Mackey topology

Remarks

Each $p_K$ is finite, because $y \mapsto \langle x, y \rangle$ is $\sigma(Y, X)$-continuous and $K$ is $\sigma(Y, X)$-compact. For a locally convex space $E$, the Mackey topology of $E$ is $\tau(E, E')$, and $E$ is called a Mackey space if its topology equals $\tau(E, E')$. Example: every metrizable locally convex space, in particular every normed space, is a Mackey space. Check for a Banach space $E$: the closed unit ball $K$ of $E'$ is convex, balanced and weak-* compact by the Banach–Alaoglu theorem, and $p_K(x) = \norm{x}$ by the Hahn–Banach theorem. So $\tau(E, E')$ is at least as fine as the norm topology. Conversely, each weak-* compact $K \subseteq E'$ is bounded at each point of $E$, hence norm bounded by the uniform boundedness principle, say by $r$. Then $p_K(x) \le r \norm{x}$, so $\tau(E, E')$ is no finer than the norm topology. Non-example: on an infinite-dimensional normed space $E$, the weak topology $\sigma(E, E')$ has the same continuous dual $E'$ but is strictly coarser than $\tau(E, E')$, so it is not the Mackey topology.