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Given a Banach space $X$ and a measure space $(\mathfrak{A}, \mu)$ One can form the Banach space $L_\infty(\mu, X)$ of all measurable, essentially bounded functions from $\mathfrak{A}$ to $X$. Is it obvious that one can identify $L_\infty(\mu, X)$ with the projective tensor product $L_\infty(\mu)\hat{\otimes}X$? If I am mistaken (that is, this is not true), can we find another tensor product to make such identification?

EDIT: Since the answer is 'no', I would appreciate any other 'reasonable' descriptions of the Banach space $L_\infty(\mu, X)$, if there are any.

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    I ask about $L_\infty(\mu, X)$. Thanks for keeping an eye on my edits.2012-05-07

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I don't know what counts as "reasonable", but here's one idea. You can form the projective tensor product $L^1(\mu) \widehat\otimes X^*$, and this is isometrically isomorphic to $L^1(\mu,X^*)$. Then $L^\infty(\mu,X)$ always forms a subspace of $L^1(\mu,X^*)^*$. But we can identify the latter as $ L^1(\mu,X^*)^* = ( L^1(\mu) \widehat\otimes X^* )^* = B(L^1(\mu), X^{**}) $ that is, bounded linear maps $L^1(\mu) \rightarrow X^{**}$. So you can always view $L^\infty(\mu,X)$ as a collection of bounded linear maps $L^1(\mu) \rightarrow X^{**}$. In fact, it's easy to see that you always get a map $L^1(\mu) \rightarrow X$. I don't know if that's any help...

Aside: It's a subtle question as to when $L^\infty(\mu,Y^*) = L^1(\mu,Y)^*$. This holds if and only if $Y^*$ has the Radon-Nikodym Property-- for example, if $Y$ is reflexive, or $Y^*$ is separable. This can be found in e.g. the Appendix of Defant+Floret's book, or the book of Diestel and Uhl.

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    Ah, yep, finite measures! Thanks... I guess if your space is "decomposable", so $L^1(\mu) = \ell^1(L^1(\mu_i))$ for some collection of finite measures $\mu_i$ (with associated measurability criteria) then you could decompose everything, and work with block matrices-- i.e. messy, but not impossible.2012-05-08