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.
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 $e_\infty(x) \neq 0$.
Yes, there is such a sequence in any non-reflexive Banach space.
It is known$^\#$ that if $Y$ is not reflexive, then $Y$ contains a basic sequence that is not boundedly complete. So if $X$ is not reflexive, then there is a semi-normalized basic sequence $(x_n^*)_{n=1}^\infty$ in $X^*$ s.t. $\sup_n \|\sum_{k=1}^n x_k^*\| < \infty$. Assume that $(x_n) \subset X$ is biorthogonal to $(x_n^*)$ (else you are done). Let $x_0^*$ be a weak$^*$ cluster point of $(\sum_{k=1}^n x_k^*)_n$. Necessarily $\langle x_0^* , x_n\rangle =1 $ for all $n\ge 1$, but $x_0^*$ is not in the normed closed span of $(x_n^*)$ because $(x_n^*)_{n=1}^\infty$ is basic, and hence $(x_n^*)_{n=0}^\infty$ is basic and not biorthogonal to any sequence in $X$.
# It is inconvenient right now for me to look for a reference for this result. It certainly is in Ivan Singer's book on bases; probably in volume one. It is almost proved in Albiac-Kalton but not stated there.
