A point $x$ belongs to closure of linear span if(f) every bounded linear functional that vanishes on the subset $\{ x_k\}$ vanishes at $x$?

Question

A point $x$ belongs to closure of linear span $\overline{<x_k>}$ if(f) every bounded linear functional that vanishes on the subset $\{ x_k\}$ vanishes at $x$?

I.e. if

$$l(x_k)=0 \space \forall x_k \in <x_k>$$ implies $$l(x)=0$$

Why is there a connection between vanishing at $\{x_k\}$ and belonging to the closure of linear span?

Or can one always draw such $l$ that one can assume that

$$l(x_k)=0 \space \forall x_k \in <x_k>$$

Perhaps it means that one takes a linear combination (a linear functional) on $x_k$ s and then sets all the coefficients to zero?

Answer 1

The link is the Hahn Banach Theorem. Suppose $x$ is not in the closure of the linear span of $\{x_k\}$ (which I name $A$), then you can find a linear functional which is zero on $A$ and $1$ at $x$.

The other direction is very easy.