Could we rigorously prove the following?
In a Hilbert Space $\mathcal{H}$, if $f, \phi_j \in \mathcal{H}$ for all $j \in J$, and $$\left<f, \phi_j\right> = 0 \text{ for all } j \in J \implies f = 0$$ then $\{\phi_j\}_{j \in J}$ spans $\mathcal{H}$.
This intuitively makes sense, but do we have formal proof?
P.S. This is from Ten Lectures on Wavelets Chapter 3, in the proof of proposition 3.2.1.
Let $V := \displaystyle\bigoplus_{j \in J} \operatorname{span}\lbrace \phi_j \rbrace$. Your assumption implies $V^\perp = \lbrace 0 \rbrace$. Let $P: \mathcal{H} \rightarrow V$ be the linear projection onto $V$, i.e. $$ (f - Pf, v) = 0 \quad \forall v \in V \quad \iff Pf = \operatorname*{argmin}_{v \in V} \lVert f - v \rVert. $$ According to the projection theorem for Hilbert Spaces, this is well defined and the equivalence stated above holds. Note that $f$ in $\ker(P)$ iff $(f, v) = 0$ for all $v \in V$. So $\ker(P) = \lbrace 0 \rbrace$ according to our assumption. Furtheremore, since $Pf \in V$, we have $PPf = Pf$ according to the $\operatorname{argmin}$ definition of the projection. Note that for any $x \in \mathcal{H}$: $$ x = \underbrace{(x - Px)}_{\in \ker(P) \implies 0} + \underbrace{Px}_{\in V} $$ Therefore $$ \mathcal{H} = \ker(P) \oplus V = \lbrace 0 \rbrace \oplus V = V $$ and we are done.
$x - Px \in \ker(P)$ holds because of $$ P(x-Px) = Px - PPx = Px - Px = 0. $$