\(\S1 \) Gaussian Measure and Hermite Polynomials

Define on \(\mathbb{R}^d\) the normalized Gaussian measure
\[ d \gamma(x)=\frac{1}{(2\pi)^{\frac{d}{2}}} e^{-\frac{|x|^2}{2}}dx\]

Consider first the case \(d= 1\). The Taylor expansion of \(e^{-\frac{1}{2}x^2}\) at the point \(x\), with increment \(t\) is
\[e^{−\frac{1}{2}(x−t)^2}=\sum_{n=0}^{\infty}a_n t^n,\]
where
\[a_n=\frac{(−1)^n}{n!}\frac{d^n}{dx}e^{-\frac{x^2}{2}}.\]
This series is convergent for all real or complex values of \(x\) and \(t\), since we are dealing with an entire function. Multiply both sides by \(e^{\frac{1}{2}x^2}\) to get
\[e^{xt−\frac{t^2}{2}}=\sum_{n=0}^{\infty}\frac{1}{n!}(-1)^n e^{\frac{x^2}{2}}\frac{d^n}{dx}e^{-\frac{x^2}{2}}.\]
It is clear that the coefficient of \(t^n\) here is a polynomial in \(x\). We define the \(n\)th Hermite polynomial \(H_n\) by
\[H_n (x)=(−1)^n e^{\frac{x^2}{2}}\frac{d^n}{dx}e^{-\frac{x^2}{2}}.\]

Then
\[e^{xt-\frac{t^2}{2}}=\sum_{n=0}^{\infty}H_n(x)t^n.\]
The functione \(e^{xt-\frac{t^2}{2}}\) is called the generating function of \((H_n)_{n=0}^{\infty}\).

Theorem 1.1. The polynomials \((H_n)_{n=0}^{\infty}\) form a complete orthogonal system in  \( L^2 (\gamma)\), and  \( ||H_n||_{L^2 (\gamma)} =\sqrt{n!}.\)

Proof. It is clear that \(H_n(x)\) is a poplynormial with degree \(n\). Let \(m\le n\). Using the definition of \(H_n\) and integrating by parts, we get, with \(D=d/dx\),
\[\begin{array}{rcl}\int H_m(x)H_n(x)d\gamma(x)&=&\sum_{n=0}^{\infty}\frac{(−1)^n}{\sqrt{2\pi}}\int H_m(x) e^{\frac{x^2}{2}}\left(D^n e^{−\frac{x^2}{2}}\right)e^{\frac{-x^2}{2}}dx\\ &=&\sum_{n=0}^{\infty}\frac{(−1)^n}{\sqrt{2\pi}}\int H_m(x)(D_n e^{−\frac{x^2}{2}})dx\\ &=&\sum_{n=0}^{\infty} \frac{(−1)^n}{\sqrt{2\pi}}\int (D^n H_m(x)) e^{−\frac{x^2}{2}}dx,\\ \end{array}\]
and this vanishes if \(m<n\). For \(m=n\) the same calculation yields
\[\frac{1}{\sqrt{2\pi}}\int (D_n H_n(x))e^{−\frac{x^2}{2}}dx=n!,\]
and thus \(||H_n||_{L^2 (\gamma)} =\sqrt{n!}\), as claimed.

It remains to prove the completeness. Since any polynomial can be expressed as linear combinations of Hermite polynomials, it suffices to show that the set of all polynomials is dense in \(L^2 (\gamma)\). Assume that \(f\in L^2(\gamma)\subset L^1(\gamma)\) is orthogonal to all polynomials. If \(f\) can be shown to be zero, completeness is proved. The product \(f(x)e^{−\frac{x^2}{2}}\) is in \(L^1(dx)\), so it has a well-defined Fourier transform. Calculating this Fourier transform, expanding \(e^{i\xi x}\) in a Taylor series and assuming that we can interchange the order of summation and integration, we get that
\[\int e^{i\xi x} f(x)d \gamma(x)=\sum_{n=0}^{\infty} \frac{i^n {\xi}^n}{n!} \int x^n f(x)d \gamma(x)=0,  \forall \xi \in \mathbb{R}.\]
We conclude that \(f=0\).

Finally, we must verify that the order of summation and integration in (1.2) can be switched. We shall majorize \(\sum_{n=0}^N\frac{|\xi|^n}{n!}|x|^n|f(x)|\) by an \(L^1(\gamma)\) function, uniformly in \(N\in \mathbb{N}\). But
\[\sum_{n=0}^N\frac{|\xi|^n}{n!}|x|^n|f(x)|\le\sum_{n=0}^{\infty}\frac{|\xi|^n}{n!}|x|^n |f(x)|=e^{|\xi||x|}|f(x)|,\]
and by the Cauchy-Schwarz inequality
\[\int e^{|\xi||x|}|f(x)|d\gamma(x)\le \left(\int |f(x)|^2d\gamma(x)\right)^{\frac{1}{2}}\left(\int e^{2|\xi||x|}d\gamma(x)\right)^{\frac{1}{2}}<\infty.\]

Remark.  Let \(X, Y\) be two random variables with joint Gaussian distribution such that \(E(X)=E(Y)=0,E(X^2)E(Y^2)=1\), then

\[E\left(e^{sX-\frac{s^2}{2}}e^{tY-\frac{t^2}{2}}\right)=e^{stE(XY)}.\]

Taking the \((n + m)\)th partial derivative \(\frac{\partial^{n+m}}{\partial s^n \partial t^m}\) at \(s = t = 0\) in both sides of the above equality yields the same result as theorem 1.1 claimed.