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

Transcription

1 Mathematics 2015, 3, ; doi: /math Article OPEN ACCESS mathematics ISSN The Schwartz Space: Tools for Quantum Mechanics and Infinite Dimensional Analysis Jeremy Becnel 1, * and Ambar Sengupta 2 1 Department of Mathematics, Stephen F. Austin State University, PO Box SFA Station, Nacogdoches, TX 75962, USA 2 Department of Mathematics, Louisiana State University, Baton Rouge, LA 70803, USA; sengupta@math.lsu.edu * Author to whom correspondence should be addressed; becneljj@sfasu.edu; Tel.: ; Fax: Academic Editor: Palle E. T. Jorgensen Received: 17 April 2015 / Accepted: 3 June 2015 / Published: 16 June 2015 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. Keywords: quantum mechanics; Schwartz space; test functions 1. Introduction In , 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 = d2 dx + x (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.

2 Mathematics 2015, 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 S(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 S(R d ), and its topology, as a decreasing intersection of subspaces S p (R d ), for p {0, 1, 2,...}: S(R d ) = p 0 S p (R d ) S 2 (R d ) S 1 (R d ) L 2 (R d ) an exact characterization of the functions in the space S p (R); summary of notions from spectral theory and quantum mechanics; Our exposition of the properties of T and of S(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,...}. (2) 2.1. The Schwartz Space The Schwartz space S(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) def = sup x a f b (x) < x R

3 Mathematics 2015, for all a, b W = {0, 1, 2,...}. The finiteness condition for all a 1 and b W, implies that x a f 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 The Schwartz Topology and The functions p a,b are semi-norms on the vector space S(R), in the sense that p a,b (f + g) p a,b (f) + p a,b (g) p a,b (zf) = z p a,b (f) for all f, g S(R), and z C. For this semi-norm, an open ball of radius r centered at some f S(R) is given by B pa,b (f; r) = {g S(R) : p a,b (g f) < r}. (3) Thus each p a,b specifies a topology τ a,b on S(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 S(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 S(R) that is very useful, which we describe in detail below The Operator T The operator T = d2 dx + x (4) 2 plays a very useful role in working with the Schwartz space. As we shall see, there is an orthonormal basis {φ n } n W of L 2 (R, dx), consisting of eigenfunctions φ n of T : T φ n = (n + 1)φ n. (5) The functions φ n, called the Hermite functions are actually in the Schwartz space S(R). Let B be the bounded linear operator on L 2 (R) given on each f L 2 (R) by It is readily checked that the right side does converge and, in fact, Bf 2 L 2 = n W Bf = n W(n + 1) 1 f, φ n φ n. (6) (n + 1) 2 f, φ n 2 L 2 n W f, φ n 2 L = 2 f 2 L2. (7) Note that B and T are inverses of each other on the linear span of the vectors φ n : T Bf = f and BT f = f for all f L, (8) where L = linear span of the vectors φ n, for n W. (9)

4 Mathematics 2015, The L 2 Approach For any p 0, the image of B p consists of all f L 2 (R) for which (n + 1) 2p f, φ n 2 <. Let n W S p (R) = B p ( L 2 (R) ). (10) This is a subspace of L 2 (R), and on S p (R) there is an inner-product, p given by f, g p def = n W(n + 1) 2p f, φ n φ n, g = B p f, B p g L 2 (11) which makes it a Hilbert space, having L, and hence also S(R), a dense subspace. We will see later that functions in S p (R) are p-times differentiable. We will prove that the intersection p W S p (R) is exactly equal to S(R). In fact, S(R) = S p (R) S 2 (R) S 1 (R) S 0 (R) = L 2 (R). (12) p W We will also prove that the topology on S(R) generated by the norms p coincides with the standard topology. Furthermore, the elements (n + 1) p φ n S p (R) form an orthonormal basis of S p (R), and n W (n + 1) (p+1) φ n 2 p = n 1 n 2p <, n2(p+1) showing that the inclusion map S p+1 (R) S p (R) is Hilbert-Schmidt. The topological vector space S(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 Coordinatization as a Sequence Space. All of the results described above follow readily from the identification of S(R) with a space of sequences. Let {φ n } n W be the orthonormal basis of L 2 (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 (R) C W : f ( f, φ n ) n W. (13) For each p W let E p be the subset of C W consisting of all (a n ) n W such that (n + 1) 2p a n 2 <. n W On E p define the inner-product, p by a, b p = n W(n + 1) 2p a n b n. (14)

