Timeline for Sufficient condition such that weak and initial topology coincide for a locally convex space
Current License: CC BY-SA 3.0
12 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| May 26, 2014 at 15:25 | vote | accept | AlexE | ||
| May 26, 2014 at 14:11 | answer | added | Sergei Akbarov | timeline score: 2 | |
| Feb 5, 2014 at 18:24 | history | edited | AlexE | CC BY-SA 3.0 |
deleted 3 characters in body
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| Feb 5, 2014 at 17:12 | answer | added | alpha | timeline score: 2 | |
| Feb 5, 2014 at 9:04 | comment | added | AlexE | @alpha: Sorry, but if your last comment was meant as an answer to my previous comment, then it does not make any sense to me. | |
| Feb 5, 2014 at 8:54 | comment | added | alpha | Not with the weak topology: with the initial locally convex topology. | |
| Feb 5, 2014 at 8:40 | comment | added | AlexE | @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? | |
| Feb 4, 2014 at 15:44 | comment | added | alpha | 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 | |
| Feb 4, 2014 at 12:34 | comment | added | AlexE | @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. | |
| Feb 3, 2014 at 15:52 | comment | added | alpha | 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). | |
| Feb 3, 2014 at 9:16 | history | asked | AlexE | CC BY-SA 3.0 |