The Topological Dual of E Equipped with σ(E, F) is F

The Topological Dual of E Equipped with σ(E, F) is F

Let $(E, F, \langle \cdot, \cdot \rangle)$ be a dual pair. Observe that we may identify $F$ as a subspace of $E^*$. Indeed, for each $y \in F$ let $f_y : E \to \mathbf{F}$ be defined for all $x \in E$ by:

(1)
\begin{align} \quad f_y(x) := \langle x, y \rangle \end{align}

Then $f_y$ is a linear form on $E$ (since $\langle \cdot, \cdot \rangle$ is a bilinear form). The map from $F$ to $E^*$ specified by $y \mapsto f_y$ is linear. It is furthermore injective, for if $y_1, y_2 \in Y$ are such that $y_1 \neq y_2$ then $y_1 - y_2 \neq 0$, and so by the properties of $\langle \cdot, \cdot \rangle$ there exists a point $x \in E$ such that $\langle x, y_1 - y_2 \rangle \neq 0$ so that $\langle x, y_1 \rangle \neq \langle x, y_2$, i.e., $f_{y_1}(x) \neq f_{y_2}(x)$ and thus $f_{y_1} \neq f_{y_2}$. So indeed $F$ may be identified as a subspace of $E^*$.

Preliminary Results

 Lemma 1: Let $E$ be a vector space. If $f_0, f_1, …, f_n$ are linear forms on $E$ then either $f_0$ is a linear combination of $f_1, f_2, …, f_n$ OR there exists a point $a \in E$ such that $f_0(a) = 1$ and $f_1(a) = f_2(a) = … = f_n(a) = 0$.
• Proof: If $\{ f_1, f_2, ..., f_n \}$ were a linearly dependent set, then $\{ f_0, f_1, ..., f_n \}$ would be a linearly dependent set too and thus $f_0$ would be a linear combination of $f_1, f_2, ..., f_n$ and we're done.
• So assume that $\{ f_1, f_2, ..., f_n \}$ is a linearly independent set. We prove the lemma by mathematical induction.
• If $n = 0$ then there is nothing to prove. Suppose that the lemma is true for the natural number $n - 1$. Since $f_1, f_2, …, f_n$ are linearly independent, for each $1 \leq i \leq n$, $f_i$ is NOT a linear combination of $f_1, f_2, …, f_{i-1}, f_{I+1}, …, f_n$. This is true for each $i$, and so by the induction hypothesis, there exists elements $a_1, a_2, …, a_n$ such that:
(2)
\begin{align} \quad f_i(a_i) = 1 \quad \mathrm{and} \quad f_i(a_j) = 0 \quad \mathrm{whenever \:} i \neq j \end{align}
• Let $x \in E$. Then observe that for each $1 \leq i \leq n$:
(3)
\begin{align} \quad f_i \left ( x - \sum_{j=1}^{n} f_j(x) a_j \right ) = f_i(x) - \sum_{j=1}^{n} f_j(x) f_i(a_j) = f_i(x) - f_i(x) f_i(a_i) = 0 \end{align}
• Therefore:
(4)
\begin{align} \quad x - \sum_{j=1}^{n} f_j(x) a_j \in \bigcap_{I=1}^{n} f_i^{-1}(0) \end{align}
• Let $\displaystyle{N := \bigcap_{i=1}^{n} f_i^{-1}(0)}$. Then for each $x \in E$ we have that:
(5)
\begin{align} \quad x \in N + \sum_{j=1}^{n} f_j(x) a_j \end{align}
• There are now two cases to consider.
• Case 1: Suppose that for all $y \in N$ we have that $f_0(y) = 0$. For each $x \in E$, write $\displaystyle{x := y + \sum_{j=1}^{n} f_j(x) a_j}$ where $y \in N$. Then:
(6)
\begin{align} \quad f_0(x) = f_0(y) + \sum_{j=1}^{n} f_j(x) f_0(a_j) = \sum_{j=1}^{n} f_j(x) f_0(a_j) \end{align}
• Hence $\displaystyle{f_0 = \sum_{j=1}^{n} f_0(a_j) f_j}$ and thus $f_0$ is a linear combination of $f_1, f_2, …, f_n$.
• Case 2: Suppose that there exists a point $y \in N$ such that $f_0(y) \neq 0$. Let $\displaystyle{a := \frac{y}{f_0(y)}}$. Then $f_0(a) = 1$. Moreover, since $a \in N$ we must have that $f_i(a) = 0$ for all $1 \leq i \leq n$.
• So by the principle of mathematical induction, the statement is true for all natural numbers $n$. $\blacksquare$
 Corollary 2: Let $(E, F)$ be a dual pair. If $f$ is a linear forms on $E$ and $y_1, y_2, ..., y_n \in F$ then either $f$ is a linear combination of $y_1, y_2, ..., y_n$ (in the sense that $f$ is a linear combination of $f_{y_1}, f_{y_2}, ..., f_{y_n}$) OR there exists a point $a \in E$ such that $f(a) = 1$ and $\langle a, y_1 \rangle = \langle a, y_2 \rangle = ... = \langle a, y_n \rangle = 0$.