5 Mathematics 2015, This makes E p a Hilbert space, essentially the Hilbert space L 2 (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 S p (R) shows that it is the set of all f L 2 (R) for which I(f) belongs to E p. We will prove in Theorem 16 that I maps S(R) exactly onto E def = p W E p. (15) This will establish essentially all of the facts mentioned above concerning the spaces S p (R). Note the chain of inclusions: E = p W 2.6. The Multi-Dimensional Setting E p E 2 E 1 E 0 = L 2 (W, µ 0 ). (16) In the multidimensional setting, the Schwartz space S(R d ) consists of all infinitely differentiable functions f on R d for which sup x k x k m 1+ +m kf(x) d d x m x m <, d d x R d for all (k 1,..., k d ) W d and m = (m 1,..., m d ) W d. standard notation: For this setting, it is best to use some k = k k d for k = (k 1,..., k d ) W d (17) D m = x k = x k 1...x k d (18) m x m x m. (19) d For the multi-dimensional case, we use the indexing set W d whose elements are d tuples j = (j 1,..., j d ), with j 1,..., j d W, and counting measure µ 0 on W d. The sequence space is replaced by C W d ; a typical element a C W d, is a map a : W d C : j = (j 1,..., j d ) a j = a j1...j d. (20) The orthonormal basis (φ n ) n W of L 2 (R) yields an orthonormal basis of L 2 (R d ) consisting of the vectors φ j = φ j1 φ jd : (x 1,..., x d ) φ j1 (x 1 )...φ jd (x d ). (21) The coordinatizing map I is replaced by the map d I d : L 2 (R d ) C W d (22) where I d (f) j = f, φ j L 2 (R d ). (23)

6 Mathematics 2015, Replace the operator T by T d def = T d = ( 2 x 2 d ) ) + x2 d ( 2 + x2 1 x (24) Then T d φ j = (j 1 + 1)...(j d + 1)φ j for all j W d. In place of B, we now have the bounded operator B d on L 2 (R d ) given by B d f = j W d [(j 1 + 1)...(j d + 1)] 1 f, φ j φ j (25) Again, T d and B d are inverses of each other on the linear subspace L d of L 2 (R d ) spanned by the vectors φ j. The space E p is now the subset of C W d consisting of all a C W d for which a 2 p def = j W d [(j 1 + 1)...(j d + 1)] 2p a j 2 <. (26) Thus, E p def = {a C W d : a 2 p < } (27) This is a Hilbert space with inner-product Again we have the chain of spaces a, b p = j W d [(j 1 + 1)...(j d + 1)] 2p a j b j. (28) E def = p W E p E 2 E 1 E 0 = L 2 (W d, µ 0 ), (29) with the inclusion E p+1 E p being Hilbert-Schmidt. To go back to functions on R d, define S p (R d ) to be the range of B d. Thus S p (R d ) is the set of all f L 2 (R d ) for which j W d [(j 1 + 1)...(j d + 1)] 2p f, φ j 2 <. The inner-product, p comes back to an inner-product, also denoted, p, on S p (R d ) and is given by f, g p = B p d f, B p d g L 2 (R d ). (30) The intersection p W S p (R d ) equals S(R d ). Moreover, the topology on S(R d ) is the smallest one generated by the inner-products obtained from, p, with p running over W.

7 Mathematics 2015, 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 V : y y + x are continuous, for every x V, and are hence also homeomorphisms since t 1 x = 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: If U is open and x U then x + U α U, for some α 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 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 1 x lies in B if t is large enough. The smallest value of t for which t 1 x 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 µ B (x) = inf{t > 0 : t 1 x 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 µ B (x + y) µ B (x) + µ 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

8 Mathematics 2015, 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} 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 µ(x + y) µ(x) + µ(y), µ(tx) = t µ(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 µ (x; r) = {y V : µ(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 µ(x) µ(y) µ(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 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 I τ 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 {W α,i } i Iα is a local base for τ α then a local base for τ is obtained by taking all finite intersections of the form W α1,i 1 W αn,i n. Proof. Let B be the collection of all sets which are of the form W α1,i 1 W αn,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: If C 1 and C 2 are sets in B, and x is a point in the intersection of the translates a + C 1 and b + C 2, (35) then x + C (a + C 1 ) (b + C 2 ) for some C in B.

