This is the opposite question to this one: Example of locally convex space such that its weak and initial topology coincide.

If we have a normed vector space $X$ than its norm topology and weak topology coincide if and only if $X$ is finite dimensional. Now I'm interested in the general case of $X$ being just locally convex.

If $X$ is a locally convex space (which is not normed) what are sufficient conditions such that the weak topology and the initial topology of $X$ do not coincide?

I think that there must be at least one quite general condition, since I think that it is rather rare that we have a locally convex space fulfilling the condition that its weak and initial topology coincide.

  • $\begingroup$ The condition is that the locally convex space support a linear functional which is not continuous. One usually needs to apply the axiom of choice (in the form of the existence of a Hamel basis) in order to apply this to concrete spaces (e.g., as you note, for infinite dimensional normed spaces). $\endgroup$ – alpha Feb 3 '14 at 15:52
  • $\begingroup$ @alpha: I don't understand your comment completely. The linear functional should be discontinuous with respect to which topology? And how do we see then that the two topologies on X do not coincide? Or do you mean that one has to find a functional which is continuous against the initial topology and discontinuous against the weak topology? But that would be just a restatement of the question. $\endgroup$ – AlexE Feb 4 '14 at 12:34
  • $\begingroup$ Your formulation postulates a locally convex space. The condition is that there should be a linear functional which is not continuous for this topology. Thus in the case of a normed, infinitely dimensional space, you have such a functional and so the initial and weak topology aee distinct as you already knew $\endgroup$ – alpha Feb 4 '14 at 15:44
  • $\begingroup$ @alpha: So you claim that given a locally convex space $X$ such that its topology coincides with the weak topology, then every linear map $X \to \mathbb{K}$ will be continuous? Can you provide a proof please, since I do not see this? $\endgroup$ – AlexE Feb 5 '14 at 8:40
  • $\begingroup$ Not with the weak topology: with the initial locally convex topology. $\endgroup$ – alpha Feb 5 '14 at 8:54

The following is mentioned in the book by H.Schaefer (IV, Exercise 6(a)), but with a little bit different formulation:

Theorem. A complete locally convex space $X$ over $\mathbb K$ ($\mathbb K=\mathbb R$ or $\mathbb C$) has weak topology (i.e. the topology of $X$ coincides with the weak topology on $X$ generated by all linear continuous functionals $f:X\to{\mathbb K}$) if and only if $X$ is isomorphic (as a locally convex space) to the direct product of a family of copies of $\mathbb K$: $$ X\cong{\mathbb K}^{\mathfrak m} $$ for some cardinality ${\mathfrak m}$.

Corollary. If $X$ is complete and $X\not\cong{\mathbb K}^{\mathfrak m}$, then the topology of $X$ can't be weak. In particular, a Fréchet space $X$ has weak topology if and only if $X\cong{\mathbb K}^{\mathbb N}$.


EDIT: Have just realised that I misunderstood your question. The first paragraph is stil valid and I am leaving the last three paragraphs to show where I misunderstood you. I thought that under initial topology you meant the finest l.c. topology on $E$.

An infinite dimensional metrisable l.c.s. is never complete for the weak topology. So any complete, infinite dimensional space wii do what you want (assuming I have now understood your question correctly). Sorry for the confusion.

THE SOLUTION TO MY UNDERSTANDING OF YOUR QUESTION. Each vector space has a finest l.c. topology. It can be defined as that l.c. structure which is defined by ALL seminorms on $E$ or as the l.c. inductive limit of the the finite dimensional subspaces. An example of an l.c. space for which this coincides with its weak topology is $\phi$, the space of finite sequences, with the natural inductive limit structure. Its topological and algebraic dual is $\omega$, the space of ALL sequences. For this see G. Köthe: Topological linear spaces.

If a l.c. space has a non continuous linear functional, then the above two topologies cannot coincide for the simple reason that they have distinct duals.

A host of such spaces can be found as follows: a separable l.c.s. whose linear dimension is uncountable automatically has a non continuous linear functional.

The last fact uses AC in the form of the existence of a Hamel basis. Work of Solovay, Schwartz and Garnir shows that without AC the situation is different.

  • $\begingroup$ Ok, I see. The misunderstanding was probably due to the fact that I had a typo in my question (I wrote "initial" were "weak" should have been) - sorry for that. I edited my question, so that it is correct now. $\endgroup$ – AlexE Feb 5 '14 at 18:27
  • $\begingroup$ Do you know a reference for the fact that the weak topology on an infinite-dimensional, metrizable locally convex space is never complete? $\endgroup$ – AlexE Feb 5 '14 at 18:38
  • $\begingroup$ This is strange for me. What is the connection between the finest and the weak topologies on $X$? $\endgroup$ – Sergei Akbarov May 26 '14 at 14:18

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