Mackey topology vs. uniform convergence on weakly compact convex sets

Let $X$ be a locally convex Hausdorff space and $X'$ its dual space. By the Mackey-Arens theorem, there is a finest locally convex topology $\tau$ on $Y$ such that $(Y, \tau)' = X$. $\tau$ is called the Mackey topology and it can be charaterized as the topology of uniform convergence on weakly compact convex balanced subsets of $X$.

In some literature, the authors sometimes refer to the topology of uniform convergence on weakly compact convex sets as the Mackey topology --- just google for "uniform convergence on weakly compact convex" "mackey".

Are these topologies really equal or are the authors just a little bit sloppy?

Edit: I think, the equality of these two topologies is at least true if $X$ is weakly complete since then the weak closure of the convex balanced hull of a weakly compact set remains weakly compact. For a general locally convex space the convex balanced hull of a compact set is only precompact and the closure need not be compact.

• But here we are starting from a (weakly compact) convex set, and that prevents the closed convex balanced hull from being incomplete. Commented Nov 14, 2016 at 14:52

The topologies are equal. If the scalar field is $\mathbb{R}$, for a (nonempty) weakly compact convex $K \subset X$, consider the convex hull of $K \cup (-K)$. Since $K$ and $-K$ are convex,

$$\operatorname{conv}(K \cup (-K)) = \{ t k_1 - (1-t)k_2 : k_1, k_2 \in K\} = f(I\times K\times (-K)),$$

where $f \colon (t,x,y) \mapsto t x + (1-t)y$ is weakly continuous, so $\operatorname{conv}(K\cup (-K))$ is weakly compact and convex. Since $\operatorname{conv}(K \cup (-K))$ is also clearly symmetric, it follows that it's balanced. Thus uniform convergence on all weakly compact balanced convex sets implies uniform convergence on $K$.

If the scalar field is $\mathbb{C}$, consider the convex hull $C$ of $\sqrt{2}\cdot(K \cup iK \cup (-K) \cup (-iK))$. By essentially the same argument as above, it is weakly compact, and therefore

$$\bigcap_{\lvert z\rvert = 1} z\cdot C$$

is a weakly compact balanced convex set. And that set contains $K$.

• I see, thank you. If I see correctly, you have proven more, namely that if $X$ is any topological vector space (and not only a space equipped with its weak topology) then every compact convex set is contained in a compact convex balanced set ($f$ remains continuous since $X$ is a TVS). In particular, the topologies of uniform convergence w.r.t. both classes are the same.
Commented Nov 14, 2016 at 15:31
• Yes, the argument doesn't use any specific property of the topology other than being a vector space topology. Commented Nov 14, 2016 at 15:33
• I found your proof for the real case in Köthe, "Topological Vector Spaces I" p. 242. The theorem there states that the absolutely convex cover of a compact convex set remains compact (and convex) in any real locally convex Hausdorff space. However, for complex spaces the absolutely convex cover is at least relatively compact and need not be closed. A counterexample is given on p. 243 with the underlying space being $\omega = \mathbb{C}^{\mathbb{N}}$. Nevertheless, the Mackey topology and the topology of uniform convergence on weakly compact convex sets coincide.
• Yes, in the real case, the absolutely convex hull of a convex set $C$ is the convex hull of the union of two convex sets ($C$ and $-C$), and since the convex hull of the union of finitely many convex sets is the image of $\Sigma \times A_1 \times \dotsc \times A_n$ (where $\Sigma$ is the $n-1$-dimensional standard simplex) under the continuous map $((t_1,\dotsc,t_n),a_1,\dotsc,a_n) \mapsto t_1a_1 + \dotsc + t_na_n$, the convex hull of finitely many compact convex sets is again compact. Since the unit circle has infinitely many points, it's not as nice over $\mathbb{C}$. Commented Nov 15, 2016 at 11:43