9 Mathematics 2015, Clearly, it suffices to consider finitely many topologies τ α. Thus, consider vector topologies τ 1,..., τ n on V. Let B n be the collection of all sets of the form B 1 B n with B i in a local base for τ i, for each i {1,..., n}. We can check that if D, D B n then there is an E 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 B 1 B n. Then there exist sets B i, with each B i being in the local base for τ i, such that z + B i B i (this follows from our earlier observation Equation (31)). Consequently, z + n B i i=1 n B i. Now consider sets C 1 an C 2, both in B n. Consider a, b V and suppose x (a + C 1 ) (b + C 2 ). Then since x a C 1 there is a set C 1 B n with x a + C 1 C 1 ; similarly, there is a C 2 B n with x b + C 2 C 2. So x + C 1 a + C 1 and x + C 2 b + C 2. So i=1 x + C (a + C 1 ) (b + C 2 ), where C B n satisfies C C 1 C 2. 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 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 B with 0 B W. Suppose B = B 1 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 i τ i, both containing 0, with D i + D i B i. Then choose C i, C i in the local base for τ i with C i D i and C i D i. Then C i +C i B i. Now let C = C 1 C n, and C = C 1 C n. Then C, C B and C + C 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 tx + W, for some W B. Using continuity of the addition map V V V V : (a, b, c) a + b + c at (0, 0, 0), we can choose W 1, W 2, W 3 B with W 1 + W 2 + W 3 W. Then we can choose W B, such that W W 1 W 2 W 3.

10 Mathematics 2015, Then W B and W + W + W W. Suppose W = B 1 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 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 α I τ µ α. An open set in this topology is a union of translates of finite intersections of balls of the form B µi (0; r i ). Thus, any open neighborhood of f contains a set of the form B µ1 (f; r 1 ) B µn (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 = {µ j} j J. Suppose each µ i is bounded above by a linear combination of the µ j. Then τ τ. Proof. Let µ M. Then there exist µ 1,..., µ n M, and real numbers c 1,..., c n > 0, such that Now consider any x, y V. Then µ c 1 µ c n µ n. µ(x) µ(y) µ(x y) n c i µ i(x y). i=1 So µ is continuous with respect to the topology generated by µ 1,..., µ n. Thus, τ µ τ. Since this is true for all µ M, we have τ τ.

11 Mathematics 2015, 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 Metrizability Suppose the topology τ on the topological vector space V is generated by a countable family of semi-norms µ 1, µ 2,... For any x, y V define d(x, y) = n 1 2 n d n (x, y) (36) where d n (x, y) = min{1, µ n (x y)}. Then d is a metric, it is translation invariant, and generates the topology τ [10]. 4. The Schwartz Space S(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 S(R) is the set of all C functions f on R for which p a,b (f) def def = f a,b = sup x R x a D b f(x < (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 S(R). We will call this the Schwartz topology on S(R). Theorem 4. The topology on S(R) generated by the family of semi-norms a,b a, b {0, 1, 2,...}, is complete. for all

12 Mathematics 2015, Proof. Let (f n ) n 1 be a Cauchy sequence on S(R). Then this sequence is Cauchy in each of the semi-norms a,b, and so each sequence of functions x a D b f n (x) is uniformly convergent. Let g b (x) = lim n D b f n (x). (38) Let f = g 0. Using a Taylor theorem argument it follows that g b is D b f. For instance, for b = 1, observe first that 1 ( ) df n (1 t)x + ty) 1 ) f n (y) = f n (x) + dt = f n (x) + f dt n( (1 t)x + ty (y x) dt, and so, letting n, we have 0 f(y) = f(x) g 1 ( (1 t)x + ty ) (y x) dt, which implies that f (x) exists and equals g 1 (x). In this way, we have x a D b f n (x) x a D b f(x) pointwise. Note that our Cauchy hypothesis implies that the sequence of functions x a D b f n (x) is Cauchy in sup-norm, and so the convergence x a D b f n (x) x a D b f(x) is uniform. In particular, the sup-norm of x a D b f(x) is finite, since it is the limit of a uniformly convergent sequence of bounded functions. Thus f S(R). Finally, we have to check that f n converges to f in the topology of S(R). We have noted above that x a D b f n (x) x a D b f(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 S(R). Now let s take a quick look at the Schwartz space S(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,..., d} {0, 1, 2,...} : j a j. If a is a multi-index, we write a to mean the sum a a d, x a to mean the product x a x a d d, and D a to mean the differential operator D a 1 x 1...D a d x d. The space S(R d ) consists of all C functions f on R d such that each function x a D b f(x) is bounded. On S(R d ) we have the semi-norms f a,b = sup x R d x a D b f(x) for each pair of multi-indices a and b. The Schwartz topology on S(R d ) is the smallest topology making each semi-norm a,b continuous. This makes S(R d ) a topological vector space. The argument for the proof of the preceding theorem goes through with minor alterations and shows that: 0 Theorem 5. The topology on S(R d ) generated by the family of semi-norms a,b a, b {0, 1, 2,...} d, is complete. for all

13 Mathematics 2015, 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 Properties of translates of p are obtained from Expanding the right side in a Taylor series we have p(x) = 1 2π e x2 /2. (39) e xy y2 p(x y) 2 =. (40) p(x) e xy y2 2 = p(x y) p(x) = n=0 1 n! H n(x)y n, (41) where the Taylor coefficients, denoted H n (x), are H n (x) = 1 p(x) ( d ) n p(x). (42) dx This is the n th Hermite polynomial and is indeed an n th degree polynomial in which x n has coefficient 1, facts which may be checked by induction. Observe the following p(x y) p(x z) p(x)dx = e y2 +z 2 2 e x(y+z) p(x) dx R p(x) p(x) R = e y2 +z 2 + (y+z)2 2 2 = e yz. 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 Thus an orthonormal set of functions is given by H n, H m L 2 (p(x)dx) = n!δ nm. (43) h n (x) = 1 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 L 2 (p(x)dx). Thus the polynomials above constitute an orthonormal basis of L 2 (p(x)dx).

