The Schwartz Space: Tools for Quantum Mechanics and Infinite Dimensional Analysis

Abstract

An account of the Schwartz space of rapidly decreasing functions as a topological vector space with additional special structures is presented in a manner that provides all the essential background ideas for some areas of quantum mechanics along with infinite-dimensional distribution theory.

1. Introduction

In 1950–1951, Laurent Schwartz published a two volumes work Théorie des Distributions [1,2], where he provided a convenient formalism for the theory of distributions. The purpose of this paper is to present a self-contained account of the main ideas, results, techniques, and proofs that underlie the approach to distribution theory that is central to aspects of quantum mechanics and infinite dimensional analysis. This approach develops the structure of the space of Schwartz test functions by utilizing the operator

$$T=-\frac{d^2}{d{x}^2}+\frac{x^2}{4}+\frac{1}{2}.$$

(1)

This operator arose in quantum mechanics as the Hamiltonian for a harmonic oscillator and, in that context as well as in white noise analysis, the operator N = T − 1 is called the number operator. The physical context provides additional useful mathematical tools such as creation and annihilation operators, which we examine in detail.

In this paper we include under one roof all the essential necessary notions of this approach to the test function space $\mathcal{S}({\mathbf{R}}^n)$. The relevant notions concerning topological vector spaces are presented so that the reader need not wade through the many voluminous available works on this subject. We also describe in brief the origins of the relevant notions in quantum mechanics.

We present

• the essential notions and results concerning topological vector spaces;
• a detailed analysis of the creation operator C, the annihilation operator A, the number operator N and the harmonic oscillator Hamiltonian T;
• a detailed account of the Schwartz space $\mathcal{S}({\mathbf{R}}^d)$, and its topology, as a decreasing intersection of subspaces ${\mathcal{S}}_p({\mathbf{R}}^d)$, for p ∈ {0, 1, 2, …}:

  $$\mathcal{S}({\mathbf{R}}^d)={\displaystyle \underset{p\ge 0}{\cap }{\mathcal{S}}_p({\mathbf{R}}^d)\subset \cdots }\subset {\mathcal{S}}_2({\mathbf{R}}^d)\subset {\mathcal{S}}_1({\mathbf{R}}^d)\subset L^2({\mathbf{R}}^d)$$
• an exact characterization of the functions in the space ${\mathcal{S}}_p(\mathbf{R})$;
• summary of notions from spectral theory and quantum mechanics;

Our exposition of the properties of T and of $\mathcal{S}(\mathbf{R})$ follows Simon’s paper [3], but we provide more detail and our notational conventions are along the lines now standard in infinite-dimensional distribution theory.

The classic work on spaces of smooth functions and their duals is that of Schwartz [1,2]. Our purpose is to present a concise and coherent account of the essential ideas and results of the theory. Of the results that we discuss, many can be found in other works such as [1–6], which is not meant to be a comprehensive list. We have presented portions of this material previously in [7], but also provide it here for convenience. The approach we take has a direct counterpart in the theory of distributions over infinite dimensional spaces [8,9].

2. Basic Notions and Framework

In this section we summarize the basic notions, notation, and results that we discuss in more detail in later sections. Here, and later in this paper, we work mainly with the case of functions of one variable and then describe the generalization to the multi-dimensional case.

We use the letter W to denote the set of all non-negative integers:

$$W=\{0,1,2,3,\dots \}.$$

(2)

2.1. The Schwartz Space

The Schwartz space$\mathcal{S}(\mathbf{R})$ is the linear space of all functions f : R → C which have derivatives of all orders and which satisfy the condition

$$p_{a,b}(f)\stackrel{\mathrm{def}}{=}\underset{x\in \mathbf{R}}{\sup }\left|x^a{f}^b(x)\right|<\infty$$

for all a, b ∈ W = {0, 1, 2, …}. The finiteness condition for all a ≥ 1 and b ∈ W, implies that x^af^b(x) actually goes to 0 as |x| → ∞, for all a, b ∈ W, and so functions of this type are said to be rapidly decreasing.

2.2. The Schwartz Topology

The functions p_a,b are semi-norms on the vector space $\mathcal{S}(\mathbf{R})$, in the sense that

$$p_{a,b}(f+g)\le p_{a,b}(f)+p_{a,b}(g)$$

and

$$p_{a,b}(zf)=\left|z\right|p_{a,b}(f)$$

for all $f,g\in \mathcal{S}(\mathbf{R})$, and z ∈ C. For this semi-norm, an open ball of radius r centered at some $f\in \mathcal{S}(\mathbf{R})$ is given by

$$B_p{}_{{}_{a,b}}(f;r)=\{g\in S(\mathbf{R}):p_{a,b}(g-f)<r\}.$$

(3)

Thus each p_a,b specifies a topology τ_a,b on $\mathcal{S}(\mathbf{R})$. A set is open according to τ_a,b if it is a union of open balls.

One way to generate the standard Schwartz topology τ on $\mathcal{S}(\mathbf{R})$ is to "combine" all the topologies τ_a,b. We will demonstrate how to generate a "smallest" topology containing all the sets of τ_a,b for all a, b ∈ W. However, there is a different approach to the topology on $\mathcal{S}(\mathbf{R})$ that is very useful, which we describe in detail below.

2.3. The Operator T

The operator

$$T=-\frac{d^2}{d{x}^2}+\frac{x^2}{4}+\frac{1}{2}$$

(4)

plays a very useful role in working with the Schwartz space. As we shall see, there is an orthonormal basis ${\left\{{\varphi }_n\right\}}_{n\in W}$ of L²(R,dx), consisting of eigenfunctions ϕ_n of T:

$$T{\varphi }_n=(n+1){\varphi }_n.$$

(5)

The functions ϕ_n, called the Hermite functions are actually in the Schwartz space $\mathcal{S}(\mathbf{R})$. Let B be the bounded linear operator on L²(R) given on each f ∈ L²(R) by

$$Bf={\displaystyle \sum _{n\in W}{\left(n+1\right)}^{-1}\langle f,{\varphi }_n\rangle {\varphi }_n.}$$

(6)

It is readily checked that the right side does converge and, in fact,

$${\Vert Bf\Vert }_{L^2}^2={\displaystyle \sum _{n\in W}{\left(n+1\right)}^{-2}{\left|\langle f,{\varphi }_n\rangle \right|}_{L^2}^2\le {\displaystyle \sum _{n\in W}{\left|\langle f,{\varphi }_n\rangle \right|}_{L^2}^2={\Vert f\Vert }_{L^2}^2}}.$$

(7)

Note that B and T are inverses of each other on the linear span of the vectors ϕ_n:

$$TBf=f\mathrm{and}BTf=f\text{for all}f\in ℒ,$$

(8)

where

$$ℒ=\text{linear span of the vector}{\varphi }_n,\mathrm{for}n\in W.$$

(9)

2.4. The L² Approach

For any p ≥ 0, the image of B^p consists of all f ∈ L²(R) for which

$${\displaystyle \sum _{n\in W}{\left(n+1\right)}^{2p}{\left|\langle f,{\varphi }_n\rangle \right|}^2<\infty .}$$

Let

$${\mathcal{S}}_p(\mathbf{R})=B^p\left(L^2(\mathbf{R})\right).$$

(10)

This is a subspace of L²(R), and on ${\mathcal{S}}_p(\mathbf{R})$ there is an inner-product 〈·, ·〉_p given by

$${\langle f,g\rangle }_p\stackrel{\mathrm{def}}{=}{\displaystyle \sum _{n\in W}{\left(n+1\right)}^{2p}\langle f,{\varphi }_n\rangle \langle {\varphi }_n,g\rangle ={\langle B^{-p}f,B^{-p}g\rangle }_{L^2}}$$

(11)

which makes it a Hilbert space, having $ℒ$, and hence also $\mathcal{S}(\mathbf{R})$, a dense subspace. We will see later that functions in ${\mathcal{S}}_p(\mathbf{R})$ are p-times differentiable.

We will prove that the intersection${\cap }_p{}_{\in W}{\mathcal{S}}_p(\mathbf{R})$ is exactly equal to S(R). In fact,

$$\mathcal{S}(\mathbf{R})={\displaystyle \underset{p\in W}{\cap }{\mathcal{S}}_p(\mathbf{R})\subset \cdots {\mathcal{S}}_2(\mathbf{R})}\subset {\mathcal{S}}_1(\mathbf{R})\subset {\mathcal{S}}_0(\mathbf{R})=L^2(\mathbf{R}).$$

(12)

We will also prove that the topology on$\mathcal{S}(\mathbf{R})$ generated by the norms ║·║_p coincides with the standard topology. Furthermore, the elements ${(n+1)}^{-p}{\varphi }_n\in {\mathcal{S}}_p(\mathbf{R})$ form an orthonormal basis of ${\mathcal{S}}_p(\mathbf{R})$, and

$${\displaystyle \sum _{n\in W}{\Vert {\left(n+1\right)}^{-(p+1)}{\varphi }_n\Vert }_p^2={\displaystyle \sum _{n\ge 1}\frac{n^{2p}}{n^{2(p+1)}}<\infty }},$$

showing that the inclusion map ${\mathcal{S}}_p{}_{+1}(\mathbf{R})\to {\mathcal{S}}_p(\mathbf{R})$ is Hilbert-Schmidt.

The topological vector space $\mathcal{S}(\mathbf{R})$ has topology generated by a complete metric [10], and has a countable dense subset given by all finite linear combinations of the vectors ϕ_n with rational coefficients.

2.5. Coordinatization as a Sequence Space

All of the results described above follow readily from the identification of $\mathcal{S}(\mathbf{R})$ with a space of sequences. Let {ϕ_n}_n∈W be the orthonormal basis of L²(R) mentioned above, where W = {0, 1, 2, …}. Then we have the set C^W; an element a ∈ C^W is a map W → C : n ⟼a_n. So we shall often write such an element a as (a_n)_n∈W.

We have then the coordinatizing map

$$I:L^2(\mathbf{R})\to {\mathbf{C}}^W:f\mapsto {(\langle f,{\varphi }_n\rangle )}_{n\in W}.$$

(13)

For each p ∈ W let E_p be the subset of C^W consisting of all (a_n)_n∈W such that

$${\displaystyle \sum _{n\in W}{\left(n+1\right)}^{2p}{\left|a_n\right|}^2}<\infty .$$

On E_p define the inner-product 〈·, ·〉_p by

$${\langle a,b\rangle }_p={\displaystyle \sum _{n\in W}{\left(n+1\right)}^{2p}{a}_n{\overline{b}}_n}.$$

(14)

This makes E_p a Hilbert space, essentially the Hilbert space L²(W, µ_p), where µ_p is the measure on W given by µ_p({n}) = (n + 1)^2p for all n ∈ W.

The definition, Equation (10), for ${\mathcal{S}}_p(\mathbf{R})$ shows that it is the set of all f ∈ L²(R) for which I(f) belongs to E_p.

We will prove in Theorem 16 that I maps$\mathcal{S}(\mathbf{R})$ exactly onto

$$E\stackrel{\mathrm{def}}{=}{\displaystyle \underset{p\in W}{\cap }{E}_p}.$$

(15)

This will establish essentially all of the facts mentioned above concerning the spaces ${\mathcal{S}}_p(\mathbf{R})$.

