Consider the real Banach space $\ell^2$ of square summable sequences and let $\mathcal{A}\subset \ell^2$ be a dense subspace. Suppose I have a bounded sequence $\psi=(\psi_n)_{n\geq 1}\in \ell^\infty$ which has the following property: $$\sum\limits_{n=1}^\infty \psi_na_n=0$$ for any $(a_n)_{n\geq 1}\in \mathcal{A}$. I would like to conclude that $\psi=0$. If I would know that $\psi\in \ell^2$, then this would be easy enough, but unfortunately I only know that $\psi\in\ell^\infty$. Does anyone have a clue how to approach this?
In my specific case I know that each sequence in $\mathcal{A}$ decays exponentially, but I do not know if this useful in this case.
This question is sort of a follow-up question of the following question Scaled series of shifted continuous function As I stated there, I encountered this problem in my research, where I consider linear differential equations with infinitely many delayed and/or forward terms.
It's not true. Consider $\psi = (1,1,1,\ldots)$ with $\mathcal A$ the set of sequences $a$ of finite support such that $\sum_i a_i = 0$. To see this is dense in $l^2$, take any $x \in l^2$ and $\epsilon > 0$. Take $N \in \mathbb N$ such that $\sum_{n > N} |x_n|^2 < \epsilon^2/2$. Let $s = \sum_{n \le N} x_n$, and take $M \in \mathbb N$ such that $M \epsilon^2 > 4 |s|^2$. Let $$ a_n = \cases{x_n & for $n \le N$\cr -s/M & for $N+1 \le n \le N+M$\cr 0 & otherwise} $$ Then $a \in \mathcal A$ and $\|x - a\| < \epsilon$.