Schwartz functions, Hermite functions, and the Hermite operator

1 Schwartz functions

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

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

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

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

$$e_{\varphi }(\omega )=\omega (\varphi ),\omega \in {𝒮}^{\prime }.$$

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

2 L² norms

For $p\ge 0$ and $\varphi ,\psi \in 𝒮$, let

$${[\varphi ,\psi ]}_p=\sum _{k=0}^p{\int }_{ℝ}{(1+u^2)}^p{\varphi }^{(k)}(u)\overline{{\psi }^{(k)}(u)}𝑑u,$$

and let

$${[\varphi ]}_p^2={[\varphi ,\varphi ]}_p=\sum _{k=0}^p{\int }_{ℝ}{(1+u^2)}^p{|{\varphi }^{(k)}(u)|}^2𝑑u.$$

Because ${(1+u^2)}^p\le {(1+u^2)}^q$ when $p\le q$, it is immediate that ${[\varphi ]}_p\le {[\varphi ]}_q$ when $p\le q$.

We relate the norms $|\cdot {|}_p$ and the norms ${[\cdot ]}_p$.

Lemma 1.

For each $p\ge 1$, for all $\varphi \in 𝒮$,

$$\frac{1}{p\sqrt{\pi }}{|\varphi |}_{p-1}\le {[\varphi ]}_p\le \sqrt{(p+1)\pi }{|\varphi |}_{p+1}.$$

Proof.

For $0\le k\le p$,

${\displaystyle {\int }_{ℝ}}{(1+u^2)}^p{|{\varphi }^{(k)}(u)|}^2𝑑u$ $\le \underset{u\in ℝ}{sup}({(1+u^2)}^{p+1}{|{\varphi }^{(k)}(u)|}^2){\displaystyle {\int }_{ℝ}}{(1+u^2)}^{-1}𝑑u$

$=\underset{u\in ℝ}{sup}({(1+u^2)}^{p+1}{|{\varphi }^{(k)}(u)|}^2)\cdot \pi$

$\le \pi {|\varphi |}_{p+1}^2,$

hence

${[\varphi ]}_p^2$ $={\displaystyle \sum _{k=0}^p}{\displaystyle {\int }_{ℝ}}{(1+u^2)}^p{|{\varphi }^{(k)}(u)|}^2𝑑u$

$\le {\displaystyle \sum _{k=0}^p}\pi {|\varphi |}_{p+1}^2$

$=(p+1)\pi {|\varphi |}_{p+1}^2.$

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

$|{(1+u^2)}^{(p-1)/2}{\varphi }^{(k)}(u)|$ $=\left|{\displaystyle {\int }_{-\mathrm{\infty }}^u}{({(1+t^2)}^{(p-1)/2}{\varphi }^{(k)}(t))}^{\prime }𝑑t\right|$

$\le {\displaystyle {\int }_{ℝ}}|(p-1)t{(1+t^2)}^{(p-1)/2-1}{\varphi }^{(k)}(t)|𝑑t$

$+{\displaystyle {\int }_{ℝ}}|{(1+t^2)}^{(p-1)/2}{\varphi }^{(k+1)}(t)|𝑑t$

$\le (p-1){\displaystyle {\int }_{ℝ}}{(1+t^2)}^{-1/2}{(1+t^2)}^{(p-1)/2}|{\varphi }^{(k)}(t)|𝑑t$

$+{\displaystyle {\int }_{ℝ}}{(1+t^2)}^{-1/2}{(1+t^2)}^{p/2}|{\varphi }^{(k+1)}(t)|𝑑t$

$\le (p-1){\left({\displaystyle {\int }_{ℝ}}{(1+t^2)}^{-1}𝑑t\right)}^{1/2}{\left({\displaystyle {\int }_{ℝ}}{(1+t^2)}^{p-1}{|{\varphi }^{(k)}(t)|}^2𝑑t\right)}^{1/2}$

$+{\left({\displaystyle {\int }_{ℝ}}{(1+t^2)}^{-1}𝑑t\right)}^{1/2}{\left({\displaystyle {\int }_{ℝ}}{(1+t^2)}^p{|{\varphi }^{(k+1)}(t)|}^2𝑑t\right)}^{1/2}$

$\le (p-1)\sqrt{\pi }{[\varphi ]}_{p-1}+\sqrt{\pi }{[\varphi ]}_p$

$\le p\sqrt{\pi }{[\varphi ]}_p,$

which shows that

$${|\varphi |}_{p-1}\le p\sqrt{\pi }{[\varphi ]}_p.$$

∎

3 Hermite functions

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

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

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

$$h_n=0,$$

to write some expressions in a uniform way.

