Let $l^2$ be the hilbert space of all complex sequences $\psi= (\psi_n)_{n=0}^{\infty}$ such that $\sum_{j=0}^{\infty} |\psi_j |^2 < \infty$. Let $\phi= (\phi_n)_{n=0}^{\infty}$ be a sequence of complex numbers, and $S$ a vector subspace of $l^2$ which is dense in $l^2$. Assume that for all $\psi = (\psi_n)_{n=0}^{\infty}$ in $S$, we have \begin{equation} \sum_{n=0}^{\infty} \bar{\phi_n} \psi_n = 0. \end{equation} Can we conclude that $\phi=0$? What if we only suppose that $S$ is a dense subset of $l^2$?
Please note that I am not assuming that $\phi$ is in $l^2$, otherwise the answer to both questions is trivially yes (take a sequence $\psi^{(N)}$ in $S$ converging to $\phi$ and use the continuity of the scalar product).
PS The motivation for this question is the following. Set $D= \{ \phi \in l^2 : \sum_{j=0}^{\infty} j |\phi_j |^2 < \infty \}$ and consider the operator $X$ on $D$ which associates to each $\phi \in D$ the vector $X \phi$ whose j-th component (j=0,1,2,...) is
$(X \phi)_j = \sqrt{j+1} \psi_{j+1} + \sqrt{j} \psi_{j-1}$,
where we set $\psi_{-1}=0$. We have for every $\phi \in l^2 , \psi \in D$ \begin{equation} \sum_{j=0}^{\infty} \bar{\phi_j} [\sqrt{j+1} \psi_{j+1} + \sqrt{j} \psi_{j-1}] = \sum_{j=0}^{\infty} \psi_j \overline{ [\sqrt{j+1} \phi_{j+1} + \sqrt{j} \phi_{j-1}]}, \end{equation} where again we set $\phi_{-1}=0$. So in particular, $X$ is a symmetric operator. But it is not self-adjoint. To see this, consider the vector $\phi$ whose j-th component is $\phi_j = (-1)^{\lfloor j/2 \rfloor} j^{-\beta}$, where $1/2 < \beta <1$. Is is easy to see that $\phi \in l^2 \backslash D$, and that the vector whose j-th component is $\sqrt{j+1} \phi_{j+1} + \sqrt{j} \phi_{j-1}$ is in $l^2$, so $\phi$ belongs to the domain of the adjoint. I conjectured that the domain of the adjoint $X^{*}$ is exactly the set of all vectors $\phi \in l^2$ such that $ \sum_{j=0}^{\infty} |\sqrt{j+1} \phi_{j+1} + \sqrt{j} \phi_{j-1} |^2 < \infty$. This is the point where my question comes into play. If the answer to the question I posted were affirmative, then it would be easy, by using riesz theorem, to prove my conjecture.
