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
\tau\circ\sigma(s)\geq s, \forall s\in S\\
\sigma\circ\tau(t)\geq t, \forall t\in T.
If $\sigma(s_1)\leq \sigma(s_2)\forall s_1\geq s_2$,
then $\sigma\circ\tau\circ\sigma=\sigma$.


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

{\Phi:} & \left. y\rightarrow\left( {X,\sigma\left( {X,Y} \right)} \right)^{\ast}\qquad \right. & & \qquad \\
& \left. y\mapsto\widehat{y}.\qquad \right. & & \qquad \\

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

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


\[\ker F \supset \bigcap\limits_{i = 1}^{n}\ker\widehat{y_{i}}\]

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 $\M$.

For each $x\in X$, define \begin{equation*}
ev_x(f)=f(x), f\in C(X),
we have

Claim. \begin{equation*}
H:X\to \M, x\to ev_x
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.