Note the chain of inclusions:

$$E={\displaystyle \underset{p\in W}{\cap }{E}_p\subset \cdots \subset E_2}\subset E_1\subset E_0=L^2(W,{\mu }_0).$$

(16)

2.6. The Multi-Dimensional Setting

In the multidimensional setting, the Schwartz space $\mathcal{S}({\mathbf{R}}^d)$ consists of all infinitely differentiable functions f on R^d for which

$$\underset{x\in {\mathbf{R}}^d}{\sup }\left|x_1^{k_1}\dots x_d^{k_d}\frac{{\partial }^{m_1+\cdots +m_k}f(x)}{\partial x_1^{m_1}\dots \partial x_d^{m_d}}\right|<\infty ,$$

for all (k₁, …, k_d) ∈ W ^d and m = (m₁, …, m_d) ∈ W^d. For this setting, it is best to use some standard notation:

$$\left|k\right|=k_1+\cdots +k_d\mathrm{for}k=(k_1,\dots ,k_d)\in W^d$$

(17)

$$x^k=x^{k_1}\dots x^{k_d}$$

(18)

$$D^m=\frac{{\partial }^{\left|m\right|}}{\partial x_1^{m_1}\dots \partial x_d^{m_d}}.$$

(19)

For the multi-dimensional case, we use the indexing set W ^d whose elements are d–tuples j = (j₁, …, j_d), with j₁, …, j_d ∈ W, and counting measure µ₀ on W^d. The sequence space is replaced by ${\mathbf{C}}^{W^d}$; a typical element $a\in {\mathbf{C}}^{W^d}$, is a map

$$a:W^d\to \mathbf{C}:j=(j_1,\dots ,j_d)\mapsto a_j=a_{j_1\dots j_d}.$$

(20)

The orthonormal basis (ϕ_n)_n∈W of L²(R) yields an orthonormal basis of L²(R^d) consisting of the vectors

$${\varphi }_j={\varphi }_{j_1}\oplus \cdots \oplus {\varphi }_{j_d}:(x_1,\dots ,x_d)\mapsto {\varphi }_{j_1}(x_1)\dots {\varphi }_{j_d}(x_d).$$

(21)

The coordinatizing map I is replaced by the map

$$I_d:L^2({\mathbf{R}}^d)\to {\mathbf{C}}^{W^d}$$

(22)

where

$$I_d{\left(f\right)}_j={\langle f,{\varphi }_j\rangle }_{L^2({\mathbf{R}}^d)}.$$

(23)

Replace the operator T by

$$T_d\stackrel{\mathrm{def}}{=}{T}^{\oplus d}=\left(-\frac{{\partial }^2}{\partial x_d^2}+\frac{x_d^2}{4}+1\right)\cdots \left(-\frac{{\partial }^2}{\partial x_1^2}+\frac{x_1^2}{4}+1\right).$$

(24)

Then

$$T_d{\varphi }_j=(j_1+1)\dots (j_d+1){\varphi }_j$$

for all j ∈ W^d.

In place of B, we now have the bounded operator B_d on L²(R^d) given by

$$B_{d}f{\displaystyle \sum _{j\in W^d}{\left[(j_1+1)\dots (j_d+1)\right]}^{-1}\langle f,{\varphi }_j\rangle {\varphi }_j}$$

(25)

Again, T_d and B_d are inverses of each other on the linear subspace $ℒ$_d of L²(R^d) spanned by the vectors ϕ_j.

The space E_p is now the subset of ${\mathbf{C}}^{W^d}$ consisting of all $a\in {\mathbf{C}}^{W^d}$ for which

$${\Vert a\Vert }_p^2\stackrel{\mathrm{def}}{=}{\displaystyle \sum _{j\in W^d}{\left[(j_1+1)\dots (j_d+1)\right]}^{2p}{\left|a_j\right|}^2<\infty .}$$

(26)

Thus,

$$E_p\stackrel{\mathrm{def}}{=}\{a\in {\mathbf{C}}^{W^d}:{\Vert a\Vert }_p^2<\infty \}$$

(27)

This is a Hilbert space with inner-product

$${\langle a,b\rangle }_p={\displaystyle \sum _{j\in W^d}{\left[(j_1+1)\dots (j_d+1)\right]}^{2p}{a}_j{\overline{b}}_j.}$$

(28)

Again we have the chain of spaces

$$E\stackrel{\mathrm{def}}{=}{\displaystyle {\cap }_{p\in W}{E}_p\subset \cdots E_2\subset E_1\subset E_0=L^2(W^d,{\mu }_0)},$$

(29)

with the inclusion E_p+1 → E_p being Hilbert-Schmidt.

To go back to functions on R^d, define ${\mathcal{S}}_p({\mathbf{R}}^d)$ to be the range of B_d. Thus ${\mathcal{S}}_p({\mathbf{R}}^d)$ is the set of all f ∈ L²(R^d) for which

$$\sum _{j\in W^d}{\left[\left(j_1+1\right)\dots \left(j_d+1\right)\right]}^{2p}{\left|\langle f,{ø}_j\rangle \right|}^2<\infty$$

The inner-product 〈·, ·〉_p comes back to an inner-product, also denoted 〈·, ·〉_p, on ${\mathcal{S}}_p({\mathbf{R}}^d)$ and is given by

$${\langle f,g\rangle }_p={\langle B_d^{-p}f,B_d^{-p}g\rangle }_{L^2({\mathbf{R}}^d)}.$$

(30)

The intersection ${\displaystyle {\cap }_{p\in W}{\mathcal{S}}_p({\mathbf{R}}^d)}$ equals $\mathcal{S}({\mathbf{R}}^d)$. Moreover, the topology on $\mathcal{S}({\mathbf{R}}^d)$ is the smallest one generated by the inner-products obtained from 〈·, ·〉_p, with p running over W.

3. Topological Vector Spaces

The Schwartz space is a topological vector space, i.e., it is a vector space equipped with a Hausdorff topology with respect to which the vector space operations (addition, and multiplication by scalar) are continuous. In this section we shall go through a few of the basic notions and results for topological vector spaces.

Let V be a real vector space. A vector topology τ on V is a topology such that addition V × V → V : (x, y) ⟼ x + y and scalar multiplication R × V → V : (t, x) ⟼ tx are continuous. If V is a complex vector space we require that C × V → V : (α, x) ⟼ αx be continuous.

It is useful to observe that when V is equipped with a vector topology, the translation maps

$$t_x:V\to V:y\mapsto y+x$$

are continuous, for every x ∈ V, and are hence also homeomorphisms since $t_x^{-1}=t_{-x}$.

A topological vector space is a vector space equipped with a Hausdorff vector topology. A local base of a vector topology τ is a family of open sets {U_α}_α∈I containing 0 such that if W is any open set containing 0 then W contains some U_α. If U is any open set and x any point in U then U − x is an open neighborhood of 0 and hence contains some U_α, and so U itself contains a neighborhood x + U_α of x:

$$IfUisopenandx\in Uthenx+U_{\alpha }\subset U,forsome\alpha \in I.$$

(31)

Doing this for each point x of U, we see that each open set is the union of translates of the local base sets U_α.

3.1. Local Convexity and the Minkowski Functional

A vector topology τ on V is locally convex if for any neighborhood W of 0 there is a convex open set B with 0 ∈ B ⊂ W. Thus, local convexity means that there is a local base of the topology τ consisting of convex sets. The principal consequence of having a convex local base is the Hahn-Banach theorem which guarantees that continuous linear functionals on subspaces of V extend to continuous linear functionals on all of V. In particular, if V ≠ {0} is locally convex then there exist non-zero continuous linear functionals on V.

Let B be a convex open neighborhood of 0. Continuity of R × V → V : (s, x) ⟼ sx at s = 0 shows that for each x the multiple sx lies in B if s is small enough, and so t⁻¹x lies in B if t is large enough. The smallest value of t for which t⁻¹x is just outside B is clearly a measure of how large x is relative to B. The Minkowski functional µ_B is the function on V given by

$${\mu }_B(x)=\inf \{t>0:t^{-1}x\in B\}.$$

Note that 0 ≤ µ_B(x) < ∞. The definition of µ_B shows that µ_B(kx) = kµ_B(x) for any k ≥ 0. Convexity of B can be used to show that

$${\mu }_B(x+y)\le {\mu }_B(x)+{\mu }_B(y).$$

If B is symmetric, i.e., B = −B, then µ_B(kx) = |k|µ_B(x) for all real k. If V is a complex vector space and B is balanced in the sense that αB = B for all complex numbers α with |α| = 1, then µ_B(kx) = |k|µ_B(x) for all complex k. Note that in general it could be possible that µ_B(x) is 0 without x being 0; this would happen if B contains the entire ray {tx : t ≥ 0}.

3.2. Semi-Norms

A typical vector topology on V is specified by a semi-norm on V, i.e., a function µ : V → R such that

$$\mu (x+y)\le \mu (x)+\mu (y),\mu (tx)=\left|t\right|\mu (x)$$

(32)

for all x, y ∈ V and t ∈ R (complex t if V is a complex vector space). Note that then, using t = 0, we have µ(0) = 0 and, using −x for y, we have µ(x) ≥ 0. For such a semi-norm, an open ball around x is the set

$$B_{\mu }(x;r)=\{y\in V:\mu (y-x)<r\},$$

(33)

and the topology τ_µ consists of all sets which can be expressed as unions of open balls. These balls are convex and so the topology τ_µ is locally convex. If µ is actually a norm, i.e., µ(x) is 0 only if x is 0, then τ_µ is Hausdorff.

A consequence of the triangle inequality Equation (32) is that a semi-norm µ is uniformly continuous with respect to the topology it generates. This follows from the inequality

$$\left|\mu (x)-\mu (y)\right|\le \mu (x-y),$$

(34)

which implies that µ, as a function on V, is continuous with respect to the topology τ_µ it generates. Now suppose µ is continuous with respect to a vector topology τ. Then the open balls {y ∈ V : µ(y − x) < r} are open in the topology τ and so τ_µ ⊂ τ.

3.3. Topologies Generated by Families of Topologies

Let {τ_α}_α∈I be a collection of topologies on a space. It is natural and useful to consider the the least upper bound topology τ, i.e., the smallest topology containing all sets of ∪_α∈Iτ_α. In our setting, we work with each τ_α a vector topology on a vector space V.

Theorem 1. The least upper bound topology τ of a collection {τ_α}_α∈I of vector topologies is again a vector topology. If${\left\{W_{\alpha }{}_{,i}\right\}}_i{}_{\in I_{\alpha }}$ is a local base for τ_α then a local base for τ is obtained by taking all finite intersections of the form$W_{\alpha }{}_{{}_1,i_1}\cap \cdots \cap W_{{\alpha }_n}{}_{,i_n}$.

Proof. Let B be the collection of all sets which are of the form $W_{\alpha }{}_{{}_1,i_1}\cap \cdots \cap W_{{\alpha }_n}{}_{,i_n}$.

Let τ′ be the collection of all sets which are unions of translates of sets in B (including the empty union). Our first objective is to show that τ′ is a topology on V. It is clear that τ′ is closed under unions and contains the empty set. We have to show that the intersection of two sets in τ′ is in τ′. To this end, it will suffice to prove the following:

$$\begin{array}{l}\text{If}{C}_1\mathrm{and}{C}_2\text{are sets in}B,\mathrm{and}x\text{is a point in}\\ \text{the intersection of the translates}a+C_1\mathrm{and}b+C_2,\\ \mathrm{then}x+C\subset (a+C_1)\cap (b+C_2)\text{for some}C\text{in}\mathcal{B}.\end{array}$$

(35)

Clearly, it suffices to consider finitely many topologies τ_α. Thus, consider vector topologies τ₁, …, τ_n on V.

