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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.

TOMILO87
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    If $f \bot h_n$ for all $n$ then $f \bot \operatorname{sp} { h_n}_n$ and since the map $g \mapsto \langle h_n, g \rangle $ is continuous, if follows that $f \bot \overline{ \operatorname{sp} { h_n}_n }$. Since $(L^2)^\bot = {0}$ you have the desired result. – copper.hat Aug 22 '22 at 21:02
  • You should replace $L^2(\mathbb{R})$ with $L^2(\mathbb{R},,e^{-x^2/2},dx).$ Another argument: $$|f|^2=\sum {|\langle f,h_n\rangle |^2\over |h_n|^2}$$ – Ryszard Szwarc Aug 22 '22 at 23:52
  • @Ryszard. Thank you. I can understand that if $(f,h_n)=0$ for all $n$ in the above expression, then that would imply that $f=0$ (a.s.) but I don't see how this necessary imply that $f$ can be written as a linear combination of $h_1, h_2, ...$ – TOMILO87 Aug 23 '22 at 09:59
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    You welcome. The function $f$ can be written as the infinite linear combination $$f=\sum \langle f,h_n\rangle h_n,$$ because $h_n $ form an orthonormal basis in $ L^2(e^{-x^2/2}dx)$ – Ryszard Szwarc Aug 23 '22 at 10:30

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