# Schwartz functions, Hermite functions, and the Hermite operator

Jordan Bell
July 17, 2015

## 1 Schwartz functions

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

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

We define $\mathscr{S}$ to be the set of those $\phi\in C^{\infty}(\mathbb{R},\mathbb{C})$ such that $|\phi|_{p}<\infty$ for all $p\geq 0$. $\mathscr{S}$ is a complex vector space and each $|\cdot|_{p}$ is a norm, and because each $|\cdot|_{p}$ is a norm, a fortiori $\{|\cdot|_{p}:p\geq 0\}$ is a separating family of seminorms. With the topology induced by this family of seminorms, $\mathscr{S}$ is a Fréchet space.11 1 Walter Rudin, Functional Analysis, second ed., p. 184, Theorem 7.4. Furthermore, $D:\mathscr{S}\to\mathscr{S}$ 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.

Let $\mathscr{S}^{\prime}$ be the collection of continuous linear maps $\mathscr{S}\to\mathbb{C}$. For $\phi\in\mathscr{S}$, define $e_{\phi}:\mathscr{S}^{\prime}\to\mathbb{C}$ by

 $e_{\phi}(\omega)=\omega(\phi),\qquad\omega\in\mathscr{S}^{\prime}.$

The initial topology for the collection $\{e_{\phi}:\phi\in\mathscr{S}\}$ is called the weak-* topology on $\mathscr{S}^{\prime}$. With this topology, $\mathscr{S}^{\prime}$ is a locally convex space whose dual space is $\{e_{\phi}:\phi\in\mathscr{S}\}$.

## 2 L2 norms

For $p\geq 0$ and $\phi,\psi\in\mathscr{S}$, let

 $[\phi,\psi]_{p}=\sum_{k=0}^{p}\int_{\mathbb{R}}(1+u^{2})^{p}\phi^{(k)}(u)% \overline{\psi^{(k)}(u)}du,$

and let

 $[\phi]_{p}^{2}=[\phi,\phi]_{p}=\sum_{k=0}^{p}\int_{\mathbb{R}}(1+u^{2})^{p}|% \phi^{(k)}(u)|^{2}du.$

Because $(1+u^{2})^{p}\leq(1+u^{2})^{q}$ when $p\leq q$, it is immediate that $[\phi]_{p}\leq[\phi]_{q}$ when $p\leq q$.

We relate the norms $|\cdot|_{p}$ and the norms $[\cdot]_{p}$.22 2 Takeyuki Hida, Brownian Motion, p. 305, Lemma A.1.

###### Lemma 1.

For each $p\geq 1$, for all $\phi\in\mathscr{S}$,

 $\frac{1}{p\sqrt{\pi}}|\phi|_{p-1}\leq[\phi]_{p}\leq\sqrt{(p+1)\pi}|\phi|_{p+1}.$
###### Proof.

For $0\leq k\leq p$,

 $\displaystyle\int_{\mathbb{R}}(1+u^{2})^{p}|\phi^{(k)}(u)|^{2}du$ $\displaystyle\leq\sup_{u\in\mathbb{R}}((1+u^{2})^{p+1}|\phi^{(k)}(u)|^{2})\int% _{\mathbb{R}}(1+u^{2})^{-1}du$ $\displaystyle=\sup_{u\in\mathbb{R}}((1+u^{2})^{p+1}|\phi^{(k)}(u)|^{2})\cdot\pi$ $\displaystyle\leq\pi|\phi|_{p+1}^{2},$

hence

 $\displaystyle[\phi]_{p}^{2}$ $\displaystyle=\sum_{k=0}^{p}\int_{\mathbb{R}}(1+u^{2})^{p}|\phi^{(k)}(u)|^{2}du$ $\displaystyle\leq\sum_{k=0}^{p}\pi|\phi|_{p+1}^{2}$ $\displaystyle=(p+1)\pi|\phi|_{p+1}^{2}.$

For $0\leq k\leq p-1$ and $u\in\mathbb{R}$, using the fundamental theorem of calculus and the Cauchy-Schwarz inequality,