Let ${\mathcal{B}}_n$ be the collection of all sets of the form B₁ ∩⋯∩ B_n with B_i in a local base for τ_i, for each i ∈ {1, …, n}. We can check that if D, $D^{\prime }\in {\mathcal{B}}_n$ then there is an $E\in {\mathcal{B}}_n$ with E ⊂ D ∩ D′.

Working with B_i drawn from a given local base for τ_i, let z be a point in the intersection B₁∩⋯∩B_n. Then there exist sets ${B^{\prime }}_i$, with each ${B^{\prime }}_i$ being in the local base for τ_i, such that $z+{B^{\prime }}_i\subset B_i$ (this follows from our earlier observation Equation (31)). Consequently,

$$z+{\displaystyle \underset{i=1}{\overset{n}{\cap }}{B^{\prime }}_i}\subset {\displaystyle \underset{i=1}{\overset{n}{\cap }}{B}_i}.$$

Now consider sets C₁ an C₂, both in ${\mathcal{B}}_n$. Consider a, b ∈ V and suppose x ∈ (a + C₁) ∩ (b + C₂). Then since x − a ∈ C₁ there is a set ${C^{\prime }}_1\in {\mathcal{B}}_n$ with $x-a+{C^{\prime }}_1\subset C_1$; similarly, there is a ${C^{\prime }}_2\in {\mathcal{B}}_n$ with $x-b+{C^{\prime }}_2\subset C_2$. So $x+{C^{\prime }}_1\subset a+C_1$ and $x+{C^{\prime }}_1\subset a+C_1$. So

$$x+C\subset (a+C_1)\cap (b+C_2),$$

where $C\in {\mathcal{B}}_n$ satisfies C ⊂ C₁ ∩ C₂. This establishes Equation (35), and shows that the intersection of two sets in τ′ is in τ′.

Thus τ′ is a topology. The definition of τ′ makes it clear that τ′ contains each τ_α. Furthermore, if any topology σ contains each τ_α then all the sets of τ′ are also open relative to σ. Thus τ′ = τ, the topology generated by the topologies τ_α.

Observe that we have shown that if W ∈ τ contains 0 then W ⊃ B for some $B\in \mathcal{B}$.

Next we have to show that τ is a vector topology. The definition of τ shows that τ is translation invariant, i.e., translations are homeomorphisms. So, for addition, it will suffice to show that addition V × V ⟼V : (x, y) ⟼x + y is continuous at (0, 0). Let W ∈ τ contain 0. Then there is a $B\in \mathcal{B}$ with 0 ∈ B ⊂ W. Suppose B = B₁ ∩⋯∩ B_n, where each B_i is in the given local base for τ_i. Since τ_i is a vector topology, there are open sets D_i, ${D^{\prime }}_i\in {\tau }_i$, both containing 0, with $D_i+{D^{\prime }}_i\subset B_i$. Then choose C_i, ${C^{\prime }}_i$ in the local base for τ_i with C_i ⊂ D_i and ${C^{\prime }}_i\subset {D^{\prime }}_i$. Then $C_i+{C^{\prime }}_i\subset B_i$. Now let C = C₁∩⋯∩C_n, and $C^{\prime }={C^{\prime }}_1\cap \cdots \cap {C^{\prime }}_n$. Then C, $C^{\prime }\in \mathcal{B}$ and $C+C^{\prime }\subset B$. Thus, addition is continuous at (0, 0).

Now consider the multiplication map R × V → V : (t, x) ⟼tx. Let (s, y), (t, x) ∈ R × V. Then

$$sy-tx=(s-t)x+t(y-x)+(s-t)(y-x).$$

Suppose F ∈ τ contains tx. Then

$$F\supset tx+W^{\prime },$$

for some $W^{\prime }\in \mathcal{B}$. Using continuity of the addition map

$$V\times V\times V\to V:(a,b,c)\mapsto a+b+c$$

at (0, 0, 0), we can choose $W_1,W_2,W_3\in \mathcal{B}$ with W₁ + W₂ + W₃ ⊂ W′. Then we can choose $W\in \mathcal{B}$, such that

$$W\subset W_1\cap W_2\cap W_3.$$

Then $W\in \mathcal{B}$ and

$$W+W+W\subset W^{\prime }.$$

Suppose W = B₁ ∩⋯∩ B_n, where each B_i is in the given local base for the vector topology τ_i. Then for s close enough to t, we have (s − t)x ∈ B_i for each i, and hence (s − t)x ∈ W. Similarly, if y is τ–close enough to x then t(y − x) ∈ W. Lastly, if s − t is close enough to 0 and y is close enough to x then (s − t)(y − x) ∈ W. So sy − tx ∈ W′, and so sy ∈ F, when s is close enough to t and y is τ–close enough to x. □

The above result makes it clear that if each τ_α has a convex local base then so is τ. Note also that if at least one τ_α is Hausdorff then so is τ.

A family of topologies {τ_α}_α∈I is directed if for any α, β ∈ I there is a γ ∈ I such that τ_α ∪ τ_β ⊂ τ_γ. In this case every open neighborhood of 0 in the generated topology contains an open neighborhood in one of the topologies τ_γ.

3.4. Topologies Generated by Families of Semi-Norms

We are concerned mainly with the topology τ generated by a family of semi-norms {µ_α}_α∈I; this is the smallest topology containing all sets of ${\displaystyle {\cup }_{\alpha \in I}{\tau }_{{\mu }_{\alpha }}}$. An open set in this topology is a union of translates of finite intersections of balls of the form $B_{{\mu }_i}(0;r_i)$. Thus, any open neighborhood of f contains a set of the form

$$B_{{\mu }_1}(f;r_1)\cap \cdots \cap B_{{\mu }_n}(f;r_n).$$

This topology is Hausdorff if for any non-zero x ∈ V there is some norm µ_α for which µ_α(x) is not zero.

The description of the neighborhoods in the topology τ shows that a sequence f_n converges to f with respect to τ if and only if µ_α(f_n − f) → 0, as n → ∞, for all α ∈ I.

We will need to examine when two families of semi-norms give rise to the same topology:

Theorem 2. Let τ be the topology on V generated by a family of semi-norms $ℳ$ = {µ_i}_i∈I, and τ′ the topology generated by a family of semi-norms${ℳ}^{\prime }={\left\{{{\mu }^{\prime }}_j\right\}}_j{}_{\in J}$. Suppose each µ_i is bounded above by a linear combination of the${{\mu }^{\prime }}_j$. Then τ ⊂ τ′.

Proof. Let µ ∈ M. Then there exist ${{\mu }^{\prime }}_j,\dots ,\in {ℳ}^{\prime }$, and real numbers c₁, …, c_n > 0, such that

$$\mu \le c_1{{\mu }^{\prime }}_1+\cdots +c_n{{\mu }^{\prime }}_n.$$

Now consider any x, y ∈ V. Then

$$\left|\mu (x)-\mu (y)\right|\le \mu (x-y)\le {\displaystyle \sum _{i=1}^n\left|c_i\right|{{\mu }^{\prime }}_i}(x-y).$$

So µ is continuous with respect to the topology generated by ${{\mu }^{\prime }}_1,\dots ,{{\mu }^{\prime }}_n$. Thus, τ_µ ⊂ τ′. Since this is true for all µ ∈ $ℳ$, we have τ ⊂ τ′. □

3.5. Completeness

A sequence (x_n)_n∈N in a topological vector space V is Cauchy if for any neighborhood U of 0 in V, the difference x_n − x_m lies in U when n and m are large enough. The topological vector space V is complete if every Cauchy sequence converges.

Theorem 3. Let {τ_α}_α∈I be a directed family of Hausdorff vector topologies on V, and τ the generated topology. If each τ_α is complete then so is τ.

Proof. Let (x_n)_n≥1 be a sequence in V, which is Cauchy with respect to τ. Then clearly it is Cauchy with respect to each τ_α. Let x_α = lim_n→∞ x_n, relative to τ_α. If τ_α ⊂ τ_γ then the sequence (x_n)_n≥1 also converges to x_γ relative to the topology τ_α, and so x_γ = x_α. Consider α, β ∈ I, and choose γ ∈ I such that τ_α ∪ τ_β ⊂ τ_γ. This shows that x_α = x_γ = x_β, i.e., all the limits are equal to each other. Let x denote the common value of this limit. We have to show that x_n → x in the topology τ. Let W ∈ τ contain x. Since the family {τ_α}_α∈I which generates τ is directed, it follows that there is a β ∈ I and a B_β ∈ τ_β with x ∈ B_β ⊂ W. Since (x_n)_n≥1 converges to x with respect to τ_β, it follows x_n ∈ B_β for large n. So x_n → x with respect to τ. □

3.6. Metrizability

Suppose the topology τ on the topological vector space V is generated by a countable family of semi-norms µ₁, µ₂, …. For any x, y ∈ V define

$$d(x,y)={\displaystyle \sum _{n\ge 1}{2}^{-n}{d}_n}(x,y)$$

(36)

where

$$d_n(x,y)=\min \{1,{\mu }_n(x-y)\}.$$

Then d is a metric, it is translation invariant, and generates the topology τ [10].

4. The Schwartz Space$\mathcal{S}(\mathbf{R})$

Our objective in this section is to show that the Schwartz space is complete, in the sense that every Cauchy sequence converges. Recall that $\mathcal{S}(\mathbf{R})$ is the set of all C^∞ functions f on R for which

$$p_{a,b}(f)\stackrel{\mathrm{def}}{=}{\Vert f\Vert }_{a,b}\underset{x\in \mathbf{R}}{\sup }\left|x^a{D}^{b}f(x\right|<\infty$$

(37)

for all a, b ∈ W = {0, 1, 2, …}. The functions p_a,b are semi-norms, with ║ · ║_0,0, being just the sup-norm. Thus the family of semi-norms given above specify a Hausdorff vector topology on $\mathcal{S}(\mathbf{R})$. We will call this the Schwartz topology on $\mathcal{S}(\mathbf{R})$.

Theorem 4. The topology on$\mathcal{S}(\mathbf{R})$ generated by the family of semi-norms ║·║_a,b for all a, b ∈ {0, 1, 2, …}, is complete.

Proof. Let (f_n)_n≥1 be a Cauchy sequence on $\mathcal{S}(\mathbf{R})$. Then this sequence is Cauchy in each of the semi-norms ║·║_a,b, and so each sequence of functions x^aD^bf_n(x) is uniformly convergent. Let

$$g_b(x)=\underset{n\to \infty }{\lim }{D}^b{f}_n(x).$$

(38)

Let f = g₀. Using a Taylor theorem argument it follows that g_b is D^bf. For instance, for b = 1, observe first that

$$f_n(y)=f_n(x)+{\displaystyle {\int }_0^1\frac{d{f}_n\left((1-t)x+ty)\right)}{dt}dt=f_n(x)+}{\displaystyle {\int }_0^1{f^{\prime }}_n\left((1-t)x+ty\right)(y-x)dt,}$$

and so, letting n → ∞, we have

$$f(y)=f(x)+{\displaystyle {\int }_0^1{g}_1\left((1-t)x+ty\right)(y-x)dt,}$$

which implies that f′(x) exists and equals g₁(x).

