Timeline for Banach space dual to $L^\infty(I,H^1(M))$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 23, 2015 at 13:49 | answer | added | corserine | timeline score: 2 | |
| Mar 20, 2015 at 11:40 | comment | added | weather | In the spirit of the remark of Nate Eldredge, it seems clear that you looking for a predual and it may be worth remarking that one can always identify the dual of a Bochner space $L^1(E)$ of integrable functions with values in a general Banach space with a space of equivalence classes of functions with values in $E'$ which are bounded and measurable in a suitable sense. The result is particularly simple to state when $E$ is separable and reflexive as in your situation---then one gets the space of equivalence classes of bounded, measurable functions. | |
| Mar 18, 2015 at 15:11 | comment | added | Nate Eldredge | @Alan: That is a very different question - you aren't looking for the dual but the predual. The answer is almost certainly going to be $L^1(I, H^1(M))$ and the proof will probably look very similar to the proof that $L^1(I, \mathbb{R})^* = L^\infty(I, \mathbb{R})$. The canonical reference for any question like "what is the dual of Banach space $X$" is Dunford and Schwartz, Linear Operators. Another possible place to look is Dinculeanu's Vector Measures. | |
| Mar 18, 2015 at 14:56 | comment | added | Alan | I need the dual for the theorem of Alaoglu, where I need to find the Banach space to which $L^\infty(I,H^1(M))$ is dual to. Please also provide me with references for proof of your claims. | |
| Mar 18, 2015 at 13:47 | comment | added | Yemon Choi | Well, "vector-valued $(L^\infty)^*$" doesn't seem well-defined to me. But I think the OP needs to clarify whether he really wants the full dual, or just something like a predual | |
| Mar 18, 2015 at 13:43 | comment | added | Hachino | @YemonChoi Okay, I indeed messed it up, sorry for that. Going back to the original question, does $(L^{\infty})^*(I, H^{-1}(M))$ sound like a reasonable dual space ? | |
| Mar 18, 2015 at 13:41 | comment | added | Yemon Choi | @Hachino The dual of $L^\infty$ is much bigger than ${\mathcal M}$. ${\mathcal M}$ would be the dual of $C(I)$ (assuming $I$ is compact). | |
| Mar 18, 2015 at 11:50 | comment | added | Hachino | Various questions on M.SE and MO asked for such results, here is one of them. (And now I wonder whether I messed up the dual of $L^{\infty}$ with that of $\mathcal{C}_0$. That's quite possible.) | |
| Mar 18, 2015 at 11:48 | comment | added | Alan | Do you have a reference for such a proof as you described? | |
| Mar 18, 2015 at 11:29 | comment | added | Hachino | My guess would be the space $\mathcal{M}(I, H^{-1}(M))$, where $\mathcal{M}$ denotes the space of bounded Radon measures and $H^{-1}$ is the dual of $H^1$. In short, "compose" the dual spaces in the same order. For a proof, I guess you could try to mimic the usual one which shows that $(L^{\infty})^* = \mathcal{M}$. (At least, it is rather clear that the suggested space acts canonically on your space.) | |
| Mar 18, 2015 at 10:49 | history | asked | Alan | CC BY-SA 3.0 |