We calculate that for $n\ge 0$,

$$D{h}_n=\sqrt{\frac{n}{2}}{h}_{n-1}-\sqrt{\frac{n+1}{2}}{h}_{n+1}.$$

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

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

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

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

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

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

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

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

$$B=D+M$$

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

$$C=-D+M,$$

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

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

(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 $\varphi ,\psi \in 𝒮$, using integration by parts,

$${(D\varphi ,\psi )}_{L^2}={\int }_{ℝ}{\varphi }^{\prime }(x)\overline{\psi (x)}𝑑x=-{\int }_{ℝ}\varphi (x)\overline{{\psi }^{\prime }(x)}𝑑x={(\varphi ,(-D)\psi )}_{L^2},$$

and

$${(M\varphi ,\psi )}_{L^2}={\int }_{ℝ}x\varphi (x)\overline{\psi (x)}𝑑x={(\varphi ,M\psi )}_{L^2}.$$

Thus,

${(B\varphi ,\psi )}_{L^2}$ $={(D\varphi ,\psi )}_{L^2}+{(M\varphi ,\psi )}_{L^2}$

$={(\varphi ,(-D)\psi )}_{L^2}+{(\varphi ,M\psi )}_{L^2}$

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

and

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

We shall use these calculations to obtain the following lemma.

Lemma 2.

For $p\ge 0$ and for $\varphi \in 𝒮$,

$$B^p\varphi =2^{p/2}\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{(n+p)!}{n!}\right)}^{1/2}{(\varphi ,h_{n+p})}_{L^2}{h}_n$$

and

$$C^p\varphi =2^{p/2}\sum _{n=0}^{\mathrm{\infty }}{(\varphi ,h_{n-p})}_{L^2}{\left(\frac{n!}{(n-p)!}\right)}^{1/2}{h}_n.$$

Proof.

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

$${(\varphi ,C^p{h}_n)}_{L^2}={(\varphi ,h_{n+p})}_{L^2}\prod _{j=n}^{n+p-1}{(2j+2)}^{1/2}={(\varphi ,h_{n+p})}_{L^2}{2}^{p/2}{\left(\frac{(n+p)!}{n!}\right)}^{1/2}.$$

With

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

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

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

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(\varphi ,C^p{h}_n)}_{L^2}{h}_n$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(\varphi ,h_{n+p})}_{L^2}{2}^{p/2}{\left({\displaystyle \frac{(n+p)!}{n!}}\right)}^{1/2}{h}_n.$

Because $B{h}_n={(2n)}^{1/2}{h}_{n-1}$, and reminding ourselves that we define $h_n=0$ for ,

$${(\varphi ,B^p{h}_n)}_{L^2}={(\varphi ,h_{n-p})}_{L^2}\prod _{j=n-p+1}^n{(2j)}^{1/2}={(\varphi ,h_{n-p})}_{L^2}{2}^{p/2}{\left(\frac{n!}{(n-p)!}\right)}^{1/2}.$$

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

$C^p\varphi$ $={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(C^p\varphi ,\psi )}_{L^2}{h}_n$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(\varphi ,B^p\psi )}_{L^2}{h}_n$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(\varphi ,h_{n-p})}_{L^2}{2}^{p/2}{\left({\displaystyle \frac{n!}{(n-p)!}}\right)}^{1/2}{h}_n.$

∎

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

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

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

$$ℱM=iDℱ,ℱD=iMℱ.$$

From these we obtain

$$ℱA=Aℱ,ℱB=iBℱ,ℱC=-iCℱ,$$

and one proves the following using the above.

Lemma 3.

For $n\ge 0$,

$$ℱh_n={(-i)}^n{h}_n.$$

We further remark that for $\varphi \in 𝒮$,

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

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

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

we get

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

4 Hermite operator

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

$${\parallel f\parallel }_p^2=\sum _{n=0}^{\mathrm{\infty }}{(2n+2)}^{2p}{|{(f,h_n)}_{L^2}|}^2.$$

We define

and for $f,g\in {𝒮}_p$ we define

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

for which

$${\parallel f\parallel }_p^2={(f,f)}_p.$$

Lemma 4.