 $\displaystyle|(1+u^{2})^{(p-1)/2}\phi^{(k)}(u)|$ $\displaystyle=\left|\int_{-\infty}^{u}((1+t^{2})^{(p-1)/2}\phi^{(k)}(t))^{% \prime}dt\right|$ $\displaystyle\leq\int_{\mathbb{R}}|(p-1)t(1+t^{2})^{(p-1)/2-1}\phi^{(k)}(t)|dt$ $\displaystyle+\int_{\mathbb{R}}|(1+t^{2})^{(p-1)/2}\phi^{(k+1)}(t)|dt$ $\displaystyle\leq(p-1)\int_{\mathbb{R}}(1+t^{2})^{-1/2}(1+t^{2})^{(p-1)/2}|% \phi^{(k)}(t)|dt$ $\displaystyle+\int_{\mathbb{R}}(1+t^{2})^{-1/2}(1+t^{2})^{p/2}|\phi^{(k+1)}(t)% |dt$ $\displaystyle\leq(p-1)\left(\int_{\mathbb{R}}(1+t^{2})^{-1}dt\right)^{1/2}% \left(\int_{\mathbb{R}}(1+t^{2})^{p-1}|\phi^{(k)}(t)|^{2}dt\right)^{1/2}$ $\displaystyle+\left(\int_{\mathbb{R}}(1+t^{2})^{-1}dt\right)^{1/2}\left(\int_{% \mathbb{R}}(1+t^{2})^{p}|\phi^{(k+1)}(t)|^{2}dt\right)^{1/2}$ $\displaystyle\leq(p-1)\sqrt{\pi}[\phi]_{p-1}+\sqrt{\pi}[\phi]_{p}$ $\displaystyle\leq p\sqrt{\pi}[\phi]_{p},$

which shows that

 $|\phi|_{p-1}\leq p\sqrt{\pi}[\phi]_{p}.$

## 3 Hermite functions

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

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

$L^{2}(\lambda)$ with the inner product $(\cdot,\cdot)_{L^{2}}$ is a separable Hilbert space. For $n\geq 0$, let

 $h_{n}(x)=(-1)^{n}(2^{n}n!\sqrt{\pi})^{-1/2}e^{x^{2}/2}D^{n}e^{-x^{2}},$

the Hermite functions, the set of which is an orthonormal basis for $L^{2}(\lambda)$. We remark that the Hermite functions belong to $\mathscr{S}$. For $n<0$ we define

 $h_{n}=0,$

to write some expressions in a uniform way.

We calculate that for $n\geq 0$,

 $Dh_{n}=\sqrt{\frac{n}{2}}h_{n-1}-\sqrt{\frac{n+1}{2}}h_{n+1}.$

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

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

$A$ is a densely defined operator in $L^{2}(\lambda)$ that is symmetric and positive, and satisfies

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

There is a unique bounded linear operator $T:L^{2}(\lambda)\to L^{2}(\lambda)$ satisfying

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

The operator norm of $T$ is $\left\|T\right\|=\frac{1}{2}$, and $T$ is self-adjoint. For $p\geq 1$, $T^{p}$ is a Hilbert-Schmidt operator with Hilbert-Schmidt norm $\left\|T^{p}\right\|_{\mathrm{HS}}=2^{-p}\sqrt{\zeta(2p)}$.

We define the creation operator $B:\mathscr{S}\to\mathscr{S}$ by

 $B=D+M$

and we define the annihilation operator $C:\mathscr{S}\to\mathscr{S}$ by

 $C=-D+M,$

which are continuous linear maps. They satisfy, for $n\geq 0$,

 $Bh_{n}=(2n)^{1/2}h_{n-1},\qquad Ch_{n}=(2n+2)^{1/2}h_{n+1}.$

(We remind ourselves that we have defined $h_{-1}=0$.) It is immediate that $BC=A$ and that $B-C=2D$. Using the creation operator, we can write the Hermite functions as

 $h_{n}=(2^{n}n!)^{-1/2}C^{n}h_{0}=\pi^{-1/4}(2^{n}n!)^{-1/2}C^{n}(e^{-x^{2}/2}).$