14 Mathematics 2015, Next, consider the derivative of H n : H n(x) = ( 1) n p(x) 1 p (n+1) (x) ( 1) n p(x) 1 p (x)p(x) 1 p n (x) = H n+1 (x) + xh n (x). So ( d ) dx + x h n (x) = n + 1 h n+1 (x). (45) The operator ( d ) dx + x is called the creation operator in L 2( R; p(x)dx ). Officially, we can take the creation operator to have domain consisting of all functions f which can be expanded in L 2 (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 n 0 n + 1 an 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 ( d dx + x) to f. Next, from the fundamental generating relation Equation (41) we have : ye xy y2 /2 1 1 = lim ɛ 0 n! ɛ [H n(x + ɛ) H n (x)] y n. (46) n 0 Using Equation (41) again on the left we have n 1 1 (n 1)! H n 1(x)y n 1 1 = lim ɛ 0 n! ɛ [H n(x + ɛ) H n (x)] y n. (47) n 0 Letting y = 0 allows us to equate the n = 0 terms, and then, successively, the higher order terms. From this we see that H n(x) = nh n 1 (x) (48) where H 1 = 0. Thus: The operator d dx h n(x) = n h n 1 (x). (49) d dx is the annihilation operator in L 2( 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.

15 Mathematics 2015, Hermite Functions, Creation and Annihilation Operators In the preceding section we studied Hermite polynomials in the setting of the Gaussian space L 2( R; p(x)dx ). Let us translate the concepts and results back to the usual space L 2( R; dx ). To this end, consider the isomorphism: U : L 2 (R, p(x)dx) L 2 (R, dx) : f pf. (50) Then the orthonormal basis polynomials h n go over to the functions φ n given by φ n (x) = ( 1) n 1 n! (2π) 1/4 e x2 /4 dn e x2 /2 dx n. (51) The family {φ n } n 0 forms an orthonormal basis for L 2 (R, dx). We now determine the annihilation and creation operators on L 2 (R, dx). If f L 2 (R, dx) is differentiable and has derivative f also in L 2 (R, dx), we have: ( U d ) dx U 1 f(x) = p(x) d [ p(x) 1/2 f(x) ] dx = f (x) + p(x) 1/2( 1/2 ) p(x) 3/2 p (x)f(x) So, on L 2 (R, dx), the annihilator operator is = f (x) xf(x). which will satisfy A = d dx x (52) Aφ n = n φ n 1 (53) where φ 1 = 0. For the moment, we proceed by taking the domain of A to be the Schwartz space S(R). Next, ( ( U d ) ) dx + x U 1 f(x) = f (x) + xf(x) 1 2 xf(x) ( = d dx + 1 ) 2 x f(x). Thus the creation operator is C = A = d dx + 1 x. (54) 2 The reason we have written A is that, as is readily checked, we have the adjoint relation ( Af, g = f, d dx + 1 ) 2 x g (55) with the inner-product being the usual one on L 2 (R, dx). Again, for the moment, we take the domain of C to be the Schwartz space S(R) (though, technically, in that case we should not write C as A, since the latter, if viewed as the L 2 adjoint operator, has a larger domain).

16 Mathematics 2015, For this we have Cφ n = n + 1 φ n+1. (56) Observe also that AC = 1 4 x2 d2 dx I and CA = 1 4 x2 d2 dx I (57) which imply: Next observe that [A, C] = AC CA = I, the identity. (58) CAφ n = n nφ n = nφ n (59) and so CA is called the number operator N: N = A A = CA = d2 dx 2 + x the number operator. (60) As noted above in Equation (59), the number operator N has the eigenfunctions φ n : Nφ n = nφ n. (61) Integration by parts (see Lemma 10) shows that f, g = f, g for every f, g S(R), and so also f, g = f, g. It follows that the operator N satisfies Nf, g = f, Ng (62) for every f, g S(R). Now consider the case of R d. For each j {1,..., d}, there are creation, annihilation, and number operators: A j = x j x j, C j = x j x j, N j = C j A j. (63) These map S(R d ) into itself and, as is readily verified, satisfy the commutation relations [A j, C k ] = δ jk I, [N j, A k ] = δ jk A j, [N j, C k ] = δ 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 L 2 (R) yields an orthonormal basis {φ m } m W d of L 2 (R d ) given by φ m = φ m1 (x 1 ) φ md (x d ). For convenience we say φ m = 0 if some m j < 0. Given its effect on the orthonormal basis {φ m } m W d, the operator C k has the form: φ m m k + 1φ m,