For $\varphi \in 𝒮$, for each $p\ge 0$, $\varphi \in {𝒮}_p$, and

$${\parallel \varphi \parallel }_p={\parallel A^p\varphi \parallel }_{L^2}.$$

Proof.

$A^p\varphi \in 𝒮$, so . Because $A$ is a symmetric operator and as $A{h}_n=(2n+2)h_n$,

${\parallel A^p\varphi \parallel }_{L^2}^2$ $={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{|{(A^p\varphi ,h_n)}_{L^2}|}^2$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{|{(\varphi ,A^p{h}_n)}_{L^2}|}^2$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(2n+2)}^{2p}{|{(\varphi ,h_n)}_{L^2}|}^2$

$={\parallel \varphi \parallel }_p^2.$

∎

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

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

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

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

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

$={(f,g)}_{L^2},$

and so ${\parallel T^{p}f\parallel }_p={\parallel f\parallel }_{L^2}$, which shows that

$$T^p{L}^2(\lambda )={𝒮}_p.$$

If $f_i\in {𝒮}_p$ is a Cauchy sequence in the norm $\parallel \cdot {\parallel }_p$, then as ${\parallel T^{-p}{f}_i-T^{-p}{f}_j\parallel }_{L^2}={\parallel f_i-f_j\parallel }_p$, $T^{-p}{f}_i$ is a Cauchy sequence in the norm $\parallel \cdot {\parallel }_{L^2}$ and so there is some $g\in L^2(\lambda )$ for which ${\parallel T^{-p}{f}_i-g\parallel }_{L^2}\to 0$. We have $T^{p}g\in {𝒮}_p$, and

$${\parallel f_i-T^{p}g\parallel }_p={\parallel T^{-p}{f}_i-g\parallel }_{L^2}\to 0,$$

thus $f_i\to T^{p}g$ in the norm $\parallel \cdot {\parallel }_p$, showing that $({𝒮}_p,{(\cdot ,\cdot )}_p)$ is a Hilbert space. Furthermore, $T^p:L^2(\lambda )\to {𝒮}_p$ is an isomorphism of Hilbert spaces, and thus $\{T^p{h}_n:n\ge 0\}$ is an orthonormal basis for $({𝒮}_p,{(\cdot ,\cdot )}_p)$.

For $p\le q$,

$${\parallel f\parallel }_p\le {\parallel f\parallel }_q,$$

so ${𝒮}_q\subset {𝒮}_p$. For $p\ge q$, let $i_{q,p}:{𝒮}_q\to {𝒮}_p$ be the inclusion map.

Theorem 5.

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

$${\parallel i_{q,p}\parallel }_{\mathrm{HS}}=2^{-q+p}\sqrt{\zeta (2q-2p)}.$$

Proof.

$\{T^q{h}_n:n\ge 0\}$ is an orthonormal basis for $({𝒮}_q,{(\cdot ,\cdot )}_q)$, and

${\parallel i_{q,p}\parallel }_{\mathrm{HS}}^2$ $={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\parallel i_{q,p}{T}^q{h}_n\parallel }_p^2$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\parallel T^q{h}_n\parallel }_p^2$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\parallel {(2n+2)}^{-q}{h}_n\parallel }_p^2$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(2n+2)}^{-2q}{(2n+2)}^{2p}$

$=2^{-2q+2p}\zeta (2q-2p).$

∎

5 The Hilbert spaces S_p

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

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

and for $N\ge 0$ we define $f_N:ℝ\to ℂ$ by

$$f_N(x)=\sum _{n=0}^N{(f,h_n)}_{L^2}{h}_n(x),x\in ℝ,$$

which belongs to $𝒮$.

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

$${\parallel f\parallel }_{C_b^k}=\sum _{j=0}^k{\parallel f^{(j)}\parallel }_{\mathrm{\infty }}$$

this is a Banach space. Because the Hermite functions belong to $𝒮$, for $f\in L^2(\lambda )$ and for any $k$ and $N$, the function $f_N$ belongs to $C_b^k(ℝ)$.

Lemma 6.