For $\phi,\psi\in\mathscr{S}$, using integration by parts,

 $(D\phi,\psi)_{L^{2}}=\int_{\mathbb{R}}\phi^{\prime}(x)\overline{\psi(x)}dx=-% \int_{\mathbb{R}}\phi(x)\overline{\psi^{\prime}(x)}dx=(\phi,(-D)\psi)_{L^{2}},$

and

 $(M\phi,\psi)_{L^{2}}=\int_{\mathbb{R}}x\phi(x)\overline{\psi(x)}dx=(\phi,M\psi% )_{L^{2}}.$

Thus,

 $\displaystyle(B\phi,\psi)_{L^{2}}$ $\displaystyle=(D\phi,\psi)_{L^{2}}+(M\phi,\psi)_{L^{2}}$ $\displaystyle=(\phi,(-D)\psi)_{L^{2}}+(\phi,M\psi)_{L^{2}}$ $\displaystyle=(\phi,C\psi)_{L^{2}}$

and

 $(C\phi,\psi)_{L^{2}}=(\phi,B\psi)_{L^{2}}.$

We shall use these calculations to obtain the following lemma.

###### Lemma 2.

For $p\geq 0$ and for $\phi\in\mathscr{S}$,

 $B^{p}\phi=2^{p/2}\sum_{n=0}^{\infty}\left(\frac{(n+p)!}{n!}\right)^{1/2}(\phi,% h_{n+p})_{L^{2}}h_{n}$

and

 $C^{p}\phi=2^{p/2}\sum_{n=0}^{\infty}(\phi,h_{n-p})_{L^{2}}\left(\frac{n!}{(n-p% )!}\right)^{1/2}h_{n}.$
###### Proof.

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

 $(\phi,C^{p}h_{n})_{L^{2}}=(\phi,h_{n+p})_{L^{2}}\prod_{j=n}^{n+p-1}(2j+2)^{1/2% }=(\phi,h_{n+p})_{L^{2}}2^{p/2}\left(\frac{(n+p)!}{n!}\right)^{1/2}.$

With

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

and because $(B\phi,\psi)_{L^{2}}=(\phi,C\psi)_{L^{2}}$, we have

 $\displaystyle B^{p}\phi$ $\displaystyle=\sum_{n=0}^{\infty}(B^{p}\phi,h_{n})_{L^{2}}h_{n}$ $\displaystyle=\sum_{n=0}^{\infty}(\phi,C^{p}h_{n})_{L^{2}}h_{n}$ $\displaystyle=\sum_{n=0}^{\infty}(\phi,h_{n+p})_{L^{2}}2^{p/2}\left(\frac{(n+p% )!}{n!}\right)^{1/2}h_{n}.$

Because $Bh_{n}=(2n)^{1/2}h_{n-1}$, and reminding ourselves that we define $h_{n}=0$ for $n<0$,

 $(\phi,B^{p}h_{n})_{L^{2}}=(\phi,h_{n-p})_{L^{2}}\prod_{j=n-p+1}^{n}(2j)^{1/2}=% (\phi,h_{n-p})_{L^{2}}2^{p/2}\left(\frac{n!}{(n-p)!}\right)^{1/2}.$

Because $(C\phi,\psi)_{L^{2}}=(\phi,B\psi)_{L^{2}}$, we have

 $\displaystyle C^{p}\phi$ $\displaystyle=\sum_{n=0}^{\infty}(C^{p}\phi,\psi)_{L^{2}}h_{n}$ $\displaystyle=\sum_{n=0}^{\infty}(\phi,B^{p}\psi)_{L^{2}}h_{n}$ $\displaystyle=\sum_{n=0}^{\infty}(\phi,h_{n-p})_{L^{2}}2^{p/2}\left(\frac{n!}{% (n-p)!}\right)^{1/2}h_{n}.$

We define the Fourier transform $\mathscr{F}:\mathscr{S}\to\mathscr{S}$ by

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

$\mathscr{F}:\mathscr{S}\to\mathscr{S}$ is a continuous linear map, and satisfies

 $\mathscr{F}M=iD\mathscr{F},\qquad\mathscr{F}D=iM\mathscr{F}.$

From these we obtain

 $\mathscr{F}A=A\mathscr{F},\qquad\mathscr{F}B=iB\mathscr{F},\qquad\mathscr{F}C=% -iC\mathscr{F},$

and one proves the following using the above.

