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location | ||
age | ||
visits | member for | 2 years, 3 months |
seen | Jun 15 '13 at 7:48 | |
stats | profile views | 21 |
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 $L^2(0,T;X)^* = L^2(0,T;X^*)$ |