# Mackey's topology on a dual space

I tried to understand this proposition and it's proof, but I don't understand, how is the Mackey's topology on a dual space defined.

Let $E$, $F$ be two locally convex Hausdorff topological vector space. Then $L(E'_{\sigma},\, F_{\sigma}) = L(E'_{\tau},\, F)$, where $\tau$ means Mackey's topology on $E'$, i.e. the topology of uniform convergence on the convex balanced weakly compact subsets of $E$.

1. The Mackey's topology $\tau(E,\, E')$ on $E$ was introduced in this book like the topology of uniform convergence on every convex balanced weakly compact subsets of $E'$. This why the Mackey's topology on $E'$ has to be defined with respect to the dual space of $E'$. But we don't have any topology given on $E'$. Do we have to put the weak topology on $E'$? Since $(E'_{\sigma})' \cong E$, it will be clear, why we can define $\tau$ like the topology of uniform convergence on the convex balanced weakly compact subsets of $E$. But how exactly is the Mackey's topology on $E'$ defined?

The first step in the proof of this proposition says:

If $u \colon E'_\tau \to F$ is continuous, its transpose $^t u \colon F'_{\sigma} \to E$ is continuous (as E is the dual of $E'_\tau$).

1. Why should be $E$ and the dual of $E'_\tau$ isomorphic? And why is $^t u$ continuous, when they are isomorphic?
• The Mackey topology on $E'$ is - in this setting - defined with respect to the dual pair $(E,E')$; thus it's the topology of uniform convergence on all $\sigma(E,E')$-compact absolutely convex subsets of $E$. It's the finest locally convex topology on $E'$ such that the topological dual of $E'$ is $E$. It is in general different from the Mackey topology on $E'$ with respect to the dual pair $(E',E'')$. The $\sigma,\tau, b$ topologies are always defined with respect to a dual pair, not a space alone. – Daniel Fischer Mar 9 '17 at 20:51
• @DanielFischer, thank you, it makes a lot of things clearer. But could you please explain to me, why is $^t u$ continuous? – Imperio Mar 10 '17 at 9:46