This is a follow up to the question: [Biorthogonal functionals](https://mathoverflow.net/questions/239471/biorthogonal-functionals). A positive answer to that question implies a negative answer to this one. 


If $X$ is a separable Banach space, can we find a basic sequence $(x^{\*}\_n)$ in $X^{\*}$ with the property that for any subsequence $(n\_k)$ we have that:
$$
\bigcap\_{k=1}^{\infty}\ker{x^{\*}\_{n\_k}}=\{0\}
$$ 


Clearly this is not possible in a reflexive space, as the previous question has a positive answer in the reflexive case. Therefore the only case of interest is when $X$ is not reflexive. 



Trivial intersection of kernels
As I understand you can answer this question in the negative using the notion of $w^\*$-basic sequence introduced by Johnson and Rosenthal (1972), see Lindenstrauss-Tzafriri, Classical Banach spaces, vol. I, p. 11. In the case where the sequence $\{x\_n^\*\}$ is weak$^\*$-convergent to $0$, we can find a $w^\*$-basic subsequence in it by the proof of Theorem 1.b.7, and get nontrivial kernel by Proposition 1.b.9. If the sequence $\{x\_n^\*\}$ is not weak$^\*$-convergent to $0$, we pick in it a subsequence weak$^\*$-convergent to some $x^\*$. We use the argument above to $\{x^\*\_{n\_k}-x^\*\}$. For a suitable subsequence the kernel will be more than one-dimensional, and hence will intersect the kernel of $x^\*$, thus we get an element in the intersection of the kernels of $\{x^\*\_{n\_k}\}$.


