Hermite functions

1 Locally convex spaces

If $V$ is a vector space and $\{p_{\alpha }:\alpha \in A\}$ is a separating family of seminorms on $V$, then there is a unique topology with which $V$ is a locally convex space and such that the collection of finite intersections of sets of the form

is a local base at $0$. We call this the topology induced by the family of seminorms. If $\{p_n:n\ge 0\}$ is a separating family of seminorms, then

$$d(v,w)=\sum _{n=0}^{\mathrm{\infty }}{2}^{-n}\frac{p_n(v-w)}{1+p_n(v-w)},v,w\in V,$$

is a metric on $V$ that induces the same topology as the family of seminorms. If $d$ is a complete metric, then $V$ is called a Fréchet space.

2 Schwartz functions

For $\varphi \in C^{\mathrm{\infty }}(ℝ,ℂ)$ and $n\ge 0$, let

$$p_n(\varphi )=\underset{0\le k\le n}{sup}\underset{u\in ℝ}{sup}{(1+u^2)}^{n/2}|{\varphi }^{(k)}(u)|.$$

We define $𝒮$ to be the set of those $\varphi \in C^{\mathrm{\infty }}(ℝ,ℂ)$ such that for all $n\ge 0$. $𝒮$ is a complex vector space and each $p_n$ is a norm, and because each $p_n$ is a norm, a fortiori $\{p_n:n\ge 0\}$ is a separating family of seminorms. With the topology induced by this family of seminorms, $𝒮$ is a Fréchet space. As well, $D:𝒮\to 𝒮$ defined by

$$(D\varphi )(x)={\varphi }^{\prime }(x),x\in ℝ$$

and $M:𝒮\to 𝒮$ defined by

$$(M\varphi )(x)=x\varphi (x),x\in ℝ$$

are continuous linear maps.

3 Hermite functions

Let $\lambda$ be Lebesgue measure on $ℝ$ and let

$${(f,g)}_{L^2}={\int }_{ℝ}f\overline{g}𝑑\lambda .$$

With this inner product, $L^2(\lambda )$ is a separable Hilbert space. We write

$${|f|}_{L^2}^2={(f,f)}_{L^2}={\int }_{ℝ}{|f|}^2𝑑\lambda .$$

For $n\ge 0$, define $H_n:ℝ\to ℝ$ by

$$H_n(x)={(-1)}^n{e}^{x^2}{D}^n{e}^{-x^2},$$

which is a polynomial of degree $n$. $H_n$ are called Hermite polynomials. It can be shown that

$$\exp (2zx-z^2)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}{H}_n(x)z^n,z\in ℂ.$$ (1)

For $m,n\ge 0$,

$${\int }_{ℝ}{H}_m(x)H_n(x)e^{-x^2}𝑑\lambda (x)=2^{n}n!\sqrt{\pi }{\delta }_{m,n}.$$

For $n\ge 0$, define $h_n:ℝ\to ℝ$ by

$$h_n(x)={(2^{n}n!\sqrt{\pi })}^{-1/2}{e}^{-x^2/2}{H}_n(x)={(-1)}^n{(2^{n}n!\sqrt{\pi })}^{-1/2}{e}^{x^2/2}{D}^n{e}^{-x^2}.$$

$h_n$ are called Hermite functions. Then for $m,n\ge 0$,

$${(h_m,h_n)}_{L^2}={\int }_{ℝ}{h}_m(x)h_n(x)𝑑\lambda (x)={\delta }_{m,n}.$$

One proves that $\{h_n:n\ge 0\}$ is an orthonormal basis for $(L^2(\lambda ),{(\cdot ,\cdot )}_{L^2})$.