If $p\ge 1$ and $f\in {𝒮}_p$, then there is some $F\in C_b^{p-1}(ℝ)$ 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 $n$, ${\parallel h_n\parallel }_{\mathrm{\infty }}\le K_0$. For , using this and the Cauchy-Schwarz inequality,

${\parallel f_N-f_M\parallel }_{C_b^0}$ $={\parallel {\displaystyle \sum _{n=M+1}^N}{(f,h_n)}_{L^2}{h}_n\parallel }_{\mathrm{\infty }}$

$\le K_0{\displaystyle \sum _{n=M+1}^N}|{(f,h_n)}_{L^2}|$

$=K_0{\displaystyle \sum _{n=M+1}^N}{(2n+2)}^{-1}(2n+2)|{(f,h_n)}_{L^2}|$

$\le {\left({\displaystyle \sum _{n=M+1}^N}{(2n+2)}^{-2}\right)}^{1/2}{\left({\displaystyle \sum _{n=M+1}^N}{(2n+2)}^2{|{(f,h_n)}_{L^2}|}^2\right)}^{1/2}$

$={\left({\displaystyle \sum _{n=M+1}^N}{(2n+2)}^{-2}\right)}^{1/2}{\parallel f_N-f_M\parallel }_1.$

Because $f\in {𝒮}_p\subset {𝒮}_1$, $f_N$ is a Cauchy sequence in ${𝒮}_1$, hence $f_N$ is a Cauchy sequence in $C_b^0(ℝ)$, so there is some $F\in C_b^0(ℝ)$ such that $f_N$ converges to $F$ in $C_b^0(ℝ)$. We assert that $F=f$ as elements of $L^2(\lambda )$.

Using

$$D{h}_n=\sqrt{\frac{n}{2}}{h}_{n-1}-\sqrt{\frac{n+1}{2}}{h}_{n+1},$$

we calculate

$f_N^{\prime }$ $=-\sqrt{{\displaystyle \frac{N}{2}}}{(f,h_{N-1})}_{L^2}{h}_N-\sqrt{{\displaystyle \frac{N+1}{2}}}{(f,h_N)}_{L^2}{h}_{N+1}$

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

hence for ,

$f_N^{\prime }-f_M^{\prime }$ $=-\sqrt{{\displaystyle \frac{N}{2}}}{(f,h_{N-1})}_{L^2}{h}_N-\sqrt{{\displaystyle \frac{N+1}{2}}}{(f,h_N)}_{L^2}{h}_{N+1}$

$+\sqrt{{\displaystyle \frac{M}{2}}}{(f,h_{M-1})}_{L^2}{h}_M+\sqrt{{\displaystyle \frac{M+1}{2}}}{(f,h_M)}_{L^2}{h}_{M+1}$

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

and for $N\ge M+2$,

${\parallel f_N^{\prime }-f_M^{\prime }\parallel }_1$ $={(2N+2)}^2{\displaystyle \frac{N}{2}}{|(f,h_{N-1})|}_{L^2}^2+{(2N+4)}^2{\displaystyle \frac{N+1}{2}}{|(f,h_{N-1})|}_{L^2}^2$

${(2M+2)}^2{\displaystyle \frac{M+1}{2}}{|(f,h_{M+1})|}_{L^2}^2+{(2M+4)}^2{\displaystyle \frac{M+2}{2}}{|(f,h_{M+1})|}_{L^2}^2$

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

$=O({\parallel f_N-f_M\parallel }_2),$

whence $f_N^{\prime }$ is a Cauchy sequence in $C_b^0(ℝ)$, and so $f_N$ is a Cauchy sequence in $C_b^1(ℝ)$. ∎

We prove that for $p\ge 1$ the derivatives of the partial sums $f_N$ are a Cauchy sequence in $L^2(\lambda )$.

Lemma 7.

For $p\ge 1$ and $f\in {𝒮}_p$, $f_N^{\prime }$ is a Cauchy sequence in $L^2(\lambda )$.

Proof.

Because $f_N\in 𝒮$,

$$f_N^{\prime }=D{f}_N=\frac{B-C}{2}{f}_N.$$

Then

$${\parallel f_N^{\prime }-f_M^{\prime }\parallel }_{L^2}\le \frac{1}{2}{\parallel B{f}_N-B{f}_M\parallel }_{L^2}+\frac{1}{2}{\parallel C{f}_N-C{f}_M\parallel }_{L^2}.$$

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

${\parallel B{f}_N-B{f}_M\parallel }_{L^2}^2$ $={\parallel B{\displaystyle \sum _{n=M+1}^N}{(f,h_n)}_{L^2}{h}_n\parallel }_{L^2}^2$

