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 $E$?

  • 1
    $\begingroup$ 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

$\sigma(H, E)$ 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 $(*)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.