Timeline for Tensor product of measure spaces
Current License: CC BY-SA 3.0
16 events
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| Jun 13, 2017 at 21:08 | comment | added | Yemon Choi | Basically, my intuition is that $M(X)$ is an $L_1$-ish kind of space, and you don't expect the injective tensor product of two $L^1$-ish things to be $L^1$-ish -- $\ell^1$ plays well with projective tensor product instead. Dually, the projective tensor product of two $C(K)$-ish things won't be $C(K)$-ish, in general | |
| Jun 13, 2017 at 21:06 | comment | added | Yemon Choi | Regarding the 2nd edit: if by Bil you mean all bilinear maps, then this space is sometimes known as the space of bimeasures. It is known in the theory of Banach spaces that not every bimeasure on $X\times Y$ is a measure (in the literature, this is often mentioned alongside results such as Grothendieck's inequality ). I don't recall if this fact is in the book of Ryan that @MatthewDaws has referred to, but that might be one place to look | |
| Jun 8, 2017 at 13:12 | comment | added | Matthias Ludewig | Thank you for your comments! Please see the edits above. | |
| Jun 8, 2017 at 13:10 | history | edited | Matthias Ludewig | CC BY-SA 3.0 |
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| Jun 8, 2017 at 12:24 | history | edited | Matthias Ludewig | CC BY-SA 3.0 |
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| Jun 8, 2017 at 12:23 | vote | accept | Matthias Ludewig | ||
| Jun 7, 2017 at 0:10 | history | edited | Yemon Choi | CC BY-SA 3.0 |
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| Jun 6, 2017 at 20:33 | answer | added | Matthew Daws | timeline score: 4 | |
| Jun 5, 2017 at 20:40 | history | edited | Matthias Ludewig | CC BY-SA 3.0 |
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| Jun 2, 2017 at 8:55 | history | edited | Matthias Ludewig | CC BY-SA 3.0 |
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| Jun 1, 2017 at 21:17 | comment | added | Matthew Daws | @YemonChoi: Yes, of course! So, correctly, the bounded bilinear maps on $C(X)\times C(Y)$ is just the bounded linear maps $C(X)\hat\otimes_\pi C(Y) \rightarrow \mathbb C$, i.e. the dual space, which we can identify with the bounded linear map $C(X)\rightarrow C(Y)^* = M(Y)$. Thus $M(X) \hat\otimes_\epsilon M(Y)$ is in general only a subspace (it gives you just the compact maps $C(X)\rightarrow M(Y)$. | |
| Jun 1, 2017 at 19:14 | comment | added | Yemon Choi | Question for the OP: I agree that $M(X)\hat\otimes_\pi M(Y)$ should embed isometrically inside $M(X\times Y)$ but I don't see why we should get equality for, say, $X=Y=[0,1]$. | |
| Jun 1, 2017 at 19:12 | comment | added | Yemon Choi | @MatthewDaws Are you sure about that last one? If $X$ and $Y$ are finite sets of size $n$ then $M(X)\hat\otimes_\pi M(Y)$ is just $\ell_1^{n^2}$. But the bilinear functionals on $\ell_\infty^n$ should not have the same norm as the linear functionals on $\ell_\infty^{n^2}$ ... | |
| Jun 1, 2017 at 14:25 | comment | added | Matthew Daws | $\epsilon$ and $\overline\epsilon$ always agree, for any (dual) Banach spaces. Good references here are Diestel and Uhl's book on Vector Measures, or a personal favourite, Ryan's book "Introduction to tensor products of Banach spaces". The proof is basically just en.wikipedia.org/wiki/Goldstine_theorem The projective tensor product linearises bilinear maps, so I think (unless I misunderstand) that $B(C(X)\times C(Y))$ is $M(X) \hat\otimes_\pi M(Y)$. | |
| Jun 1, 2017 at 14:17 | comment | added | Nate Eldredge | In your first paragraph, you need a word like "Radon" or "regular" or maybe "metrizable"... | |
| Jun 1, 2017 at 13:44 | history | asked | Matthias Ludewig | CC BY-SA 3.0 |