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Let $X$ be a Banach space. Let $Y$ be a closed subspace. Suppose that the normed spaces (in fact Banach spaces) $Y$ and $X/Y$ are both reflexive. I need to show that $X$ is reflexive.

I cannot show this but I feel that I could use the fact that a Banach space is reflexive if and only if its closed unit ball is weakly compact. So in this case we know $B_Y$ and $B_{X/Y}$ are weakly compact. Let $\mathcal{U}$ be a weakly open cover for $B_X$. I feel that a finite subcover can be obtained by considering the sets of the form $a+Y\cap B_X\subset X$, which are certainly compact, as $a+Y\cap B_X\subset a+nB_Y\cap B_X$, which is w-closed in the weakly compact set $a+nB_Y$ for some sufficiently large $n$, and the corresponding $a+Y\in X/Y$.

Could anybody suggest anything? Thanks.

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  • $\begingroup$ Isn't this equivalent to prove that if $X$ and $Y$ are reflexive, so is $X\times Y$? $\endgroup$ – Emanuele Paolini Jan 16 '14 at 20:41
  • $\begingroup$ @EmanuelePaolini I'm not sure, but wouldn't it be $X+Y$ $\endgroup$ – Tim Seguine Jan 16 '14 at 20:46
  • $\begingroup$ @EmanuelePaolini You only have an isomorphism $Y\times (X/Y) \cong X$ if $Y$ is a complemented subspace. The premises don't immediately yield that. $\endgroup$ – Daniel Fischer Jan 16 '14 at 20:50
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A property of a Banach space is a three-space property if whenever $E$ is a Banach space, $F\subseteq E$ is a closed linear subspace and two of the spaces $E$, $F$ and $E/F$ have the property, then all three of the spaces $E$, $F$ and $E/F$ necessarily have the property.

Reflexivity is a three-space property. For the proof see theorem 1.11.19 in An Introduction to Banach Space Theory Graduate Texts in Mathematics. Robert E. Megginson.

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  • $\begingroup$ Thank you very much for the answer. Thank you for mentioning the three-space property. That is very interesting! I'll read around it. $\endgroup$ – Spook Jan 16 '14 at 21:48
  • $\begingroup$ You are wellcome! $\endgroup$ – Norbert Jan 16 '14 at 21:48

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