Normalized Hermite Fourier eigenfunctions

Put $a=\sqrt{\pi }x+(2\sqrt{\pi }{)}^{-1}d\mathrm{/}dx$ and $u_m=(a^{†}{)}^m{h}_0$. Differentiation gives $a{h}_0=0$ and $a{a}^{†}-a^{†}a=I$, since $(d\mathrm{/}dx)(xf)-x{f}^{\mathrm{\prime }}=f$. Induction gives $a{u}_m=m{u}_{m-1}$ for $m\ge 1$, and $a^{†}a{u}_m=m{u}_m$ for $m\ge 0$. If $u_m=Q_m{h}_0$, then $Q_{m+1}=2\sqrt{\pi }x{Q}_m-(2\sqrt{\pi }{)}^{-1}{Q}_m^{\mathrm{\prime }}$. Thus $Q_m$ is real of degree $m$ with leading coefficient $(2\sqrt{\pi }{)}^m$, and [F1] makes every $u_m$ Schwartz.

For polynomial Gaussians $v,w$, [F4] gives $⟨a^{†}v,w⟩=⟨v,aw⟩$: derivative products are polynomial Gaussians and integrable, and their endpoint products vanish. Consequently $N=a^{†}a$ is symmetric on these functions, since $⟨Nv,w⟩=⟨av,aw⟩=⟨v,Nw⟩$. Step 1.1 implies $(m-l)⟨u_m,u_l⟩=0$. Also $\mathrm{\parallel }{u}_m{\mathrm{\parallel }}_2^2=⟨a^{†}{u}_{m-1},u_m⟩=m\mathrm{\parallel }{u}_{m-1}{\mathrm{\parallel }}_2^2$. The base norm is $\mathrm{\parallel }{h}_0{\mathrm{\parallel }}_2^2=\sqrt{2}\int e^{-2\pi x^2}dx=1$ by [F2]. Hence $\mathrm{\parallel }{u}_m{\mathrm{\parallel }}_2^2=m!$ and the normalized functions are orthonormal, including $m=0$.

The derivative identities [F3] give $\mathcal{F}(a^{†}v)=(i\mathrm{/}(2\sqrt{\pi }))(\widehat{v}{)}^{\mathrm{\prime }}-i\sqrt{\pi }\xi \widehat{v}=-i{a}^{†}\widehat{v}$. Since [F2] gives ${\widehat{h}}_0=h_0$, induction yields ${\widehat{u}}_m=(-i{)}^m{u}_m$ and the asserted normalized identity. All operations are on Schwartz functions, so this also holds for their Plancherel classes.

Suppose $f\in L^2$ is orthogonal to all $h_m$. By the nonzero real leading coefficients in step 1.1, triangular induction expresses each monomial $x^r$ as a real linear combination of $Q_0,\dots ,Q_r$. Therefore $\int f(x)h_0(x)x^{r}dx=0$ for every $r$; these integrals exist by [F5], since $x^r{h}_0\in L^2$. Put $v=f{h}_0\in L^1$ by [F5]. For fixed real $\xi$, the exponential Taylor partial sums are bounded by $e^{2\pi \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}$. The majorant $\mathrm{\mid }f\mathrm{\mid }{h}_0{e}^{2\pi \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}$ is integrable by [F5]: its second factor has finite square integral, because $-2\pi x^2+4\pi \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }\le -\pi x^2+4\pi {\xi }^2$. Thus [F7] integrates the exponential series termwise, all terms being the zero moments. It gives $\widehat{v}(\xi )=0$ for every $\xi$. By [F6], $v=0$ a.e.; positivity of $h_0$ gives $f=0$ a.e.

For arbitrary $f\in L^2$, put $s_N={\sum }_{m=0}^N⟨f,h_m⟩h_m$. Finite orthogonality in step 2.1 gives $\mathrm{\parallel }f-s_N{\mathrm{\parallel }}_2^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2-{\sum }_{m=0}^N\mathrm{\mid }⟨f,h_m⟩{\mathrm{\mid }}^2\ge 0$. Hence the coefficient-square partial sums are bounded increasing and converge; their tails give $\mathrm{\parallel }{s}_M-s_N{\mathrm{\parallel }}_2^2\to 0$. Completeness in [F5] supplies $s\in L^2$ with $s_N\to s$. Pairing continuity shows $⟨f-s,h_m⟩=0$ for every $m$, so step 2.3 gives $f=s$. This proves the promised expansion, not merely orthogonality. All sequences are specified; countable choice is inherited from the complex integral and completeness interfaces.