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Given a complex Banach space $X$ and its antidual $\hat X^*$, it is possible in some cases to apply the complex interpolation method, and get as $(X,\hat X^*)_{1/2}$ a Hilbert space. See [F. Watbled, Complex interpolation of a Banach space and its dual, Math. Scand. 87 (2000), 200-210].

I am trying to understand the case when $X$ has a Schauder basis, in which $X$ and $\hat X^*$ can be seen as spaces of complex sequences. Following Watbled's paper, the key to get $(X,\hat X^*)_{1/2}=\ell_2$ is to show that the spaces satisfy the following property:

(H)$\quad$ $\ell_2$ is continuously embedded in $X+\hat X^*$.

I would like to know under which conditions the space $X$ satisfies (H); in particular if it does when the basis is unconditional.

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  • $\begingroup$ What do you mean by "continuously embedded"? There is a bounded one to one linear operator from $\ell_2$ into any infinite dimensional Banach space. On the other hand, $\ell_2$ is linearly homeomorphic to a subspace of $X\oplus Y$ iff $\ell_2$ is linearly homeomorphic to either a subspace of $X$ or a subsapce of $Y$. $\endgroup$ Oct 29, 2013 at 17:54
  • $\begingroup$ @Bill Johnson: I mean that $X+\hat X^*$ algebraically contains $\ell_2$ as spaces of sequences, and that the natural inclusion is continuous. $\endgroup$ Oct 29, 2013 at 18:09
  • $\begingroup$ @Bill Johnson: Clearly $X\cap \hat X^*$ is continuously embedded in $\ell_2$. Hence $\ell_2$ is continuosly embedded in $X^*+\hat X^{**}$, and (I think) the result is clear in the reflexive case. $\endgroup$ Oct 29, 2013 at 18:18
  • $\begingroup$ I see the problem, but have no idea how to solve it. $\endgroup$ Oct 29, 2013 at 18:31

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