Mackey–Arens theorem

Remarks

The weak topology is compatible by construction, and it is the coarsest topology for which all the forms $\langle \cdot, y \rangle$ are continuous. If $\mathcal{T}$ is compatible, each $\mathcal{T}$-neighbourhood of $0$ contains a closed convex balanced neighbourhood $U$ of $0$. Its polar $U^\circ$ in $Y$ is convex, balanced and $\sigma(Y, X)$-compact by the Banach–Alaoglu theorem, and $U = U^{\circ\circ}$ by the bipolar theorem. So $U$ is a $\tau(X, Y)$-neighbourhood of $0$, which gives $\mathcal{T} \subseteq \tau(X, Y)$. That $\tau(X, Y)$ itself is compatible is the harder half; it uses the bipolar theorem and a compactness argument in the algebraic dual of $Y$. Example: for a normed space $E$, the weak topology $\sigma(E, E')$ and the norm topology are both compatible with the pair $(E, E')$, so they have the same continuous linear forms. A consequence: all compatible topologies have the same closed convex sets. Indeed, by the Hahn–Banach separation theorem, a closed convex set is an intersection of closed half-spaces defined by continuous forms.

Going deeper

For a student

In duality theory one often starts from a dual pair and asks which topologies on one side give the other side as dual. The Mackey–Arens theorem answers this completely. It explains why closed convex sets and bounded sets depend only on the dual pair: they are the same for all compatible topologies. For bounded sets this is Mackey's theorem. It also shows that the Mackey topology is natural: metrizable and barrelled locally convex spaces carry their Mackey topology. The proof rests on the bipolar theorem and on Banach–Alaoglu compactness.