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May 3, 2013 at 14:57 vote accept Chris Gerig
Mar 12, 2013 at 16:59 answer added Connor Mooney timeline score: 20
Mar 11, 2013 at 23:12 comment added Spencer I'm not sure I quite appreciate exactly what is being asked. You lay out two cases and then say that what is essentially just a third case is an example of the other two not working. You say that the Sobolev embedding theorem "fails" or "goes wrong" when $k=n/p$, but one might say that it is simply neither of the two cases you lay out at the start. Nothing "fails", it just happens to be its own special case.
Mar 11, 2013 at 22:28 history edited Chris Gerig CC BY-SA 3.0
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Mar 9, 2013 at 20:03 history edited Chris Gerig CC BY-SA 3.0
added 314 characters in body; edited title
Mar 9, 2013 at 7:45 comment added Denis Serre actually, the embedding holds true in one critical case, namely $p=1$, $k=n$.
Mar 8, 2013 at 22:09 history asked Chris Gerig CC BY-SA 3.0