Completeness of coefficient functionnals 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 $X^*$? 
 A: 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).
A: 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.
