Timeline for Can you tell whether a space is Banach from the unit ball?
Current License: CC BY-SA 2.5
26 events
| when toggle format | what | by | license | comment | |
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| S May 28, 2015 at 6:11 | history | suggested | Ali Taghavi |
I add atag
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| May 28, 2015 at 5:53 | review | Suggested edits | |||
| S May 28, 2015 at 6:11 | |||||
| Oct 23, 2012 at 8:05 | answer | added | jbc | timeline score: 4 | |
| Mar 3, 2011 at 4:42 | vote | accept | Jim Belk | ||
| Mar 3, 2011 at 4:42 | vote | accept | Jim Belk | ||
| Mar 3, 2011 at 4:42 | |||||
| Mar 3, 2011 at 4:40 | vote | accept | Jim Belk | ||
| Mar 3, 2011 at 4:42 | |||||
| Mar 3, 2011 at 4:38 | comment | added | Jim Belk | Cool, so Bill Johnson's answer solves the case I asked about, and makes me understand that completeness is a much more general concept than I had realized. Thanks! | |
| Mar 3, 2011 at 4:20 | comment | added | Jim Belk | Ah, I see -- every topological vector space has an obvious uniform structure. | |
| Mar 2, 2011 at 22:16 | comment | added | TaQ | @Jim. Every product of complete topological vector spaces is complete. See, e.g. Jarchow's Locally Convex Spaces, Proposition 3.3.6, page 59. | |
| Mar 2, 2011 at 20:58 | comment | added | Jim Belk | I'm a bit confused by this -- $\mathbb{R}^\Omega$ is not metrizable in the product topology, so I don't see how it can be complete. | |
| Mar 2, 2011 at 16:57 | comment | added | Bill Johnson | Your ambient space, Jim, with the product topology is a complete locally convex space. If $B$ is closed in the product topology, the norm induced by $B$ is complete. | |
| Mar 2, 2011 at 16:05 | comment | added | Jim Belk | I'm picturing $V$ as something like $\mathbb{R}^\Omega$, where $\Omega$ is an uncountable set. It doesn't come with a topology, other than perhaps the infinite product topology. I find it very puzzling that there can be something so different about different convex sets in a vector space like this, and I'm wondering if there's any way to understand this difference from an external point of view. For example, suppose you are given a set $S\subset\mathbb{R}^\Omega$, and you let $B$ be the convex hull of $S\cup-S$. Under what conditions on $S$ will the norm induced by $B$ be complete? | |
| Mar 2, 2011 at 10:52 | comment | added | Andrew Stacey | Jim: How much are we allowed to assume about $V$? The problem is (as you know) that completeness is about filling in holes so I feel that I would want to assume that $V$ itself had no holes. Since I'm not allowed to specify a topology on $V$, but "holes" is definitely a topological condition, maybe I could get away with: "There is some norm on $V$ that makes $V$ a Banach space". My thinking is that then one might be able to show that if $B_1$ is such that $(V,B_1)$ is complete then $B$ and $B_1$ must be equivalent (note I say "might", I haven't thought this through). | |
| Mar 2, 2011 at 4:23 | comment | added | Jim Belk | The total convexity condition is certainly interesting, as is the nested intersection condition, though neither of these seems that easy to check. These are good conditions because they manage to avoid discussing the topology induced by $B$, which is something relatively complicated. I guess my question is whether someone living in $V$ can "see" whether a given convex set will lead to a complete norm. | |
| Mar 1, 2011 at 14:14 | comment | added | Andrew Stacey | Another silly condition is that B must be an algebra for the monad "$\ell^1$" (that sends a set to the unit ball of the $\ell^1$-space on it). Aka, $B$ must be totally convex. (silly for the same reason) | |
| Mar 1, 2011 at 14:13 | comment | added | Andrew Stacey | One (silly) condition that feels a little more "shape like" is that it is possible to draw "infinite polygons". The reason why this is silly is that it is easily equivalent to completeness. (An infinite polygon is a curve (domain [0,1]) which is piecewise linear with breaks at 1/n.) | |
| Mar 1, 2011 at 12:37 | answer | added | TaQ | timeline score: 8 | |
| Mar 1, 2011 at 4:18 | comment | added | Deane Yang | I don't see why the shape of the ball has anything to do with completeness. It is just the completeness of the ball (the lack of holes) that matters. In finite dimensions, the convexity of the ball forces completeness, but this is not true in infinite dimensions. | |
| Feb 28, 2011 at 23:11 | comment | added | Jim Belk | These observations are all of course correct, and have made me realize that "simple" isn't quite what I meant. I have edited the question to clarify what I'm looking for. | |
| Feb 28, 2011 at 23:03 | history | edited | Jim Belk | CC BY-SA 2.5 |
Clarified Question
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| Feb 28, 2011 at 20:27 | comment | added | Deane Yang | Another way to say what's been said already: First, if the vector space is finite-dimensional, then you get a topology free of charge and it is always both consistent with the norm defined by $B$ and complete. If the vector space is infinite-dimensional, then $B$ defines a topology (via the norm). It seems to me that the space is complete if and only if the completion of $B$ with respect to this topology lies in $V$. | |
| Feb 28, 2011 at 19:49 | answer | added | Matthew Daws | timeline score: 13 | |
| Feb 28, 2011 at 19:48 | comment | added | Pietro Majer | Translating the notion of 'converging sequence' and of 'Cauchy sequence' in terms of $B$ in place of the norm is quite immediate. The resulting formulation of completeness in terms of $B$ is not that different form the usual one. Do you expect a simpler way than that? | |
| Feb 28, 2011 at 18:46 | comment | added | Mark Meckes | More directly to the point, if the dimension is finite all norms give $V$ a complete metric structure. An easy necessary condition for completeness is that $B$ satisfies a version of the nested interval property: any nested sequence of translates of dilates of $B$ has a nonempty intersection. | |
| Feb 28, 2011 at 17:46 | comment | added | Adam Hughes | If the dimension is finite all norms induce the same topology, so that's something. | |
| Feb 28, 2011 at 17:35 | history | asked | Jim Belk | CC BY-SA 2.5 |