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My questions is about Schauder bases and more specifically about coefficient functionals.

Let $(x_n)$ be a Schauder basis of a Banach space $X$. Thus for all $x$ in $X$, $x = \sum f_n(x) x_n$. The $f_n$ are called coefficient functionals. They are continuous. If $X$ is reflexive, they form a basis of $X^*$ (with Hahn–Banach theorem). My question is : if $X$ is not reflexive, but if we suppose $X^*$ separable, is $(f_n)$ a basis of $X^*$?

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2 Answers 2

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M. Zippin showed that for a Banach space $X$ with a basis, if every basis of $X$ is boundedly complete or if every basis of $X$ is shrinking, then $X$ is reflexive.

The result of Zippin answers you question in the negative (or, rather, the answer is, "not necessarily".) However to complete the picture let us recall the earlier result of R.C. James (alluded to in your question) asserting the following: For a reflexive Banach space $X$ with a basis $(x_n)$, the basis $(x_n)$ is both shrinking and boundedly complete.

Putting the results of Zippin and James together yields the following:

Let $X$ be a Banach space with a basis. The following are equivalent:

  • $X$ is reflexive;
  • every basis of $X$ is shrinking; and
  • every basis of $X$ is boundedly complete.
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No, there are counterexamples. For instance, there exists a space $X$ with a basis, $X^*$ separable, and yet $X^*$ fails approximation property. See Lindenstrauss-Tzafriri's book Theorem 1.e.7 for a discussion of this (it is an existence proof).

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    $\begingroup$ The dual space $J^*$ of James' space, with the basis formed by the coefficient functionals of the (usual) shrinking basis of $J$, is another example. $\endgroup$
    – M.González
    Commented Mar 31, 2020 at 16:50
  • $\begingroup$ Yes, in fact, that is a better answer:) But you mean boundedly complete basis (i.e., `the summing basis') of $J$ $\endgroup$
    – Bunyamin Sari
    Commented Mar 31, 2020 at 18:35
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    $\begingroup$ And Zippin showed that for a Banach space $X$ with a basis, if every basis of $X$ is boundedly complete or if every basis of $X$ is shrinking, then $X$ is reflexive: link.springer.com/article/10.1007/BF02771607 $\endgroup$
    – Philip Brooker
    Commented Mar 31, 2020 at 22:07
  • $\begingroup$ @ Philip Brooker Nice. This should be the answer to the question. $\endgroup$
    – Bunyamin Sari
    Commented Apr 1, 2020 at 18:02
  • $\begingroup$ @BunyaminSari Thanks, Bunyamin, I've just cobbled together an answer :) $\endgroup$
    – Philip Brooker
    Commented Apr 2, 2020 at 22:53

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