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I remember a professor remarking a while back that the double dual of the sequence space $l_1^{\infty}(\mathbb{R})$ is a very complicated space. I understand it must contain a copy of the original space. Past that how much is understood about the double dual? I am looking for sources on this topic, books or papers included.

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    $\begingroup$ First dual is $\ell_{\infty}$, and the second dual consists of finitely additive finite signed measures on $\mathbb{N}$. In particular, it contains ultrafilters: any ultrafilter $A$ on $\mathbb{N}$ send a sequence to its $A$-limit. $\endgroup$ Dec 17, 2015 at 10:14
  • $\begingroup$ Thank you. This is the type of answer I was looking for. $\endgroup$
    – Joe
    Dec 17, 2015 at 11:38
  • $\begingroup$ What is $\ell_1^\infty(\mathbb R)$? $\endgroup$ Dec 17, 2015 at 14:47
  • $\begingroup$ $l_1^{\infty}(\mathbb{R})$ is the space of all countably infinite sequences of real numbers which converge absolutely when summed as a series. $\endgroup$
    – Joe
    Dec 17, 2015 at 17:17
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    $\begingroup$ But the standard notion then is $\ell_1$ or $\ell^1$. $\endgroup$ Dec 18, 2015 at 7:22

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The space about whose double dual you are asking is usually denoted $\ell_1$. It is a very well-understood space, and so is its dual $\ell_1^*=\ell_\infty$. Many of the properties of $\ell_1^{**}=\ell_\infty^*$ are therefore inherited by virtue of it being the dual of a well-understood space. For instance, it is well-known that every separable space embeds into $\ell_\infty$.

A nice introduction to $\ell_\infty^*$ is given in Diestel's book Sequences and Series in Banach Spaces, p76cc. A very nice characterization is given there as follows.

Let $\Sigma$ be a $\sigma$-field of subsets of a set $\Omega$, and denote by $B(\Sigma)$ the set of bounded, $\Sigma$-measurable scalar-valued functions on $\Sigma$, and note that $B(\Sigma)$ is a Banach space when endowed with the sup norm $\|\cdot\|_\infty$. Also denote by $ba(\Sigma)$ the set of all finitely additive scalar-valued signed measures with bounded total variation, and note that $ba(\Sigma)$ is also a Banach space, as long as it is endowed with the total variation norm $\|\cdot\|_1$. We can now make the isometric identification $B(\Sigma)^*=ba(\Sigma)$ via the action $\mu(f)=\int f\,d\mu$ for all $\mu\in ba(\Sigma)$ and $f\in B(\Sigma)$. If $\Omega=\mathbb{N}$ and $\Sigma$ is the counting measure then $B(\Sigma)=\ell_\infty$, and hence $\ell_\infty^*$ is precisely the space of finitely additive signed measures on $\mathbb{N}$ with bounded total variation, denoted $ba$ for short. More information on this identification can be found in Dunford-Schwartz's Linear Operators I (on p296, it seems).

Note that, according to Diestel, you should read this paper to get a better understanding of the space $ba$. However, I have not read it myself, so beware : )

There is another natural identification, due to the fact that $\ell_\infty=C(\beta\mathbb{N})$, where $\beta\mathbb{N}$ denotes the Stone-Cech compactification of $\mathbb{N}$. Thus, $\ell_\infty^*$ can be identified with $C(\beta\mathbb{N})^*$, which in turn is identifiable with the space of regular Borel measures on $\beta\mathbb{N}$. See chapter 15 of Carother's A Short Course In Banach Space Theory on this, especially remark 2 on p152.

Also, note that this question has been asked before on MO, here. If you're wondering about the title of the question, it's because $\ell_\infty\cong L_\infty[0,1]$, and so $\ell_\infty^*\cong L_\infty[0,1]^*$ (isomorphically, not isometrically).

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    $\begingroup$ When you say $\ell_\infty=L_\infty$ is the one on the left atomic and the one on the right continuous? Because if so, then the Banach spaces are isomorphic but not isometrically so $\endgroup$
    – Yemon Choi
    Jan 3, 2016 at 4:04
  • $\begingroup$ Also, it seems off to say that every separable space is a quotient of $\ell_\infty^*$ -- that does not follow immediately from the sentence which precedes it. It is of course true that every separable Banach space is a quotient of $\ell_1$ $\endgroup$
    – Yemon Choi
    Jan 3, 2016 at 4:05
  • $\begingroup$ Oops about the quotient thing. I was thinking of about the duality between bounded below operators and surjective operators. So, the dual of an isomorphic embedding is a surjection. However, I forgot that we need it to be open, too, if we want a quotient map. Anyway, I have removed the quotient bit and clarified the $L_\infty$ bit. $\endgroup$
    – Ben W
    Jan 3, 2016 at 4:16
  • $\begingroup$ It's worth remarking that the Carothers book, while it has some interesting parts, is a bit uneven when it comes to quality and accuracy; see the remarks in www.math.leidenuniv.nl/~mdejeu/reading_guide_carothers.pdf However, I guess that the part you refer to in your answer is Garling's use of $\beta (X_d)$ to prove the Riesz representation theorem for $C(X)^*$ ? $\endgroup$
    – Yemon Choi
    Jan 3, 2016 at 14:08

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