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It is known that there exist some isomorphism between $L_\infty$ and $\ell_\infty$, which is not explicit at all.

Could someone tell me whether there exist an isometric isomorphism between $L_\infty$ and $\ell_\infty$?

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    Side note (sorry): existence of this isomorphism depends essentially on the Axiom of Choice. Which is why you found it to be "not explicit at all".2012-10-05

1 Answers 1

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They aren't isometric.

We have $\ell_\infty = C(K)$ and $L_\infty = C(L)$ for $K = \beta \mathbb N$ and $L$ the Stone space of the Lebesgue algebra on $[0,1]$. By the Banach-Stone theorem an isometry would involve a homeomorphism between $K$ and $L$. Such a homeomorphism can't exist: For example, $K = \beta\mathbb{N}$ is separable while $L$ isn't separable, being the Stone space of an atomless probability algebra. In fact, every countable subset of $L$ is nowhere dense.


A more explicit and probably more elementary approach is outlined in these two exercises from page 268 in M. Fabian, P. Habala et al. Functional Analysis and Infinite-Dimensional Geometry, Springer 2001, depending on recognizing points of differentiability of the norms:

Exercises 8.30 and 8.31

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    @commenter thank you this is much more easier approach!2013-06-29