Is there a Banach space with an unconditional basis which contains an isomorphic copy of $\ell_p$ (for some $p\in (1,\infty)$) but such that no copy of $\ell_p$ is complemented?
Copies of $\ell_p$ in spaces with unconditional basis
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0Do you have such an example for spaces without an unconditional basis? – 2012-10-04
1 Answers
The answer is yes; a good reference for what follows is Chapter 6 of Albiac and Kalton's book Topics in Banach space theory (see in particular p.130 and p.160).
The Banach spaces $L_p[0,1]$, $1 , have an unconditional basis. Moreover, if $1\leq p<2$ and $p\leq q\leq 2$, then $\ell_q$ embeds isometrically into $L_p[0,1]$. On the other hand, if $1 then $\ell_q$ embeds complementably into $L_p[0,1]$ if and only if $p=q$ or $q=2$. Thus the spaces $L_p[0,1]$, $1 , provide examples desired by the OP. P.S. With regards to the question posed in Theo's comment above, the answer is also yes in that case by considering $L_1[0,1]$, which does not have an uncondtional basis (p.144 of Albiac-Kalton) and does not have any infinite-dimensional reflexive complemented subspaces (since it has the Dunford-Pettis property).