Timeline for Can $L^p([0,1])$ be built up from countably infnite copies of $l^p({F})$ , where $F$ is a finite set or $\mathbb{N}$?
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| Jun 5 at 22:09 | comment | added | David Gao | I retract my previous comment suggesting maybe $L^p$ tensor product norm can be defined via that specific inf formula. I just realized that formula does not even work when $p = 2$. But still, it is interesting whether the $L^p$ tensor product norm could be defined in some intrinsic way, without writing down the measure spaces. (After all, it can be done in case of $p = 1$ and $p = 2$. The former projective tensor norm, the latter Hilbert space tensor norm. Maybe when $1 \leq p \leq 2$, one can do this via some interpolation? Then for $p > 2$ by dualizing from the $p < 2$ case?) | |
| Jun 3 at 19:55 | comment | added | David Gao | @JohnDepp You need to fix a unit vector for each space (in this case the constant function $1$) and then consider simple tensors with all but finitely many components $1$. Equip the space of finite linear combinations of such simple tensors with the $L^p$ tensor product norm and then complete it. | |
| Jun 3 at 12:11 | comment | added | John Depp | @ DavidGao : How does one define an 'infinite tensor product' rigorously? | |
| Jun 2 at 14:56 | comment | added | John Depp | @BillJohnson : How does one define an 'infinite tensor product' rigorously ? | |
| Jun 2 at 5:21 | comment | added | David Gao | @BillJohnson Maybe an interesting question to ask is whether the $L^p$ tensor product norm can be defined abstractly. I'd assume maybe the following norm on the algebraic tensor product $L^p(X, \mu) \otimes L^p(Y, \nu)$ matches the $L^p$ tensor norm: $\|x\|_p^p = \inf\{\sum_{i=1}^n \|f_i\|^p\|g_i\|^p: x = \sum_{i=1}^n f_i \otimes g_i\}$. (It is at least true when $p = 1$, if I recall correctly.) | |
| Jun 1 at 20:58 | comment | added | Bill Johnson | The $L^p$ tensor product of $L^p(X,\mu)$ with $L^p(Y,\nu)$ is usually DEFINED to be $L^p(X\times Y,\mu\times \nu)$. If you extend this to infinite products you get just what you know; namely, the $L^p$ infinite tensor product of $\ell^p(2)$ is $ L^p(\{0,1\}^{\aleph_0})$ with the product measure when $\{0,1\}$ has the uniform probability measure. That is, you end up with $L^p$ of the Cantor group, which is isometrically isomorphic to $L^p(0,1)$. | |
| Jun 1 at 20:56 | comment | added | David Gao | It is correct if you define “infinite tensor product” in a correct way and equip it with the appropriate norm. | |
| Jun 1 at 14:04 | history | edited | John Depp | CC BY-SA 4.0 |
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| Jun 1 at 13:50 | history | asked | John Depp | CC BY-SA 4.0 |