Completeness of normalized probabilists’ Hermite polynomials in Gaussian L²

Let $r\in L^2(\gamma )$ be orthogonal to every $h_m$. Each monomial $x^j$ is a finite linear combination of $H_0,\dots ,H_j$ by [F2], and each fixed polynomial is square-integrable under $\gamma$ because a polynomial times $e^{-x^2\mathrm{/}2}$ has an integrable tail. Therefore $\int r(x)x^{j}d\gamma (x)=0$ for every $j\ge 0$.

Put $v(x)=r(x)\varphi (x)$, so $v\in L^1(\mathbb{R})$ by [F1] and [F6]. For every $a\ge 0$, Cauchy–Schwarz and completion of the square give $\int \mathrm{\mid }r(x)\mathrm{\mid }{e}^{a\mathrm{\mid }x\mathrm{\mid }}d\gamma (x)\le \mathrm{\parallel }r{\mathrm{\parallel }}_2(\int e^{2a\mathrm{\mid }x\mathrm{\mid }}d\gamma {)}^{1\mathrm{/}2}<\mathrm{\infty }$, since each half-line integral is at most $e^{2a^2}$.

Fix $\xi \in \mathbb{R}$. The partial sums $S_M(x)={\sum }_{j=0}^M(-2\pi i\xi x{)}^j\mathrm{/}j!$ converge pointwise to $e^{-2\pi i\xi x}$ and satisfy $\mathrm{\mid }{S}_M(x)\mathrm{\mid }\le e^{2\pi \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}$ by [F5]. Applying [F7] with the integrable majorant from step 1.2 gives $\widehat{v}(\xi )={\lim }_{M\to \mathrm{\infty }}\int r(x)S_M(x)d\gamma (x)=0$, because every term is a scalar multiple of a zero moment from step 1.1.

By [F4], $\widehat{v}=0$ implies $v=0$ Lebesgue-almost everywhere. Since $\varphi (x)>0$ for all $x$, $r=0$ $\gamma$-almost everywhere. Thus the only $L^2(\gamma )$ function orthogonal to all the $h_m$ is zero.

For arbitrary real $f\in L^2(\gamma )$, set $a_m=\int f{h}_{m}d\gamma$ and $S_M={\sum }_{m=0}^M{a}_m{h}_m$. Orthogonality in [F2] gives $\mathrm{\parallel }f-S_M{\mathrm{\parallel }}_2^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2-{\sum }_{m=0}^M{a}_m^2\ge 0$ and, for $M>K$, $\mathrm{\parallel }{S}_M-S_K{\mathrm{\parallel }}_2^2={\sum }_{K<m\le M}{a}_m^2$; hence the partial sums are Cauchy. View them in complex $L^2(\gamma )$; by [F3] and [A1] they converge there, and their limit is real because every $S_M$ is real. The residual $f-S$ is orthogonal to each $h_m$, hence zero by step 3.1. Therefore $f=S$ in $L^2(\gamma )$ and taking norms yields Parseval’s identity.

If only finitely many $a_m$ are nonzero, step 4.1 makes $f$ equal almost everywhere to a polynomial $p$. If $f$ is continuous and differs from $p$ at some point, it differs on a nonempty open interval, which has positive $\gamma$ measure by [F1]; hence $f=p$ everywhere, contrary to nonpolynomiality. Thus a continuous nonpolynomial $f$ has unbounded coefficient support. A globally Lipschitz $f$ obeys $\mathrm{\mid }f(x)\mathrm{\mid }\le \mathrm{\mid }f(0)\mathrm{\mid }+L\mathrm{\mid }x\mathrm{\mid }$; [F8] gives $\int \mathrm{\mid }x{\mathrm{\mid }}^{2}d\gamma <\mathrm{\infty }$, so this growth bound implies $f\in L^2(\gamma )$.