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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 $E$?

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    $\begingroup$ 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 $H^\perp$. $\endgroup$
    – Jochen
    Jan 10 '20 at 15:27
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$\sigma(H, E)$ 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 $(*)$.

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