17 Mathematics 2015, where m i = m i for all i {1,..., d} except when i = k, in which case m k = m k + 1. The domain of C k is the set D(C k ) given by { } D(C k ) = f L 2 (R) (m k + 1) a m 2 < where a m = f, φ m. m W d The operator C k is then officially defined by specifying its action on a typical element of its domain: ( ) C k a m φ m = a m mk + 1φ m, (65) m W d m W d 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 h T is a closed operator. Suppose x n x. Since T is a bounded linear operator, T x n T x. Now suppose also that M h (T x n ) y. Since M h is closed, it follows then that T x D(M h ) and y = M h T x. The operators A k and N k are defined analogously. Proposition 6. Let L 0 be the vector subspace of L 2 (R d ) spanned by the basis vectors {φ m } m 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 Gr(C k L 0 ) (see to subsection A.1 for the notion of graph). It is clear that Gr(C k L 0 ) Gr(C k ). Using this and the fact that C k is a closed operator, we have Gr(C k L 0 ) Gr(C k ) = Gr(C k ). Going in the other direction, take (f, C k f) Gr(C k ). Now f = m W a d m φ m where a m = f, φ m. Let f N be given by f N = a m φ m where WN d = { m W d 0 m 1 N,..., 0 m d N}. m W d N Observe that f N L 0. Moreover lim f N = f and N lim C kf N = lim N N m WN d (m k + 1)a m φ m = C k f in L 2 (R d ). Thus (f, C k f) Gr(C k L 0 ) and so we have Gr(C k ) 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 S(R d ) then C k f = f x k + x k 2 f, and A kf = f x k + x k 2 f.

18 Mathematics 2015, Proof. Let g = f x k + x k 2 f. Since f S(R d ), we have g L 2 (R d ). So we can write g as g = j W a d j φ j where a j = g, φ j. Let us examine these a j s more closely. Observe a j = g, φ j = f + x k x k 2 f, φ j = f, φ j + x k x k 2 φ j = f, j k φ j 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 g = j W d = m W d jk f, φ j φ j mk + 1 f, φ m φ m = C k f by (65). where m is as defined above 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 S p (R) Our aim here is to obtain a complete characterization of the functions in S p (R). We will prove that S p (R) consists of all square-integrable functions f for which all derivatives f (k) exist for k {1, 2,..., p} and sup x a f (b) (x) < x R for all a, b {0, 1,..., p 1} with a + b p 1. A significant tool we will use is the Fourier transform: ˆf(p) = Ff(p) = (2π) 1/2 R e ipx f(x) dx. (66) This is meaningful whenever f is in L 1 (R), but we will work mainly with f in S(R). We will use the following standard facts: F maps S(R) onto itself and satisfies the Plancherel identity: f(x) 2 dx = ˆf(p) 2 dp (67) R R for any f S(R), if f S(R) then f(x) = (2π) 1/2 e ipx ˆf(p) dp (68) R p ˆf(p) = if(f )(p). (69)

19 Mathematics 2015, Consequently, we have f sup (2π) 1/2 ˆf(p) dp R = (2π) 1/2 (1 + p 2 ) 1/2 ˆf(p) (1 + p 2 ) 1/2 dp R [ 1/2 = (2π) 1/2 (1 + p 2 ) dp] ˆf(p) 2 π 1/2 by Cauchy-Schwartz R 2 [ 1/2 ˆf L 2 + p ˆf ] L 2 (R,dp) 2 1/2 ( f L 2 + f L 2) by Plancherel and Equation (69). (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 L 2 (R), with common domain D(C) = D(A) = {f L 2 (R) n f, φ n 2 < } n 0 and Cf = n 0 f, φ n n + 1φ n+1, Af = n 0 f, φ n nφ n 1. and Moreover, define operators C 1 and A 1 on the common domain C 1 f(x) = D 1 = all differentiable f L 2 with f L 2 and xf(x) L 2 (dx) (71) [ d dx + 1 ] [ d 2 x f(x), A 1 f(x) = dx + 1 ] 2 x f(x) for all f D 1. (72) We will prove below that C and C 1 (and A and A 1 ) are, in fact, equal. For a function f = n 0 a n φ n L 2 (R), we will use the notation f N for the partial sum: f N = Observe the following about the derivatives f N : N a n φ n. n=0 Lemma 8. If f D(C), then {f N } is Cauchy in L2 (R). Proof. Note that ( A C f N = 2 ) f N. So f N f M L Af N Af M L Cf N Cf M L 2.

