What is the dual of the space L-infinity ($L^\infty$)?

Asked Jun 05 '12 at 21:09

Active Jun 05 '12 at 21:09

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Possible Duplicate:
The Duals of $l^\infty$ and $L^{\infty}$

In learning real analysis, I do understand that the dual of $L^\infty$ cannot be $L^1$ because the latter is separable, whereas the former is not separable. So the double dual of $L^1$ is strictly larger than $L^1$ (non-reflexivity). See for example this question:

Dual of $\ell_{\infty}$ is not $\ell_1$

However, I am not able to find in my textbooks etc. the precise space which is the dual of $L^\infty(\Bbb R)$.

I want to know exactly what it is. I would very much appreciate help.

edited Apr 13 '17 at 12:21

Community

asked Jun 05 '12 at 21:09

Gandhi Viswanathan

The short answer is that the nontrivial elements of the dual aren't describable without some form of the axiom of choice. See http://mathoverflow.net/questions/22661/explicit-element-of-ell-infty-ell1 . – Qiaochu Yuan Jun 05 '12 at 21:16
It's described in IV.8.16 of Dunford and Schwartz' Linear Operators, Part I as a certain space of bounded, scalar-valued set functions. – David Mitra Jun 05 '12 at 21:27
1

See here. – David Mitra Jun 05 '12 at 21:36
weaker statement than what quaochu said : it contains elements that cannot reasonnably be represented as functions. – Albert Jun 05 '12 at 21:39
Thank you very much for all these helpful answers. – Gandhi Viswanathan Jun 06 '12 at 17:34

What is the dual of the space L-infinity ($L^\infty$)?

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