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


*

*every $A \in \mathcal{A}$ is bounded (w.r.t. to the duality)

*every point of $y$ is contained in some $A \in \mathcal{A}$: $\ \bigcup \{ A \mid A \in \mathcal{A} \} = Y$

*$\mathcal{A}$ is directed by inclusion: $\ \forall B, C \in \mathcal{A} \, \exists A \in \mathcal{A}: \, B \cup C \subseteq A$

*$\mathcal{A}$ 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?
 A: This is an old question but perhaps others will benefit from it being answered anyways. 
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 $n$. 