20 Mathematics 2015, Now for M < N we have Likewise, Af N Af M 2 L 2 = Cf N Cf M 2 L 2 = = = N n=m+1 N n=m+1 N n=m+1 N n=m+1 a n nφn 1 2 a n 2 n. L 2 a n n + 1φn+1 2 a n 2 (n + 1). Since f D(C), we know N n=m+1 a n 2 (n + 1) tends to 0 as M goes to infinity. Thus {f N } is Cauchy in L 2 (R). Lemma 9. If f 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 M f N sup 0 as M, N. Note that f M f N sup 2 1/2 ( f N f M L 2 + f N f M L 2) by Equation (70). Since f L 2 (R) we have f N f M L 2 0 as M, N and by Lemma 8 we have that f N f M L 2 0 as M, N. Therefore {f N } converges uniformly to f. Next we establish an integration-by-parts formula: Lemma 10. If f, g L 2 (R) are differentiable with derivatives also in L 2 (R) then f (x)g(x) dx = f(x)g (x) dx. (73) R Proof. The derivative of fg, being f g + fg, is in L 1. So the fundamental theorem of calculus applies to give: b for all real numbers a < b. Now fg L 1, and so a f (x)g(x) dx + lim N b Consequently, there exist a N < N < N < b N with N a R f(x)g (x) dx = f(b)g(b) f(a)g(a) (74) N f(x)g(x) dx = lim f(x)g(x) dx = 0. N f(a N )g(a N ) 0 and f(b N )g(b N ) 0 as N. Plugging into Equation (74) we obtain the desired result. L 2

21 Mathematics 2015, Next we have the first step to showing that C 1 equals C: Lemma 11. If f is in the domain of C 1 then f is in the domain of C and Cf = C 1 f and Af = A 1 f. (75) Proof. Let f be in the domain of C 1. Then we may assume that f is differentiable and both f and the derivative f are in L 2 (R). We have then [ C 1 f, φ n = φ n (x) d R dx + 1 ] 2 x f(x) dx [ d = f(x) dx + 1 ] 2 x φ n (x) dx by Equation (73) R = n f, φ n 1 by Equation (53). (76) Then C 1 f 2 = n 0 C 1 f, φ n 2 = n 1 n f, φ n 1 2. (77) Because this sum is finite, it follows that f is in the domain D(C) of C. Moreover, Cf = n 0 n + 1 f, φn φ n+1 = m 0 C 1 f, φ m φ m by Equation (76) = C 1 f. The argument showing Af = A 1 f is similar. We can now prove: Theorem 12. The operators C and C 1 are equal, and the operators A and A 1 are equal. Thus, a function f L 2 (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 L 2 (R) and with R xf(x) 2 dx <. Proof. In view of Lemma 11, it will suffice to prove that D(C) D(C 1 ). Let f D(C). Then n f, φ n 2 <. n 0 This implies that the sequences {C 1 f N } N 0 and {A 1 f N } N 0 are Cauchy, where f N is the partial sum f N = N f, φ n φ n. n=0 Now 1 2 (A 1 C 1 )f N = f N, and 1 2 (A 1 + C 1 )f N (x) = xf N (x).

22 Mathematics 2015, So the sequences of functions {f N } N 0 and {h N } N 0, where h N (x) = xf N (x), are also L 2 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 is Cauchy in L2 (R). Let g = lim N f N in L2 (R). Observe that f N (y) = f N (x) f N( (x + t(y x) ) (y x) dt. (78) Now 1 f 0 N (x + t(y x))(y x) g(x + t(y x))(y x) dt y x f N g L 2 by the Cauchy-Schwartz inequality. Since f N g L2 0 as N, we have 1 0 f N(x + t(y x))(y x) dt 1 0 g(x + t(y x))(y x) dt. Because {f N } N 0 converges to f uniformly by Lemma 9, taking the limit as N in Equation (78) we obtain f(y) = f(x) g ( (x + t(y x) ) (y x) dt. Therefore f = g L 2 (R). Lastly, we have, by Fatou s Lemma: xf(x) 2 dx lim inf xf N (x) 2 dx <, N R because the sequence {g N } N 0 is convergent. Thus we have established that f 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 sup x a f (b) (x) < x R 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 a f (b) (x) are in L 2 (R) for every a, b {0, 1, 2,...} with a + b 2p. Proof. Consider f S 1 (R). Then R f = n 0 a n φ n with n 0 n2 a n 2 <. (79) In particular, f D(C). Moreover, Cf = n 0 a n n + 1φn+1 Af = n 0 a n nφn 1. From these expressions and Equation (79) it is clear that Cf and Af both belong to D(C). Thus, B 1 B 2 f L 2 (R) for all B 1, B 2 {C, A, I}.

