# Orthogonality of Hermite functions

I would like to prove to myself that Hermite functions, defined by $\varphi_n(x)=(-1)^n e^{x^2/2}\frac{d^n e^{-x^2}}{dx^n}$, $n\in\mathbb{N}$ are an orthogonal system in $L^2(\mathbb{R})$, i.e. that, for any $m\ne n$, $$\int_{-\infty}^\infty e^{x^2} \frac{d^m e^{-x^2}}{dx^m} \frac{d^n e^{-x^2}}{dx^n}dx=0$$ I have tried by integrating by parts as the book where I find such a statement, Kolmogorov-Fomin's Элементы теории функций и функционального анализа, says that it can be proved, but I am landing nowhere since I have no experience in manipulating such integrals... Thank you so much for any answer!

It is actually easier to prove the orthogonality of the Hermite polynomials $$H_n=(-1)^n \exp(x^2)\frac{d^n}{dx^n}\exp(-x^2)$$ with respect to the weight $d\omega=\exp(-x^2)dx$. It should be obvious that both claims are the same.
First notice, that each $H_n$ is obviously a polynomial of degree $n$ (this is a simple proof by induction). Let $k$ be any nonnegative integer smaller than $n$, then there holds $$\int_\mathbb{R}x^kH_n(x)d\omega=\int_\mathbb{R}x^kH_n(x)\exp(-x^2)dx=\int_\mathbb{R}x^k\frac{d^n}{dx^n}\exp(-x^2)dx=0$$From the last expression, you can actually see that this integral is zero, as you can partially integrate and shift the derivation to $x^k$ until the term vanishes (remember $k<n$). The boundary terms obviously vanish as $\exp(-x^2)$ decays rapidly. As $H_m$ is only a linear combination of $x^k$, this directly implies, that for any $m<n$,$$\int_\mathbb{R}H_m(x)H_n(x)d\omega=\int_\mathbb{R}\varphi_m(x)\varphi_n(x)dx=0$$ which proves the claim.
• At the first shift I get $\int^b_a x^k\frac{d^n}{dx^n}e^{-x^2}dx=x^k\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}|^b_a-\int_a^b kx^{k-1}\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}dx$, at the $j$-th step I have the "left" term outside the integral in the form $x^{k-j+1}\frac{d^{n-j}}{dx^{n-j}}e^{-x^2}|^b_a$, but I'm not able to prove that it approaches 0 as $b\to\infty$... Thank you so much again! – Self-teaching worker Nov 13 '14 at 16:41
• That expression is just of the type $P(x)\exp(-x^2)$ with some polynomial $P$, which vanishes at $\pm\infty$ because $\exp(-x^2)$ decays faster than any polynomial. This can be proven using elementary properties of the exponential function. – Daniel Nov 13 '14 at 19:09