# proof for a basis in $L^2$

I know, correct me if I am wrong, that the functions $H_n(x)\exp(-x^2/2)$ form a complete basis in $L^2(\mathbb{R},dx)$, where $H_n(x)$ is the $n$th Hermite polynomial. This must be true also for $x^n\exp(-x^2/2)$ with $n\in\mathbb{N}_0$. Does someone know a proof of the latter or can give me a reference? Since I am not a mathematician, I will really appreciate it if the proof contains all the details that might puzzle a non-mathematician.

I also have the problem to choose functions $f_k$ such that

$$\int_{-\infty}^{+\infty}f_k(x)x^n\exp(-x^2)dx = \delta_{kn} \mbox{.}$$

Does anyone know a method to construct the $f_k$s?

• "This must be true also for...": can you clarify this sentence? Are you asking whether the functions $x^n \exp(-x^2/2)$ also form a complete basis for $L^2(\mathbb{R}, dx)$? Note that $x^n$ (or any other polynomial) can be written as a linear combination of Hermite polynomials. – Nate Eldredge Apr 9 '13 at 14:00
• Yes, $x^n$ can be written as superposition of Hermite polynomials. Therefore, I wrote "This must be true also for...". However, I don't know how the proof that this is a complete basis in $L^2$ looks like, neither for $x^n$ nor for $H_n$. That is why I asked for a detailed proof or a reference where such a proof can be found. – Physicist Apr 9 '13 at 15:12

Wikipedia has the following proof of completeness. Suppose $f\in L^2(\mathbb R,\exp(-x^2))$ is orthogonal to all $H_n$, that is $$\int_{\mathbb R} f(x)x^n \exp(-x^2)\,dx=0\quad \text{ for all }\ n\tag1$$ Consider the entire function $$F(z)= \int_{\mathbb R} f(x)\exp(zx-x^2)\,dx \tag2$$ where the integral converges because both $f$ and $e^{zx}$ are in $L^2(\mathbb R,\exp(-x^2))$. Calculating the $n$th derivative $F^{(n)}(0)$, we obtain
$$F^{(n)}(0)= \int_{\mathbb R} f(x)x^n\exp( -x^2)\,dx =0 \tag3$$ Thus $F\equiv 0$. In particular, $F(iy)\equiv 0$, which says precisely that the Fourier transform of $f(x)\exp(-x^2)$ is zero. By the Fourier inversion formula, $f(x)\exp(-x^2)\equiv 0$, completing the proof.