On the proof of the Hermitian functions being a basis of $L^2(R)$ space. Define $h_k(x) = H_k(x) e^{- x^2/2} $ where $H_k(x)$ is the k-th Hermitian polynomial.
Assuming already having shown that the $h_k(x)$ form an orthonormal system in $L^2(R)$ with the usual inner product how can one prove that they are a basis?
 A: To complement @MartinArgerami's answer: although it is certainly traditional to invoke something like Weierstrass approximation (for polynomials on compacts) to know that Hermite polynomials times Gaussians are dense in $L^2(\mathbb R)$, it is also possible to give a more structural argument that can apply in situations where a Stone-Weierstrass argument is much less clear. Namely, the functions $h_k$ of the question are eigenfunctions for the so-called quantum harmonic oscillator $-\Delta+x^2$ (studied by many people long before quantum theory, e.g., Hermite, Mehler (1868), and others). It is an interesting calculus exercise to show that the resolvent is compact (self-adjoint) (e.g., using Friedrichs' extension construction), and for general reasons this implies that the spectrum of the operator is the collection of inverses of the spectrum of the resolvent, in particular purely discrete. 
Then the issue becomes to show that there are no other eigenfunctions than the $h_k$. At this point, the "ladder" or "raising/lowering" operators are relevant, looking at the factorization of $-\Delta+x^2$ via the "Dirac operator", in this case simply $i{\partial\over \partial x}$. One shows that all eigenfunctions are obtained by "raising" the bottom eigenfunction, $e^{-x^2/2}$, and these are exactly the $h_k$...
A: You need to show that $\{h_k\}$ is total: that is,  $$\tag{1}\langle f,h_k\rangle=0,\ \ \  \text{ for all } k,\ \ \implies f=0.$$ Since the orthogonal of a set is the same as the orthogonal of the closure of a set, it is enough to show the above on a dense set. So now you have to show $(1)$ when $f$ is a polynomial times $e^{-x^2}$, or a characteristic function. 
