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Let $p\in (1,\infty)$ and let $(E_\alpha)_{\alpha<\omega_1}$ be a family of Banach spaces. Set $E=\left(\bigoplus_{\alpha<\omega_1}E_\alpha\right)_{\ell_p(\omega_1)}$. Must $E$ be isomorphic to $\ell_p(E)$?

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If I am understanding your notation correctly, consider the trivial case $E_\alpha = 0$ for all $\alpha$. Then $E = 0$ whereas $\ell^p(E)$ is one-dimensional. –  Nate Eldredge Mar 3 '12 at 15:27
$\ell_p(E)=\left(\bigoplus_{n\in \mathbb{N}} X_n\right)_{\ell_p}$, where $X_n = E$, $n\in \mathbb{N}$. I see no reason why it would be 1-dimensional. On the other hand, I'd like to exclude the 'trivial case'. –  Jan Veselý Mar 3 '12 at 15:35
Yes, I see that I did not understand your notation correctly after all. Disregard my comment. –  Nate Eldredge Mar 3 '12 at 15:38
Try this. $E_\alpha$ has dimension 1 for a single $\alpha$ and dimension $0$ for all the others. –  GEdgar Mar 3 '12 at 17:05
@GEdgar Jan Vesely asked to exclude trivial cases. –  Norbert Mar 3 '12 at 17:29

1 Answer 1

up vote 1 down vote accepted

I do not know if this is nontrivial enough for the OP, but I will post it just in case. The main fact we use here to give a counterexample is the fact that $\ell_p(\ell_2)$ is not isomorphic to a subspace of $\ell_p \oplus \ell_2$; there may be a more 'formal' reference for this result, but since I do not have institutional journal access let me just point the interested reader towards Proposition 23 of Ted Odell's lecture notes on $L_p$ spaces at http://congreso.us.es/cidama/activos/cursos/EOdellfull.pdf (and the references contained therein.)

Now, let us consider the OP's situation where $E_0=\ell_2$ and $E_\alpha=\ell_p$ for $0<\alpha<\omega_1$. In this case we have that $\ell_p(E)$ contains a subspace isomorphic to $\ell_p(\ell_2)$. So let us suppose by way of contraposition that in this case we do have that $E$ is isomorphic to $\ell_p(E)$. There there exists a subspace $X$ of $E$ isomorphic to $\ell_p(\ell_2)$. As $X$ is separable there exists a countably infinite set $S\subseteq \omega_1$ such that $(x_\alpha)_{\alpha<\omega_1}\in X$ and $x_\beta\neq 0$ implies $\beta \in S$; we may assume, moreover, that $0\in S$. In particular, $\ell_p(\ell_2)$ embeds isomorphically into $(\bigoplus_{\alpha\in S}E_\alpha)_{\ell_p}$, which in turn is isomorphic to $\ell_p\oplus \ell_2$ - contradicting the assertion from the previous paragraph.

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