I am confused about something really basic that's kind of embarrassing me. Maybe I'm just sleepy, who knows. I will ask my question in generality.
Suppose I have a Hilbert space $\mathcal{H}$ and I have a closed subspace $F$. I have a basis for $F$ which I label $\{f_n\}_{n \in \mathbb{N}}$ and I am trying to find its dual basis. I have found a collection of vectors in $\mathcal{H}$ , $\{g_n\}_{n \in \mathbb{N} }$ such that
$$\langle f_n , g_m\rangle = \delta_{m,n} $$
I thought that this was enough to claim that my vectors $ \{g_n \}$ must be the dual basis to $\{f_n \}$. However, I know as an irrefutable fact that the vectors $\{g_n\}$ which I found cannot span my subspace $F$ (you have to just take my word for this). This is confusing me though because I thought that the dual basis to $\{f_n\}$ must span the same space as $\overline{\text{span}}\{f_n \} = F $, so I see that one of the following must be true:
$1.)$ The vectors I computed $\{g_n\}$ must not actually satisfy that $\langle f_n , g_m\rangle = \delta_{m,n} $
$2.)$ The vectors $\{g_n\}$ do satisfy $\langle f_n , g_m\rangle = \delta_{m,n} $; however, that does not make $g_n = f_n^*$.
$3.)$ I am not making a mistake
From the information I provided, could someone tell me which of those 2 mistakes (or perhaps another one I am not seeing) I must be making?
The $g_i$ do not form the dual basis if they do not lie in $F$. The reason is that $g_i$ is not uniquely determined by the equations $\langle f_j, g_i\rangle = \delta_{j, i}$. You can change such $g_i$ by any element from $F^\perp$ without disturbing these equations.
You can find the actual dual basis of $(f_i)$ by taking your $(g_i)$ and projecting orthogonally onto $F$.