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Apr 19, 2015 at 6:21 review Reopen votes
Apr 19, 2015 at 15:59
Apr 18, 2015 at 21:48 history closed Michael Renardy
Dima Pasechnik
coudy
Alex Degtyarev
Joonas Ilmavirta
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Apr 18, 2015 at 16:03 review Close votes
Apr 18, 2015 at 21:48
Apr 18, 2015 at 15:54 comment added Alan Can you elaborate, I am not sure I follow. BTW, I edited and changed $L^2(M)$ to $H^1(M)$ in the space $L^\infty(I,L^2(M))\cap Lip(I,L^2(M))$.
Apr 18, 2015 at 15:51 history edited Alan CC BY-SA 3.0
edited body
Apr 18, 2015 at 15:46 comment added Michael Renardy Because that is the same as Lip.
Apr 18, 2015 at 15:07 history asked Alan CC BY-SA 3.0