###### Lemma 3.

For $n\geq 0$,

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

We further remark that for $\phi\in\mathscr{S}$,

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

Finally, there is a unique Hilbert space isomorphism $\mathscr{F}:L^{2}(\lambda)\to L^{2}(\lambda)$ whose restriction to $\mathscr{S}$ is equal to $\mathscr{F}$ as already defined. Thus for $f\in L^{2}(\lambda)$, as

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

we get

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

## 4 Hermite operator

For $p\geq 0$ and $f\in L^{2}(\lambda)$, we define

 $\left\|f\right\|_{p}^{2}=\sum_{n=0}^{\infty}(2n+2)^{2p}|(f,h_{n})_{L^{2}}|^{2}.$

We define

 $\mathscr{S}_{p}=\{f\in L^{2}(\lambda):\left\|f\right\|_{p}<\infty\},$

and for $f,g\in\mathscr{S}_{p}$ we define

 $(f,g)_{p}=\sum_{n=0}^{\infty}(2n+2)^{2p}(f,h_{n})_{L^{2}}\overline{(g,h_{n})_{% L^{2}}},$

for which

 $\left\|f\right\|_{p}^{2}=(f,f)_{p}.$
###### Lemma 4.

For $\phi\in\mathscr{S}$, for each $p\geq 0$, $\phi\in\mathscr{S}_{p}$, and

 $\left\|\phi\right\|_{p}=\left\|A^{p}\phi\right\|_{L^{2}}.$
###### Proof.

$A^{p}\phi\in\mathscr{S}$, so $\left\|A^{p}\phi\right\|_{L^{2}}<\infty$. Because $A$ is a symmetric operator and as $Ah_{n}=(2n+2)h_{n}$,

 $\displaystyle\left\|A^{p}\phi\right\|_{L^{2}}^{2}$ $\displaystyle=\sum_{n=0}^{\infty}|(A^{p}\phi,h_{n})_{L^{2}}|^{2}$ $\displaystyle=\sum_{n=0}^{\infty}|(\phi,A^{p}h_{n})_{L^{2}}|^{2}$ $\displaystyle=\sum_{n=0}^{\infty}(2n+2)^{2p}|(\phi,h_{n})_{L^{2}}|^{2}$ $\displaystyle=\left\|\phi\right\|_{p}^{2}.$

For $f,g\in L^{2}(\lambda)$, because $T$ is self-adjoint,

 $\displaystyle(T^{p}f,T^{p}g)_{p}$ $\displaystyle=\sum_{n=0}^{\infty}(2n+2)^{2p}(T^{p}f,h_{n})_{L^{2}}\overline{(T% ^{p}f,h_{n})_{L^{2}}}$ $\displaystyle=\sum_{n=0}^{\infty}(2n+2)^{2p}(f,T^{p}h_{n})_{L^{2}}\overline{(g% ,T^{p}h_{n})_{L^{2}}}$ $\displaystyle=\sum_{n=0}^{\infty}(2n+2)^{2p}(f,(2n+2)^{-p}h_{n})_{L^{2}}% \overline{(g,(2n+2)^{-p}h_{n})_{L^{2}}}$ $\displaystyle=\sum_{n=0}^{\infty}(f,h_{n})_{L^{2}}\overline{(g,h_{n})_{L^{2}}}$ $\displaystyle=(f,g)_{L^{2}},$

and so $\left\|T^{p}f\right\|_{p}=\left\|f\right\|_{L^{2}}$, which shows that

 $T^{p}L^{2}(\lambda)=\mathscr{S}_{p}.$

If $f_{i}\in\mathscr{S}_{p}$ is a Cauchy sequence in the norm $\left\|\cdot\right\|_{p}$, then as $\left\|T^{-p}f_{i}-T^{-p}f_{j}\right\|_{L^{2}}=\left\|f_{i}-f_{j}\right\|_{p}$, $T^{-p}f_{i}$ is a Cauchy sequence in the norm $\left\|\cdot\right\|_{L^{2}}$ and so there is some $g\in L^{2}(\lambda)$ for which $\left\|T^{-p}f_{i}-g\right\|_{L^{2}}\to 0$. We have $T^{p}g\in\mathscr{S}_{p}$, and