We remind ourselves that for $x\in ℝ$,

$$e^{-x^2}=2^{-1}{\pi }^{-1/2}{\int }_{ℝ}{e}^{-y^2/4}{e}^{-ixy}𝑑y,$$

and by the dominated convergence theorem this yields

$$D^n{e}^{-x^2}=2^{-1}{\pi }^{-1/2}{\int }_{ℝ}{(-iy)}^n{e}^{-y^2/4}{e}^{-ixy}𝑑y,$$

and so

$$h_n(x)={(2^{n}n!\sqrt{\pi })}^{-1/2}{e}^{x^2/2}\cdot 2^{-1}{\pi }^{-1/2}{\int }_{ℝ}{(iy)}^n{e}^{-y^2/4}{e}^{-ixy}𝑑y.$$ (2)

4 Mehler’s formula

We now prove Mehler’s formula for the Hermite functions.

Theorem 1 (Mehler’s formula).

For $z\in ℂ$ with and for $x,y\in ℝ$,

$$\sum _{n=0}^{\mathrm{\infty }}{h}_n(x)h_n(y)z^n={\pi }^{-1/2}{(1-z^2)}^{-1/2}\exp \left(-\frac{1}{2}\cdot \frac{1+z^2}{1-z^2}(x^2+y^2)+\frac{2z}{1-z^2}xy\right).$$

Proof.

Using (2),

Now, writing $a=\frac{iy}{\pi }+\xi z$, we calculate

${\displaystyle {\int }_{ℝ}}{e}^{-{\pi }^2{\zeta }^2-2\pi i\zeta y-2{\pi }^2\xi \zeta z}𝑑\zeta$ $={\displaystyle {\int }_{ℝ}}{e}^{-{\pi }^2{(\zeta +a)}^2+{\pi }^2{a}^2}𝑑\zeta$

$={\displaystyle \frac{1}{\sqrt{\pi }}}{e}^{{\pi }^2{a}^2}$

$={\displaystyle \frac{1}{\sqrt{\pi }}}\exp \left(-y^2+2\pi iy\xi z+{\pi }^2{\xi }^2{z}^2\right).$

Then, for $\alpha =(1-z^2){\pi }^2$,

∎

5 The Hermite operator

We define $A:𝒮\to 𝒮$ by

$$(A\varphi )(x)=-{\varphi }^{\prime \prime }(x)+(x^2+1)\varphi (x),x\in ℝ,$$

i.e.,

$$A=-D^2+M^2+1,$$

which is a continuous linear map $𝒮\to 𝒮$, which we call the Hermite operator. $𝒮$ is a dense linear subspace of the Hilbert space $L^2(\lambda )$, and $A:𝒮\to 𝒮$ is a linear map, so $A$ is a densely defined operator in $L^2(\lambda )$. For $\varphi ,\psi \in 𝒮$, integrating by parts,

${(A\varphi ,\psi )}_{L^2}$ $={\displaystyle {\int }_{ℝ}}(-{\varphi }^{\prime \prime }(x)+(x^2+1)\varphi (x))\overline{\psi (x)}𝑑\lambda (x)$

$={\displaystyle {\int }_{ℝ}}-{\varphi }^{\prime \prime }(x)\overline{\psi (x)}d\lambda (x)+{\displaystyle {\int }_{ℝ}}(x^2+1)\varphi (x)\overline{\psi (x)}𝑑\lambda (x)$

$={\displaystyle {\int }_{ℝ}}-\varphi (x)\overline{{\psi }^{\prime \prime }(x)}d\lambda (x)+{\displaystyle {\int }_{ℝ}}(x^2+1)\varphi (x)\overline{\psi (x)}𝑑\lambda (x)$

$={(\varphi ,A\psi )}_{L^2},$

showing that $A:𝒮\to 𝒮$ is symmetric. Furthermore, also integrating by parts,

$${(A\varphi ,\varphi )}_{L^2}={\int }_{ℝ}({\varphi }^{\prime }(x)\overline{{\varphi }^{\prime }(x)}+(x^2+1)\varphi (x)\overline{\varphi (x)})𝑑\lambda (x)\ge 0,$$

so $A$ is a positive operator.

