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Let $X,Y$ be Banach spaces and denote $Z = X \cap Y$. It is easy to show that $Z$ is a Banach space with norm $$\|x\|_Z:= \|x\|_X+\|x\|_Y.$$

Assume that $X$ is reflexive and $Y$ is non-reflexive. Can we conclude anything about the reflexivity of $Z$?

Marry Mag
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  • What does $X \cap Y$ even mean if $X$ and $Y$ are abstract Banach spaces? Surely they must be linear subspaces of some larger vector space $\Omega \supseteq X,Y$, or something like that? – Josse van Dobben de Bruyn Dec 13 '17 at 21:02
  • The part with "it is easy to show" is wrong. Indeed, if you have a Cauchy sequence in $Z$, it is a Cauchy sequence in $X$ and in $Y$. However, you do not have any guarantee that the limits in $X$ and $Y$ coincide. – gerw Nov 24 '18 at 18:13

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The intersection can be either reflexive or non-reflexive. For example, $\ell^2\cap \ell^\infty = \ell^2$ is reflexive while $\ell^1\cap \ell^2 = \ell^1$ is non-reflexive.

  • You still have to prove that the intersection norm (as defined by OP) is equivalent to the usual one. Luckily this is easy: for all $p\geq 1$ we have $\lVert x\rVert_p \leq \lVert x\rVert_p + \lVert x\rVert_\infty \leq 2\cdot \lVert x\rVert_p$. Then, note that reflexivity is preserved when passing to an equivalent norm. – Josse van Dobben de Bruyn Dec 13 '17 at 21:06