For a Dual Pair (E, F), the Topological Dual of E Equipped with σ(E, F) is F

 Theorem 3: Let $(E, F)$ be a dual pair. Then $(E^{\sigma(E, F)})' = F$.

Here the notation $(E^{\sigma(E, F)})'$ means the dual of $E$ when $E$ is equipped with the topology $\sigma (E, F)$.

• Proof: Let $f$ be a linear form on $E$ that is $\sigma(E, F)$-continuous. Since $E$ equipped with $\sigma(E, F)$ is a locally convex topological vector space, we know that $f$ is $\sigma(E, F)$-continuous if and only if it is bounded on a $\sigma(E, F)$-neighbourhood of the origin (see the proposition on the Continuous Linear Forms on a TVS and its Continuous Dual page). We already know what a base of $\sigma(E, F)$-closed neighbourhoods of the origin look like though (see the page on The Weak Topology on E Determined by F) and so there exists a set $U$ of the form:
(7)
\begin{align} U := \{ x : \sup_{1 \leq i \leq n} |\langle x, y_i \rangle| \leq 1 \} \end{align}
• with $y_1, y_2, ..., y_n \in F$, and a constant $\alpha$ with $0 < \alpha < 1$ such that $|f(x)| \leq \alpha$ for all $x \in U$.
• By the previous corollary for the collection of linear forms $\{ f, y_1, y_2, ..., y_n \}$ we have that either $f$ is a linear combination of $y_1, y_2, ..., y_n$ OR there exists a point $a \in E$ such that $f(a) = 1$ and $\langle a, y_i \rangle = 0$ for all $1 \leq i \leq n$.
• However, if this latter case holds, then $\displaystyle{\sup_{1 \leq i \leq n} |\langle a, y_i \rangle| = 0 \leq 1}$ and so $a \in U$. But then also $|f(a)| = 1 > \alpha$ which implies that $a \not \in U$, a contradiction.
• Thus $f$ is indeed a linear combination of $y_1, y_2, ..., y_n$ and so there exists scalars $\lambda_1, \lambda_2, …, \lambda_n \in \mathbf{F}$ such that:
(8)
\begin{align} \quad f = \sum_{I=1}^{n} \lambda_i y_i \in F \end{align}
• Thus every $\sigma(E, F)$-continuous linear form $f$ on $E$ is contained in $F$, and so:
(9)
\begin{align} (E^{\sigma(E, F)})' \subseteq F \end{align}
• On the other hand, if $y \in F$, then $y$ is certainly $\sigma(E, F)$ continuous, for $|f_y(x)| = |\langle x, f \rangle| = p_f(x)$, which shows that $y$ is a linear form bounded by the $\sigma (E, F)$-continuous $p_f$ seminorm, and hence:
(10)
\begin{align} (E^{\sigma(E, F)})' \supseteq F \end{align}
• So $(E^{\sigma(E, F)})' = F$. $\blacksquare$