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I know that $L^p([0,1])$ is not isometrically isomorphic to $l^p(\mathbb{N})$ when $p\neq 2$? But, there is an isometric copy of $l^p(\mathbb{N})$ inside $L^p([0,1])$. My question is that whether $L^p([0,1])$ can be written as an infinite direct/tensor product of $l^p(\mathbb{N})$ or $L^p([0,1])$ can be built up from countably infinite copies of $l^p(\mathbb{N})$?

Since $[0,1]$ is measurably isomorphic to $2^\mathbb{N}$ with product measure, I am feeling that $L^p([0,1])$ is isometrically isomorphic to infinite tensor product of $ \mathbb{C}^2$ with $l^p$-norm. Is this statement correct?

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  • $\begingroup$ It is correct if you define “infinite tensor product” in a correct way and equip it with the appropriate norm. $\endgroup$
    – David Gao
    Commented Jun 1 at 20:56
  • $\begingroup$ 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 $L^p(0,1)$. $\endgroup$
    – Bill Johnson
    Commented Jun 1 at 20:58
  • $\begingroup$ @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.) $\endgroup$
    – David Gao
    Commented Jun 2 at 5:21
  • $\begingroup$ @BillJohnson : How does one define an 'infinite tensor product' rigorously ? $\endgroup$
    – John Depp
    Commented Jun 2 at 14:56
  • $\begingroup$ @ DavidGao : How does one define an 'infinite tensor product' rigorously? $\endgroup$
    – John Depp
    Commented Jun 3 at 12:11

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