# Relations between different definitions of polar topologies

Let $\langle X, Y \rangle$ be a separated dual pair of (real) vector spaces. A non-empty family $\mathcal{A}$ of non-empty subsets of $Y$ is called polar if

1. every $A \in \mathcal{A}$ is bounded (w.r.t. to the duality)
2. every point of $y$ is contained in some $A \in \mathcal{A}$: $\ \bigcup \{ A \mid A \in \mathcal{A} \} = Y$
3. $\mathcal{A}$ is directed by inclusion: $\ \forall B, C \in \mathcal{A} \, \exists A \in \mathcal{A}: \, B \cup C \subseteq A$
4. $\mathcal{A}$ is closed under scalar multiplication: $\ \forall \lambda \in \mathbb{R} \, \forall A \in \mathcal{A}: \, \lambda A \in \mathcal{A}$.

This is the definition as in Wikipedia and many other books in the literature. A polar family $\mathcal{A}$ on $Y$ induces a locally convex topology (the polar topology) on $X$, namely the topology of uniform convergence on $\mathcal{A}$. Due to property 2. this topology is Hausdorff. The weak topology $\sigma(X, Y)$ is the weakest polar topology ($\mathcal{A}$ consists of all finite sets) and the strong topology $\beta(X,Y)$ the strongest polar topology ($\mathcal{A}$ consists of all bounded sets).

Wilansky, "Modern Methods of Topological Vector Spaces" uses a slightly more general definition which uses 1., 3. and

4'. $\mathcal{A}$ is directed under doubling: $\ \forall A \in \mathcal{A} \, \exists B \in \mathcal{A}: 2A \subseteq B$.

So, Wilansky omits property 2. and weakens property 4. He then considers a locally convex topology on $X$ generated by the set of all polars $A^o$ where $A \in \mathcal{A}$. Let us call this a W-polar topology. He shows that this topology is just the topology of uniform convergence on $\mathcal{A}$. Every W-polar topology is weaker than $\beta(X,Y)$. A W-polar topology is then called admissible if it is also finer than $\sigma(X,Y)$. He then shows that a W-polar topology is admissible iff 2'. holds where

2'. $\bigcup \{ A^{oo} \mid A \in \mathcal{A} \} = Y$

is a weakening of property 2.

Question 1: Am I right to say that admissible topologies are just the usual polar topologies as defined above?

Question 2: I was surprised to see Exercise 8-5-105:

"Give an example of a separated nonadmissible W-polar topology."

So the topology in question must be weaker than $\sigma(X,Y)$ or equivalently not satisfying 2'., but still be Hausdorff and W-polar. Do you have some ideas?

Question 1: Yes, the polar topology is unchanged by taking the saturated hull of $$\mathcal A$$ (that is, the collection $$\{B | B \subseteq A^{\circ \circ} \text{ for some } A \in \mathcal A\}$$). C.f. Jarchow Section 8.4, Prop. 1 or Bourbaki Chapter 3, Section 3, No. 1, Prop. 2 for instance. (In these references, for $$E, F$$ locally convex, the topology on $$\mathscr L(E,F)$$ of uniform convergence on some collection $$\mathfrak S$$ of bounded subsets of $$E$$ is discussed; this contains the situation of $$\mathcal A$$-convergence described in your question).
Question 2: Here is an example. Let $$E= \mathcal D((0,1))$$ with its usual inductive limit topology, and consider the duality $$\langle E, E' \rangle$$. Let $$\mathcal B = \{ \{T_{\varphi}\} | \varphi \in E \}$$ where $$T_{\varphi}$$ is the distribution defined by $$\langle T_{\varphi}, \psi \rangle = \int \varphi \psi dm$$, $$\psi \in E$$. Let $$\mathcal A$$ be the saturated hull of $$\mathcal B$$. Then the $$\mathcal A$$-topology $$\tau_{\mathcal A}$$ is a W-polar topology and $$\tau_{\mathcal A}$$ is just the topology of pointwise convergence in $$\mathcal D$$ (regarding $$\mathcal D$$ as a subspace of $$\mathcal D'$$), and this topology is Hausdorff because $$\bigcup \mathcal A$$ is $$\sigma(\mathcal D', D)$$-dense in $$\mathcal D'$$. But the identity map $$(\mathcal D, \tau_{\mathcal A}) \to (\mathcal D, \sigma(\mathcal D, \mathcal D'))$$ is not even sequentially continuous: choose test functions $$\varphi_n \in E$$ with $$0 \leq \varphi_n \leq 1$$ and which are identically 1 on $$[\frac{1}{2n}, \frac{1}{n}]$$ and supported in $$[\frac{1}{3n}, \frac{2}{n}]$$. Then, $$\varphi_n \to 0$$ in $$\tau_{\mathcal A}$$ but $$\langle 1/x, \varphi_n \rangle = \int_0^1 \frac{1}{x} \varphi_n(x) dm(x) \geq \log 2$$ for all $$n$$.
• Very nice example. Just to recapitulate: $\varphi_n \to 0$ in $\tau_{\mathcal{A}}$ iff $\langle T_\varphi, \varphi_n \rangle = \int_0^1 \varphi(x) \varphi_n(x) \, dx \to 0$ for all $\varphi \in \mathcal{D}$, whereas $\varphi_n \to 0$ in $\sigma(\mathcal{D}, \mathcal{D}')$ iff $\langle T, \varphi_n \rangle \to 0$ for all $T \in \mathcal{D}'$. $T= "\frac{1}{x}"$ (i.e. $T \psi:= \int_0^1 \frac{1}{x} \psi(x) \, dx$) is a distribution that is not of the form $T = T_\varphi$ for some $\varphi \in \mathcal{D}$. So, weak convergence is strictly stronger than $\tau_\mathcal{A}$-convergence.