In the context of Hermite Polynomials (here) it is stated that $\overline{\text{span}(h_n(x))} = L^2(\mathbf{R}) \iff$ if $f \in L^2(\mathbf{R})$ satisfies $(f,h_n)=0$ for all $n$, then $f=0$. How does this equivalence follow? Here $h_n$ are the Hermite polynomials, but I think I have seen similar arguments for other families of functions as well so the particular form of $h_n$ is probably not central for my question.
Attempt: I considered a finite space to get some intuition. Let $V$ be a vector space with dimension $\text{dim}(V)=n$ and $v \in V$ be some element. Further let $H \subseteq V$. It seems like the inner product condition ($(f,h_k)=0, \forall k$) would translate to
$va + h_k a_k = 0 => a=a_k=0, \forall k$.
where $h_k \in H$ and $a, a_k$ are elements of some field (say the complex numbers). If this in turn implies that $v=0$, this should somehow, be equivalent to, for all $u \neq 0$ in $V$, it exists complex coefficients $b_1, ..., b_n$ such that:
$ v' = b_1h_1 + ... + b_nh_n. $
But I couldn't make a clear argument.