The complete orthonormal basis on $L^2(-\infty,+\infty)$ about Hermite polynomials.

Question

Prove that $\{\varphi_n(t)=(2^n n! \sqrt{\pi})^{-\frac{1}{2}}e^{-\frac{t^2}{2}}H_n(t)\}_{n=0}^{\infty}$ are the complete orthonormal basis on $L^2(-\infty,+\infty)$ where $H_n(t)$ are the Hermite polynomials.

I have aleady proved $\{\varphi_n(t)\}$ are orthonormal by the trick "Integration by parts". But I have difficulties on proving the completeness. I want to use the denseness of the continuous functions with a compact supported set in $L^2(-\infty,+\infty)$ and use polynomials to approximate those continuous functions, and the use the properties of integral $\int_{-\infty}^{+\infty} e^{-\frac{t^2}{2}}p(t)dt$ where $p(t)$ is a polynomial. But I faied.

I want to know if I am on the right track. And if it isn't correct, what's the right way to do it?

Any kind of help is appreciated and thank you very much in advance!

Answer 1

There is a nice machinery for proving that polynomials are dense on certain weighted $L^2$-spaces, which involves some beautiful complex analysis tools. In this particular case, the proof is rather simple and follows from standard Hilbert space theory on $L^2(\mathbb{R})$.

Let $\mathcal{P}$ denote set of polynomials with complex coefficients. According to the comments, we came to agreement that it suffices to prove that $$ S = \left\{e^{-t^2 /2}p(t): p \in \mathcal{P} \right\}. $$ forms a dense subspace in $L^{2}(\mathbb{R})$. This is equivalent to proving that the orthogonal complement of $S$ in $L^2(\mathbb{R})$, denoted by $S^\perp$ only consists of the zero function, that is $S^\perp = \{0\}$. Let $f\in S^\perp$, that is $$ \int_\mathbb{R} f(t)\,t^n \, e^{-t^2 /2}dt = 0 \qquad ,\forall n\geq 0. $$ Consider the Fourier transform of $f(t)e^{-t^2 /2}$, defined by $$ F(\xi) := \int_{\mathbb{R}}f(t) e^{-t^2 /2 } \, e^{-it\xi} dt \qquad, \, \xi \in \mathbb{R}. $$ A simple series expansion (easily justified by the dominated convergence theorem, for instance) shows that $$ F(\xi) = \sum_{n=0}^{\infty} \frac{(-i\xi)^n}{n!} \underbrace{\int_{\mathbb{R}}f(t)\,t^n\, e^{-t^2 /2} dt}_{=0} = 0 \qquad, \, \forall \xi \in \mathbb{R}. $$ Since $\varphi(t)=f(t)e^{-t^2 /2} \in L^2$, it follows that by Parseval's theorem that $$ \int_{\mathbb{R}} \left| f(t)e^{-t^2/2} \right|^2 dt = \int_{\mathbb{R}} \left|F(\xi) \right|^2 d\xi = 0. $$ This implies that $f(t) \equiv 0$, thus we conclude that $S$ is dense in $L^2(\mathbb{R})$.