 $\left\|f_{i}-T^{p}g\right\|_{p}=\left\|T^{-p}f_{i}-g\right\|_{L^{2}}\to 0,$

thus $f_{i}\to T^{p}g$ in the norm $\left\|\cdot\right\|_{p}$, showing that $(\mathscr{S}_{p},(\cdot,\cdot)_{p})$ is a Hilbert space. Furthermore, $T^{p}:L^{2}(\lambda)\to\mathscr{S}_{p}$ is an isomorphism of Hilbert spaces, and thus $\{T^{p}h_{n}:n\geq 0\}$ is an orthonormal basis for $(\mathscr{S}_{p},(\cdot,\cdot)_{p})$.

For $p\leq q$,

 $\left\|f\right\|_{p}\leq\left\|f\right\|_{q},$

so $\mathscr{S}_{q}\subset\mathscr{S}_{p}$. For $p\geq q$, let $i_{q,p}:\mathscr{S}_{q}\to\mathscr{S}_{p}$ be the inclusion map.33 3 Hui-Hsiung Kuo, White Noise Distribution Theory, p. 18, Lemma 3.3.

###### Theorem 5.

For $p, the inclusion map $i_{q,p}:\mathscr{S}_{q}\to\mathscr{S}_{p}$ is a Hilbert-Schmidt operator, with Hilbert-Schmidt norm

 $\left\|i_{q,p}\right\|_{\mathrm{HS}}=2^{-q+p}\sqrt{\zeta(2q-2p)}.$
###### Proof.

$\{T^{q}h_{n}:n\geq 0\}$ is an orthonormal basis for $(\mathscr{S}_{q},(\cdot,\cdot)_{q})$, and

 $\displaystyle\left\|i_{q,p}\right\|_{\mathrm{HS}}^{2}$ $\displaystyle=\sum_{n=0}^{\infty}\left\|i_{q,p}T^{q}h_{n}\right\|_{p}^{2}$ $\displaystyle=\sum_{n=0}^{\infty}\left\|T^{q}h_{n}\right\|_{p}^{2}$ $\displaystyle=\sum_{n=0}^{\infty}\left\|(2n+2)^{-q}h_{n}\right\|_{p}^{2}$ $\displaystyle=\sum_{n=0}^{\infty}(2n+2)^{-2q}(2n+2)^{2p}$ $\displaystyle=2^{-2q+2p}\zeta(2q-2p).$

## 5 The Hilbert spaces Sp

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

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

and for $N\geq 0$ we define $f_{N}:\mathbb{R}\to\mathbb{C}$ by

 $f_{N}(x)=\sum_{n=0}^{N}(f,h_{n})_{L^{2}}h_{n}(x),\qquad x\in\mathbb{R},$

which belongs to $\mathscr{S}$.

For $k\geq 0$, we define $C_{b}^{k}(\mathbb{R})$ to be the set of those functions $\mathbb{R}\to\mathbb{C}$ that are $k$-times differentiable and such that for each $0\leq j\leq k$, $f^{(j)}$ is continuous and bounded. With the norm

 $\left\|f\right\|_{C_{b}^{k}}=\sum_{j=0}^{k}\left\|f^{(j)}\right\|_{\infty}$

this is a Banach space. Because the Hermite functions belong to $\mathscr{S}$, for $f\in L^{2}(\lambda)$ and for any $k$ and $N$, the function $f_{N}$ belongs to $C_{b}^{k}(\mathbb{R})$.

###### Lemma 6.

If $p\geq 1$ and $f\in\mathscr{S}_{p}$, then there is some $F\in C_{b}^{p-1}(\mathbb{R})$ such that $f$ is equal almost everywhere to $F$.

