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.

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 $K$.