In this way, we have x^aD^bf_n(x) → x^aD^bf(x) pointwise. Note that our Cauchy hypothesis implies that the sequence of functions x^aD^bf_n(x) is Cauchy in sup-norm, and so the convergence

$$x^a{D}^b{f}_n(x)\to x^a{D}^{b}f(x)$$

is uniform. In particular, the sup-norm of x^aD^bf(x) is finite, since it is the limit of a uniformly convergent sequence of bounded functions. Thus $f\in \mathcal{S}(\mathbf{R})$.

Finally, we have to check that f_n converges to f in the topology of $\mathcal{S}(\mathbf{R})$. We have noted above that x^aD^bf_n(x) → x^aD^bf(x) uniformly. Thus f_n → f relative to the semi-norm ║·║_a,b. Since this holds for every a, b ∈ {0, 1, 2, 3, …}, we have f_n → f in the topology of $\mathcal{S}(\mathbf{R})$. □

Now let’s take a quick look at the Schwartz space $\mathcal{S}({\mathbf{R}}^d)$. First some notation. A multi-index a is an element of {0, 1, 2, …}^d, i.e., it is a mapping

$$a:\{1,\dots ,d\}\to \{0,1,2,\dots \}:j\mapsto a_j.$$

If a is a multi-index, we write |a| to mean the sum a₁+⋯+a_d, x^a to mean the product $x_1^{a_1}\dots x_d^{a_d}$, and D^a to mean the differential operator $x_{x_1}^{a_1}\dots x_{x_d}^{a_d}$. The space $\mathcal{S}({\mathbf{R}}^d)$ consists of all C^∞ functions f on Rd such that each function x^aD^bf(x) is bounded. On $\mathcal{S}({\mathbf{R}}^d)$ we have the semi-norms

$${\Vert f\Vert }_{a,b}=\underset{x\in {\mathbf{R}}^d}{\sup }\left|x^a{D}^{b}f(x)\right|$$

for each pair of multi-indices a and b. The Schwartz topology on $\mathcal{S}({\mathbf{R}}^d)$ is the smallest topology making each semi-norm ║·║_a,b continuous. This makes $\mathcal{S}({\mathbf{R}}^d)$ a topological vector space.

The argument for the proof of the preceding theorem goes through with minor alterations and shows that:

Theorem 5. The topology on$\mathcal{S}({\mathbf{R}}^d)$ generated by the family of semi-norms ║·║_a,b for all a, b ∈ {0, 1, 2, …}^d, is complete.

5. Hermite Polynomials, Creation and Annihilation Operators

We shall summarize the definition and basic properties of Hermite polynomials (our approach is essentially that of Hermite’s original [11]). We repeat for convenience of reference much of the presentation in Section 2.1 of [7].

A central role is played by the Gaussian kernel

$$p(x)=\frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}.$$

(39)

Properties of translates of p are obtained from

$$e^{xy-\frac{y^2}{2}}=\frac{p(x-y)}{p(x)}.$$

(40)

Expanding the right side in a Taylor series we have

$$e^{xy-\frac{y^2}{2}}=\frac{p(x-y)}{p(x)}={\displaystyle \sum _{n=0}^{\infty }\frac{1}{n!}{H}_n(x)y^n,}$$

(41)

where the Taylor coefficients, denoted H_n(x), are

$$H_n(x)=\frac{1}{p(x)}{\left(-\frac{d}{dx}\right)}^{n}p(x).$$

(42)

This is the n–th Hermite polynomial and is indeed an n–th degree polynomial in which xⁿ has coefficient 1, facts which may be checked by induction.

Observe the following

$$\begin{array}{c}{\displaystyle {\int }_{\mathbf{R}}\frac{p(x-y)}{p(x)}\frac{p(x-z)}{p(x)}p(x)dx=}{e}^{-\frac{y^2+z^2}{2}}{\displaystyle {\int }_{\mathbf{R}}{e}^{x(y+z)}p(x)dx}\\ =e^{-\frac{y^2+z^2}{2}+\frac{{\left(y+z\right)}^2}{2}}\\ =e^{yz}.\end{array}$$

Going over to the Taylor series and comparing the appropriate Taylor coefficients (differentiation with respect to y and z can be carried out under the integral) we have

$${\langle H_n,H_m\rangle }_{L^2(p(x)dx)}=n!{\delta }_{nm}.$$

(43)

Thus an orthonormal set of functions is given by

$$h_n(x)=\frac{1}{\sqrt{n!}}{H}_n(x).$$

(44)

Because these are orthogonal polynomials, the n–th one being exactly of degree n, their span contains all polynomials. It can be shown that the span is in fact dense in L²(p(x)dx). Thus the polynomials above constitute an orthonormal basis of L²(p(x)dx).

Next, consider the derivative of H_n:

$$\begin{array}{c}{H^{\prime }}_n(x)={(-1)}^{n}p{(x)}^{-1}{p}^{(n+1)}(x)-{\left(-1\right)}^{n}p{(x)}^{-1}{p}^{\prime }(x)p{(x)}^{-1}{p}^n(x)\\ =-H_{n+1}(x)+x{H}_n(x).\end{array}$$

So

$$\left(-\frac{d}{dx}+x\right)h_n(x)=\sqrt{n+1}{h}_{n+1}(x).$$

(45)

The operator

$$\left(-\frac{d}{dx}+x\right)$$

is called the creation operator in L²(R;p(x)dx).

Officially, we can take the creation operator to have domain consisting of all functions f which can be expanded in L²(p(x)dx) as ∑_n≥0 a_nh_n, with each a_n a complex number, and satisfying the condition ∑_n≥0(n + 1)|a_n|² < ∞; the action of the operator on f yields the function ${\displaystyle {\sum }_{n\ge 0}\sqrt{n+1}{a}_n{h}_{n+1}}$. This makes the creation operator unitarily equivalent to a multiplication operator (in the sense discussed later in subsection A.5) and hence a closed operator (see A.1 for definition). For the type of smooth functions f we will mostly work with, the effect of the operator on f will in fact be given by application of $\left(-\frac{d}{dx}+x\right)$ to f.

Next, from the fundamental generating relation Equation (41) we have :

$$y{e}^{xy-y^2/2}=\underset{ϵ\downarrow 0}{\lim }{\displaystyle \sum _{n\ge 0}\frac{1}{n!}\frac{1}{ϵ}[H_n(x+ϵ)-H_n(x)]y^n.}$$

(46)

Using Equation (41) again on the left we have

$${\displaystyle \sum _{n\ge 1}\frac{1}{(n-1)!}{H}_{n-1}(x)y^n}=\underset{ϵ\downarrow 0}{\lim }{\displaystyle \sum _{n\ge 0}\frac{1}{n!}\frac{1}{ϵ}[H_n(x+ϵ)-H_n(x)]y^n.}$$

(47)

Letting y = 0 allows us to equate the n = 0 terms, and then, successively, the higher order terms. From this we see that

$${H^{\prime }}_n(x)=n{H}_{n-1}(x)$$

(48)

where H₋₁ = 0. Thus:

$$\frac{d}{dx}{h}_n(x)=\sqrt{n}{h}_{n-1}(x).$$

(49)

The operator

$$\frac{d}{dx}$$

is the annihilation operator in L²(R;p(x)dx). As with the creation operator, we may define it in a more specific way, as a closed operator on a specified domain.

6. Hermite Functions, Creation and Annihilation Operators

In the preceding section we studied Hermite polynomials in the setting of the Gaussian space L²(R;p(x)dx). Let us translate the concepts and results back to the usual space L²(R; dx).

To this end, consider the isomorphism:

$$\mathcal{U}:L^2(\mathbf{R},p(x)dx)\to L^2(\mathbf{R},dx):f\mapsto \sqrt{p}f.$$

(50)

Then the orthonormal basis polynomials h_n go over to the functions ϕ_n given by

$${\varphi }_n(x)={\left(-1\right)}^n\frac{1}{\sqrt{n!}}{\left(2\pi \right)}^{-1/4}{e}^{x^2/4}\frac{d^n{e}^{-x^2/2}}{d{x}^n}.$$

(51)

The family {ϕ_n}_n≥0 forms an orthonormal basis for L²(R, dx).

We now determine the annihilation and creation operators on L²(R, dx). If f ∈ L²(R, dx) is differentiable and has derivative f′ also in L²(R, dx), we have:

$$\begin{array}{c}\left(\mathcal{U}\frac{d}{dx}{\mathcal{U}}^{-1}\right)f(x)=\sqrt{p(x)}\frac{d}{dx}[p{\left(x\right)}^{-1/2}f(x)]\\ =f^{\prime }(x)+p{\left(x\right)}^{1/2}(-1/2)p{\left(x\right)}^{-3/2}{p}^{\prime }\left(x\right)f(x)\\ =f^{\prime }(x)+\frac{1}{2}xf(x).\end{array}$$

So, on L²(R, dx), the annihilator operator is

$$A=\frac{d}{dx}+\frac{1}{2}x$$

(52)

which will satisfy

$$A{\varphi }_n=\sqrt{n}{\varphi }_{n-1}$$

(53)

where ϕ₋₁ = 0. For the moment, we proceed by taking the domain of A to be the Schwartz space $\mathcal{S}(\mathbf{R})$.

Next,

$$\begin{array}{c}\left(\mathcal{U}\left(-\frac{d}{dx}+x\right){\mathcal{U}}^{-1}\right)f(x)=f^{\prime }(x)+xf(x)-\frac{1}{2}xf(x)\\ =\left(-\frac{d}{dx}+\frac{1}{2}x\right)f(x)\end{array}$$

Thus the creation operator is

$$C=A^{*}=-\frac{d}{dx}+\frac{1}{2}x.$$

(54)

The reason we have written A* is that, as is readily checked, we have the adjoint relation

$$\langle Af,g\rangle =\langle f,\left(-\frac{d}{dx}+\frac{1}{2}x\right)g\rangle$$

(55)

with the inner-product being the usual one on L²(R, dx). Again, for the moment, we take the domain of C to be the Schwartz space $\mathcal{S}(\mathbf{R})$ (though, technically, in that case we should not write C as A^∗, since the latter, if viewed as the L²–adjoint operator, has a larger domain).

For this we have

$$C{\varphi }_n=\sqrt{n+1}{\varphi }_{n+1}.$$

(56)

Observe also that

$$AC=\frac{1}{4}{x}^2-\frac{d^2}{d{x}^2}+\frac{1}{2}I\mathrm{and}CA=\frac{1}{4}{x}^2-\frac{d^2}{d{x}^2}-\frac{1}{2}I$$

(57)

which imply:

$$[A,C]=AC-CA=I,\text{the identity}.$$

(58)

Next observe that

$$CA{\varphi }_n=\sqrt{n}\sqrt{n}{\varphi }_n=n{\varphi }_n$$

(59)

and so CA is called the number operator N:

$$N=A*A=CA=-\frac{d^2}{d{x}^2}+\frac{x^2}{4}-\frac{1}{2}\mathrm{the}numberoperater.$$

(60)

As noted above in Equation (59), the number operator N has the eigenfunctions ϕ_n:

$$N{\varphi }_n=n{\varphi }_n.$$

(61)

Integration by parts (see Lemma 10) shows that

$$\langle f,g^{\prime }\rangle =-\langle f^{\prime },g\rangle$$

for every $f,g\in \mathcal{S}(\mathbf{R})$, and so also

$$\langle f,g^{″}\rangle =-\langle f^{″},g\rangle .$$

It follows that the operator N satisfies

$$\langle Nf,g\rangle =\langle f,Ng\rangle$$

(62)

for every $f,g\in \mathcal{S}(\mathbf{R})$.