###### Proof.

Cramér’s inequality states that there is a constant $K_{0}$ such that for all $n$, $\left\|h_{n}\right\|_{\infty}\leq K_{0}$. For $M, using this and the Cauchy-Schwarz inequality,

 $\displaystyle\left\|f_{N}-f_{M}\right\|_{C_{b}^{0}}$ $\displaystyle=\left\|\sum_{n=M+1}^{N}(f,h_{n})_{L^{2}}h_{n}\right\|_{\infty}$ $\displaystyle\leq K_{0}\sum_{n=M+1}^{N}|(f,h_{n})_{L^{2}}|$ $\displaystyle=K_{0}\sum_{n=M+1}^{N}(2n+2)^{-1}(2n+2)|(f,h_{n})_{L^{2}}|$ $\displaystyle\leq\left(\sum_{n=M+1}^{N}(2n+2)^{-2}\right)^{1/2}\left(\sum_{n=M% +1}^{N}(2n+2)^{2}|(f,h_{n})_{L^{2}}|^{2}\right)^{1/2}$ $\displaystyle=\left(\sum_{n=M+1}^{N}(2n+2)^{-2}\right)^{1/2}\left\|f_{N}-f_{M}% \right\|_{1}.$

Because $f\in\mathscr{S}_{p}\subset\mathscr{S}_{1}$, $f_{N}$ is a Cauchy sequence in $\mathscr{S}_{1}$, hence $f_{N}$ is a Cauchy sequence in $C_{b}^{0}(\mathbb{R})$, so there is some $F\in C_{b}^{0}(\mathbb{R})$ such that $f_{N}$ converges to $F$ in $C_{b}^{0}(\mathbb{R})$. We assert that $F=f$ as elements of $L^{2}(\lambda)$.

Using

 $Dh_{n}=\sqrt{\frac{n}{2}}h_{n-1}-\sqrt{\frac{n+1}{2}}h_{n+1},$

we calculate

 $\displaystyle f_{N}^{\prime}$ $\displaystyle=-\sqrt{\frac{N}{2}}(f,h_{N-1})_{L^{2}}h_{N}-\sqrt{\frac{N+1}{2}}% (f,h_{N})_{L^{2}}h_{N+1}$ $\displaystyle+\sum_{n=0}^{N-1}\left(\sqrt{\frac{n+1}{2}}(f,h_{n+1})_{L^{2}}-% \sqrt{\frac{n}{2}}(f,h_{n-1})_{L^{2}}\right)h_{n},$

hence for $M,

 $\displaystyle f_{N}^{\prime}-f_{M}^{\prime}$ $\displaystyle=-\sqrt{\frac{N}{2}}(f,h_{N-1})_{L^{2}}h_{N}-\sqrt{\frac{N+1}{2}}% (f,h_{N})_{L^{2}}h_{N+1}$ $\displaystyle+\sqrt{\frac{M}{2}}(f,h_{M-1})_{L^{2}}h_{M}+\sqrt{\frac{M+1}{2}}(% f,h_{M})_{L^{2}}h_{M+1}$ $\displaystyle+\sum_{n=M}^{N-1}\left(\sqrt{\frac{n+1}{2}}(f,h_{n+1})_{L^{2}}-% \sqrt{\frac{n}{2}}(f,h_{n-1})_{L^{2}}\right)h_{n},$

and for $N\geq M+2$,

 $\displaystyle\left\|f_{N}^{\prime}-f_{M}^{\prime}\right\|_{1}$ $\displaystyle=(2N+2)^{2}\frac{N}{2}|(f,h_{N-1})|_{L^{2}}^{2}+(2N+4)^{2}\frac{N% +1}{2}|(f,h_{N-1})|_{L^{2}}^{2}$ $\displaystyle(2M+2)^{2}\frac{M+1}{2}|(f,h_{M+1})|_{L^{2}}^{2}+(2M+4)^{2}\frac{% M+2}{2}|(f,h_{M+1})|_{L^{2}}^{2}$ $\displaystyle+\sum_{n=M+2}^{N-1}(2n+2)^{2}\left|\sqrt{\frac{n+1}{2}}(f,h_{n+1}% )_{L^{2}}-\sqrt{\frac{n}{2}}(f,h_{n-1})_{L^{2}}\right|^{2}$ $\displaystyle=O(\left\|f_{N}-f_{M}\right\|_{2}),$

whence $f_{N}^{\prime}$ is a Cauchy sequence in $C_{b}^{0}(\mathbb{R})$, and so $f_{N}$ is a Cauchy sequence in $C_{b}^{1}(\mathbb{R})$. ∎

We prove that for $p\geq 1$ the derivatives of the partial sums $f_{N}$ are a Cauchy sequence in $L^{2}(\lambda)$.44 4 Jeremy J. Becnel and Ambar N. Sengupta, The Schwartz space: a background to white noise analysis, https://www.math.lsu.edu/~preprint/2004/as20041.pdf, Lemma 7.1.

