Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\ell_p(E)$?

share|cite|improve this question
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
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 (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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.