It is straightforward to check that each $h_n$ belongs to $𝒮$. For $n\ge 0$, we calculate that

$$h_n^{\prime \prime }(x)+(2n+1-x^2)h_n(x)=0,$$

and hence

$$(A{h}_n)(x)=(2n+1-x^2)h_n(x)+x^2{h}_n(x)+h_n(x)=(2n+2)h_n(x),$$

i.e.

$$A{h}_n=(2n+2)h_n.$$

Therefore, for each $h_n$, $A^{-1}{h}_n=\frac{1}{2n+2}{h}_n$, and it follows that there is a unique bounded linear operator $T:L^2(\lambda )\to L^2(\lambda )$ such that

$$T{h}_n=A^{-1}{h}_n={(2n+2)}^{-1}{h}_n,n\ge 0.$$ (3)

The operator norm of $T$ is

$$\parallel T\parallel =\underset{n\ge 0}{sup}\frac{1}{2n+2}=\frac{1}{2}.$$

The Hermite functions are an orthonormal basis for $L^2(\lambda )$, so for $f\in L^2(\lambda )$,

$$f=\sum _{n=0}^{\mathrm{\infty }}{(f,h_n)}_{L^2}{h}_n.$$

For $f,g\in L^2(\lambda )$,

${(Tf,g)}_{L^2}$ $={({\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(f,h_n)}_{L^2}T{h}_n,{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(g,h_n)}_{L^2}{h}_n)}_{L^2}$

$={({\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(f,h_n)}_{L^2}{(2n+2)}^{-1}{h}_n,{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(g,h_n)}_{L^2}{h}_n)}_{L^2}$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(2n+2)}^{-1}{(f,h_n)}_{L^2}\overline{{(g,h_n)}_{L^2}},$

from which it is immediate that $T$ is self-adjoint.

For $p\ge 0$,

$${|T^p{h}_n|}_{L^2}^2={|{(2n+2)}^{-p}{h}_n|}_{L^2}^2={(2n+2)}^{-2p}{|h_n|}_{L^2}^2={(2n+2)}^{-2p}.$$

Therefore for $p\ge 1$,

$$\sum _{n=0}^{\mathrm{\infty }}{|T^p{h}_n|}_{L^2}^2=\sum _{n=0}^{\mathrm{\infty }}{(2n+2)}^{-2p}=2^{-2p}\sum _{m=1}^{\mathrm{\infty }}{m}^{-2p}=2^{-2p}\zeta (2p).$$

This means that for $p\ge 1$, $T^p$ is a Hilbert-Schmidt operator with Hilbert-Schmidt norm

$${\parallel T^p\parallel }_{\mathrm{HS}}=2^{-p}\sqrt{\zeta (2p)}.$$

6 Creation and annihilation operators

Taking the derivative of (1) with respect to $x$ gives

$$2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}{H}_n(x)z^{n+1}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}{H}_n^{\prime }(x)z^n,$$

so $H_0^{\prime }=0$ and for $n\ge 1$, $\frac{1}{n!}{H}_n^{\prime }(x)=\frac{1}{(n-1)!}2H_{n-1}(x)$, i.e.

$$H_n^{\prime }=2n{H}_{n-1},$$

and so

$$h_n^{\prime }(x)={(2n)}^{1/2}{h}_{n-1}(x)-x{h}_n(x),$$

i.e.

$$D{h}_n={(2n)}^{1/2}{h}_{n-1}-M{h}_n.$$

Furthermore, from its definition we calculate

$$h_n^{\prime }(x)=x{h}_n(x)-{(2n+2)}^{1/2}{h}_{n+1}(x),$$

i.e.

$$D{h}_n=M{h}_n-{(2n+2)}^{1/2}{h}_{n+1}.$$

We define $B:𝒮\to 𝒮$, called the annihilation operator, by

$$(B\varphi )(x)={\varphi }^{\prime }(x)+x\varphi (x),x\in ℝ,$$

i.e.

$$B=D+M,$$

which is a continuous linear map $𝒮\to 𝒮$. For $n\ge 1$, we calculate

$$B{h}_n={(2n)}^{1/2}{h}_{n-1},$$

and $h_0(x)={\pi }^{-1/4}{e}^{-x^2/2}$, so $B{h}_0=0$.

We define $C:𝒮\to 𝒮$, called the creation operator, by

$$(C\varphi )(x)=-{\varphi }^{\prime }(x)+x\varphi (x),x\in ℝ,$$

i.e.

$$C=-D+M,$$

which is a continuous linear map $𝒮\to 𝒮$. For $n\ge 0$, we calculate

$$C{h}_n={(2n+2)}^{1/2}{h}_{n+1}.$$

Thus,

$$h_n={(2^{n}n!)}^{-1/2}{C}^n{h}_0={\pi }^{-1/4}{(2^{n}n!)}^{-1/2}{C}^n(e^{-x^2/2}).$$ (4)

For $\varphi \in 𝒮$,

$$B-C=2D.$$

Furthermore,

$$BC=-D^2+M^2+1=A$$

and

$$CB=-D^2+M^2-1=A-2.$$

7 The Fourier transform

Define $ℱ:𝒮\to 𝒮$, for $\varphi \in 𝒮$, by

$$(ℱ\varphi )(\xi )=\widehat{\varphi }(\xi )={\int }_{ℝ}\varphi (x)e^{-i\xi x}\frac{dx}{{(2\pi )}^{1/2}},\xi \in ℝ.$$

For $\xi \in ℝ$, by the dominated convergence theorem we have

$$\underset{h\to 0}{lim}\frac{\widehat{\varphi }(\xi +h)-\widehat{\varphi }(\xi )}{h}={\int }_{ℝ}(-ix)\varphi (x)e^{-i\xi x}\frac{dx}{{(2\pi )}^{1/2}},,$$

i.e.

$$\widehat{x\varphi (x)}(\xi )=-i^{-1}D\widehat{\varphi }(\xi )=iD\widehat{\varphi }(\xi ),$$

in other words,

$$ℱ(M\varphi )=iD(ℱ\varphi ).$$ (5)

Also, by the dominated convergence theorem we obtain

$$\widehat{D\varphi }(\xi )=i\xi \widehat{\varphi }(\xi ),$$

in other words,

$$ℱ(D\varphi )=iM(ℱ\varphi ).$$ (6)

For $\varphi \in 𝒮$,

$$\varphi (x)={\int }_{ℝ}\widehat{\varphi }(\xi )e^{ix\xi }\frac{d\xi }{{(2\pi )}^{1/2}},x\in ℝ.$$ (7)

$\varphi \mapsto \widehat{\varphi }$ is an isomorphism of locally convex spaces $𝒮\to 𝒮$. Using (7) and the Cauchy-Schwarz inequality

${\parallel \varphi \parallel }_{\mathrm{\infty }}$ $\le {\displaystyle {\int }_{ℝ}}{(1+{\xi }^2)}^{1/2}{(1+{\xi }^2)}^{-1/2}|\widehat{\varphi }(\xi )|{\displaystyle \frac{d\xi }{{(2\pi )}^{1/2}}}$

$\le {(2\pi )}^{-1/2}{\left({\displaystyle {\int }_{ℝ}}{(1+{\xi }^2)}^{-1}𝑑\xi \right)}^{1/2}{\left({\displaystyle {\int }_{ℝ}}(1+{\xi }^2){|\widehat{\varphi }(\xi )|}^2𝑑\xi \right)}^{1/2}$

$=2^{-1/2}{\left({\displaystyle {\int }_{ℝ}}(1+{\xi }^2){|\widehat{\varphi }(\xi )|}^2𝑑\xi \right)}^{1/2},$

and using (6) and the fact that ${|\widehat{\varphi }|}_{L^2}={|\varphi |}_{L^2}$,

${\parallel \varphi \parallel }_{\mathrm{\infty }}^2$ $\le 2^{-1}{\displaystyle {\int }_{ℝ}}{|\widehat{\varphi }(\xi )|}^2𝑑\xi +2^{-1}{\displaystyle {\int }_{ℝ}}{\xi }^2{|\widehat{\varphi }(\xi )|}^2𝑑\xi$

$=2^{-1}{\displaystyle {\int }_{ℝ}}{|\widehat{\varphi }(\xi )|}^2𝑑\xi +2^{-1}{\displaystyle {\int }_{ℝ}}{|(ℱ{\varphi }^{\prime })(\xi )|}^2𝑑\xi$

$=2^{-1}{|\varphi |}_{L^2}^2+2^{-1}{|{\varphi }^{\prime }|}_{L^2}^2,$

and therefore

$${\parallel \varphi \parallel }_{\mathrm{\infty }}\le 2^{-1/2}({|\varphi |}_{L^2}+{|{\varphi }^{\prime }|}_{L^2}).$$ (8)

We remind ourselves that

$$A=-D^2+M^2+1,B=D+M,C=-D+M.$$

Using

$$ℱD=iMℱ,Dℱ=\frac{1}{i}ℱM,$$

we get

$ℱA$ $=ℱ(-D^2+M^2+1)$

$=-(iMℱ)D+(iDℱ)M+ℱ$

$=-iM(iMℱ)+iD(iDℱ)+ℱ$

$=M^2ℱ-D^2ℱ+ℱ$

$=Aℱ,$

and

$$ℱB=ℱ(D+M)=iMℱ+iDℱ=iBℱ$$

and

$$ℱC=ℱ(-D+M)=-iMℱ+iDℱ=-iCℱ.$$

We now determine the Fourier transform of the Hermite functions.

Theorem 2.

For $n\ge 0$,

$$ℱh_n={(-i)}^n{h}_n.$$

Proof.

For $n\ge 0$, by induction, from $ℱC=-iCℱ$ we get

$$ℱC^n={(-iC)}^nℱ.$$

From (4),

$$h_n={\pi }^{-1/4}{(2^{n}n!)}^{-1/2}{C}^n(e^{-x^2/2}).$$

Writing $g(x)=e^{-x^2/2}$, it is a fact that

$$ℱg=g,$$

and using this with the above yields

$ℱh_n$ $={\pi }^{-1/4}{(2^{n}n!)}^{-1/2}ℱC^{n}g$

$={\pi }^{-1/4}{(2^{n}n!)}^{-1/2}{(-iC)}^nℱg$

$={\pi }^{-1/4}{(2^{n}n!)}^{-1/2}{(-iC)}^{n}g$

$={\pi }^{-1/4}{(2^{n}n!)}^{-1/2}{(-i)}^n\cdot {\pi }^{1/4}{(2^{n}n!)}^{1/2}{h}_n$

$={(-i)}^n{h}_n.$

∎

There is a unique Hilbert space isomorphism $ℱ:L^2(\lambda )\to L^2(\lambda )$ such that $ℱf=\widehat{f}$ for all $f\in 𝒮$. For $f\in L^2(\lambda )$,

$$f=\sum _{n=0}^{\mathrm{\infty }}{(f,h_n)}_{L^2}{h}_n,$$

and then

$$ℱf=\sum _{n=0}^{\mathrm{\infty }}{(f,h_n)}_{L^2}ℱh_n=\sum _{n=0}^{\mathrm{\infty }}{(f,h_n)}_{L^2}{(-i)}^n{h}_n.$$

8 Asymptotics

For $x=0$, (1) reads

$$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}{H}_n(0)z^n=\exp (-z^2)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-z^2)}^n}{n!},$$

thus

$$H_{2n}(0)={(-1)}^n\frac{(2n)!}{n!},H_{2n+1}(0)=0.$$

Similarly, taking the derivative of (1) with respect to $x$ yields

$$H_{2n}^{\prime }(0)=0,H_{2n+1}^{\prime }(0)=2{(-1)}^n\frac{(2n+1)!}{n!}.$$

For $u(x)=e^{-x^2/2}{H}_n(x)$,

$$u^{\prime }(x)=-xu+e^{-x^2/2}{H}_n^{\prime }(x),u^{\prime \prime }(x)=-u-x{u}^{\prime }-x{e}^{-x^2/2}{H}_n^{\prime }(x)+e^{-x^2/2}{H}_n^{\prime \prime }(x).$$

Using

$$H_n^{\prime }(x)=2x{H}_n(x)-H_{n+1}(x),H_n^{\prime }(x)=2n{H}_{n-1}(x)$$

we get

$$H_n^{\prime \prime }(x)-2x{H}_n^{\prime }(x)+2n{H}_n(x)=0,$$

and thence

$$u^{\prime \prime }=-u+x^{2}u-2nu.$$

Thus, writing $f(x)=x^{2}u(x)$, $u$ satisfies the initial value problem

$$v^{\prime \prime }+(2n+1)v=f,v(0)=H_n(0),v^{\prime }(0)=H_n^{\prime }(0).$$ (9)

Now, for $\lambda >0$, two linearly independent solutions of $v^{\prime \prime }+\lambda v=0$ are $v_1(x)=\cos ({\lambda }^{1/2}x)$ and $v_2(x)=\sin ({\lambda }^{1/2}x)$. The Wronskian of $(v_1,v_2)$ is $W={\lambda }^{1/2}$, and using variation of parameters, if $v$ satisfies $v^{\prime \prime }+\lambda v=g$ then there are $c_1,c_2$ such that

$$v(x)=c_1{v}_1+c_2{v}_2+A{v}_1+B{v}_2,$$

where

$$A(x)=-{\int }_0^x\frac{1}{W}{v}_2(t)g(t)𝑑t,B(x)={\int }_0^x\frac{1}{W}{v}_1(t)g(t)𝑑t.$$

We calculate that the unique solution of the initial value problem $v^{\prime \prime }+\lambda v=g$, $v(0)=a$, $v^{\prime }(0)=b$, is

$v(x)$ $=a{v}_1(x)+b{\lambda }^{-1/2}{v}_2(x)$

$-{\lambda }^{-1/2}{v}_1(x){\displaystyle {\int }_0^x}{v}_2(t)g(t)𝑑t+{\lambda }^{-1/2}{v}_2(x){\displaystyle {\int }_0^x}{v}_1(t)g(t)𝑑t$

$=a\cos ({\lambda }^{1/2}x)+b{\lambda }^{-1/2}\sin ({\lambda }^{1/2}x)$

$+{\lambda }^{-1/2}{\displaystyle {\int }_0^x}(\cos ({\lambda }^{1/2}t)\sin ({\lambda }^{1/2}x)-\sin ({\lambda }^{1/2}t)\cos ({\lambda }^{1/2}x))g(t)𝑑t$

$=a\cos ({\lambda }^{1/2}x)+b{\lambda }^{-1/2}\sin ({\lambda }^{1/2}x)+{\lambda }^{-1/2}{\displaystyle {\int }_0^x}\sin ({\lambda }^{1/2}(x-t))g(t)𝑑t.$

Therefore the unique solution of the initial value problem (9) is

$v(x)$ $=H_n(0)\cos ({(2n+1)}^{1/2}x)+H_n^{\prime }(0){(2n+1)}^{-1/2}\sin ({(2n+1)}^{1/2}x)$

$+{(2n+1)}^{-1/2}{\displaystyle {\int }_0^x}\sin ({(2n+1)}^{1/2}(x-t))\cdot t^{2}u(t)𝑑t,$

where $u(x)=e^{-x^2/2}{H}_n(x)$. If $n=2k$ then

$v(x)$ $={(-1)}^k{\displaystyle \frac{(2k)!}{k!}}\cos ({(4k+1)}^{1/2}x)$

$+{(4k+1)}^{-1/2}{\displaystyle {\int }_0^x}\sin ({(4k+1)}^{1/2}(x-t))\cdot t^{2}u(t)𝑑t$

$={(-1)}^k{\displaystyle \frac{(2k)!}{k!}}\cos ({(4k+1)}^{1/2}x)+{(4k+1)}^{-1/2}{r}_{2k}(x).$

We calculate

${|r_{2k}(x)|}^2$ $\le \left({\displaystyle {\int }_0^{|x|}}{t}^4𝑑t\right)\left({\displaystyle {\int }_0^{|x|}}{|u(t)|}^2𝑑t\right)$

$\le {\displaystyle \frac{{|x|}^5}{10}}\cdot {\displaystyle {\int }_{ℝ}}{e}^{-t^2}{|H_{2k}(t)|}^2𝑑t$

$={\displaystyle \frac{{|x|}^5}{10}}\cdot 2^{2k}(2k)!\sqrt{\pi },$

i.e.

$$|r_{2k}(x)|\le {\pi }^{1/4}\frac{{|x|}^{5/2}}{\sqrt{10}}{2}^k\sqrt{(2k)!}.$$

By Stirling’s approximation,

${\displaystyle \frac{2^k\sqrt{(2k)!}}{\frac{(2k)!}{k!}}}$ $={\displaystyle \frac{2^{k}k!}{\sqrt{(2k)!}}}\sim {\displaystyle \frac{2^k{(2\pi k)}^{1/2}{k}^k{e}^{-k}}{{({(4\pi k)}^{1/2}{(2k)}^{2k}{e}^{-2k})}^{1/2}}}={\pi }^{1/4}{k}^{1/4}.$

Thus for ${\alpha }_{2k}=\frac{(2k)!}{k!}$,

$$\frac{|r_{2k}(x)|}{{\alpha }_{2k}}=O({|x|}^{5/2}\cdot k^{1/4}\cdot k^{-1/2})=O({|x|}^{5/2}{k}^{-1/4}).$$

Thangavelu states the following inequality and asymptotics without proof, and refers to Szegő and Muckenhoupt.

Lemma 3.

There are $\gamma ,C,ϵ>0$ such that for $N=2n+1$,

$|h_n(x)|$ $\le C{(N^{1/3}+|x^2-N|)}^{-1/4},x^2\le 2N$

$\le C{e}^{-\gamma x^2},x^2>2N,$

and

$$|h_n(x)|\le N^{-1/8}{(x-N^{1/2})}^{-1/4}{e}^{-ϵN^{1/4}{(x-N^{1/2})}^{3/2}}$$

for $N^{1/2}+N^{-1/6}\le x\le {(2N)}^{1/2}$.

Lemma 4.

For $N=2n+1$, $0\le x\le N^{\frac{1}{2}}-N^{-\frac{1}{6}}$, and $\theta =\arccos (x{N}^{-\frac{1}{2}})$,

$$h_n(x)={\left(\frac{2}{\pi }\right)}^{1/2}{(N-x^2)}^{-1/4}\cos \left(\frac{N(2\theta -\sin \theta )-\pi }{4}\right)+O(N^{1/2}{(N-x^2)}^{-7/4}).$$

Theorem 5.

1. ${\parallel h_n\parallel }_p\asymp n^{\frac{1}{2p}-\frac{1}{4}}$ for .

2. ${\parallel h_n\parallel }_p\asymp n^{-\frac{1}{8}}\log n$ for $p=4$.

3. ${\parallel h_n\parallel }_p\asymp n^{-\frac{1}{6p}-\frac{1}{12}}$ for .

Rather than taking the $p$th power of $h_n$, one can instead take the $p$th power of $H_n$ and integrate this with respect to Gaussian measure. Writing $d\gamma (x)={(2\pi )}^{-1/2}{e}^{-x^2/2}dx$ and taking $H_n$ to be the Hermite polynomial that is monic, now write

$${\parallel H_n\parallel }_p^p={\int }_{ℝ}{|H_n|}^p𝑑\gamma .$$

Larsson-Cohn proves that for there is an explicit $c(p)$ such that

$${\parallel H_n\parallel }_p=\frac{c(p)}{n^{1/4}}\sqrt{n!}(1+O(n^{-1})),$$

and for there is an explicit $c(p)$ such that

$${\parallel H_n\parallel }_p=\frac{c(p)}{n^{1/4}}\sqrt{n!}{(p-1)}^{n/2}(1+O(n^{-1})).$$

This uses the asymptotic expansion of Plancherel and Rotach.