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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.

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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 $e_\infty(x) \neq 0$.

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  • $\begingroup$ Could you please expand a little? I don't understand very well how $l_1(\mathbb{N}^{*})$ is defined and what is $e_\infty$. What would be the basic sequence in classical $l_1(\mathbb{N})$? $\endgroup$ – Markus May 21 '16 at 22:58
  • $\begingroup$ You know that if $\Omega$ is a compact Hausdorff space then $C(\Omega)^* \cong M(\Omega)$, right? Here $\Omega$ is the one-point compactification of $\mathbb{N}$, the continuous functions on $\Omega$ can be identified with the convergent sequences, and the Borel measures on $\Omega$ can be identified with the functions from $\Omega$ to the scalars whose values are summable. $\endgroup$ – Nik Weaver May 21 '16 at 23:02
  • $\begingroup$ $l^1(\mathbb{N}^*)$ is isometrically isomorphic to $l^1$ by simply rearranging the natural basis vectors $e_n$, putting $e_\infty$ first and shifting all the others up. $\endgroup$ – Nik Weaver May 21 '16 at 23:03
  • $\begingroup$ Thank you. Can you find such a sequence in any non-reflexive Banach space? Seems like a harder question. $\endgroup$ – Markus May 22 '16 at 0:09
  • $\begingroup$ I don't know. That does sound harder. $\endgroup$ – Nik Weaver May 22 '16 at 0:50
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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.

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