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Let $(E_n)$ be a sequence of Banach spaces and $(w_n)$ be a sequence of positive real numbers. For $1\leq p <\infty$ define $\bigoplus\limits_P E_n:=\{(x_n):x_n\in E_n,\sum\limits_n\lVert x_n\rVert^pw_n<\infty\}$ and $\lVert (x_n)\rVert_p:=(\sum\limits_n\lVert x_n\rVert^pw_n)^{1/p}$. Is $\bigoplus\limits_P E_n$ a Banach space with the given norm? If so can you describe its dual in terms of dual of $E_n$?

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I think that your question reduces to the case $w_n=1$ for all $n$, that is, the usual $\ell_p$-spaces, because for each sequence $(x_n)\in\prod E_n$ you can define $y_n=(w_n)^{1/p}x_n$, and so $\sum\|x_n\|^pw_n=\sum\|y_n\|^p$. – Matemáticos Chibchas Jan 23 '13 at 16:20
But how to identify $E_n$ with $\mathbb{R}$? What do you think about its dual? – user41481 Jan 23 '13 at 16:27
The “obvious” expectation (with $w_n=1$ for simplicity) is yes, it's a Banach space, and the dual is the corresponding $L^q$ direct sum of the duals (so in case $p=1$, $q=\infty$ you get a supremum norm). I'd be very surprised if the usual proof that $\ell^q$ is the dual of $\ell^p$ doesn't extend to this case almost verbatim. – Harald Hanche-Olsen Jan 23 '13 at 16:35
@HaraldHanche-Olsen Indeed, I did once these calculations and the proof is pretty much identical. – Theo Jan 23 '13 at 17:30
@MatemáticosChibchas, that's not accurate, because $\prod E_n$ contains sequences that are not summable. And changing the weights changes which sequences are summable. – Martin Argerami Jan 25 '13 at 15:08

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