Now consider the case of R^d. For each j ∈ {1, …, d}, there are creation, annihilation, and number operators:

$$A_j=\frac{\partial }{\partial x_j}+\frac{1}{2}{x}_j,C_j=-\frac{\partial }{\partial x_j}+\frac{1}{2}{x}_j,N_j=C_j{A}_j.$$

(63)

These map $\mathcal{S}({\mathbf{R}}^d)$ into itself and, as is readily verified, satisfy the commutation relations

$$[A_j,C_k]={\delta }_{jk}I,[N_j,A_k]=-{\delta }_{jk}{A}_j,[N_j,C_k]={\delta }_{jk}{C}_j.$$

(64)

Now let us be more specific about the precise definition of the creation and annihilation operators. The basis {ϕ_n}_n≥0 for L²(R) yields an orthonormal basis ${\left\{{\varphi }_m\right\}}_m{}_{\in W^d}$ of L²(R^d) given by ${\varphi }_m={\varphi }_{m_1}(x_1)\cdots {\varphi }_{m_d}(x_d)$. For convenience we say ϕ_m=0 if some m_j <0. Given its effect on the orthonormal basis ${\left\{{\varphi }_m\right\}}_m{}_{\in W^d}$ the operator C_k has the form:

$${\varphi }_m\mapsto \sqrt{m_k+1{\varphi }_{m^{\prime }}},$$

where ${m^{\prime }}_i=m_i$ for all i ∈ {1, …, d} except when i = k, in which case ${m^{\prime }}_k=m_k+1$. The domain of C_k is the set $\mathcal{D}(C_k)$ given by

$$\mathcal{D}(C_k)=\left\{f\in L^2(\mathbf{R})|{\displaystyle \sum _{m\in W^d}(m_k+1){\left|a_m\right|}^2<\infty }\mathrm{where}{a}_m=\langle f,{\varphi }_m\rangle \right\}.$$

The operator C_k is then officially defined by specifying its action on a typical element of its domain:

$$C_k\left({\displaystyle \sum _{m\in W^d}{a}_m{\varphi }_m}\right)={\displaystyle \sum _{m\in W^d}{a}_m\sqrt{m_m+1}}{\varphi }_{m^{\prime }},$$

(65)

where m′ is as before. The operator C_k is essentially the composite of a multiplication operator and a bounded linear map taking ϕ_m → ϕ_m′ where m′ is as defined above. (See subsection A.5 for precise formulation of a multiplication operator.) Noting this, it can be readily checked that C_k is a closed operator using the following argument: Let T be a bounded linear operator and M_h a multiplication operator (any closed operator will do); we show that the composite M_hT is a closed operator. Suppose x_n → x. Since T is a bounded linear operator, Tx_n → Tx. Now suppose also that M_h(Tx_n) → y. Since M_h is closed, it follows then that $Tx\in \mathcal{D}(M_h)$ and y = M_hTx.

The operators A_k and N_k are defined analogously.

Proposition 6. Let $ℒ$₀ be the vector subspace of L²(R^d) spanned by the basis vectors${\left\{{\varphi }_m\right\}}_m{}_{\in W^d}$. Then for k ∈ {1, 2, …, d}, C_k|$ℒ$₀ and A_k|$ℒ$₀ have closures given by C_k and A_k, respectively (see subsection A.4 for the notion of closure).

Proof. We need to show that the graph of C_k, denoted Gr(C_k), is equal to the closure of the graph of C_k|$ℒ$₀, i.e., to $\overline{Gr(C_k|{ℒ}_0)}$ (see to subsection A.1 for the notion of graph). It is clear that $Gr(C_k|{ℒ}_0)\subseteq Gr(C_k)$. Using this and the fact that Ck is a closed operator, we have

$$\overline{Gr(C_k|{ℒ}_0)}\subseteq \overline{Gr(C_k)}=Gr(C_k).$$

Going in the other direction, take (f, C_kf) ∈ Gr(C_k). Now ${\displaystyle {\sum }_{m\in W^d}{a}_m{\varphi }_m}$ where a_m = 〈f, ϕ_m〉. Let f_N be given by

$$f_N={\displaystyle \sum _{m\in W_N^d}{a}_m{\varphi }_m}\mathrm{where}{W}_N^d=\{m\in W^d|0\le m_1\le N,\dots ,0\le m_d\le N\}.$$

Observe that f_N ∈ $ℒ$₀. Moreover

$$\underset{N\to \infty }{\lim }{f}_N=f\mathrm{and}\underset{N\to \infty }{\lim }{C}_k{f}_N=\underset{N\to \infty }{\lim }{\displaystyle \sum _{m\in W_N^d}(m_k+1)a_m{\varphi }_m=C_k}f$$

in L²(R^d). Thus $(f,C_{k}f)\in \overline{Gr(C_k|L_0)}$ and so we have $Gr(C_k)\subseteq \overline{Gr(C_k|L_0)}$.

The proof for A_k follows similarly. □

Linking this new definition for C_k with our earlier formulas Equation (63) we have:

Proposition 7. If$f\in \mathcal{S}({\mathbf{R}}^d)$ then

$$C_{k}f=-\frac{\partial f}{\partial x_k}+\frac{x_k}{2}f,and{A}_{k}f=\frac{\partial f}{\partial x_k}+\frac{x_k}{2}f.$$

Proof. Let $g=\frac{\partial f}{\partial x_k}+\frac{x_k}{2}f$. Since $f\in \mathcal{S}({\mathbf{R}}^d)$, we have g ∈ L²(R^d). So we can write g as $g={\displaystyle {\sum }_{j\in W^d}{a}_j{\varphi }_j}$ where a_j=〈g, ϕ_j〉. Let us examine these a_j’s more closely. Observe

$$\begin{array}{c}{a}_j=\langle g,{\varphi }_j\rangle =\langle -\frac{\partial f}{\partial x_k}+\frac{x_k}{2}f,{\varphi }_j\rangle \\ =\langle f,\frac{\partial {\varphi }_j}{\partial x_k}+\frac{x_k}{2}{\varphi }_j\rangle \\ =\langle f,\sqrt{jk{\varphi }_{j^{″}}}\rangle \end{array}$$

where ${j^{″}}_i=j_i$ for all i ∈ {1, …, d} except when i = k, in which case ${j^{″}}_k=j_k-1$.

Bringing this information back to our expression for g we see that

$$\begin{array}{c}g={\displaystyle \sum _{j\in W^d}\sqrt{jk\langle f,{\varphi }_{j^{″}}\rangle }}{\varphi }_j\\ ={\displaystyle \sum _{m\in W^d}\sqrt{m_k+1\langle f,{\varphi }_m\rangle }}{\varphi }_{m^{″}}\mathrm{where}{m}^{″}\text{is as defined above}\\ =C_{k}f\mathrm{by}(65).\end{array}$$

The second equality is obtained by letting m = j″ and noting that ϕ_j″ = 0 when ${j^{″}}_k$ is −1. The proof follows similarly for A_k. □

7. Properties of the Functions in ${\mathcal{S}}_p(\mathbf{R})$

Our aim here is to obtain a complete characterization of the functions in ${\mathcal{S}}_p(\mathbf{R})$. We will prove that ${\mathcal{S}}_p(\mathbf{R})$ consists of all square-integrable functions f for which all derivatives f^(k) exist for k ∈ {1, 2, …, p} and

$$\underset{x\in \mathbf{R}}{\sup }\left|x^a{f}^{(b)}(x)\right|<\infty$$

for all a, b ∈ {0, 1, …, p − 1} with a + b ≤ p − 1.

A significant tool we will use is the Fourier transform:

$$\widehat{f}(p)=ℱf(p)={\left(2\pi \right)}^{-1/2}{\displaystyle {\int }_{\mathbf{R}}{e}^{-ipx}f(x)dx}.$$

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This is meaningful whenever f is in L¹(R), but we will work mainly with f in $\mathcal{S}(\mathbf{R})$. We will use the following standard facts:

• $ℱ$ maps $\mathcal{S}(\mathbf{R})$ onto itself and satisfies the Plancherel identity:

  $${\displaystyle {\int }_{\mathbf{R}}{\left|f(x)\right|}^{2}dx}={\displaystyle {\int }_{\mathbf{R}}{\left|\widehat{f}(p)\right|}^{2}d}p$$

  (67)
• for any $f\in \mathcal{S}(\mathbf{R})$,

  $$f(x)={\left(2\pi \right)}^{-1/2}{\displaystyle {\int }_{\mathbf{R}}{e}^{ipx}\widehat{f}(p)dp}$$

  (68)
• if $f\in \mathcal{S}(\mathbf{R})$ then

$$p\widehat{f}(p)=-iℱ(f^{\prime })(p).$$

(69)

Consequently, we have

$$\begin{array}{l}{\Vert f\Vert }_{\sup }\le {\left(2\pi \right)}^{-1/2}{\displaystyle {\int }_{\mathbf{R}}\left|\widehat{f}(p)\right|dp}\\ ={\left(2\pi \right)}^{-1/2}{\displaystyle {\int }_{\mathbf{R}}{\left(1+p^2\right)}^{1/2}\left|\widehat{f}(p)\right|}{\left(1+p^2\right)}^{-1/2}dp\\ ={\left(2\pi \right)}^{-1/2}{\left[{\displaystyle {\int }_{\mathbf{R}}\left(1+p^2\right){\left|\widehat{f}(p)\right|}^{2}dp}\right]}^{1/2}{\pi }^{1/2}\text{by Cauchy-Schwartz}\\ \le 2^{-1/2}\left[{\Vert \widehat{f}\Vert }_{L^2}+{\Vert p\widehat{f}\Vert }_{L^2(\mathbf{R},dp)}\right]\\ \le 2^{-1/2}\left[{\Vert f\Vert }_{L^2}+{\Vert f^{\prime }\Vert }_{L^2}\right]\text{by Plancherel and Equation}\left(69\right).\end{array}$$

(70)

For the purposes of this section it is necessary to be precise about domains. So we take now A and C to be closed operators in L²(R), with common domain

$$\mathcal{D}(C)=\mathcal{D}(A)=\{f\in L^2(\mathbf{R})|{\displaystyle \sum _{n\ge 0}n{\left|\langle f,{\varphi }_n\rangle \right|}^2<\infty }\}$$

and

$$Cf={\displaystyle \sum _{n\ge 0}\langle f,{\varphi }_n\rangle \sqrt{n+1}{\varphi }_{n+1}},Af={\displaystyle \sum _{n\ge 0}\langle f,{\varphi }_n\rangle \sqrt{n+1}{\varphi }_{n-1}}.$$

Moreover, define operators C₁ and A₁ on the common domain

$${\mathcal{D}}_1=\text{all differentiable}f\in L^2\mathrm{with}{f}^{\prime }\in L^2\mathrm{and}xf(x)\in L^2(dx)$$

(71)

and

$$C_{1}f(x)=\left[-\frac{d}{dx}+\frac{1}{2}x\right]f(x),A_{1}f(x)=\left[\frac{d}{dx}+\frac{1}{2}x\right]f(x)\text{for all}f\in {\mathcal{D}}_1.$$

(72)

We will prove below that C and C₁ (and A and A₁) are, in fact, equal.

For a function

$$f={\displaystyle \sum _{n\ge 0}{a}_n{\varphi }_n}\in L^2(\mathbf{R}),$$

we will use the notation f_N for the partial sum:

$$f_N{\displaystyle \sum _{n=0}^N{a}_n{\varphi }_n}.$$

Observe the following about the derivatives ${f^{\prime }}_N$:

Lemma 8. If$f\in \mathcal{D}(C)$, then$\left\{{f^{\prime }}_N\right\}$ is Cauchy in L²(R).

