I have come across the following statement a couple of times, but cannot figure out quite how to justify it:
If $A$ is not a dense subset of $\ell^2$ then its closure is not all of $\ell^2$ so that we can find a nonzero functional that disappears on $A$.
The first part follows from the definition of being dense, but the fact about the linear functional that follows is a bit mysterious. I am guessing it has to be some implication of either the Open Mapping Theorem or Closed Graph Theorem, but I can't quite tease it out. Any suggestions?
UPDATE: So since $A$ is not dense we have that $\overline{A}$ is not all of $\ell^2$. So $\overline{A}^{\perp}\ne\{0\}$ contains nonzero elements and for each of these nonzero elements $y\in\overline{A}^{\perp}$ we have $$ \langle x, y \rangle =0, $$ for all $x\in\overline{A}$. Then each inner product corresponds to a nontrivial bounded linear functional which disappears over $\overline{A}$.
That is a standard application of the Hahn-Banach theorem. If $A$ is a closed subspace of a Banach space $X$ and $a \notin A$, there exists a bounded linear functional $L$ such that $La = 1$ but $Lx = 0$ for all $x \in A$.