If $X$ is a separable Banach space and $(x\_n)$ is a basic sequence, then we can define biorthogonal functionals $(x^{\*}\_n)$ in $X^{\*}$ such that $x^{\*}\_n(x\_k)=\delta\_{nk}$. 


What about conversely? If $(x^{\*}\_n)$ is a basic sequence in $X^{\*}$, can we always find vectors $(x\_n)$ in $X$ such that $x^{\*}\_n(x\_k)=\delta\_{nk}$? This is true when $X$ is reflexive, but I cannot see a proof or a counterexample when $X$ is not reflexive.



Biorthogonal functionals
No, that's not true. Let $\mathbb{N}^\* = \mathbb{N} \cup \{\infty\}$ and set $X = C(\mathbb{N}^\*) \cong c$ and $X^\* = l^1(\mathbb{N}^\*) \cong l^1$. Take as the basic sequence of $X^\*$ the vectors $e\_n$ for $n \in \mathbb{N}^\*$. There is no vector $x$ in $c$ with $e\_n(x) = 0$ for all $n \in \mathbb{N}$ but $e\_\infty(x) \neq 0$.


