lollypop

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seen Jun 15 '13 at 7:48

May
30
revised $H^1(\Omega) \subset L^2(\Omega)$ dense for $\Omega$ a $C^l$ hypersurface with boundary?
edited body
May
30
asked $H^1(\Omega) \subset L^2(\Omega)$ dense for $\Omega$ a $C^l$ hypersurface with boundary?
May
24
awarded  Scholar
May
18
comment Proof that $L^2(0,T;X)^* = L^2(0,T;X^*)$
@AndrasBatkai Is it as simple as: $L^2(0,T;X)' = L^2(0,T;X) = L^2(0,T;X')$? Where the first equality is by Riesz representation theorem (RRT) for the Bochner space (which is Hilbert), and the second is by RRT for $X$ which is also Hilbert. Would this be a good proof?? (The longer way to do this requires us to show that the map is isometric, which seems difficult to do even in this case.)
May
17
accepted Proof that $L^2(0,T;X)^* = L^2(0,T;X^*)$
May
17
comment Proof that $L^2(0,T;X)^* = L^2(0,T;X^*)$
@Andras you are right. But I am interested in more general $X$, but your answers caters for this. Thanks.
May
16
awarded  Student
May
16
awarded  Editor
May
16
comment Proof that $L^2(0,T;X)^* = L^2(0,T;X^*)$
@Wlodzimierz I added some details. These are Bochner spaces.
May
16
revised Proof that $L^2(0,T;X)^* = L^2(0,T;X^*)$
added 102 characters in body; added 4 characters in body; deleted 2 characters in body
May
16
asked Proof that $L^2(0,T;X)^* = L^2(0,T;X^*)$