Examples for topological vector space

Question

I'm looking for examples for a topological vector space that is $T_1$, locally convex but not metrizable.
I know that a topological vector space is metrizable if, and only if, it is both $T_1$ and $C_1$ (first countable), but I don't know of certain examples that are not first countable.

Answer 1

A very frequently used example is the space under "weak topology". For a Banach space $X$ the following statements are equivalent.

(1) $X$ is finite dimensional.

(2) The weak topology is metrizable.

(3) The weak$^*$ topology is metrizable.

We recall that $X$ under weak topology and $X^*$ under weak$^*$ topology are always locally convex. So if $X$ is of infinite dimensional, then $X$ under weak topology $\sigma(X,X^*)$ is not metrizable.

On the other hand, a normed space $X$ is separable if and only if the closed unit ball $B^*$ of its dual space is $w^*$-metrizable. That is, if $X$ is not separable, then $B^*$ under weak$^*$ topology is NOT metrizable.

For the sources of the statements above: The first statement is from Theorem 3.33 in "Positive Operators" by Charalambos D. Aliprantis and Owen Burkinshaw. The second statement is from Theorem 6.30 in "Infinite-Dimensional-Analysis" by by Charalambos D. Aliprantis.

Answer 2

In analysis, a very important example of these types of spaces is that of Test Functions which is the base for the theory of distributions (generalized functions).

Answer 3

Let $\Bbb R$ have the usual topology. Let $X=\Bbb R^{\Bbb N}$ with the box topology $T_B$ on X. A base for $T_B$ is the family of all $\prod_{n\in \Bbb N}S_n$ where each $S_n$ is open in $\Bbb R.$ The points of $X$ are the functions from $\Bbb N$ to $\Bbb R.$ Define $f+rg$ by $(f+rg)(n)=f(n)+rg(n)$ for $f,g \in X, \;r\in \Bbb R,$ and $n\in \Bbb N.$ You may confirm that $(X,T_B)$ is a locally convex topological vector space over $\Bbb R$ and that it is a Hausdorff space (In fact it is a completely regular (Tychonoff) space.)

Let $\Bbb U=\{U_j: j\in \Bbb N\}$ be a family of neighborhoods of $f\in X.$ For each $j$ take $g_j:\Bbb N\to \Bbb R^+$ such that $U_j\supset \{h\in X: \forall n\in \Bbb N\;(|h(n)-f(n)|<g_j(n))\}.$

Let $V=\{h\in X: \forall n\in \Bbb N\;(|h(n)-f(n)|<g_n(n)/3)\}.$

Then $V$ is a nbhd of $f$ and $\forall j\in \Bbb N\; (\neg (U_j \subset V)).$ Because if $h_j(n)=f(n)$ when $n\ne j$ and $h_j(j)=f(j)+2g_j(j)/3$ then $h_j\in U_j$ \ $V.$

So $\Bbb U$ cannot be a nbhd base at $f.$

Answer 4

Let $X$ be any Tychonoff topological space, and define $C_p(X)$ as the space of all continuous functions $f:X \to \mathbb{R}$, and give this the topology inherited from the product topology on $\mathbb{R}^X$. This makes $C_p(X)$ into a topological vector space, locally convex, as basic open neighbourhoods are of the form $$U(f, x_1, \ldots,x _n, \varepsilon) = \{g \in C_p(X): |g(x_i ) - f(x_i)| < \varepsilon\}, \text{ where }: \\ f \in C_p(X), n \in \mathbb{N}, x_1 ,\ldots x_n \in X, \varepsilon >0$$

and these are convex.

And here I show that $C_p(X)$ is first countable iff $X$ is countable. So e.g. $C_p([0,1])$ is an example of such a vector space. This is a well-studied and interesting class of TVS's.