# Mackey–Arens theorem and the weak topology

Let $X$ be a topological vector space and let $\mathcal {T}$ be a locally convex Hausdorff topological vector space topology on $X$. Let $X^{*}$ denote the continuous dual space of $X$ and let $X_{\mathcal {T}}$ denote $X$ with the topology $\mathcal {T}$. Then the following are equivalent:

1. The continuous dual of $X_{{{\mathcal {T}}}}$ is identical to $X^{*}$.

2. $\mathcal {T}$is identical to a $\mathcal {G'}$-topology on $X$, where $\mathcal {G'}$ is a covering of $X^{*}$ consisting of convex, balanced, $\sigma (X^{*},X)$-compact sets with the properties that:

1. If $G_{1}',G_{2}'\in {\mathcal {G'}}$ then there exists a $G'\in {\mathcal {G'}}$ such that $G_{1}'\cup G_{2}'\subseteq G'$, and
2. If $G_{1}'\in {\mathcal {G'}}$ and $\lambda$ is a scalar then there exists a $G'\in {\mathcal {G'}}$ such that $\lambda G_{1}'\subseteq G'$.

I know, that the dual of $X$ with the weak topology $\sigma(X, X^{*})$ is identical to $X^*$, but I don't understand, how this theorem implies this fact. For the weak topology we have that ${\mathcal {G'}}$ is the set of all finite subsets of $X'$, and finite subsets of $X'$ aren't convex and balanced.

Is there is another one $G'$-topology, which consiste of convex, balanced and $\sigma (X^{*},X)$-compact sets with this two properties and which coincide with the weak topology?

The point is that - for continuous linear maps to a locally convex space, in particular for continuous linear functionals - uniform convergence on a set $S$ is equivalent to uniform convergence on the closed absolutely convex hull of $S$. Let $\Gamma(S)$ denote the absolutely convex (or convex and balanced) hull of $S$. Then since $S \subset \Gamma(S) \subset \overline{\Gamma(S)}$, clearly uniform convergence on the latter implies uniform convergence on $S$. If $Y,Z$ are topological vector spaces, $f\colon Y \to Z$ is a continuous linear map, and $W\subset Z$ a closed absolutely convex set, then $f^{-1}(W)$ is a closed absolutely convex set. Thus $S \subset f^{-1}(W)$ implies $\overline{\Gamma(S)} \subset f^{-1}(W)$. In a locally convex space, the closed absolutely convex neighbourhoods of $0$ form a neighbourhood basis, so for a family $\mathcal{G}$ of subsets of $Y$, and the family $\overline{\Gamma}(\mathcal{G}) = \{\overline{\Gamma(S)} : S \in \mathcal{G}\}$ we have

$$L_{\mathcal{G}}(Y;Z) = L_{\overline{\Gamma}(\mathcal{G})}(Y;Z).$$

It is easy to verify that if $\mathcal{G}$ is a covering of $Y$ with properties 1. and 2., then so is $\overline{\Gamma}(\mathcal{G})$.

Letting $Y = X^{\ast}$, endowed with $\sigma(X^{\ast},X)$, $Z = \mathbb{C}$ (or $\mathbb{R}$; endowed with the standard topology), and $\mathcal{G}$ the family of finite subsets of $X^{\ast}$, we see that $\sigma(X,X^{\ast})$ is the $\overline{\Gamma}(\mathcal{G})$-topology. The only thing that remains is that for a finite $S = \{ \varphi_1,\dotsc, \varphi_n\} \subset X^{\ast}$ the closed absolutely convex hull is $\sigma(X^{\ast},X)$-compact. But

$$\overline{\Gamma(S)} = F(K),$$

where

$$F \colon \mathbb{C}^n \to X^{\ast};\quad \alpha \mapsto \sum_{k = 1}^n \alpha_k \varphi_k$$

is continuous since $\sigma(X^{\ast},X)$ is a vector space topology, and

$$K = \Biggl\{\alpha \in \mathbb{C}^n : \sum_{k = 1}^n \lvert \alpha_k\rvert \leqslant 1\Biggr\}$$

is compact.

• It‘s a really nice property of $\mathfrak{G}$-Topology. Thank you very much for your explanation! – Imperio Mar 8 '17 at 20:02