23 Mathematics 2015, d dx Similarly, we can check that if f S p (R), where p 2, then Thus, inductively, we see that B 1 B 2 f S p 1 (R) for all B 1, B 2 {C, A, I}. B 1 B 2p f L 2 (R) for all B 1,..., B 2p {C, A, I}. (This really means that f is in the domain of each product operator B 1 B 2p.) Now the operators 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 ( d dx )b as a linear combination of operators B 1...B 2p with B 1,..., B 2p {C, A, I}. Conversely, suppose f is 2p times differentiable and the functions x x a f (b) (x) are in L 2 (R) for every a, b {0, 1, 2,...} with a + b 2p. Then f is in the domain of C 2p and so f, φ n 2 n 2p <. n 0 Thus f S p (R). The preceding facts show that if f S p (R) then for every B 1,..., B 2p B 1 B 2p 1 f is in the domain of C, and so, in particular, is bounded. Thus, {C, A, I}, the element for all a, b {0, 1, 2,...} with a + b 2p 1. sup x a f (b) (x) < x R 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 f, g p = f, T 2p d g 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 L 2 (R), define { } 1/2 f t = (n + 1) 2t f, φ n 2 for every t > 0. More generally, define n 0 (80) f, g t = n 0(n + 1) 2t f, φ n L 2 φ n, g L 2, (81) for all f, g in the subspace of L 2 (R) consisting of functions F for which F t <.

24 Mathematics 2015, 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 = n 0 n m f, φ n φ n, (82) where on the left N m is the differential operator d2 dx 2 series is taken in the sense of L 2 (R, dx). Furthermore, + x This result will be strengthened and a converse proved later. applied n times, and on the right the f 2 m/2 = f, (N + 1) m f. (83) Proof. Let m 0 be an integer. Since f S(R), it is readily seen that Nf is also in S(R), and thus, inductively, so is N m f. Then we have f, N m f = n 0 f, φ n φ n, N m f = n 0 f, φ n N m φ n, f by Equation (62) = n 0 f, φ n n m φ n, f = n 0 n m f, φ n 2. Thus we have proven the relation f, N m f = n 0 n m f, φ n 2. (84) An exactly similar argument shows f, (N + 1) m f = n 0(n + 1) m f, φ n 2 = f 2 m/2. (85) So if t > 0, choosing any integer m t we have f 2 t/2 f 2 m/2 = f, (N + 1) m f <. Observe that the series is convergent in L 2 (R, dx) since n m f, φ n φ n (86) n 0 n 2m f, φ n 2 = N 2m f, f <. n 0

25 Mathematics 2015, So for any g L 2 (R, dx) we have, by an argument similar to the calculations done above: N m f, g = n m f, φ n φ n, g n 0 = n m f, φ n φ n, g n 0 = n m f, φ n φ n, g. This proves the statement about N m f. n 0 We have similar observations concerning C m f and A m f. operators involving d and x, they map S(R) into itself. Also, dx First observe that since C and A are Af, g = f, Cg, for all f, g S(R), as already noted. Using this, for f S(R), we have φ n+m, C m f = A m φ n+m, f = (n + m)(n + m 1) (n + 1) φ n, f. Therefore, [ ] 1/2 (n + m)! f, φ n φ n+m. (87) C m f = n 0 n! Similarly, A m f = [ ] 1/2 n! f, φ n φ n m. (88) (n m)! n 0 More generally, if B 1,..., B k are such that each B i is either A or C then B 1...B k f = n 0 θ n,k f, φ n φ n+r, (89) where the integer r is the excess number of C s over the A s in the sequence B 1,..., B k, and θ n,k is a real number determined by n and k. We do have the upper bound Note also that θ 2 n,k (n + k) k [(n + 1)k] k = (n + 1) k k k. (90) B 1...B k f 2 = (B 1...B k ) B 1...B k f, f = n 0 θ 2 n,k f, φ n 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 L 2 (R d ) by φ a (x) = φ a1 (x 1 )...φ ad (x d ).

