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Hermite polynomials form a complete orthonormal basis of the weighted $L^2(\mathbb R, w \; dx)$ space, with inner product $$ \langle f, g \rangle_w = \int_\mathbb R f(x) g(x) \; w(x) dx. $$

A short Wikipedia proof shows that if for any $f \in L^2(\mathbb R, \exp (-x^2) \; dx)$ we have $\forall n \geq 0$ $$ \int_\mathbb R f(x) x^n e^{-x^2} dx = 0 \; \Longrightarrow \; f = 0 \; a.e. $$ then these weighted polynomials form a complete basis (orthonormal after Gram-Schmidt). Wikipedia then says "Variants of the above completeness proof apply to other weights with exponential decay."

My question is: What are the precise conditions on $w(x)$ with respect to which an inner product is defined to preserve completeness of the polynomial basis? Is exponential decay necessary, or merely sufficient? For example, are all of these heavy-tailed densities ruled out? Thanks.

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Note that the Hermite polynomials already form a complete orthogonal family (and can be easily normalized), so Gram-Schmidt isn't needed. –  JohnD Jan 7 '13 at 19:54
    
Oh, sorry, I was referring to the polynomials in the integral above, $x^n$. –  covstat Jan 7 '13 at 20:10

1 Answer 1

I can think of a few potential interpretations of that (admittedly vague) comment:

  1. They are referring to the fact that the Hermite polynomials are sometimes defined in terms of the weight function $\exp(-x^2/2)$ (used by probabilists) instead of $\exp(-x^2)$ (used by physicists); see the different orthogonality statements here.

  2. Maybe they were referring to orthogonal families other than Hermites which have an exponential weight function, such as the Laguerre polynomials on $(0,\infty)$ with weight $e^{-x}$, or the associated Laguerre polynomials on $(0,\infty)$ with weight $x^\alpha e^{-x}$.

  3. This could be an obtuse reference to the associated Hermite functions here, although as the link indicates, the typical definition builds the exponential into the family of functions itself so that the weight function is simply $1$.

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