## 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 $\sigma\circ\tau\circ\sigma=\sigma$.

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

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.