Proof. Note that

$$\begin{array}{l}{f^{\prime }}_N=\left(\frac{A-C}{2}\right)f_N.\\ \mathrm{So}{\Vert {f^{\prime }}_N-{f^{\prime }}_M\Vert }_{L^2}\le \frac{1}{2}{\Vert A{f}_N-A{f}_M\Vert }_{L^2}+{\Vert C{f}_N-C{f}_M\Vert }_{L^2}.\end{array}$$

Now for M < N we have

$$\begin{array}{c}{\Vert A{f}_N-A{f}_M\Vert }_{L^2}^2={\Vert {\displaystyle \sum _{n=M+1}^N{a}_n\sqrt{n}{\varphi }_{n-1}}\Vert }_{L^2}^2\\ ={\displaystyle \sum _{n=M+1}^N{\left|a_n\right|}^{2}n.}\end{array}$$

Likewise,

$$\begin{array}{c}{\Vert C{f}_N-C{f}_M\Vert }_{L^2}^2={\Vert {\displaystyle \sum _{n=M+1}^N{a}_n\sqrt{n+1}{\varphi }_{n+1}}\Vert }_{L^2}^2\\ ={\displaystyle \sum _{n=M+1}^N{\left|a_n\right|}^2(n+1).}\end{array}$$

Since $f\in \mathcal{D}(C)$, we know ${\displaystyle {\sum }_{n=M+1}^N|a_n{|}^2(n+1)}$ tends to 0 as M goes to infinity. Thus $\left\{{f^{\prime }}_N\right\}$ is Cauchy in L²(R). □

Lemma 9. If$f\in \mathcal{D}(C)$ then f is, up to equality almost everywhere, bounded, continuous and {f_N} converges uniformly to f, i.e., ║f−f_N║_sup→0 as N→∞.

Proof. It is enough to show that ║f−f_N║_sup→0 as M,N→∞. Note that

$$║f_M-f_N{║}_{\sup }\le 2^{-1/2}\left(║f_N-f_M{║}_{L^2}+║{f^{\prime }}_N-{f^{\prime }}_M{║}_{L^2}\right)$$

by Equation (70). Since f ∈ L²(R) we have $║f_N-f_M{║}_{L^2}\to 0$ as M, N → ∞ and by Lemma 8 we have that $║{f^{\prime }}_N-{f^{\prime }}_M{║}_{L^2}\to 0$ as M, N → ∞ Therefore {f_N} converges uniformly to f. □

Next we establish an integration-by-parts formula:

Lemma 10. If f,g ∈ L²(R) are differentiable with derivatives also in L²(R) then

$${\displaystyle {\int }_{\mathbf{R}}{f}^{\prime }(x)g(x)dx}=-{\displaystyle {\int }_{\mathbf{R}}f(x)g^{\prime }(x)dx}.$$

(73)

Proof. The derivative of fg, being f′g + fg′, is in L¹. So the fundamental theorem of calculus applies to give:

$${\displaystyle {\int }_a^b{f}^{\prime }(x)g(x)dx+}{\displaystyle {\int }_a^{b}f(x)g^{\prime }(x)dx}=f(b)g(b)-(a)g(a)$$

(74)

for all real numbers a < b.

Now fg ∈ L¹, and so

$$\underset{N\to \infty }{\lim }{\displaystyle {\int }_N^{\infty }f(x)g(x)dx}=\underset{N\to \infty }{\lim }{\displaystyle {\int }_{-\infty }^{-N}f(x)g(x)dx}=0.$$

Consequently, there exist a_N < −N < N < b_N with

$$f(a_N)g(a_N)\to 0\mathrm{and}f(b_N)g(b_N)\to 0\mathrm{as}N\to \infty .$$

Plugging into Equation (74) we obtain the desired result. □

Next we have the first step to showing that C₁ equals C:

Lemma 11. If f is in the domain of C₁ then f is in the domain of C and

$$Cf=C_{1}fandAf=A_{1}f.$$

(75)

Proof. Let f be in the domain of C₁. Then we may assume that f is differentiable and both f and the derivative f′ are in L²(R). We have then

$$\begin{array}{l}\langle C_{1}f,{\varphi }_n\rangle ={\displaystyle {\int }_{\mathbf{R}}{\varphi }_n(x)\left[-\frac{d}{dx}+\frac{1}{2}x\right]f(x)}dx\\ ={\displaystyle {\int }_{\mathbf{R}}f(x)\left[\frac{d}{dx}+\frac{1}{2}x\right]{\varphi }_n(x)dx}\text{by Equation}\left(73\right)\\ =\sqrt{n}\langle f,{\varphi }_{n-1}\rangle \text{by Equation}\left(53\right).\end{array}$$

(76)

Then

$$\left|\right|C_{1}f|{|}^2={\displaystyle \sum _{n\ge 0}{\left|\langle C_{1}f,{\varphi }_n\rangle \right|}^2={\displaystyle \sum _{n\ge 1}n{\left|\langle f,{\varphi }_{n-1}\rangle \right|}^2.}}$$

(77)

Because this sum is finite, it follows that f is in the domain $\mathcal{D}\left(C\right)$ of C. Moreover,

$$\begin{array}{l}Cf={\displaystyle \sum _{n\ge 0}\sqrt{n+1}\langle f,{\varphi }_{n+1}\rangle }\\ ={\displaystyle \sum _{m\ge 0}\langle C_{1}f,{\varphi }_m\rangle {\varphi }_m}\text{by Equation}\left(76\right)\\ =C_{1}f.\end{array}$$

The argument showing Af = A₁f is similar. □

We can now prove:

Theorem 12. The operators C and C₁ are equal, and the operators A and A₁ are equal. Thus, a function f ∈ L²(R) is in the domain of C (which is the same as the domain of A) if and only if f is, up to equality almost everywhere, a differentiable function with derivative f′ also in L²(R) and with${\displaystyle {\int }_{\mathbf{R}}{\left|xf(x)\right|}^{2}dx<\infty }$.

Proof. In view of Lemma 11, it will suffice to prove that $\mathcal{D}(C)\subset \mathcal{D}(C_1)$. Let f ∈ $\mathcal{D}\left(C\right)$. Then

$${\displaystyle \sum _{n\ge 0}n{\left|\langle f,{\varphi }_n\rangle \right|}^2<\infty .}$$

This implies that the sequences {C₁f_N}_N≥0 and {A₁f_N}_N≥0 are Cauchy, where f_N is the partial sum

$$f_N={\displaystyle \sum _{n=0}^N\langle f,{\varphi }_n\rangle {\varphi }_n}.$$

Now

$$\frac{1}{2}\left(A_1-C_1\right)f_N=f_N^{\prime },\mathrm{and}\frac{1}{2}\left(A_1+C_1\right)f_N(x)=x{f}_N(x).$$

So the sequences of functions ${\left\{f_N^{\prime }\right\}}_{N\ge 0}$ and {h_N}_N≥0, where

$$h_N(x)=x{f}_N(x),$$

are also L²–Cauchy. Now, as shown in Lemma 9, we can take f to be the uniformly convergent pointwise limit of the sequence of continuous functions f_N.

By Lemma 8, the sequence of derivatives $f_N^{\prime }$ is Cauchy in L²(R). Let $g={\lim }_{N\to \infty }{f}_N^{\prime }$ in L²(R). Observe that

$$f_N(y)=f_N(x)+{\displaystyle {\int }_0^1{f}_N^{\prime }\left((x+t(y-x)\right)(y-x)dt.}$$

(78)

Now ${{\displaystyle {\int }_0^1|f_N^{\prime }(x+t(y-x))(y-x)-g(x+t(y-x))(y-x)|dt\le \sqrt{\left|y-x\right|}\Vert f_N^{\prime }-g\Vert }}_{L^2}$ by the Cauchy-Schwartz inequality. Since ${\Vert f_N^{\prime }-g\Vert }_{L^2}\to 0$ as N → ∞, we have

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

Because ${\left\{f_N\right\}}_{N\ge 0}$ converges to f uniformly by Lemma 9, taking the limit as N → ∞ in Equation (78) we obtain

$$f(y)=f(x)+{\displaystyle {\int }_0^{1}g\left((x+t(y-x)\right)(y-x)dt.}$$

Therefore f′ = g ∈ L²(R). Lastly, we have, by Fatou’s Lemma:

$${\displaystyle {\int }_{\mathbf{R}}{\left|xf(x)\right|}^{2}dx\le }\underset{N\to \infty }{\lim \inf }{\displaystyle {\int }_{\mathbf{R}}{\left|x{f}_N(x)\right|}^{2}dx<\infty ,}$$

because the sequence {g_N}_N≥0 is convergent. Thus we have established that $f\in \mathcal{D}(C_1)$. □

Finally we can characterize the space S_p(R):

Theorem 13. Suppose f ∈ S_p(R), where p ≥ 1. Then f is (up to equality almost every where) a 2p times differentiable function and

$$\underset{x\in \mathbf{R}}{\sup }\left|x^a{f}^{(b)}(x)\right|<\infty$$

for every a, b ∈ {0, 1, 2, …} with a + b < 2p. Moreover, S_p(R) consists of all 2p times differentiable functions for which the functions x ⟼ x^af^(b)(x) are in L²(R) for every a, b ∈ {0, 1, 2, …} with a + b ≤ 2p.

Proof. Consider f ∈ S₁(R). Then

$$f={\displaystyle \sum _{n\ge 0}{a}_n{\varphi }_n}\mathrm{with}{\sum }_{n\ge 0}{n}^2{\left|a_n\right|}^2<\infty .$$

(79)

In particular, $f\in \mathcal{D}(C)$. Moreover,

$$Cf={\displaystyle \sum _{n\ge 0}{a}_n\sqrt{n+1}{\varphi }_{n+1}}Af={\displaystyle \sum _{n\ge 0}{a}_n\sqrt{n{\varphi }_{n-1}}}.$$

From these expressions and Equation (79) it is clear that Cf and Af both belong to $\mathcal{D}(C)$. Thus,

$$B_1{B}_{2}f\in L^2(\mathbf{R})\text{for all}{B}_2\in \{C,A,I\}.$$

Similarly, we can check that if f ∈ S_p(R), where p ≥ 2, then

$$B_1{B}_{2}f\in S_{p-1}(\mathbf{R})\text{for all}{B}_1,B_2\in \{C,A,I\}.$$

Thus, inductively, we see that

$$B_1\cdots B_{2p}f\in L^2(\mathbf{R})\text{for all}{B}_1,\dots ,B_{2p}\in \{C,A,I\}.$$

(This really means that f is in the domain of each product operator B₁ ···B_2p.) Now the operators $\frac{d}{dx}$ and multiplication by x are simple linear combinations of A and C. So for any a, b ∈ {0, 1, 2, …} with a + b ≤ 2p we can write the operator $x^a{\left(\frac{d}{dx}\right)}^b$ as a linear combination of operators B₁…B_2p with B₁, …, B_2p ∈ {C, A, I}.

Conversely, suppose f is 2p times differentiable and the functions x ⟼ x^af^(b)(x) are in L²(R) forevery a, b ∈ {0, 1, 2, … } with a + b ≤ 2p. Then f is in the domain of C^2p and so

$${\displaystyle \sum _{n\ge 0}{\left|\langle f,{\varphi }_n\rangle \right|}^2{n}^{2p}}<\infty .$$

Thus f ∈ S_p(R).

