0
$\begingroup$

Let $E$ be a Hausdorff locally convex (complex or real) topological vector space, and let $E'$ be the dual of $E$, that is the vector space of all continuous linear functionals on $E$. Consider $E'$ with the weak*-topology. Can we characterize the metrizability of $E'$?

In the following answer, I show that the characterization is very simple, but since many people ignore it, I decided to share this well established result with the community of math.stackexchange.com.

$\endgroup$
1

1 Answer 1

1
$\begingroup$

We have the following result.

Theorem

Let $E$ be a Hausdorff locally convex (complex or real) topological vector space, and let $E'$ be the vector space of all continuous linear functionals on $E$. Then $E'$ with the weak*-topology is metrizable if and only if $E$ has a Hamel basis which is at most countable.

Proof. First of all note that the weak*-topology is induced by the family of seminorms $(p_x)_{x \in E}$, where for every $x \in E$ and $\Lambda \in E'$ \begin{equation} p_{x}(\Lambda)= \left| \Lambda(x) \right|. \end{equation}

Now assume that $(x_j)_{j=1}^{m}$ is a finite ($m$ positive integer) or countable ($m=\infty$) Hamel basis of $E$. Since if $y=c_1 x_1 + \dots + c_j x_j$, we have for each $\Lambda \in E'$ \begin{equation} p_{y}(\Lambda) \leq \sum_{k=1}^{j} \left| c_k \right| p_{x_k}(\Lambda) , \end{equation} the weak*-topology is then equal to that induced by the family of seminorms $(p_{x_j})_{j=1}^{m}$, so that $E'$ is metrizable (see e.g. Rudin, Functional Analysis, Remark (1.38)(c)).

Conversely, assume that $E'$ is metrizable. Then it has countable local base at $0$. Since the collection of all finite intersections of sets of the form \begin{equation} W_{x,n} = \{ \Lambda \in E' : p_x(\Lambda) < 1/n \}, \end{equation} where $x$ ranges over $E$ and $n$ over the set of positive integers, is a local base at $0$, we conclude that there exists $(x_j)_{j=1}^{m}$, where $m$ is a positive integer or $m=\infty$ and $x_j \in E$ for all positive integer $j < m+1$, such that the collection of all finite intersection of sets of the form \begin{equation} W_{x_j,n} = \{ \Lambda \in E' : p_{x_j}(\Lambda) < 1/n \}, \end{equation} where $n$ ranges over the set of positive integers and $j$ ranges over the set of positive integers less than $m + 1$, is a local base at $0$. Moreover, since as we already noted if $y=c_1 x_{1} + \dots + c_j x_{j}$ we have for each $\Lambda \in E'$ \begin{equation} p_{y}(\Lambda) \leq \sum_{k=1}^{m} \left| c_k \right| p_{x_k}(\Lambda) \end{equation} by eventually deleting some of the $x_j$'s, we can suppose that the $x_j$'s are linearly independent. Now, since $E$ does not have an a finite or countable Hamel basis, there exists $x \in E$, such that the set $\{ x \} \cup \{ x_j : j \quad \textrm{is a positive integer less than m + 1} \}$ is still a linearly independent set.

Since any finite-dimensional linear subspace $V$ of $E$ is closed (see e.g. Rudin, Functional Analysis, Theorem 1.21), from the Hahn-Banach Theorem (in the form given e.g. in Rudin, Functional Analysis, Theorem 3.5) we deduce that for each positive integer $j < m+1$ there exists $\Lambda \in E'$ such that $\Lambda(x)=1$ and $\Lambda(x_k)=0$ for $1 \leq k \leq j$. Clearly, we have for each positive integer $n$ that $\Lambda \in W_{x_1,n} \cap \dots \cap W_{x_j,n}$, but $\Lambda \notin W_{x,1}$. So we conclude that $W_{x,1}$ is not included in $W_{x_1,n} \cap \dots \cap W_{x_j,n}$. Since this is true for every positive integer $n$ and every positive integer $j <m + 1$, we conclude that the collection of all finite intersection of sets of the form \begin{equation} W_{x_j,n} = \{ \Lambda \in E' : p_{x_j}(\Lambda) < 1/n \}, \end{equation} where $n$ ranges over the set of positive integers and $j$ ranges over the set of positive integers less than $m + 1$, cannot be a local base at $0$, a contradiction.

QED

PS For the characterization of the metrizability of (what in the language of Bourbaki is called) the strong dual of $E$, see my answer in the post The Space of Distributions.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .