# Hermite functions

Jordan Bell
September 9, 2015

## 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

 $\{v\in V:p_{\alpha}(v)<\epsilon\},\qquad\alpha\in A,\quad\epsilon>0$

is a local base at $0$.11 1 http://individual.utoronto.ca/jordanbell/notes/holomorphic.pdf, Theorem 1 and Theorem 4. We call this the topology induced by the family of seminorms. If $\{p_{n}:n\geq 0\}$ is a separating family of seminorms, then

 $d(v,w)=\sum_{n=0}^{\infty}2^{-n}\frac{p_{n}(v-w)}{1+p_{n}(v-w)},\qquad v,w\in V,$

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 $\phi\in C^{\infty}(\mathbb{R},\mathbb{C})$ and $n\geq 0$, let

 $p_{n}(\phi)=\sup_{0\leq k\leq n}\sup_{u\in\mathbb{R}}(1+u^{2})^{n/2}|\phi^{(k)% }(u)|.$

We define $\mathscr{S}$ to be the set of those $\phi\in C^{\infty}(\mathbb{R},\mathbb{C})$ such that $p_{n}(\phi)<\infty$ for all $n\geq 0$. $\mathscr{S}$ is a complex vector space and each $p_{n}$ is a norm, and because each $p_{n}$ is a norm, a fortiori $\{p_{n}:n\geq 0\}$ is a separating family of seminorms. With the topology induced by this family of seminorms, $\mathscr{S}$ is a Fréchet space.22 2 Walter Rudin, Functional Analysis, second ed., p. 184, Theorem 7.4. As well, $D:\mathscr{S}\to\mathscr{S}$ defined by

 $(D\phi)(x)=\phi^{\prime}(x),\qquad x\in\mathbb{R}$

and $M:\mathscr{S}\to\mathscr{S}$ defined by

 $(M\phi)(x)=x\phi(x),\qquad x\in\mathbb{R}$

are continuous linear maps.

## 3 Hermite functions

Let $\lambda$ be Lebesgue measure on $\mathbb{R}$ and let

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

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

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

For $n\geq 0$, define $H_{n}:\mathbb{R}\to\mathbb{R}$ 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

 $\exp(2zx-z^{2})=\sum_{n=0}^{\infty}\frac{1}{n!}H_{n}(x)z^{n},\qquad z\in% \mathbb{C}.$ (1)

For $m,n\geq 0$,

 $\int_{\mathbb{R}}H_{m}(x)H_{n}(x)e^{-x^{2}}d\lambda(x)=2^{n}n!\sqrt{\pi}\delta% _{m,n}.$

For $n\geq 0$, define $h_{n}:\mathbb{R}\to\mathbb{R}$ 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 $m,n\geq 0$,

 $(h_{m},h_{n})_{L^{2}}=\int_{\mathbb{R}}h_{m}(x)h_{n}(x)d\lambda(x)=\delta_{m,n}.$

One proves that $\{h_{n}:n\geq 0\}$ is an orthonormal basis for $(L^{2}(\lambda),(\cdot,\cdot)_{L^{2}})$.3

We remind ourselves that for $x\in\mathbb{R}$,4

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

and by the dominated convergence theorem this yields

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

and so

 $h_{n}(x)=(2^{n}n!\sqrt{\pi})^{-1/2}e^{x^{2}/2}\cdot 2^{-1}\pi^{-1/2}\int_{% \mathbb{R}}(iy)^{n}e^{-y^{2}/4}e^{-ixy}dy.$ (2)

## 4 Mehler’s formula

We now prove Mehler’s formula for the Hermite functions.55 5 Sundaram Thangavelu, An Introduction to the Uncertainty Principle: Hardy’s Theorem on Lie Groups, p. 8, Proposition 1.2.1.

###### Theorem 1 (Mehler’s formula).

For $z\in\mathbb{C}$ with $|z|<1$ and for $x,y\in\mathbb{R}$,

 $\sum_{n=0}^{\infty}h_{n}(x)h_{n}(y)z^{n}=\pi^{-1/2}(1-z^{2})^{-1/2}\exp\left(-% \frac{1}{2}\cdot\frac{1+z^{2}}{1-z^{2}}(x^{2}+y^{2})+\frac{2z}{1-z^{2}}xy% \right).$
###### Proof.

Using (2),

 $\begin{split}&\displaystyle\sum_{n=0}^{\infty}h_{n}(x)h_{n}(y)z^{n}\\ \displaystyle=&\displaystyle\sum_{n=0}^{\infty}\frac{\sqrt{\pi}}{2^{n}n!}e^{(x% ^{2}+y^{2})/2}z^{n}\left(\int_{\mathbb{R}}(2\pi i\xi)^{n}e^{-\pi^{2}\xi^{2}}e^% {-2\pi ix\xi}d\xi\right)\left(\int_{\mathbb{R}}(2\pi i\zeta)^{n}e^{-\pi^{2}% \zeta^{2}}e^{-2\pi iy\zeta}d\zeta\right)\\ &\displaystyle=\sqrt{\pi}e^{(x^{2}+y^{2})/2}\int_{\mathbb{R}}\int_{\mathbb{R}}% e^{-\pi^{2}\xi^{2}-\pi^{2}\zeta^{2}-2\pi ix\xi-2\pi i\zeta y}\sum_{n=0}^{% \infty}\frac{(-2\pi^{2}\xi\zeta z)^{n}}{n!}d\xi d\zeta\\ &\displaystyle=\sqrt{\pi}e^{(x^{2}+y^{2})/2}\int_{\mathbb{R}}\int_{\mathbb{R}}% e^{-\pi^{2}\xi^{2}-\pi^{2}\zeta^{2}-2\pi ix\xi-2\pi i\zeta y}e^{-2\pi^{2}\xi% \zeta z}d\xi d\zeta.\end{split}$

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

 $\displaystyle\int_{\mathbb{R}}e^{-\pi^{2}\zeta^{2}-2\pi i\zeta y-2\pi^{2}\xi% \zeta z}d\zeta$ $\displaystyle=\int_{\mathbb{R}}e^{-\pi^{2}(\zeta+a)^{2}+\pi^{2}a^{2}}d\zeta$ $\displaystyle=\frac{1}{\sqrt{\pi}}e^{\pi^{2}a^{2}}$ $\displaystyle=\frac{1}{\sqrt{\pi}}\exp\left(-y^{2}+2\pi iy\xi z+\pi^{2}\xi^{2}% z^{2}\right).$

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

 $\begin{split}&\displaystyle\sum_{n=0}^{\infty}h_{n}(x)h_{n}(y)z^{n}\\ \displaystyle=&\displaystyle e^{(x^{2}+y^{2})/2}\int_{\mathbb{R}}e^{-\pi^{2}% \xi^{2}-2\pi ix\xi-y^{2}+2\pi iy\xi z+\pi^{2}\xi^{2}z^{2}}d\xi\\ \displaystyle=&\displaystyle e^{(x^{2}-y^{2})/2}\int_{\mathbb{R}}e^{-\alpha\xi% ^{2}-2\pi i(x-yz)\xi}d\xi\\ \displaystyle=&\displaystyle e^{(x^{2}-y^{2})/2}\sqrt{\frac{\pi}{\alpha}}\exp% \left(-\frac{\pi^{2}}{\alpha}(x-yz)^{2}\right)\\ \displaystyle=&\displaystyle\pi^{-1/2}e^{(x^{2}-y^{2})/2}(1-z^{2})^{-1/2}\exp% \left(-\frac{(x-yz)^{2}}{1-z^{2}}\right)\\ \displaystyle=&\displaystyle\pi^{-1/2}(1-z^{2})^{-1/2}\exp\left(-\frac{x^{2}}{% 1-z^{2}}+\frac{2xyz}{1-z^{2}}-\frac{y^{2}z^{2}}{1-z^{2}}+\frac{x^{2}}{2}-\frac% {y^{2}}{2}\right)\\ \displaystyle=&\displaystyle\pi^{-1/2}(1-z^{2})^{-1/2}\exp\left(-\frac{1}{2}% \frac{1+z^{2}}{1-z^{2}}(x^{2}+y^{2})+\frac{2z}{1-z^{2}}xy\right).\end{split}$

## 5 The Hermite operator

We define $A:\mathscr{S}\to\mathscr{S}$ by

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

i.e.,

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

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

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

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

 $(A\phi,\phi)_{L^{2}}=\int_{\mathbb{R}}(\phi^{\prime}(x)\overline{\phi^{\prime}% (x)}+(x^{2}+1)\phi(x)\overline{\phi(x)})d\lambda(x)\geq 0,$

so $A$ is a positive operator.

It is straightforward to check that each $h_{n}$ belongs to $\mathscr{S}$. For $n\geq 0$, we calculate that

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

and hence

 $(Ah_{n})(x)=(2n+1-x^{2})h_{n}(x)+x^{2}h_{n}(x)+h_{n}(x)=(2n+2)h_{n}(x),$

i.e.

 $Ah_{n}=(2n+2)h_{n}.$

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 that6

 $Th_{n}=A^{-1}h_{n}=(2n+2)^{-1}h_{n},\qquad n\geq 0.$ (3)

The operator norm of $T$ is

 $\left\|T\right\|=\sup_{n\geq 0}\frac{1}{2n+2}=\frac{1}{2}.$

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

 $f=\sum_{n=0}^{\infty}(f,h_{n})_{L^{2}}h_{n}.$

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

 $\displaystyle(Tf,g)_{L^{2}}$ $\displaystyle=\left(\sum_{n=0}^{\infty}(f,h_{n})_{L^{2}}Th_{n},\sum_{n=0}^{% \infty}(g,h_{n})_{L^{2}}h_{n}\right)_{L^{2}}$ $\displaystyle=\left(\sum_{n=0}^{\infty}(f,h_{n})_{L^{2}}(2n+2)^{-1}h_{n},\sum_% {n=0}^{\infty}(g,h_{n})_{L^{2}}h_{n}\right)_{L^{2}}$ $\displaystyle=\sum_{n=0}^{\infty}(2n+2)^{-1}(f,h_{n})_{L^{2}}\overline{(g,h_{n% })_{L^{2}}},$

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

For $p\geq 0$,

 $|T^{p}h_{n}|_{L^{2}}^{2}=|(2n+2)^{-p}h_{n}|_{L^{2}}^{2}=(2n+2)^{-2p}|h_{n}|_{L% ^{2}}^{2}=(2n+2)^{-2p}.$

Therefore for $p\geq 1$,

 $\sum_{n=0}^{\infty}|T^{p}h_{n}|_{L^{2}}^{2}=\sum_{n=0}^{\infty}(2n+2)^{-2p}=2^% {-2p}\sum_{m=1}^{\infty}m^{-2p}=2^{-2p}\zeta(2p).$

This means that for $p\geq 1$, $T^{p}$ is a Hilbert-Schmidt operator with Hilbert-Schmidt norm7

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

## 6 Creation and annihilation operators

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

 $2\sum_{n=0}^{\infty}\frac{1}{n!}H_{n}(x)z^{n+1}=\sum_{n=0}^{\infty}\frac{1}{n!% }H_{n}^{\prime}(x)z^{n},$

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

 $H_{n}^{\prime}=2nH_{n-1},$

and so

 $h_{n}^{\prime}(x)=(2n)^{1/2}h_{n-1}(x)-xh_{n}(x),$

i.e.

 $Dh_{n}=(2n)^{1/2}h_{n-1}-Mh_{n}.$

Furthermore, from its definition we calculate

 $h_{n}^{\prime}(x)=xh_{n}(x)-(2n+2)^{1/2}h_{n+1}(x),$

i.e.

 $Dh_{n}=Mh_{n}-(2n+2)^{1/2}h_{n+1}.$

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

 $(B\phi)(x)=\phi^{\prime}(x)+x\phi(x),\qquad x\in\mathbb{R},$

i.e.

 $B=D+M,$

which is a continuous linear map $\mathscr{S}\to\mathscr{S}$. For $n\geq 1$, we calculate

 $Bh_{n}=(2n)^{1/2}h_{n-1},$

and $h_{0}(x)=\pi^{-1/4}e^{-x^{2}/2}$, so $Bh_{0}=0$.

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

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

i.e.

 $C=-D+M,$

which is a continuous linear map $\mathscr{S}\to\mathscr{S}$. For $n\geq 0$, we calculate

 $Ch_{n}=(2n+2)^{1/2}h_{n+1}.$

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 $\phi\in\mathscr{S}$,

 $B-C=2D.$

Furthermore,

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

and

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

## 7 The Fourier transform

Define $\mathscr{F}:\mathscr{S}\to\mathscr{S}$, for $\phi\in\mathscr{S}$, by

 $(\mathscr{F}\phi)(\xi)=\hat{\phi}(\xi)=\int_{\mathbb{R}}\phi(x)e^{-i\xi x}% \frac{dx}{(2\pi)^{1/2}},\qquad\xi\in\mathbb{R}.$

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

 $\lim_{h\to 0}\frac{\hat{\phi}(\xi+h)-\hat{\phi}(\xi)}{h}=\int_{\mathbb{R}}(-ix% )\phi(x)e^{-i\xi x}\frac{dx}{(2\pi)^{1/2}},,$

i.e.

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

in other words,

 $\mathscr{F}(M\phi)=iD(\mathscr{F}\phi).$ (5)

Also, by the dominated convergence theorem we obtain

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

in other words,

 $\mathscr{F}(D\phi)=iM(\mathscr{F}\phi).$ (6)

For $\phi\in\mathscr{S}$,

 $\phi(x)=\int_{\mathbb{R}}\hat{\phi}(\xi)e^{ix\xi}\frac{d\xi}{(2\pi)^{1/2}},% \qquad x\in\mathbb{R}.$ (7)

$\phi\mapsto\hat{\phi}$ is an isomorphism of locally convex spaces $\mathscr{S}\to\mathscr{S}$.88 8 Walter Rudin, Functional Analysis, second ed., p. 186, Theorem 7.7. Using (7) and the Cauchy-Schwarz inequality

 $\displaystyle\left\|\phi\right\|_{\infty}$ $\displaystyle\leq\int_{\mathbb{R}}(1+\xi^{2})^{1/2}(1+\xi^{2})^{-1/2}|\hat{% \phi}(\xi)|\frac{d\xi}{(2\pi)^{1/2}}$ $\displaystyle\leq(2\pi)^{-1/2}\left(\int_{\mathbb{R}}(1+\xi^{2})^{-1}d\xi% \right)^{1/2}\left(\int_{\mathbb{R}}(1+\xi^{2})|\hat{\phi}(\xi)|^{2}d\xi\right% )^{1/2}$ $\displaystyle=2^{-1/2}\left(\int_{\mathbb{R}}(1+\xi^{2})|\hat{\phi}(\xi)|^{2}d% \xi\right)^{1/2},$

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

 $\displaystyle\left\|\phi\right\|_{\infty}^{2}$ $\displaystyle\leq 2^{-1}\int_{\mathbb{R}}|\hat{\phi}(\xi)|^{2}d\xi+2^{-1}\int_% {\mathbb{R}}\xi^{2}|\hat{\phi}(\xi)|^{2}d\xi$ $\displaystyle=2^{-1}\int_{\mathbb{R}}|\hat{\phi}(\xi)|^{2}d\xi+2^{-1}\int_{% \mathbb{R}}|(\mathscr{F}\phi^{\prime})(\xi)|^{2}d\xi$ $\displaystyle=2^{-1}|\phi|_{L^{2}}^{2}+2^{-1}|\phi^{\prime}|_{L^{2}}^{2},$

and therefore

 $\left\|\phi\right\|_{\infty}\leq 2^{-1/2}(|\phi|_{L^{2}}+|\phi^{\prime}|_{L^{2% }}).$ (8)

We remind ourselves that

 $A=-D^{2}+M^{2}+1,\qquad B=D+M,\qquad C=-D+M.$

Using

 $\mathscr{F}D=iM\mathscr{F},\qquad D\mathscr{F}=\frac{1}{i}\mathscr{F}M,$

we get

 $\displaystyle\mathscr{F}A$ $\displaystyle=\mathscr{F}(-D^{2}+M^{2}+1)$ $\displaystyle=-(iM\mathscr{F})D+(iD\mathscr{F})M+\mathscr{F}$ $\displaystyle=-iM(iM\mathscr{F})+iD(iD\mathscr{F})+\mathscr{F}$ $\displaystyle=M^{2}\mathscr{F}-D^{2}\mathscr{F}+\mathscr{F}$ $\displaystyle=A\mathscr{F},$

and

 $\mathscr{F}B=\mathscr{F}(D+M)=iM\mathscr{F}+iD\mathscr{F}=iB\mathscr{F}$

and

 $\mathscr{F}C=\mathscr{F}(-D+M)=-iM\mathscr{F}+iD\mathscr{F}=-iC\mathscr{F}.$

We now determine the Fourier transform of the Hermite functions.

###### Theorem 2.

For $n\geq 0$,

 $\mathscr{F}h_{n}=(-i)^{n}h_{n}.$
###### Proof.

For $n\geq 0$, by induction, from $\mathscr{F}C=-iC\mathscr{F}$ we get

 $\mathscr{F}C^{n}=(-iC)^{n}\mathscr{F}.$

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

 $\mathscr{F}g=g,$

and using this with the above yields

 $\displaystyle\mathscr{F}h_{n}$ $\displaystyle=\pi^{-1/4}(2^{n}n!)^{-1/2}\mathscr{F}C^{n}g$ $\displaystyle=\pi^{-1/4}(2^{n}n!)^{-1/2}(-iC)^{n}\mathscr{F}g$ $\displaystyle=\pi^{-1/4}(2^{n}n!)^{-1/2}(-iC)^{n}g$ $\displaystyle=\pi^{-1/4}(2^{n}n!)^{-1/2}(-i)^{n}\cdot\pi^{1/4}(2^{n}n!)^{1/2}h% _{n}$ $\displaystyle=(-i)^{n}h_{n}.$

There is a unique Hilbert space isomorphism $\mathscr{F}:L^{2}(\lambda)\to L^{2}(\lambda)$ such that $\mathscr{F}f=\hat{f}$ for all $f\in\mathscr{S}$.99 9 Walter Rudin, Functional Analysis, second ed., p. 188, Theorem 7.9. For $f\in L^{2}(\lambda)$,

 $f=\sum_{n=0}^{\infty}(f,h_{n})_{L^{2}}h_{n},$

and then

 $\mathscr{F}f=\sum_{n=0}^{\infty}(f,h_{n})_{L^{2}}\mathscr{F}h_{n}=\sum_{n=0}^{% \infty}(f,h_{n})_{L^{2}}(-i)^{n}h_{n}.$

## 8 Asymptotics

For $x=0$, (1) reads

 $\sum_{n=0}^{\infty}\frac{1}{n!}H_{n}(0)z^{n}=\exp(-z^{2})=\sum_{n=0}^{\infty}% \frac{(-z^{2})^{n}}{n!},$

thus

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

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

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

For $u(x)=e^{-x^{2}/2}H_{n}(x)$,1010 10 N. N. Lebedev, Special Functions and Their Applications, p. 66, §4.14.

 $u^{\prime}(x)=-xu+e^{-x^{2}/2}H_{n}^{\prime}(x),\qquad u^{\prime\prime}(x)=-u-% xu^{\prime}-xe^{-x^{2}/2}H_{n}^{\prime}(x)+e^{-x^{2}/2}H_{n}^{\prime\prime}(x).$

Using

 $H_{n}^{\prime}(x)=2xH_{n}(x)-H_{n+1}(x),\qquad H_{n}^{\prime}(x)=2nH_{n-1}(x)$

we get

 $H_{n}^{\prime\prime}(x)-2xH_{n}^{\prime}(x)+2nH_{n}(x)=0,$

and thence

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

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

 $v^{\prime\prime}+(2n+1)v=f,\qquad v(0)=H_{n}(0),\qquad v^{\prime}(0)=H_{n}^{% \prime}(0).$ (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

 $v(x)=c_{1}v_{1}+c_{2}v_{2}+Av_{1}+Bv_{2},$

where

 $A(x)=-\int_{0}^{x}\frac{1}{W}v_{2}(t)g(t)dt,\qquad B(x)=\int_{0}^{x}\frac{1}{W% }v_{1}(t)g(t)dt.$

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

 $\displaystyle v(x)$ $\displaystyle=av_{1}(x)+b\lambda^{-1/2}v_{2}(x)$ $\displaystyle-\lambda^{-1/2}v_{1}(x)\int_{0}^{x}v_{2}(t)g(t)dt+\lambda^{-1/2}v% _{2}(x)\int_{0}^{x}v_{1}(t)g(t)dt$ $\displaystyle=a\cos(\lambda^{1/2}x)+b\lambda^{-1/2}\sin(\lambda^{1/2}x)$ $\displaystyle+\lambda^{-1/2}\int_{0}^{x}(\cos(\lambda^{1/2}t)\sin(\lambda^{1/2% }x)-\sin(\lambda^{1/2}t)\cos(\lambda^{1/2}x))g(t)dt$ $\displaystyle=a\cos(\lambda^{1/2}x)+b\lambda^{-1/2}\sin(\lambda^{1/2}x)+% \lambda^{-1/2}\int_{0}^{x}\sin(\lambda^{1/2}(x-t))g(t)dt.$

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

 $\displaystyle v(x)$ $\displaystyle=H_{n}(0)\cos((2n+1)^{1/2}x)+H_{n}^{\prime}(0)(2n+1)^{-1/2}\sin((% 2n+1)^{1/2}x)$ $\displaystyle+(2n+1)^{-1/2}\int_{0}^{x}\sin((2n+1)^{1/2}(x-t))\cdot t^{2}u(t)dt,$

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

 $\displaystyle v(x)$ $\displaystyle=(-1)^{k}\frac{(2k)!}{k!}\cos((4k+1)^{1/2}x)$ $\displaystyle+(4k+1)^{-1/2}\int_{0}^{x}\sin((4k+1)^{1/2}(x-t))\cdot t^{2}u(t)dt$ $\displaystyle=(-1)^{k}\frac{(2k)!}{k!}\cos((4k+1)^{1/2}x)+(4k+1)^{-1/2}r_{2k}(% x).$

We calculate

 $\displaystyle|r_{2k}(x)|^{2}$ $\displaystyle\leq\left(\int_{0}^{|x|}t^{4}dt\right)\left(\int_{0}^{|x|}|u(t)|^% {2}dt\right)$ $\displaystyle\leq\frac{|x|^{5}}{10}\cdot\int_{\mathbb{R}}e^{-t^{2}}|H_{2k}(t)|% ^{2}dt$ $\displaystyle=\frac{|x|^{5}}{10}\cdot 2^{2k}(2k)!\sqrt{\pi},$

i.e.

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

By Stirling’s approximation,

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

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.1111 11 Sundaram Thangavelu, Lectures on Hermite and Laguerre Expansions, pp. 26–27, Lemma 1.5.1 and Lemma 1.5.2; Gábor Szegő, Orthogonal Polynomials; Benjamin Muckenhoupt, Mean convergence of Hermite and Laguerre series. II, Trans. Amer. Math. Soc. 147 (1970), 433–470, Lemma 15.

###### Lemma 3.

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

 $\displaystyle|h_{n}(x)|$ $\displaystyle\leq C(N^{1/3}+|x^{2}-N|)^{-1/4},\qquad x^{2}\leq 2N$ $\displaystyle\leq Ce^{-\gamma x^{2}},\qquad x^{2}>2N,$

and

 $|h_{n}(x)|\leq N^{-1/8}(x-N^{1/2})^{-1/4}e^{-\epsilon N^{1/4}(x-N^{1/2})^{3/2}}$

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

###### Lemma 4.

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

 $h_{n}(x)=\left(\frac{2}{\pi}\right)^{1/2}(N-x^{2})^{-1/4}\cos\left(\frac{N(2% \theta-\sin\theta)-\pi}{4}\right)+O(N^{1/2}(N-x^{2})^{-7/4}).$
###### Theorem 5.
1. 1.

$\left\|h_{n}\right\|_{p}\asymp n^{\frac{1}{2p}-\frac{1}{4}}$ for $1\leq p<4$.

2. 2.

$\left\|h_{n}\right\|_{p}\asymp n^{-\frac{1}{8}}\log n$ for $p=4$.

3. 3.

$\left\|h_{n}\right\|_{p}\asymp n^{-\frac{1}{6p}-\frac{1}{12}}$ for $4.

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

 $\left\|H_{n}\right\|_{p}^{p}=\int_{\mathbb{R}}|H_{n}|^{p}d\gamma.$

Larsson-Cohn1212 12 Lars Larsson-Cohn, $L^{p}$-norms of Hermite polynomials and an extremal problem on Wiener chaos, Ark. Mat. 40 (2002), 134–144. proves that for $0 there is an explicit $c(p)$ such that

 $\left\|H_{n}\right\|_{p}=\frac{c(p)}{n^{1/4}}\sqrt{n!}(1+O(n^{-1})),$

and for $2 there is an explicit $c(p)$ such that

 $\left\|H_{n}\right\|_{p}=\frac{c(p)}{n^{1/4}}\sqrt{n!}(p-1)^{n/2}(1+O(n^{-1})).$

This uses the asymptotic expansion of Plancherel and Rotach.1313 13 M. Plancherel and W. Rotach, Sur les valeurs asymptotiques des polynomes d’Hermite, Commentarii mathematici Helvetici 1 (1929), 227–254.