Dual of a subspace problem - identification of weak topologies

Let $$\langle E, F \rangle$$ be a dual pair (separating the points in both components). For a linear subspace $$H \subseteq F$$ we can consider the following two pairings:

• $$\langle E, H \rangle$$ as the restriction of $$\langle E, F \rangle$$ (this need not be a dual pair) and
• $$\langle E / H^\perp, H \rangle$$ by $$\langle x + H^\perp, y \rangle := \langle x, y \rangle$$ (this is well-defined and a dual pair).

This provides us with the following three topologies on $$H$$:

$$\sigma(F, E)|_H, \quad \sigma(H, E) \quad \textrm{and} \quad \sigma(H, E / H^\perp).$$

It is well known that $$\sigma(F, E)|_H = \sigma(H, E/H^\perp)$$ (see e.g. [Kelley-Namioka, "Linear Topological Spaces", 16.11]) and this is a Hausdorff topology. In particular, the dual $$(H, \sigma(F, E)|_H)'$$ of a subspace $$H$$ can be identified with the quotient $$E / H^\perp$$. Also $$\sigma(H, E)$$ is Hausdorff, because $$E$$ is separating the points of $$H$$.

[Wilansky, "Modern methods in topological vector spaces", problem 8-2-3] asks to show that $$\sigma(H, E) = \sigma(F, E)|_H$$. This would imply that the dual of $$(H, \sigma(H, E))$$ would be $$E$$ and $$E/H^\perp$$, which is obscure. Is the problem here that $$H$$ is not necessarily separating the points of $$E$$?

• One should be careful with wordings like the dual of $(H,\sigma(H,E))$ is $E$. The precise satement is: Every $\sigma(H,E)$-continuous linear functional on $H$ is of the form $\langle \cdot,e\rangle$ for some $e\in E$. One has thus a surjective linear map $E \to (H,\sigma(H,E))'$ whose kernel is $H^\perp$. Jan 10 '20 at 15:27

$$\sigma(H, E)$$ is the weak topology of the pairing $$\langle E, H \rangle$$ but its dual is not $$E$$, it is $$E / H^\perp$$. In general, $$\langle E, H \rangle$$ is not a dual pair. Therefore, $$\sigma(H, E)$$ is compatible with $$\langle E/H^\perp, H \rangle$$ and moreover $$\sigma(H, E) = \sigma(F,E)|_H = \sigma(H, E/H^\perp) \tag{*}.$$
More generally, let $$\langle X, Y \rangle$$ be an arbitrary pairing of vector spaces $$X$$ and $$Y$$ (not necessarily separating the points in $$X$$ or $$Y$$). Then $$\langle X, Y \rangle$$ defines the linear mappings $$\Phi : X \to Y^*, x \mapsto (y \mapsto \langle x, y \rangle) \quad \textrm{and} \quad \Psi : Y \to X^*, y \mapsto (x \mapsto \langle x, y \rangle).$$ These mappings need not be injective or surjective. Let us focus on $$\Psi$$. $$\Psi$$ induces the linear isomorphism $$Y / ker(\Psi) \to im(\Psi)$$. It holds $$ker(\Psi) = X^\perp = \{ y \in Y \mid \langle x, y \rangle = 0 \textrm{ for all } x \in X \}$$. To characterize $$im(\Psi)$$, look at $$\Phi$$, equip $$Y^* \subseteq \mathbb{R}^Y$$ with the product topology and denote by $$\sigma(X, Y)$$ the initial topology on $$X$$ induced by $$\Phi$$, called the weak topology. Then $$im(\Psi) = X' := (X, \sigma(X,Y))' \subseteq X^*$$, see e.g. [Horvath, "Topological vector spaces and distributions", Section 3-2]). We then get a linear isomorphism $$Y / X^\perp \to X'$$ with which we identify $$X' = Y / X^\perp$$. Now, the pairing $$\langle X, Y \rangle$$ induces the pairings
• $$\langle X, Y / X^\perp \rangle$$, $$\langle x, y + X^\perp \rangle := \langle x, y \rangle$$ (this is well-defined by the definition of $$X^\perp$$) and
• $$\langle X, X' \rangle$$, $$\langle x, f \rangle := f(x)$$.
We then similarly get the two weak topologies $$\sigma(X, Y / X^\perp)$$ and $$\sigma(X, X')$$ on $$X$$. It holds $$\sigma(X, Y) = \sigma(X, X') = \sigma(X, Y / X^\perp).$$
In particular, for the pairing $$\langle X, Y \rangle = \langle E, H \rangle$$ we get the identification $$(H, \sigma(H, E))' = E / H^\perp$$ and $$(*)$$.