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The bounded coarse structure on a metric space is the topological coarse structure associated to its Higson compactification.

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.

Solutions to Some Exercises in Murphy’s book (Chapter 3)

$\newcommand{\zhang}[1]{{\Large{\text{#1}}}} \def \A{\mathscr A} \def \F{\mathbb{F}} \def \N{\mathbb{N}} \def \R{\mathbb{R}} \def \C{\mathbb{C}}\def \H{\mathscr{H}}\def \d{\mathrm{d}} \def \Inv{\mathrm{Inv}} \def \ni{^{-1}} \def \jx{\lim_{n\to\infty}} \def \yb{\frac{1}{2}} \newcommand{\x}{\times} \newcommand{\wt}{\textbf{Question.}} \newcommand{\bh}[1]{\textbf{#1.}} \newcommand{\xl}[2]{#1_1, #1_2, \cdots, #1_{#2}} \newcommand{\xl}[2]{#1_1, #1_2, \cdots, #1_{#2}} \newcommand{\tm}[1]{\textbf{ #1.}} \renewcommand{\l}[1]{\lVert #1\rVert}\newcommand{\zm}{Proof.} \newcommand{\zb}{}$ $\bh{3.5}$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.

left maximal modular ideal in $M_n$

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 $$\left( \begin{array} & 0 & a\\ 0 & b \end{array}\right). $$

a projection orthoisomorphism of q-closed projections can be extended to a projection orthoisomorphism of all projections

$\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/

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.

Borel Functional Caculus on Orthogonal Hermitian Elements

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.

a projection orthoisomorphism of closed projections of $C(X,\mathcal M_n)$ can be implemented by a Jordan *-isomorphism

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/

A Jordan-isomorphism $J:A^{**}_a\to B^{**}_a$ that preserves q-open projections maps $A$ onto $B$

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.

$Y=(X,\sigma(X,Y))^*$

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 \(A\).

A compact Hausdorff space $X$ is homeomorphic to the maximal ideal space of $C(X)$

$\newcommand{\M}{\operatorname{Max}}\newcommand{\C}{\mathbb{C}}$**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.