# Are there spaces “smaller” than $c_0$ whose dual is $\ell^1$?

It is well known that both the sequence spaces $c$ and $c_0$ have duals which are isometrically isomorphic to $\ell^1$. Now, $c_0$ is a subspace of $c$. My question - is there an even smaller subspace of $c$ whose dual is isometrically isomorphic to $\ell^1$? More generally, is there a characterization of preduals of $\ell^1$? References are most welcome.

• I'm not sure if this has an answer but it is extremely interesting (I've only read the first page or so for now): arxiv.org/pdf/1102.4325.pdf . $\ell^1$ does not have a unique predual which is really quite surprising to me. – Cameron Williams Oct 10 '14 at 3:37
• Your question is not well posed because you have not given a rigorous definition of smaller. Indeed as sets $c_0$ is smaller that $c$, but in fact $c_0$ is isomorphic to $c$ as Banach spaces. – Norbert Oct 10 '14 at 6:38
• @noidentity yes. it's not well-posed, since I didn't define what exactly I mean by smaller. I assume that smaller means a strictly smaller subspace, but then realized that it could be given different meanings, so I just left it within quotes. – Nirakar Neo Oct 11 '14 at 19:07
• @CameronWilliams yeah. That seems an interesting read, and $\ell^1$ is indeed pathological in this sense. – Nirakar Neo Oct 11 '14 at 19:07

Preduals of $\ell_1$ are very interesting creatures. The question is slightly ill-posed but let me comment on that anyway.

We have quite a lot of isometric preduals of $\ell_1$. Indeed, for any countably infinite ordinal number $\alpha$ (endowed with the order topology) we have $C_0(\alpha)^* \cong \ell_1$. This essentially follows from the Riesz–Markov–Kakutani representation theorem as every measure on such a space has to be atomic.

Note that $c_0 = C_0(\omega)$ and $c=C(\omega+1)$ are isomorphic. Actually for each $\alpha\in [\omega, \omega^\omega)$ we have $C_0(\alpha) \cong c_0$. However, there exist $\aleph_1$ many countable ordinals which give pair-wise non-isomorphic Banach spaces. To be more precise, if $\alpha, \beta$ are countable ordinals then $C(\omega^{\omega^\alpha}+1)$ and $C(\omega^{\omega^\beta}+1)$ are isomorphic if and only if $\alpha = \beta$.

Historically, the first example different from the above-mentioned ones was due to Y. Benyamini and J. Lindenstrauss:

Y. Benyamini and J. Lindenstrauss. A predual of $\ell_1$ which is not isomorphic to a $C(K)$ space, Israel J. Math. 13 (1972), 246–254.

Johnson and Zippin proved that every isometric predual is a quotient of $C(\Delta)$, where $\Delta$ is the Cantor set.

W.B. Johnson and M. Zippin, Every separable predual of an $L_1$-space is a quotient of $C(\Delta)$, Israel J. Math. 16 (1973), 198–202.

This result combined with an old result of Pełczyński

A. Pełczyński, Banach spaces on which every unconditionally converging operator is weakly compact, Bull. Acad. Pol. Sci. Ser. Math. Astr. Phys. 10 (1962), 265–270.

asserting that an operator $T\colon C(K)\to X$ is weakly compact if and only if it is not bounded below on any isomorphic copy of $c_0$ yields the following corollary:

Corollary. Each isometric predual of $\ell_1$ contains a subspace isomorphic to $c_0$.

In some sense this answers OP's question.

One may wonder whether the Cantor set in the statement of the Johnson–Zippin theorem may be replaced by a countable compact Hausdorff space. This is not the case as shown by D. Alspach:

D. E. Alspach, A $\ell_1$-predual which is not isometric to a quotient of $C(\alpha)$, Contemp. Math. 144, Amer. Math. Soc., (1993), 9–14.

On the other hand, Gasparis constructed some new preduals of $\ell_1$ which are quotients of $C(\alpha)$:

I. Gasparis, A class of $\ell_1$-preduals which are isomorphic to quotients of $C(\omega^\omega)$, Studia Math. 133 (2) (1999), 131–143.

Since there is no characterisation of complemented subspaces of $C(K)$-spaces, this result is also noteworthy:

I. Gasparis, A new isomorphic $\ell_1$ predual not isomorphic to a complemented subspace of a $C(K)$ space, Bull. London Math. Soc. 45 (2013), 789–799.

If you are interested in spaces whose dual space is only isomorphic to $\ell_1$ then the situation is even more exciting. J. Bourgain and F. Delbaen constructed isomorphic preduals without subspaces isomorphic to $c_0$:

J. Bourgain and F. Delbaen, A class of special $\mathscr{L}_\infty$-spaces, Acta Math. 145 (1980), 155–176.

A recent variation of the BD construction is the famous Argyros–Haydon space which is an isomorphic $\ell_1$-predual with the property that each operator on this space is of the form $cI + K$, where $c$ is a scalar and $K$ is a compact operator.

S.A. Argyros and R.G. Haydon, A Hereditarily Indecomposable $\mathscr{L}_\infty$-space that solves the scalar-plus-compact problem, Acta Math. 206 (2011), no. 1, 1–54.

Even more recently, Spiros A. Argyros, Ioannis Gasparis and Pavlos Motakis constructed a $c_0$-asymptotic $\ell_1$-predual without copies of $c_0$:

S.A. Argyros, I. Gasparis and P. Motakis, On the structure of separable $\mathcal{L}_\infty$-spaces, Mathematika 62 (2016), no. 3, 685–700.

A nice overview of the BD-spaces (and some new variants of the Argyros–Haydon space) can be found in Matt Tarbard's PhD thesis:

M. Tarbard, Operators on Banach Spaces of Bourgain–Delbaen Type, PhD Thesis, University of Oxford, 2012.

I would also recommend the following paper by M. Daws, R. Haydon, Th. Schlumprecht and S. White

M. Daws, R. Haydon, Th. Schlumprecht, and S. White, Shift invariant preduals of $\ell_1(\mathbb{Z})$, Israel Journal of Mathematics, 192 (2012), 541–585.

which explains these subtleties very well.

Laustsen and myself have exhibited a version of the Argyros–Haydon space whose algebra of bounded operators has certain peculiar properties:

T. Kania and N.J. Laustsen, Ideal structure of the algebra of bounded operators acting on a Banach space, Indiana University Mathematics Journal 66 (2017), 1019–1043.

Let me finish with an open problem (I think) which I like very much:

Problem. We can easily construct $\aleph_1$ many isometric preduals of $\ell_1$. Can we construct in $\mathsf{ZFC}$ continuum many isometric preduals of $\ell_1$?

• Thank you for the excellent answer. That helps a lot. I guess I'll have to go a bit deeper in to measure theory. – Nirakar Neo Oct 11 '14 at 19:09