###### Lemma 7.

For $p\geq 1$ and $f\in\mathscr{S}_{p}$, $f_{N}^{\prime}$ is a Cauchy sequence in $L^{2}(\lambda)$.

###### Proof.

Because $f_{N}\in\mathscr{S}$,

 $f_{N}^{\prime}=Df_{N}=\frac{B-C}{2}f_{N}.$

Then

 $\left\|f_{N}^{\prime}-f_{M}^{\prime}\right\|_{L^{2}}\leq\frac{1}{2}\left\|Bf_{% N}-Bf_{M}\right\|_{L^{2}}+\frac{1}{2}\left\|Cf_{N}-Cf_{M}\right\|_{L^{2}}.$

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

 $\displaystyle\left\|Bf_{N}-Bf_{M}\right\|_{L^{2}}^{2}$ $\displaystyle=\left\|B\sum_{n=M+1}^{N}(f,h_{n})_{L^{2}}h_{n}\right\|_{L^{2}}^{2}$ $\displaystyle=\left\|\sum_{n=M+1}^{N}(f,h_{n})_{L^{2}}(2n)^{1/2}h_{n-1}\right% \|_{L^{2}}^{2}$ $\displaystyle=\sum_{n=M+1}^{N}|(f,h_{n})_{L^{2}}|^{2}(2n)$ $\displaystyle\leq\sum_{n=M+1}^{N}(2n+2)^{2}|(f,h_{n})_{L^{2}}|^{2},$

and as $Ch_{n}=(2n+2)^{1/2}h_{n+1}$,

 $\displaystyle\left\|Cf_{N}-Cf_{M}\right\|_{L^{2}}^{2}$ $\displaystyle=\left\|C\sum_{n=M+1}^{N}(f,h_{n})_{L^{2}}h_{n}\right\|_{L^{2}}^{2}$ $\displaystyle=\left\|\sum_{n=M+1}^{N}(f,h_{n})_{L^{2}}(2n+2)^{1/2}h_{n+1}% \right\|_{L^{2}}^{2}$ $\displaystyle=\sum_{n=M+1}^{N}|(f,h_{n})_{L^{2}}|^{2}(2n+2)$ $\displaystyle\leq\sum_{n=M+1}^{N}(2n+2)^{2}|(f,h_{n})_{L^{2}}|^{2}.$

Thus

 $\left\|f_{N}^{\prime}-f_{M}^{\prime}\right\|_{L^{2}}\leq\frac{1}{2}\left\|f_{N% }-f_{M}\right\|_{1}+\frac{1}{2}\left\|f_{N}-f_{M}\right\|_{1}=\left\|f_{N}-f_{% M}\right\|_{1}.$

Because $f\in\mathscr{S}_{p}$ and $p\geq 1$, the series $\sum_{n=0}^{\infty}(2n+2)^{2}|(f,h_{n})_{L^{2}}|^{2}$ converges, from which the claim follows. ∎

Now we establish that if $p\geq 1$ and $f\in\mathscr{S}_{p}$ then there is some $F\in C_{b}^{0}(\mathbb{R})$ such that $f$ is equal almost everywhere to $F$, $F$ is differentiable almost everywhere, and $F^{\prime}\in\mathscr{S}_{p-1}$.55 5 Jeremy J. Becnel and Ambar N. Sengupta, The Schwartz space: a background to white noise analysis, https://www.math.lsu.edu/~preprint/2004/as20041.pdf, Theorem 7.3.

