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The following theorem, Theorem, is of utmost importance. But don't forget the other stuff in the second file. Namely AA and for example AA, AA and AA. Here is a link pretending to be in this text: link.
Let $\psi_\alpha$, $\alpha\in\N$, be Hermite functions. Then \[ \psi_\alpha'(x) = \sqrt{\frac{\alpha}{2}}\psi_{\alpha-1}(x) - \sqrt{\frac{\alpha+1}{2}}\psi_{\alpha+1}(x) \] and \[ x\psi_\alpha(x) = \sqrt{\frac{\alpha}{2}}\psi_{\alpha-1}(x) + \sqrt{\frac{\alpha+1}{2}}\psi_{\alpha+1}(x). \] Moreover the Hermite functions are eigenfunctions to the quantum mechanical oscillator \begin{equation} \label{eigenFunction} (x^2 - \partial_x^2) \psi_\alpha(x) = (2\alpha + 1)\psi_\alpha(x). \end{equation}
In this section we will first show that the Hermite functions form an orthonormal set in $L^2(\R)$. After that we will show that the set is complete and that leads to the Hermite transform.
Let $\psi_\alpha:\R\to\R$ be Hermite functions. Then \[ \int_{-\infty}^\infty \psi_\alpha(x)\psi_\beta(x) dx = \delta_{\alpha\beta} \] i.e. the sequence $(\psi_\alpha)_{\alpha=0}^\infty$ is orthonormal in $L^2(\R)$.
Note first the identity $(x+\partial_x)(x-\partial_x) = (x^2-\partial_x^2) + 1$. Combine it with the fact that $\psi_n$ is an eigenfunction for the quantum oscillator \eqref{eigenFunction} to get \[ (x+\partial_x)(x-\partial_x)\psi_n = 2(n+1)\psi_n \] for any $n\in\N$. Note also that the transpose of $x-\partial_x$ in the $L^2$ inner product is $x+\partial_x$.
Hence we get \begin{align*} &\int \psi_\alpha \psi_\beta dx = \frac{1}{2\sqrt{\alpha\beta}} \int (x-\partial_x)\psi_{\alpha-1}(x+\partial_x)\psi_{\beta-1} dx \\ &\qquad = \frac{1}{2\sqrt{\alpha\beta}} \int \psi_{\alpha-1} (x+\partial_x)(x-\partial_x)\psi_{\beta-1} \\ &\qquad = \sqrt{\frac{\beta}{\alpha}} \int \psi_{\alpha-1}\psi_{\beta-1} dx. \end{align*}
Using the previous equation we see that \[ \int \psi_\alpha \psi_\alpha dx = \int \psi_0^2 dx = \pi^{-1/2}\int_{-\infty}^\infty e^{-x^2} dx = 1. \]
If, on the other hand for example $\alpha\lt\beta$, then \[ \int\psi_\alpha\psi_\beta dx = \ldots = c_{\alpha,\beta} \int \psi_0\psi_{\beta-\alpha} dx = c'_{\alpha,\beta} \int_{-\infty}^\infty \partial_x^{\beta-\alpha} e^{-x^2} dx = 0 \] since all the derivatives of Gaussians vanish at infinity.
Let $\mathscr{O}=\{x, \partial_x\}$ be the set of operators containing the multiplication by $x$ and differentiation by $x$ operators. Let $L^0_\alpha = \{\psi_\alpha\}$ and \[ L^{k+1}_\alpha = \{ Sf \mid S\in\mathscr{O}, \, f\in L^k_\alpha \}. \] Then \[ \norm{f}_{L^2(\R)} \leq 2^{k/2} \prod_{\ell=1}^k \sqrt{\alpha+\ell} = \sqrt{ 2^k \frac{(a+k)!}{a!} } \] for any $f \in L^k_\alpha$.
The sequence $(\psi_\alpha)_\alpha$ is orthonormal by Lemma . Hence we have $\norm{\psi_\alpha}_2=1$. Assume that the claim is true for $L^k_\alpha$ for any $\alpha\in\N$. We will prove it for $L^{k+1}_\alpha$. Let that $f\in L^k_\alpha$. Then there are operators $S_1,\ldots,S_k \in \mathscr{O}$ such that $f = S_1S_2\cdots S_k \psi_\alpha$.
Consider the case $S_k = \partial_x$. Let $O\in\mathscr{O}$ and use one of the basic identities to get \begin{align*} &\norm{Of}_2 = \norm{OS_1\cdots S_{k-1} \partial_x \psi_\alpha}_2 \\ &\qquad = \norm{OS_1\cdots S_{k-1} \left( \sqrt{\tfrac{\alpha}{2}}\psi_{\alpha-1} - \sqrt{\tfrac{\alpha+1}{2}}\psi_{\alpha+1}\right)}_2 \\ &\qquad \leq \sqrt{\tfrac{\alpha}{2}} \norm{OS_1\cdots S_{k-1} \psi_{\alpha-1}}_2 + \sqrt{\tfrac{\alpha+1}{2}} \norm{OS_1\cdots S_{k-1} \psi_{\alpha+1}}_2 \\ &\qquad \leq \sqrt{\tfrac{\alpha+1}{2}} 2^{k/2} \left( \prod_{\ell=1}^k \sqrt{\alpha-1+\ell} + \prod_{\ell=1}^k \sqrt{\alpha+1+\ell} \right) \\ &\qquad \leq 2^{(k-1)/2} \sqrt{\alpha+1} \cdot 2 \prod_{\ell=1}^k\sqrt{\alpha+1+\ell} \\ &\qquad \leq 2^{(k+1)/2} \prod_{\ell=1}^{k+1} \sqrt{\alpha+\ell} \end{align*} since $OS_1\cdots S_{k-1} \psi_{\alpha-1}$ and $OS_1\cdots S_{k-1} \psi_{\alpha+1}$ are in $L^k_{\alpha-1}$ and $L^k_{\alpha+1}$, respectively. The same kind of deduction works in the case when $S_k$ is multiplication by $x$. Hence the claim follows by induction.