $X$ onto $\M$.
*Proof*.
For each $x\in X$, define \begin{equation*}
ev_x(f)=f(x), f\in C(X),
\end{equation*}
we have
**Claim. ** \begin{equation*}
H:X\to \M, x\to ev_x
\end{equation*}
is a homeomorphism.
(a)Obviously, $H$ is well-defined.
(b)By Urysohn's lemma, $H$ is an injection.
(c)If $\{x_\lambda\}$ is a net in $X$ converges to $x\in X$, then $f(x_\lambda)\to f(x)$ for all $f\in C(X)$, i.e., $H(x_\lambda)\xrightarrow{weak*}H(x)$. Hence $H$ is continuous.
(d)Let $I$ be a none-trivial ideal of $C(X)$ and denote all the zero points of $f\in C(X)$ by $Z(f)$. For any finite continuous functions $f_i\in I, i=1,2,\cdots, n$, $$f:=\bar{f_1}f_1+\bar{f_2}f_2+\cdots \bar{f_n}f_n\in I,$$ therefore $ f$ is not invertible, and thus $Z(f)\neq\emptyset$.
Since $$Z(f_1)\cap Z(f_2)\cap\cdots\cap Z(f_n)\supset Z(\bar{f_1}f_1+\bar{f_2}f_2+\cdots \bar{f_n}f_n),$$
$Z(f_1)\cap Z(f_2)\cap\cdots\cap Z(f_n)\neq\emptyset.$
Hence we have $$\bigcap_{f\in I}Z(f)\neq\emptyset.$$ Note that $Z(f)$ is closed for every continuous function and that $X$ is compact.
In particular, take $I=\ker\tau, \tau\in \M$. Suppose $x\in \bigcap_{f\in I}Z(f)$, then $\ker\tau=I\subset \{f\in C(X)|f(x)=0\}=\ker ev_x$, so $\tau=ev_x$.
Therefore, $H$ is a surjection.
(e) Since $X$ is compact and $\M$ is Hausdorff, $H$ is a homeomorphism.