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Mar 25, 2020 at 20:31 history became hot network question
Mar 25, 2020 at 15:26 vote accept ABIM
Mar 25, 2020 at 15:26 answer added Jochen Wengenroth timeline score: 2
Mar 25, 2020 at 15:25 comment added ABIM Oh my brain filtered the word "steps" sorry. I've never seen this terminology, you mean the finite "sub-colimits?". In that case definitely.
Mar 25, 2020 at 15:21 comment added Jochen Wengenroth Indeed, the $L^p_{comp}$ topology is stricly finer than the $L^p$-topology, but on the steps they coincide. There is no contradiction.
Mar 25, 2020 at 15:15 comment added ABIM But in this post: mathoverflow.net/questions/347318/… don't we find that the topology on $L^p_{comp}$ is strictly finer?
Mar 25, 2020 at 15:05 comment added Jochen Wengenroth Why should this be true? The $L^p_{comp}$-topology gives nothing for the distributional derivatives! You probably know that because of the stricness of the inductive system the relative topology of $L^p_{comp}$ on the steps is just the $L^p$-topology.
S Mar 25, 2020 at 13:56 history suggested ABIM
added the pde tag
Mar 25, 2020 at 13:56 review Suggested edits
S Mar 25, 2020 at 13:56
Mar 25, 2020 at 13:43 history edited YCor CC BY-SA 4.0
formatting, partly retagged
Mar 25, 2020 at 12:23 history asked ABIM CC BY-SA 4.0