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Jul 8, 2013 at 5:08 comment added Dan Lee To answer my own question, I suppose that the Schauder fixed point theorem (being an infinite dimensional analog of the Brouwer fixed point theorem) provides some understanding of the case of the unit ball.
Jul 8, 2013 at 4:46 history edited Dan Lee CC BY-SA 3.0
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Jul 8, 2013 at 4:43 comment added Dan Lee Given how embarrassingly wrong my original premise was, it's a bit awkward to ask this, but I still wonder if something interesting can be said about the unit ball in Banach space.
Jul 8, 2013 at 4:19 vote accept Dan Lee
Jul 7, 2013 at 11:00 review Close votes
Jul 7, 2013 at 16:50
Jul 7, 2013 at 10:40 comment added Ryan Budney Dan, what you claim as a "simple fact" is not true. Take for example $S^1 \times [0,1]$, this has an inward-pointing everywhere non-zero vector field. The Poincare-Hopf index theorem is what tells you when you have to have a zero, and that's given in terms of the Euler characteristic of the manifold.
Jul 7, 2013 at 10:21 answer added Vivek Shende timeline score: 5
Jul 7, 2013 at 0:35 comment added Dan Lee I know that the answer can't be super simple since there is no compactness, but maybe there are hypotheses that make it work. (For example, something like the Palais-Smale condition.) Even one specific situation in the literature where such an argument was used would be interesting to me.
Jul 6, 2013 at 20:58 comment added Matthias Ludewig This depends on the topology though. In general, closed balls can be compact if the model space is not Banach.
Jul 6, 2013 at 20:52 comment added Ben McKay I would bet against because compactness isn't available. Closed balls aren't compact.
Jul 6, 2013 at 20:41 history asked Dan Lee CC BY-SA 3.0