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Jan 10 at 10:20 history edited YCor CC BY-SA 4.0
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Jun 15, 2020 at 7:27 history edited CommunityBot
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Nov 30, 2018 at 15:21 vote accept Matt R.
Nov 28, 2018 at 23:01 answer added Dirk Werner timeline score: 2
Nov 28, 2018 at 20:21 comment added Matt R. @DirkWerner: Those are good comments, and a useful reference. I'll accept them as an answer, if you make one.
Nov 28, 2018 at 6:13 comment added Dirk Werner In my previous comment I meant to say closed, not compact. There are subspaces of $L_1$ with a unit ball that is compact in measure; see G. Godefroy, N. Kalton, D. Li, J. Reine Angew. Math. 471, 43-75 (1996).
Nov 27, 2018 at 22:45 comment added Dirk Werner I guess the authors (who are they anyway?) are talking about the space $S$, also known as $L_0(\mu)$, equipped with the topology of convergence in measure, which can be generated by the metric $d(f,g)=\inf\{\varepsilon>0: \mu\{|f-g|\ge\varepsilon\}\le\varepsilon\}$. Reference to this topology is often made by saying ``in measure'', e.g., one can say that the unit ball of $L_1(\mu)$ is compact in this topology by saying it's compact in measure.
Nov 27, 2018 at 16:22 comment added Nate Eldredge Then I'm confused how this is a norm. If $V$ has measure, say, 3, then no function could have norm greater than 3 (taking $s \to 0$). So it can't be homogeneous. Am I missing something?
Nov 27, 2018 at 15:49 comment added Matt R. Yes, the inf is over $s$; I've fixed that. This topic comes up in the section on $L^p(V)$, where $V$ is a set of finite Lebesgue measure in a finite-dimensional space.
Nov 27, 2018 at 15:46 history edited Matt R. CC BY-SA 4.0
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Nov 27, 2018 at 15:36 comment added Nate Eldredge Is the inf supposed to be over $s$? Are we working on any particular measure space, or a general one? I guess it has to be an infinite measure space?
Nov 27, 2018 at 15:35 review First posts
Nov 27, 2018 at 15:44
Nov 27, 2018 at 15:32 history asked Matt R. CC BY-SA 4.0