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Jul 30, 2017 at 6:09 vote accept CSH
Jul 30, 2017 at 6:09
Jul 24, 2017 at 14:54 comment added Matthew Daws No: you just need that if $X_0$ is the linear span of $(x_n)$ (so $X_0$ is not closed, and it's closure is separable) then $(f_n)$ should separate the points of $X_0$. Hahn-Banach does this.
Jul 24, 2017 at 14:33 comment added CSH But it seems that the argument holds only when $X^*$ is separable, isn't it?
Jul 24, 2017 at 14:30 comment added Matthew Daws @CSH: Yes, pretty much. You start with $(x_n)$ being just linearly independent, and $(f_n)$ being linearly independent separating the points of the linear span of the $(x_n)$. Then adjust $f_1$ so $f_1(x_1)=1$, adjust $x_2$ so $f_1(x_2)=0$, adjust $f_2$ with $f_2(x_1)=0, f_2(x_2)=1$, and so on....
Jul 24, 2017 at 14:17 comment added CSH I've tried to thinking all day but can not found the series. So thank you very much about the answer and more intuition about that biorthogonal systems(I think it is just as the Gram-Schmidt process?).
Jul 24, 2017 at 8:25 history answered Matthew Daws CC BY-SA 3.0