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A few years ago, Schechtman showed that $\ell_p(\ell_q)$ fails to admit a greedy basis whenever $1\leq p\neq q<\infty$. This furnishes an example of a Banach space with an unconditional basis but not a greedy one. However, I am also interested in the following:

Question 1. Does there exist a Banach space $X$ admitting an unconditional basis but not a quasi-greedy basis?

It was shown in Dilworth/Kalton/Kutzarova ("On the existence of almost greedy bases") that $c_0$ is the unique (up to isomorphism) $\mathcal{L}_\infty$-space admitting a quasi-greedy basis. Thus, to find a positive answer to question 1, it suffices to find a negative answer to the following:

Question 2. Is $c_0$ the unique $\mathcal{L}_\infty$-space admitting an unconditional basis?

It is proved in Albiac/Kalton that if $K$ is metrizable then $C(K)$ admits an unconditional basis if and only if $C(K)\approx c_0$. The fact that they used metrizability in their proof makes me think perhaps there is a lurking counter-example when $K$ is not metrizable. Hence, we could ask:

Question 3. Is $c_0$ the unique $C(K)$ space (where $K$ is compact Hausdorff) admitting an unconditional basis?

As every $C(K)$ space is an $\mathcal{L}_\infty$-space, a negative answer to question 3 would give a negative answer to question 2, and hence a positive answer to question 1.

Recently, Argyros/Gasparis/Motakis showed that $X$ is a separable $\mathcal{L}_\infty$-space if and only if it is isomorphic to a so-called Bourgain Delbaen space (whose definition is complicated but can be found in their paper). Hence, question 2 is equivalent to the following:

Question 4. Is $c_0$ the unique Bourgain Delbaen space admitting an unconditional basis?

Again, a negative answer here would mean a positive answer to question 1, which is what I hope to find.

Thanks in advance guys!

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2 Answers 2

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In their Absolutely Summing Operators paper, Lindenstrauss and Pelczynski gave positive answers to questions 2, 3, and 4. You can find a proof on p. 29 of the Basic Concepts article Joram and I wrote for the Handbook of the Geometry of Banach Spaces. (This proof uses Khintchine's inequality rather than Grothendieck's inequality.) To see that you get 2 and 4, observe that the proof gives that if $P$ is a projection on $\ell_\infty^n$ and the range of $P$ has a normalized $\lambda$-unconditional basis, then the basis is $f(\lambda, \|P\|)$- equivalent to the unit vector basis of $\ell_\infty^k$, where $k$ is the rank of $P$.

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  • $\begingroup$ Thank you very much. It might have been nice to have at least one negative answer to q's #2/3/4. However, I suppose this will make answering question 1 a lot more interesting! $\endgroup$
    – Ben W
    Commented Oct 12, 2016 at 21:37
  • $\begingroup$ I'm going to "accept" this answer even though Q1 is still open. As long as the answer to Q1 is neither known nor trivial, I should probably just work on it myself. (However, if anyone has hints/ideas, feel free to say so.) $\endgroup$
    – Ben W
    Commented Oct 12, 2016 at 21:42
  • $\begingroup$ Unconditional bases are quasi-greedy. Should be immediate from the definition. $\endgroup$
    – Bunyamin Sari
    Commented Oct 13, 2016 at 4:44
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Notice that any unconditional basis is quasi-greedy.

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  • $\begingroup$ This answer is a duplicate of the comment of Bunyamin Sari. $\endgroup$
    – Stefan Kohl ♦
    Commented Mar 7, 2017 at 11:03
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    $\begingroup$ Actually, this does appear to be an attempt to answer question 1, and it's in the spirit of SE not to relegate important and relevant information to comments. However, the OP ought to acknowledge the comment, and also provide more detail besides a single declarative (or imperative) sentence. $\endgroup$
    – Todd Trimble
    Commented Mar 7, 2017 at 11:26

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