The preceding facts show that if f ∈ S_p(R) then for every B₁, …, B_2p ∈ {C, A, I}, the element B₁ ····B_2p−1f is in the domain of C, and so, in particular, is bounded. Thus,

$$\underset{x\in \mathbf{R}}{\sup }\left|x^a{f}^{(b)}(x)\right|<\infty$$

for all a, b ∈ {0, 1, 2, …} with a + b ≤ 2p − 1. □

We do not carry out a similar study for S_p(R^d), but from the discussions in the following sections, it will be clear that:

• S_p(R^d) is a Hilbert space with inner-product given by

  $${\langle f,g\rangle }_p=\langle f,T_d^{2p}g\rangle$$
• as a Hilbert space, S_p (R^d) is the d–fold tensor product of S_p(R) with itself.

8. Inner-Products on S(R) from N

For f ∈ L²(R), define

$${\Vert f\Vert }_t={\left\{{\displaystyle \sum _{n\ge 0}{\left(n+1\right)}^{2t}{\left|\langle f,{\varphi }_n\rangle \right|}^2}\right\}}^{1/2}$$

(80)

for every t > 0. More generally, define

$${\langle f,g\rangle }_t={\displaystyle \sum _{n\ge 0}{\left(n+1\right)}^{2t}}{\langle f,{\varphi }_n\rangle }_{L^2}{\langle {\varphi }_n,g\rangle }_{L^2},$$

(81)

for all f, g in the subspace of L²(R) consisting of functions F for which ||F ||_t < ∞.

Theorem 14. Let f ∈ S(R). Then for every t > 0 we have ||f||_t < ∞. Moreover, for every integer m ≥ 0, we also have

$$N^{m}f={\displaystyle \sum _{n\ge 0}{n}^m}\langle f,{\varphi }_n\rangle {\varphi }_n,$$

(82)

where on the left N^m is the differential operator$-\frac{d^2}{d{x}^2}+\frac{x^2}{4}-\frac{1}{2}$ applied n times, and on the right the series is taken in the sense of L²(R, dx). Furthermore,

$${\Vert f\Vert }_{m/2}^2=\langle f,{\left(N+1\right)}^{m}f\rangle .$$

(83)

This result will be strengthened and a converse proved later.

Proof. Let m ≥ 0 be an integer. Since f ∈ S(R), it is readily seen that N f is also in S(R), and thus, inductively, so is N^mf. Then we have

$$\begin{array}{l}\langle f,N^{m}f\rangle ={\displaystyle \sum _{n\ge 0}\langle f,{\varphi }_n\rangle \langle {\varphi }_n,N^{m}f\rangle }\\ ={\displaystyle \sum _{n\ge 0}\langle f,{\varphi }_n\rangle \langle N^m{\varphi }_n,f\rangle }\text{by Equation}\left(62\right)\\ ={\displaystyle \sum _{n\ge 0}\langle f,{\varphi }_n\rangle \langle n^m{\varphi }_n,f\rangle }\\ ={\displaystyle \sum _{n\ge 0}{n}^m}{\left|\langle f,{\varphi }_n\rangle \right|}^2.\end{array}$$

Thus we have proven the relation

$$\langle f,N^{m}f\rangle ={\displaystyle \sum _{n\ge 0}{n}^m}{\left|\langle f,{\varphi }_n\rangle \right|}^2.$$

(84)

An exactly similar argument shows

$$\langle f,{\left(N+1\right)}^{m}f\rangle ={\displaystyle \sum _{n\ge 0}{\left(n+1\right)}^m}{\left|\langle f,{\varphi }_n\rangle \right|}^2={\Vert f\Vert }_{m/2}^2.$$

(85)

So if t > 0, choosing any integer m ≥ t we have

$${\Vert f\Vert }_{t/2}^2\le {\Vert f\Vert }_{m/2}^2=\langle f,{\left(N+1\right)}^{m}f\rangle <\infty .$$

Observe that the series

$${\displaystyle \sum _{n\ge 0}{n}^m\langle f,{\varphi }_n\rangle {\varphi }_n}$$

(86)

is convergent in L²(R, dx) since

$${\displaystyle \sum _{n\ge 0}{n}^{2m}{\left|\langle f,{\varphi }_n\rangle \right|}^2=\langle N^{2m}f,f\rangle }<\infty .$$

So for any g ∈ L²(R, dx) we have, by an argument similar to the calculations done above:

$$\begin{array}{l}\langle N^{m}f,g\rangle ={\displaystyle \sum _{n\ge 0}{n}^m\langle f,{\varphi }_n\rangle \langle {\varphi }_n,g\rangle }\\ ={\displaystyle \sum _{n\ge 0}\langle n^m\langle f,{\varphi }_n\rangle {\varphi }_n,g\rangle }\\ =\langle {\displaystyle \sum _{n\ge 0}{n}^m\langle f,{\varphi }_n\rangle {\varphi }_n,g}\rangle .\end{array}$$

This proves the statement about N^mf. □

We have similar observations concerning C^mf and A^mf. First observe that since C and A are operators involving $\frac{d}{dx}$ and x, they map S(R) into itself. Also,

$$\langle Af,g\rangle =\langle f,Cg\rangle ,$$

for all f, g ∈ S(R), as already noted. Using this, for f ∈ S(R), we have

$$\begin{array}{l}\langle {\varphi }_{n+m},C^{m}f\rangle =\langle A^m{\varphi }_{n+m},f\rangle \\ =\sqrt{(n+m)(n+m-1)\cdots (n+1)}\langle {\varphi }_n,f\rangle .\end{array}$$

Therefore,

$$C^{m}f={\displaystyle \sum _{n\ge 0}{\left[\frac{(n+m)!}{n!}\right]}^{1/2}\langle f,{\varphi }_n\rangle {\varphi }_{n+m}.}$$

(87)

Similarly,

$$A^{m}f={\displaystyle \sum _{n\ge 0}{\left[\frac{n!}{(n-m)!}\right]}^{1/2}\langle f,{\varphi }_n\rangle {\varphi }_{n-m}.}$$

(88)

More generally, if B₁, …, B_k are such that each B_i is either A or C then

$$B_{1\cdots }{B}_{k}f={\displaystyle \sum _{n\ge 0}{\theta }_{n,k}\langle f,{\varphi }_n\rangle {\varphi }_{n+r},}$$

(89)

where the integer r is the excess number of C’s over the A’s in the sequence B₁, …, B_k, and θ_n,k is a real number determined by n and k. We do have the upper bound

$${\theta }_{n,k}^2\le {\left(n+k\right)}^k\le {\left[\left(n+1\right)k\right]}^k={\left(n+1\right)}^k{k}^k.$$

(90)

Note also that

$${\Vert B_{1\cdots }{B}_{k}f\Vert }^2=\langle {\left(B_{1\cdots }{B}_k\right)}^{*}{B}_{1\cdots }{B}_{k}f,f\rangle ={\displaystyle \sum _{n\ge 0}{\theta }_{n,k}^2{\left|\langle f,{\varphi }_n\rangle \right|}^2}.$$

(91)

Let’s look at the case of R^d. The functions ϕ_n generate an orthonormal basis by tensor products. In more detail, if a ∈ W^d is a multi-index, define ϕ_a ∈ L²(R^d) by

$${\varphi }_a(x)={\varphi }_{a_1}(x_1)\dots {\varphi }_{a_d}(x_d).$$

Now, for each t > 0, and f ∈ L²(R^d), define

$${\Vert f\Vert }_t\stackrel{\mathrm{def}}{=}{\left\{{{\displaystyle \sum _{a\in W^d}\left[\left(a_1+1\right)\dots \left(a_d+1\right)\right]}}^{2t}{\left|\langle f,{\varphi }_a\rangle \right|}^2\right\}}^{1/2},$$

(92)

and then define

$${\langle f,g\rangle }_t={\displaystyle \sum _{a\in W^d}{\left[\left(a_1+1\right)\dots \left(a_d+1\right)\right]}^{2t}\langle f,{\varphi }_a\rangle \langle {\varphi }_a,g\rangle },$$

(93)

for all f, g in the subspace of L²(R^d) consisting of functions F for which ║F║_t < ∞.

Let T_d be the operator on $\mathcal{S}\left({\mathbf{R}}^d\right)$ given by

$$T_d=\left(N_d+1\right)\dots \left(N_1+1\right).$$

Then, for every non-negative integer m, we have

$${\Vert f\Vert }_{m/2}^2=\langle f,T_d^{m}f\rangle .$$

The other results of this section also extend in a natural way to R^d.

9. L²–Type Norms on$\mathcal{S}\left(\mathbf{R}\right)$

For integers a, b ≥ 0, and f ∈ $\mathcal{S}\left(\mathbf{R}\right)$, define

$${\Vert f\Vert }_{a,b,2}={\Vert x^a{D}^{b}f\left(x\right)\Vert }_{L^2\left(\mathbf{R},dx\right)}.$$

(94)

Recall the operators

$$A=\frac{d}{dx}+\frac{1}{2}x,C=-\frac{d}{dx}+\frac{1}{2}x,N=CA$$

and the norms

$${\Vert f\Vert }_m=\langle f,{\left(N+1\right)}^{m}f\rangle .$$

The purpose of this section is to prove the following:

Theorem 15. The system of semi-norms given by ║f║_a,b,2 and the system given by the norms ║f║_m generate the same topology on$\mathcal{S}\left(\mathbf{R}\right)$.

Proof. Let a, b be non-negative integers. Then

$$\begin{array}{l}{\Vert f\Vert }_{a,b,2}={\Vert {\left(A+C\right)}^a{2}^{-b}{\left(A-C\right)}^{b}f\Vert }_{L^2}\\ \le \mathrm{a}\mathrm{linear}\mathrm{combination}\mathrm{of}\mathrm{terms}\\ \text{of the}\mathrm{form}{\Vert B_1\dots B_{k}f\Vert }_{L^2},\end{array}$$

where each B_i is either A or C, and k = a + b. Writing c_n = ⟨f, ϕ_n⟩, we have where

$$\begin{array}{l}{\Vert B_1\dots B_{k}f\Vert }_{L^2}^2=\Vert {\displaystyle \sum _{n\ge 0}{c}_n{\theta }_{n,k}{\varphi }_{n+r}}\Vert \\ ={\displaystyle \sum _{n\ge 0}{\left|c_n\right|}^2{\theta }_{n,k}^2},\end{array}$$

where

$$r=\#\left\{j:B_j=C\right\}-\#\left\{j:B_j=A\right\},$$

and, as noted earlier in Equation (90),

$${\theta }_{n,k}^2\le \left(n+1\right){}^{k}k{}^k.$$

So

$${\Vert B_1\dots B_{k}f\Vert }_{L^2}^2\le {{\displaystyle \sum _{n\ge 0}\left|c_n\right|}}^2\left(n+1\right){}^{k}k{}^k=k^k{\Vert f\Vert }_{k/2}^2\le k^k{\Vert f\Vert }_k^2.$$

(95)

Thus ║f║_a,b,2 is bounded above by a multiple of the norm ║f║_a+b.

It follows, that the topology generated by the semi-norms || · ||_a,b,2 is contained in the topology generated by the norms || · ||_k.

