Another orthonormal basis: Hermite functions

This is an orthonormal basis for L^2(\mathbb R). Since the measure of \mathbb R is infinite, functions will have to decay at infinity in order to be in L^2. The Hermite functions are
\displaystyle \Phi_n(x)=(2^n n! \sqrt{\pi})^{-1/2} H_n(x)e^{-x^2/2}
where H_n is the nth Hermite polynomial, defined by
\displaystyle H_n(x)=(-1)^n e^{x^2} \left(\frac{d}{dx}\right)^n e^{-x^2}.
The goal is to prove that the functions \Phi_n can be obtained from x^n e^{-x^2/2} via the Gram-Schmidt process. (They actually form a basis, but I won’t prove that.)

One can observe that the term e^{-x^2/2} would be unnecessary if we considered the weighted space L^2(\mathbb R, w) with weight w(x)=e^{-x^2} and the inner product \langle f,g\rangle=\int_{\mathbb R} fg\,w\,dx. In this language, we orthogonalize the sequence of monomials \lbrace x^n\rbrace\subset L^2(\mathbb R, w) and get the ON basis of polynomials \{c_n H_n\} with c_n = (2^n n! \sqrt{\pi})^{-1/2} being a normalizing constant. But since weighted spaces were never introduced in class, I’ll proceed with the original formulation. First, an unnecessary graph of \Phi_0,\dots,\Phi_4; the order is red, green, yellow, blue, magenta.

Hermite Functions

Claim 1. H_n is a polynomial of degree n with the leading term 2^n x^n. Proof by induction, starting with H_0=1. Observe that

\displaystyle H_{n+1}=- e^{x^2} \frac{d}{dx}\left(e^{-x^2} H_n\right) =2x H_n - H_n'

where the first term has degree n+1 and the second n-1. So, their sum has degree exactly n+1, and the leading coefficient is 2^{n+1}. Claim 1 is proved.

In particular, Claim 1 tells us that the span of the \Phi_0,\dots,\Phi_n is the same as the span of \lbrace x^k e^{-x^2/2}\colon 0\le k\le n\rbrace.

Claim 2. \Phi_m\perp \Phi_n for m\ne n. We may assume m<n. Must show \int_{\mathbb R} H_m(x) H_n(x) e^{-x^2}\,dx=0. Since H_m is a polynomial of degree m<n, it suffices to prove

(*) \displaystyle \int_{\mathbb R} x^k H_n(x) e^{-x^2}\,dx=0 for integers 0\le k<n.

Rewrite (*) as \int_{\mathbb R} x^k \left(\frac{d}{dx}\right)^n e^{-x^2} \,dx=0 and integrate by parts repeatedly, throwing the derivatives onto x^k until the poor guy can't handle it anymore and dies. No boundary terms appear because e^{-x^2} decays superexponentially at infinity, easily beating any polynomial factors. Claim 2 is proved.

Combining Claim 1 and Claim 2, we see that \Phi_n belongs to the (n+1)-dimensional space \mathrm{span}\,\lbrace x^k e^{-x^2/2}\colon 0\le k\le n\rbrace, and is orthogonal to the n-dimensional subspace \mathrm{span}\,\lbrace x^k e^{-x^2/2}\colon 0\le k\le n-1\rbrace. Since the “Gram-Schmidtization'' of x^n e^{-x^2/2} has the same properties, we conclude that \Phi_n agrees with this “Gram-Schmidtization'' up to a scalar factor.

It remains to prove that the scalar factor is unimodular (\pm 1 since we are over reals).

Claim 3. \langle \Phi_n, \Phi_n\rangle=1 for all n. To this end we must show \int_{\mathbb R} H_n(x)H_n(x)e^{-x^2}\,dx =2^n n! \sqrt{\pi}. Expand the first factor H_n into monomials, use (*) to kill the degrees less than n, and recall Claim 1 to obtain
\int_{\mathbb R} H_n(x)H_n(x)e^{-x^2}\,dx = 2^n \int_{\mathbb R} x^n H_n(x)e^{-x^2}\,dx = (-1)^n 2^n\int_{\mathbb R} x^n \left(\frac{d}{dx}\right)^n e^{-x^2} \,dx.
As in the proof of Claim 2, we integrate by parts throwing the derivatives onto x^n. After n integrations the result is
2^n \int_{\mathbb R} n! e^{-x^2} \,dx = 2^n n! \sqrt{\pi}, as claimed.

P.S. According to Wikipedia, these are the “physicists’ Hermite polynomials”. The “probabilists’ Hermite polynomials” are normalized to have the leading coefficient 1.