$={\parallel {\displaystyle \sum _{n=M+1}^N}{(f,h_n)}_{L^2}{(2n)}^{1/2}{h}_{n-1}\parallel }_{L^2}^2$

$={\displaystyle \sum _{n=M+1}^N}{|{(f,h_n)}_{L^2}|}^2(2n)$

$\le {\displaystyle \sum _{n=M+1}^N}{(2n+2)}^2{|{(f,h_n)}_{L^2}|}^2,$

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

${\parallel C{f}_N-C{f}_M\parallel }_{L^2}^2$ $={\parallel C{\displaystyle \sum _{n=M+1}^N}{(f,h_n)}_{L^2}{h}_n\parallel }_{L^2}^2$

$={\parallel {\displaystyle \sum _{n=M+1}^N}{(f,h_n)}_{L^2}{(2n+2)}^{1/2}{h}_{n+1}\parallel }_{L^2}^2$

$={\displaystyle \sum _{n=M+1}^N}{|{(f,h_n)}_{L^2}|}^2(2n+2)$

$\le {\displaystyle \sum _{n=M+1}^N}{(2n+2)}^2{|{(f,h_n)}_{L^2}|}^2.$

Thus

$${\parallel f_N^{\prime }-f_M^{\prime }\parallel }_{L^2}\le \frac{1}{2}{\parallel f_N-f_M\parallel }_1+\frac{1}{2}{\parallel f_N-f_M\parallel }_1={\parallel f_N-f_M\parallel }_1.$$

Because $f\in {𝒮}_p$ and $p\ge 1$, the series ${\sum }_{n=0}^{\mathrm{\infty }}{(2n+2)}^2{|{(f,h_n)}_{L^2}|}^2$ converges, from which the claim follows. ∎

Now we establish that if $p\ge 1$ and $f\in {𝒮}_p$ then there is some $F\in C_b^0(ℝ)$ such that $f$ is equal almost everywhere to $F$, $F$ is differentiable almost everywhere, and $F^{\prime }\in {𝒮}_{p-1}$.

Theorem 8.

For $p\ge 1$ and $f\in {𝒮}_p$, there is some $F\in C_b^0(ℝ)$ such that $f$ is equal almost everywhere to $F$, $F$ is differentiable almost everywhere, $f_N^{\prime }$ converges to $F^{\prime }$ in the norm $\parallel \cdot {\parallel }_{L^2}$, and $F^{\prime }\in {𝒮}_{p-1}$.

Proof.

Lemma 7 tells us that $f_N^{\prime }$ is a Cauchy sequence in the norm $\parallel \cdot {\parallel }_{L^2}$, and hence there is some $g\in L^2(\lambda )$ to which $f_N^{\prime }$ converges in the norm $\parallel \cdot {\parallel }_{L^2}$. For $x\le y$, 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,

Because ${\parallel f_N^{\prime }-g\parallel }_{L^2}\to 0$ as $N\to \mathrm{\infty }$,

$${\int }_0^1{f}_N^{\prime }(x+t(y-x))\cdot (y-x)𝑑t\to {\int }_0^{1}g(x+t(y-x))\cdot (y-x)𝑑t.$$

Then by Lemma 6, taking $N\to \mathrm{\infty }$, 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 $x\in ℝ$,

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

Therefore for almost all $x\in ℝ$,

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

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

$F^{\prime }$ $=\underset{N\to \mathrm{\infty }}{lim}{f}_N^{\prime }$

$=\underset{N\to \mathrm{\infty }}{lim}\left({\displaystyle \frac{B-C}{2}}\right){\displaystyle \sum _{n=0}^N}{(f,h_n)}_{L^2}{h}_n$

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

$={\displaystyle \frac{1}{2}}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}\left({(2n+2)}^{1/2}{(f,h_n)}_{L^2}-{(2n)}^{1/2}{(f,h_{n-1})}_{L^2}\right)h_n,$

for which

${\parallel F^{\prime }\parallel }_{p-1}^2$ $={\displaystyle \frac{1}{4}}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(2n+2)}^{2p-2}{\left|{(2n+2)}^{1/2}{(f,h_n)}_{L^2}-{(2n)}^{1/2}{(f,h_{n-1})}_{L^2}\right|}^2$

$\le {\displaystyle \frac{1}{2}}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(2n+2)}^{2p-2}\left((2n+2){|{(f,h_n)}_{L^2}|}^2+2n{|{(f,h_{n-1})}_{L^2}|}^2\right),$

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