Now we show the converse inclusion. From

$${\Vert f\Vert }_k^2={\langle f,{\left(N+1\right)}^{2k}f\rangle }_{L^2}\le {\Vert f\Vert }_{L^2}^2+{\Vert {\left(N+1\right)}^{2k}f\Vert }_{L^2}^2$$

and the expression of N as a differential operator we see that ${\Vert f\Vert }_k^2$ is bounded above by a linear combination of ${\Vert f\Vert }_{a,b,2}^2$ for appropriate a and b. It follows then that the topology generated by the norms ║·║_k is contained in the topology generated by the semi-norms ║·║_a,b,2. □

Now consider R^d. Let a, b ∈ W^d be multi-indices, where W = {0, 1, 2, …}. Then for f ∈ S(R^d) define

$${\Vert f\Vert }_{a,b,2}=\left\{{\displaystyle \int {}_{R^d}}{\left|x^a{D}^{b}f\left(x\right)\right|}^{2}dx\right\}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$$

These specify semi-norms and they generate the same topology as the one generated by the norms || · ||_m, with m ∈ W. The argument is a straightforward modification of the one used above.

10. Equivalence of the Three Topologies

We will demonstrate that the topology generated by the family of norms || · ||_k, or, equivalently, by the semi-norms ║ · ║_a,b,2, is the same as the Schwartz topology on. $\mathcal{S}\left(\mathbf{R}\right)$.

Recall from Equation (70) that we have, for f ∈ $\mathcal{S}\left(\mathbf{R}\right)$

$${\Vert f\Vert }_{\sup }\le 2^{-1/2}\left({\Vert f\Vert }_{L^2}+{\Vert f^{\prime }\Vert }_{L^2}\right).$$

Putting in x^aD^bf(x) in place of f(x) we then have

$${\Vert f\Vert }_{a,b}\le \mathrm{a}\mathrm{linear}\mathrm{combination}\mathrm{of}{\Vert f\Vert }_{a,b,2},\mathrm{and}{\Vert f\Vert }_{a,b+1,2}.$$

(96)

Next we bound the semi-norms ║f║_a,b,2 by the semi-norms ║f║_a,b. To this end, observe first

$$\begin{array}{l}{\Vert f\Vert }_{L^2}^2={{\displaystyle {\int }_{\mathbf{R}}\left(1+x^2\right)}}^{-1}\left(1+x^2\right){\left|f\left(x\right)\right|}^{2}dx\\ \le \pi {\Vert \left(1+x^2\right){\left|f\left(x\right)\right|}^2\Vert }_{\sup }\\ \le \pi \left({\Vert f\Vert }_{\sup }^2+{\Vert xf\left(x\right)\Vert }_{\sup }^2\right)\\ \le \pi {\left({\Vert f\Vert }_{\sup }+{\Vert xf\left(x\right)\Vert }_{\sup }\right)}^2.\end{array}$$

So for any integers a, b ≥ 0, we have

$${\Vert f\Vert }_{a,b,2}={\Vert x^a{D}^{b}f\Vert }_{L^2}\le {\pi }^{1/2}\left({\Vert f\Vert }_{a,b}+{\Vert f\Vert }_{a+1,b}\right).$$

(97)

Thus, the topology generated by the semi-norms ║ · ║_a,b,2 coincides with the Schwartz topology.

Now lets look at the situation for R^d. The same result holds in this case and the arguments are similar. The appropriate Sobolev inequalities require using (1 + |p|²)^d instead of 1 + p². For $f\in \mathcal{S}\left({\mathbf{R}}^d\right)$, we have the Fourier transform given by

$$ℱ\left(f\right)\left(p\right)=\widehat{f}\left(p\right)={\left(2\pi \right)}^{-d/2}{\displaystyle {\int }_{{\mathbf{R}}^d}{e}^{-i\langle p,x\rangle }}f\left(x\right)dx.$$

(98)

Again, this preserves the L² norm, and transforms derivatives into multiplications:

$$p_i\widehat{f}\left(p\right)=-iℱ\left(\frac{\partial f}{\partial x_i}\right)(p).$$

Repeated application of this shows that

$${\left|p\right|}^2\widehat{f}\left(p\right)=-ℱ\left(\mathrm{\Delta }f\right)\left(p\right),$$

(99)

where

$$\mathrm{\Delta }={\displaystyle \sum _{j=1}^n\frac{{\partial }^2}{\partial x_j^2}}$$

is the Laplacian. Iterating this gives, for each r ∈ {0, 1, 2, …} and f ∈ $\mathcal{S}\left({\mathbf{R}}^d\right)$,

$${\left|p\right|}^{2r}\widehat{f}\left(p\right)={\left(-1\right)}^rℱ\left({\mathrm{\Delta }}^{r}f\right)\left(p\right),$$

(100)

which in turn implies, by the Plancherel formula Equation (67), the identity:

$${{\displaystyle {\int }_{{\mathbf{R}}^d}\left|{\left|p\right|}^{2r}\left|\widehat{f}\left(p\right)\right|\right|}}^{2}dp={{\displaystyle {\int }_{{\mathbf{R}}^d}\left|{\mathrm{\Delta }}^{r}f\left(x\right)\right|}}^{2}dx.$$

(101)

Then we have, for any m > d/4,

$$\begin{array}{l}{\Vert f\Vert }_{\sup }\le {\left(2\pi \right)}^{-d/2}{\displaystyle {\int }_{{\mathbf{R}}^d}\left|\widehat{f}\left(p\right)\right|}dp\\ ={\left(2\pi \right)}^{-d/2}{\displaystyle {\int }_{{\mathbf{R}}^d}{\left(1+{\left|p\right|}^2\right)}^m}\left|\widehat{f}\left(p\right)\right|{\left(1+{\left|p\right|}^2\right)}^{-m}dp\\ =K{\left[{\displaystyle {\int }_{{\mathbf{R}}^d}{\left(1+{\left|p\right|}^2\right)}^{2m}}{\left|\widehat{f}\left(p\right)\right|}^{2}dp\right]}^{1/2}\mathrm{by}\text{Cauchy-Schwartz},\end{array}$$

where $K={\left(2\pi \right)}^{-d/2}{\left[{\displaystyle {\int }_{{\mathbf{R}}^d}\frac{dp}{{\left(1+|p{|}^2\right)}^{2m}}}\right]}^{1/2}<\infty$ The function (1 + s)ⁿ/(1 + sⁿ), for s ≥ 0, attains a maximum value of 2ⁿ⁻¹, and so we have the inequality (1 + s)^2m ≤ 2^2m−1(1 + s^2m), which leads to

$${\left(1+{\left|p\right|}^2\right)}^{2m}\le 2^{2m-1}\left(1+{\left|p\right|}^{4m}\right).$$

Then, from Equation (102), we have

$${\Vert f\Vert }_{\sup }^2\le K^2{2}^{2m-1}\left({\Vert f\Vert }_{L^2}^2+{\Vert {\mathrm{\Delta }}^{m}f\Vert }_{L^2}^2\right).$$

(102)

This last quantity is clearly bounded above by a linear combination of ║f║_0,b,2 for certain multi-indices b. Thus ║f║_sup is bounded above by a linear combination of ║f║_0,b,2 for certain multi-indices b. It follows that ║x^a D^bf║_sup is bounded above by a linear combination of ║f║_a′,_b′,₂ for certain multi-indices a′, b′.

For the inequality going the other way, the reasoning used above for Equation (97) generalizes readily, again with (1 + x²) replaced by (1 + |x|²)^d. Thus, on $\mathcal{S}\left({\mathbf{R}}^d\right)$ the topology generated by the family of semi-norms ║ · ║_a,b,2 coincides with the Schwartz topology.

Now we return to Equation (102) for some further observations. First note that

$$\mathrm{\Delta }=\frac{1}{4}{\displaystyle \sum _{j=1}^d{\left(C_j-A_j\right)}^2}$$

and so Δ^m consists of a sum of multiples of (3d)^m terms each a product of 2m elements drawn from the set {A₁, C₁,…, A_d, C_d}. Consequently, by Equation (95)

$${\Vert {\mathrm{\Delta }}^{m}f\Vert }_{L^2}^2\le c_{d,m}^2{\Vert f\Vert }_m^2,$$

(103)

for some positive constant c_d,m. Combining this with Equation (102), we see that for m > d/4, there is a constant k_d,m such that

$${\Vert f\Vert }_{\sup }\le k_{d,m}{\Vert f\Vert }_m$$

(104)

holds for all $f\in \mathcal{S}\left({\mathbf{R}}^d\right)$.

Now consider $f\in {\mathcal{S}}_p\left({\mathbf{R}}^d\right)$, with p > d/4. Let

$$f_N={\displaystyle \sum _{j\in W^d,\left|j\right|\le N}\langle f,{\varphi }_j\rangle }{\varphi }_j.$$

Then f_N → f in L² and so a subsequence ${\left\{f_{N_k}\right\}}_{k\ge 1}$ converges pointwise almost everywhere to f. It follows then that the essential supremum ║f║_∞ is bounded above as follows:

$${\Vert f\Vert }_{\infty }\le \underset{N\to \infty }{\lim \sup }{\left|f_N\right|}_{\sup }.$$

Note that f_N → f also in the ║ · ║_p–norm. It follows then from Equation (104) that

$${\Vert f\Vert }_{\infty }\le k_{d,p}{\Vert f\Vert }_p$$

(105)

holds for all $f\in {\mathcal{S}}_p\left({\mathbf{R}}^d\right)$ with p > d/4. Replacing f by the difference f − f_N in Equation (105), we see that f is the L^∞–limit of a sequence of continuous functions which, being Cauchy in the sup-norm, has a continuous limit; thus f is a.e. equal to a continuous function, and may thus be redefined to be continuous.

11. Identification of $\mathcal{S}\left(\mathbf{R}\right)$ with a Sequence Space

Suppose a₀, a₁, … form a sequence of complex numbers such that

$${{\displaystyle \sum _{n\ge 0}{\left(n+1\right)}^m\left|a_n\right|}}^2<\infty ,\mathrm{for}\mathrm{every}\mathrm{integer}m\ge 0.$$

(106)

We will show that the sequence of functions given by

$$s_n={\displaystyle \sum _{j=0}^n{a}_j{\varphi }_j}$$

converges in the topology of $\mathcal{S}\left(\mathbf{R}\right)$ to a function $f\in \mathcal{S}\left(\mathbf{R}\right)$ for which a_n = ⟨f, ϕ_n⟩ for every n ≥ 0.

All the hard work has already been done. From Equation (106) we see that (s_n)_n≥0 is Cauchy in each norm ║·║_m. So it is Cauchy in the Schwartz topology of $\mathcal{S}\left(\mathbf{R}\right)$, and hence convergent to some $f\in \mathcal{S}(\mathbf{R})$. In particular, s_n → f in L². Taking inner-products with ϕ_j we see that a_j = ⟨f, ϕ_j⟩.

Thus we have

Theorem 16. Let W = {0, 1, 2, …}, and define

$$F:L^2\left(\mathbf{R}\right)\to {\mathbf{C}}^W$$

by requiring that

$$F{\left(f\right)}_n={\langle f,{\varphi }_n\rangle }_{L^2}$$

for all n ∈ W. Then the image of$\mathcal{S}\left(\mathbf{R}\right)$ under F is the set of all a ∈ C^W for which${\Vert a\Vert }_m^2\stackrel{\mathrm{def}}{=}{{\displaystyle {\sum }_{n\ge 0}{\left(n+1\right)}^{2m}\left|a_n\right|}}^2<\infty$ for every integer m ≥ 0. Moreover, if$F\left(\mathcal{S}\left(\mathbf{R}\right)\right)$ is equipped with the topology generated by the norms ║ · ║_m then F is a homeomorphism.