Trivial intersection of kernels

Question

This is a follow up to the question: 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.

Answer 1

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 $\{x^*_{n_k}\}$.