MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\tau'$?

share|cite|improve this question
Is $V$ already a topological vector space? – Giuseppe 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

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 $\tau$.

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.