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Jan 8, 2021 at 13:41 comment added Bill Johnson Maybe type $P^*$ is needed. I think $\ell^+$ only gives a linear functional that is bounded below away from zero on the sequence while you want it to be one at each term of the sequence.
Jan 8, 2021 at 1:34 vote accept erz
Jan 8, 2021 at 1:20 comment added erz When looking up $l^+$ basic sequences, I came across an even narrower class of $P^*$ basic sequences, whose existence is also equivalent to non-reflexivity. To be honest I don't understand why the fact that $(x_n)$ is of type $l^+$ implies that $(x_1-x_n)$ is basic, but I understand how to get there from the assumption that $(x_n)$ is of type $P^*$,
Jan 7, 2021 at 18:46 history answered Bill Johnson CC BY-SA 4.0