Giorgos Petsoulas, in his paper "A class of $\ell^p$ saturated Banach spaces," has constructed for each $1<p<\infty$ a space $\mathfrak{X}_p$ which is complementably $\ell_p$-saturated but admits no unconditional basis. I was wondering if such an example has been proved for the case $p=1$.

I have managed to construct a space which is complementably $\ell_1$-saturated but admits no symmetric basis. However, it does admit a subsymmetric one---in particular, an unconditional one. I was going to adapt my technique to see if I could also kill off unconditionality. But, it would be more motivating to me to know whether this has been done before ; )

Just to be clear, when I ask for a space $X$ to be "complementably $\ell_p$-saturated" I mean that for every infinite-dimensional closed subspace $Y$ of $X$ we can find a continuous linear projection $P:X\to X$ such that $PX\subseteq Y$ and $PX\cong\ell_p$.

Thank you!

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    $\begingroup$ Take the $\ell_1$ sum of any sequence of finite dimensional spaces whose GL constants tend to infinity. $\endgroup$ Dec 7, 2015 at 14:22
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    $\begingroup$ J. Lindenstrauss showed $\ell_1$ has a subspace with no unconditional basis in On a certain subspace of $\ell_1$, Bull. Acad. Polon. Sci., 12, 539-542, 1964. $\endgroup$
    – user19038
    Dec 7, 2015 at 22:17

1 Answer 1


Szankowski (Subspaces without the approximation property. Israel J. Math. 30 (1978), no. 1-2, 123–129) for each $1\le p<2$ constructed a subspace $X$ of $\ell_p$ without the approximation property and thus without an unconditional basis. On the other hand, as a subspace of $\ell_p$, $1\le p<\infty$, the space $X$ is complementably $\ell_p$-saturated.


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