###### Theorem 8.

For $p\geq 1$ and $f\in\mathscr{S}_{p}$, there is some $F\in C_{b}^{0}(\mathbb{R})$ such that $f$ is equal almost everywhere to $F$, $F$ is differentiable almost everywhere, $f_{N}^{\prime}$ converges to $F^{\prime}$ in the norm $\left\|\cdot\right\|_{L^{2}}$, and $F^{\prime}\in\mathscr{S}_{p-1}$.

###### Proof.

Lemma 7 tells us that $f_{N}^{\prime}$ is a Cauchy sequence in the norm $\left\|\cdot\right\|_{L^{2}}$, and hence there is some $g\in L^{2}(\lambda)$ to which $f_{N}^{\prime}$ converges in the norm $\left\|\cdot\right\|_{L^{2}}$. For $x\leq y$, by the fundamental theorem of calculus,

 $f_{N}(y)=f_{N}(x)+\int_{0}^{1}f_{N}^{\prime}(x+t(y-x))\cdot(y-x)dt.$

By the Cauchy-Schwarz inequality,

 $\begin{split}&\displaystyle\int_{0}^{1}|f_{N}^{\prime}(x+t(y-x))\cdot(y-x)-g(x% +t(y-x))\cdot(y-x)|dt\\ \displaystyle=&\displaystyle\int_{x}^{y}|f_{N}^{\prime}(u)-g(u)|du\\ \displaystyle\leq&\displaystyle\sqrt{y-x}\left\|f_{N}^{\prime}-g\right\|_{L^{2% }}.\end{split}$

Because $\left\|f_{N}^{\prime}-g\right\|_{L^{2}}\to 0$ as $N\to\infty$,

 $\int_{0}^{1}f_{N}^{\prime}(x+t(y-x))\cdot(y-x)dt\to\int_{0}^{1}g(x+t(y-x))% \cdot(y-x)dt.$

Then by Lemma 6, taking $N\to\infty$, for any $y>x$ we have

 $F(y)=F(x)+\int_{0}^{1}g(x+t(y-x))\cdot(y-x)dt=F(x)+\frac{1}{y-x}\int_{x}^{y}g(% s)ds.$

By the Lebesgue differentiation theorem, for almost all $x\in\mathbb{R}$,

 $\frac{1}{y-x}\int_{x}^{y}g(s)ds\to g(x),\qquad y\to x.$

Therefore for almost all $x\in\mathbb{R}$,

 $F^{\prime}(x)=g(x).$

Thus $F^{\prime}=g$ in $L^{2}(\lambda)$, and as $f_{N}^{\prime}\to g$ in $L^{2}(\lambda)$,

 $\displaystyle F^{\prime}$ $\displaystyle=\lim_{N\to\infty}f_{N}^{\prime}$ $\displaystyle=\lim_{N\to\infty}\left(\frac{B-C}{2}\right)\sum_{n=0}^{N}(f,h_{n% })_{L^{2}}h_{n}$ $\displaystyle=\frac{1}{2}\sum_{n=0}^{\infty}(f,h_{n})_{L^{2}}((2n)^{1/2}h_{n-1% }-(2n+2)^{1/2}h_{n+1})$ $\displaystyle=\frac{1}{2}\sum_{n=0}^{\infty}\bigg{(}(2n+2)^{1/2}(f,h_{n})_{L^{% 2}}-(2n)^{1/2}(f,h_{n-1})_{L^{2}}\bigg{)}h_{n},$

for which

 $\displaystyle\left\|F^{\prime}\right\|_{p-1}^{2}$ $\displaystyle=\frac{1}{4}\sum_{n=0}^{\infty}(2n+2)^{2p-2}\bigg{|}(2n+2)^{1/2}(% f,h_{n})_{L^{2}}-(2n)^{1/2}(f,h_{n-1})_{L^{2}}\bigg{|}^{2}$ $\displaystyle\leq\frac{1}{2}\sum_{n=0}^{\infty}(2n+2)^{2p-2}\bigg{(}(2n+2)|(f,% h_{n})_{L^{2}}|^{2}+2n|(f,h_{n-1})_{L^{2}}|^{2}\bigg{)},$

which is finite because $f\in\mathscr{S}_{p}$. Therefore $F^{\prime}\in\mathscr{S}_{p-1}$. ∎