26 Mathematics 2015, Now, for each t > 0, and f L 2 (R d ), define and then define f t def = { f, g t = a W d [(a 1 + 1)...(a d + 1)] 2t f, φ a 2 } 1/2, (92) a W d [(a 1 + 1)...(a d + 1)] 2t f, φ a φ a, g, (93) for all f, g in the subspace of L 2 (R d ) consisting of functions F for which F t <. Let T d be the operator on S(R d ) given by Then, for every non-negative integer m, we have T d = (N d + 1)...(N 1 + 1). f 2 m/2 = f, T m d f. The other results of this section also extend in a natural way to R d. 9. L 2 Type Norms on S(R) For integers a, b 0, and f S(R), define f a,b,2 = x a D b f(x) L 2 (R,dx). (94) Recall the operators and the norms A = d dx x, C = d dx x, f m = f, (N + 1) m f. N = CA 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 S(R). Proof. Let a, b be non-negative integers. Then f a,b,2 = (A + C) a 2 b (A C) b f L 2 a linear combination of terms of the form B 1...B k f L 2, where each B i is either A or C, and k = a + b. Writing c n = f, φ n, we have B 1...B k f 2 L = c 2 n θ n,k φ n+r n 0 = c n 2 θn,k, 2 n 0

27 Mathematics 2015, where r = #{j : B j = C} #{j : B j = A}, and, as noted earlier in Equation (90), θ 2 n,k (n + 1) k k k. So B 1...B k f 2 L 2 n 0 c n 2 (n + 1) k k k = k k f 2 k/2 k k f 2 k. (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 f 2 k = f, (N + 1) 2k f L 2 f 2 L 2 + (N + 1)2k f 2 L 2 and the expression of N as a differential operator we see that f 2 k is bounded above by a linear combination of f 2 a,b,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 { } 1/2 f a,b,2 = x a D b f(x) 2 dx. R d 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 S(R). Recall from Equation (70) that we have, for f S(R) f sup 2 1/2 ( f L 2 + f L 2). Putting in x a D b f(x) in place of f(x) we then have f a,b a linear combination of f a,b,2, f a 1,b,2, and f 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 f 2 L = (1 + x 2 ) 1 (1 + x 2 ) f(x) 2 dx 2 R π (1 + x 2 ) f(x) 2 sup π ( ) f 2 sup + xf(x) 2 sup π ( f sup + xf(x) sup ) 2.

28 Mathematics 2015, So for any integers a, b 0, we have f a,b,2 = x a D b f L 2 π 1/2 ( f a,b + f a+1,b ). (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 + p 2. For f S(R d ), we have the Fourier transform given by F(f)(p) = ˆf(p) = (2π) d/2 R d e i p,x f(x) dx. (98) Again, this preserves the L 2 norm, and transforms derivatives into multiplications: ( ) f p j ˆf(p) = if (p). x j Repeated application of this shows that where p 2 ˆf(p) = F ( f) (p), (99) = n 2 x 2 j=1 j is the Laplacian. Iterating this gives, for each r {0, 1, 2,...} and f S(R d ), p 2r ˆf(p) = ( 1) r F ( r f) (p), (100) which in turn implies, by the Plancherel formula Equation (67), the identity: p 2r ˆf(p) 2 dp = r f(x) 2 dx. (101) R d R d Then we have, for any m > d/4, f sup (2π) d/2 R d ˆf(p) dp = (2π) d/2 R d (1 + p 2 ) m ˆf(p) (1 + p 2 ) m dp [ ] 1/2 = K (1 + p 2 ) 2m ˆf(p) 2 dp R d by Cauchy-Schwartz, [ ] 1/2 where K = (2π) d/2 dp R d (1+ p 2 ) <. The function (1 + s) n /(1 + s n ), for s 0, attains a 2m maximum value of 2 n 1, and so we have the inequality (1 + s) 2m 2 2m 1 (1 + s 2m ), which leads to Then, from Equation (102), we have (1 + p 2 ) 2m 2 2m 1 (1 + p 4m ). f 2 sup K 2 2 2m 1 ( f 2 L 2 + m f 2 L 2 ). (102)

29 Mathematics 2015, 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 b f 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 + x 2 ) replaced by (1 + x 2 ) d. Thus, on S(R d ) 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 = 1 4 d (C j A j ) 2 j=1 and so m consists of a sum of multiples of (3d) m terms each a product of 2m elements drawn from the set {A 1, C 1,..., A d, C d }. Consequently, by Equation (95) m f 2 L 2 c2 d,m f 2 m, (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 f sup k d,m f m (104) holds for all f S(R d ). Now consider f S p (R d ), with p > d/4. Let f N = j W d, j N f, φ j φ j. Then f N f in L 2 and so a subsequence {f Nk } k 1 converges pointwise almost everywhere to f. It follows then that the essential supremum f is bounded above as follows: f lim sup f N sup. N Note that f N f also in the p norm. It follows then from Equation (104) that f k d,p f p (105) holds for all f S p (R d ) 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 S(R) with a Sequence Space Suppose a 0, a 1,... form a sequence of complex numbers such that (n + 1) m a n 2 <, for every integer m 0. (106) n 0