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 $X^\*$? 



Completeness of coefficient functionnals
M. Zippin [showed](http://link.springer.com/article/10.1007/BF02771607) 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.*


