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Let $V$ be a real vector space and let $V'$ be the algebraic dual of $V$, i.e. the space of all the linear functionals $V\to\mathbb{R}$. Then there exists the weakest topology $\tau$ which makes all the elements of $V'$ continuous, and $\tau$ is locally convex and Hausdorff. For example, if the dimension on $V$ is finite, then $\tau$ is the usual Euclidean topology.

I have two questions:

  1. Is there a commonly used name for $\tau$?

  2. Let $\tau'$ be the maximal locally convex topology on $V$, i.e. the weakest topology that makes all the seminorms on $V$ continuous. Of course $\tau'$ is finer than $\tau$, but $\tau'$ and $\tau$ could coincide in some cases (for example, for finite dimensional spaces). What is the exact relationship between $\tau$ and $\tau'$?

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Is $V$ already a topological vector space? –  Giuseppe Tortorella Apr 23 '12 at 15:11
No, it isn't. The topology tau only depends on the underlying linear structure of V. –  Roberto Frigerio Apr 23 '12 at 15:33
Just a wee addendum to the answer below. The topology $\tau'$ is the Mackey topology on $V$, i.e., the finest locally convex topology for the duality $(V,V')$, aka the topology of uniform convergence on the weakly (i.e., $\sigma(V',V)$) compact subsets of $V'$. Thus, in a certain sense, they are as far apart as possible---the finest and the weakest locally convex topologies for this duality. –  jbc Feb 16 '13 at 12:06
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2 Answers

I got into trouble in following prof. Johnson's suggestion, and now I am quite convinced that the topologies $\tau$ and $\tau'$ coincide only when $V$ is finite-dimensional.

In fact, let $\{x_i\}_{i\in I}$ be a Hamel basis for $V$, and let us consider the seminorm

$p(\sum_{i\in I} v_i x_i)=\sum_{i\in I} |v_i|$

(this seminorm is well-defined since every element in $V$ admits a unique representation as a finite linear combination of elements of the basis). Then the set $U=p^{-1}(-1,1)$ is open in $\tau'$, and does not contain any nontrivial linear subspace of $V$.

On the other hand, since $\tau$ is the weak topology with respect to a family of linear functionals (in fact, with respect to all the linear functionals), every $\tau$-neighbourhood of $0\in V$ must contain a finite-codimensional linear subspace of $V$. Therefore, if the dimension of $V$ is not finite, then the set $U$ introduced above is open with respect to $\tau'$, and not open with respect to $\tau$.

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The two topologies are the same. See e.g. Problem 20G in Kelley-Namioka "Linear Topological Spaces".

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BTW, they call the topology the "strongest locally convex topology on $V$". –  Bill Johnson Apr 23 '12 at 15:35
Bill, are you sure? this seems strange. –  Pietro Majer Apr 24 '12 at 16:10
Indeed it is. I did not think about the problem but just gave the reference and misquoted it. In K-L problem 20G, the statement is that all admissible topologies on $V^*$ (rather than on $V$ itself) are the same, which is pretty clear because $V^*$ is a product of copies of the scalar field. Roberto'a answer is the correct one. –  Bill Johnson Apr 24 '12 at 17:02
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