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Let $X$ be a Banach space embedded in $X^{**}$ in the usual way. We consider the set $$ O_X := \{ x^{**}\in X^{**} : \|x^{**}-x\|\geq \|x\| \textrm{ for all }x\in X\}. $$ I think this is the orthogonal set of the subspace $X$ in the sense of Birkhoff. It is a closed cone in $X^{**}$, but not a subspace in general.

Suppose that $X$ is (isometrically) a dual space with unique predual $X_*$. Hence $X^{**} = X \oplus X_*^\perp$ and, since the associated projection $X^{**}\to X$ has norm one, $X_*^\perp\subset O_X$.

QUESTION: Is is true in this special case that $X_*^\perp= O_X$?

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  • $\begingroup$ I think you can find the answer in this survey $\endgroup$
    – Norbert
    Commented Mar 1, 2014 at 15:11
  • $\begingroup$ @Norbert: I have a copy of that survey. I studied it some time ago, and I thought then that the answer to my question was obviously positive. However I have recently realized that I was mistaken, or cannot remember the argument. Could you be more precise in your observation? $\endgroup$
    – M.González
    Commented Mar 1, 2014 at 15:32
  • $\begingroup$ I' just scrolled this survey few month ago that is why I thought it might containt the answer. Unfortunately I do not have full access, so I can't help you with exact reference. $\endgroup$
    – Norbert
    Commented Mar 1, 2014 at 16:03

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