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I am asked to show that the $\ell^p$-direct sum of a sequence of Banach Spaces $X_n$ is isometrically isomorphic to the $\ell^q$ direct sum of $X_n^*$ where $X_n^*$ is the dual of $X_n$ for each $n$ respectively. (Here $p$, $q$ are conjugate indices with $1\leq p<\infty$). So far I have tried the case $p>1$, and I am trying to adapt the proof of $(\ell^p)^*$ being isometrically isomorphic to $\ell^q$ but this isn't getting anywhere. Could someone please give any hints?

EDITED by idonknow: Fix $1\leq p<\infty.$ The $\ell^p$-direct sum of a sequence of Banach spaces $X_n,$ denoted as $\left( \bigoplus_{n=1}^\infty X_n\right)_{\ell^p},$ means that for every $x=(x_n)_{n=1}^\infty\in \left( \bigoplus_{n=1}^\infty X_n\right)_{\ell^p},$ we have $$\|x\| = \left(\sum_{n=1}^\infty\|x_n\|^p \right)^{\frac{1}{p}}.$$

Idonknow
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Jack
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    There are two parts of the proof: easy ($\oplus X_n^*$ acts on $\oplus X_n$) and hard (there are no other functionals). Any problems with the first part? –  Dec 21 '14 at 02:43
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    What is the $l^p$ direct sum of the $X_n?$ – zhw. Oct 02 '17 at 18:37
  • Could you define what you mean by $\ell^p$ direct sum? – Guy Fsone Oct 03 '17 at 09:30
  • Maybe each $X_n$ should be a reflexive Banach space ? – Guy Fsone Oct 03 '17 at 09:38
  • @Idonknow But in this problem we have an infinite sequence of Banach spaces, right? The definition of the norm would then be for those $x$ for which the sum converges I suppose. – zhw. Oct 03 '17 at 14:58
  • I think there something missing in this problem. Assume that $X_n{0}$ for all $n>1$ except then , $X\equiv X_1$ from this problem we have that $X_1$ and $X_1^*$ must be isometrics. which is curious. Since given a random Banach space it is not always clear how to define a map from the space to its dual – Guy Fsone Oct 03 '17 at 15:05
  • in the above comment Asume that $X_n\equiv {0}$ for all $n>1$ – Guy Fsone Oct 03 '17 at 15:50
  • @GuyFsone I don't see how it is saying that. – zhw. Oct 03 '17 at 17:49
  • @Idonknow It should be $\left( \bigoplus_{n=1}^\infty X_n\right)_{\ell^p},$ right? – zhw. Oct 03 '17 at 17:52
  • YES But what if we Assum all the spaces are trivial except one of them.? – Guy Fsone Oct 03 '17 at 18:47
  • @GuyFsone I don't see the problem. Why do you say $X_1,X_1^*$ are isometric? – zhw. Oct 03 '17 at 19:00
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    @Idonknow A proof is here. [Assuming that the intention here is, as indicated by the question title, to show that the dual of the $\ell^p$-direct sum is isometrically isomorphic to the $\ell^q$-direct sum of the duals. As written in the first sentence of the question body, the assertion is easily seen to be wrong.] – Daniel Fischer Oct 03 '17 at 20:01
  • @zhw. In the case whre $X_n\equiv {0}$ for $n>1$ $X\equiv X_1 $ and $X^\equiv X^_1 $ and the $\ell^p$ norm reduce to the norm of $X_1$ . Same for the $\ell^q$ norm on $X^$ reduce norm of $X^_1$. Is that not correct? – Guy Fsone Oct 04 '17 at 08:26
  • @DanielFischer I saw your answer and it confirms my doubt . as part of assumptions one should assum $X_n$ to reflexive and the asked isometry from the body question seem to be wrong. Thank you – Guy Fsone Oct 04 '17 at 08:36
  • @GuyFsone I see what you mean now, sorry for the confusion. I hadn't even read the first sentence carefully, I had just assumed the result to be proved was what "it should be", as Daniel Fischer said. – zhw. Oct 04 '17 at 14:52

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