A functional Hilbert space approach to frame transforms and wavelet transforms

Transcription

1 Eindhoven University of Technology MASTER A functional Hilbert space approach to frame transforms and wavelet transforms Duits, M. Award date: 2004 Link to publication Disclaimer This document contains a student thesis bachelor's or master's), as authored by a student at Eindhoven University of Technology. Student theses are made available in the TU/e repository upon obtaining the required degree. The grade received is not published on the document as presented in the repository. The required complexity or quality of research of student theses may vary by program, and the required minimum study period may vary in duration. General rights Copyright and moral rights for the publications made accessible in the public portal are retained by the authors and/or other copyright owners and it is a condition of accessing publications that users recognise and abide by the legal requirements associated with these rights. Users may download and print one copy of any publication from the public portal for the purpose of private study or research. You may not further distribute the material or use it for any profit-making activity or commercial gain

2 TECHNISCHE UNVERSITEIT EINDHOVEN Department of Mathematics and Computing Science MASTER S THESIS A functional Hilbert space approach to frame transforms and wavelet transforms by Maurice Duits Supervisor: Prof. dr. ir. J. de Graaf Eindhoven: September 2004

3 Contents Introduction 3 2 Frame transforms and functional Hilbert spaces 5 2. Introduction Construction of a frame transform The inverse frame transform 9 3. Inversion using projections Inversion using Gelfand triples Frame transforms constructed from generating functions 4 4. An expansion theorem Bargmann-transform Laguerre polynomials Gegenbauer polynomials Introduction The semi-group generated by the Euler operator The classical Fourier and Laplace transforms The Fourier transform as a frame transform on a Gelfand triple The Laplace transform Sampling Theorems A general sampling theorem A special subspace of L 2 R) A space of q-functions Operators on G q A generalization of Lagrange interpolation Wavelet transforms Construction of V using group representations Unitary representations Topological conditions Cyclic representations Orthogonal sums of functional Hilbert spaces An example: diffusion on a sphere Irreducible representations 60

4 0 A representation of a semi-direct product S T on L 2 S) Introduction The wavelet transform Alternative description of C S T for admissible wavelets L 2 R 2 ) and the Euclidean motion group 70. The wavelet transform Admissible wavelets Generalized admissible wavelets An example of an admissible generalized wavelet A transform by Sherman revisited Preliminaries The transform by Sherman An application of Theorem An approach using a special orthonormal basis A Schur s lemma 86 B Gelfand triples 88 C An open problem 88 C. Introduction C.2 q-functions D Acknowledgements 90 2

5 Introduction A frame in a Hilbert space H is defined to be a subset V = {φ x x E} H, with E an index set, such that the span V of V is dense in H. Mostly, the set E is a subset of C n or a group. Starting from a frame V we introduce the frame transform W : H C E : f W f, where W f : E C : x φ x, f) H..) The underlying master s thesis is mainly focused on the questions how and when the above transform defines a unitary map. We succeeded answering these questions in the most general way by using the theory of functional Hilbert spaces = theory of reproducing kernels). The idea of working with these kind of spaces is inspired by the identity W f ) x) ψx H f H..2) This identity states that if the frame transform defines a unitary map from H onto a Hilbert subspace of C E, then point evaluation δ x : f fx) in the latter space is a continuous linear functional for all x E. This means that the Hilbert subspace has a reproducing kernel and we denote it by C E. The central observation of this thesis is Theorem 2.3. It states that W defines a unitary map from H onto the functional Hilbert space C E where is the function of positive defined by x, x ) = ψ x, ψ x ) H. This is a new and important result. Since the functional Hilbert space C E is constructed in a rather abstract fashion, we are challenged to find a more tangible alternative description of this space. As shown in chapter 4, an alternative description comes within sight in case the frame V is constructed from a generating function for orthogonal polynomials. The space C E then typically consist of analytic functions on a subset of) the complex plain. The Bargmanntransform is a famous illustration of this phenomenon. In chapter 5 the well-known Laplace and Fourier transforms are looked upon as frame transforms. Note that the Fourier transform is not a frame transform itself, but with the aid of Gelfand triples we can construct a frame transform which leads to the Fourier transform. Theorem 2.3 can also be used to construct sampling theorems. A famous example of a sampling theorem concerns the space of functions f L 2 R) for which Ff has support within, ). Then fx) = n= fn) sinπx n))..3) πx n) In chapter 6 a general sampling theorem is proved by a simple argument, which covers most of the classical cases. As an excursion, we also mention a construction of functional Hilbert spaces that admit a sampling theorem. In Appendix D an open problem is formulated which is inspired by these results. 3

6 A special kind of frame transforms is the wavelet transform. In this case V is constructed from a vector ψ H and a group representation of a group G in H V ψ = {U g ψ g G}..4) We denote a wavelet transform by W ψ, where ψ H is called wavelet. In the last twenty years a lot has been written about these kind of transformations. In 985 Grossmann, Morlet and Paul published a basic paper [GMP] which can be seen as the fundament of the theory of wavelet transforms. Their main result is that the wavelet transform W ψ defines a unitary map from a Hilbert space H onto L 2 G) for a suitable vector ψ H, where G is a locally compact group with a unitary, irreducible and square integrable representation U of G in H. A square integrable representation is a representation for which a ψ exist such that C ψ = U ψ 2 g ψ, ψ) H dµ G g) <,.5) H G where µ G is a left invariant Haar measure. The irreducibility condition is a very strong one. However, many representations of practical interest are not irreducible at all. Therefore it is often suggested, to replace the condition of irreducibility by the condition that the representation is cyclic, i.e. it has a cyclic vector, i.e. a vector for which the span of the orbit under U is dense in the Hilbert space. But no really successful unitarity results were obtained. For a nice survey of some posed suggestions, see [FM]. Noticeably, our Theorem 2.3 states that W ψ defines a unitary map from V ψ onto C G. The conditions we impose on the representation are quite simple: none! Note that V ψ = H if and only if ψ is a cyclic vector. Although the above solves the unitarity questions, the functional Hilbert space, as mentioned before, is not easily characterized. We managed to give an easy to grasp description of the functional Hilbert space in the case H = L 2 S) and G = S T for an abelian group S and an arbitrary group T which acts on S. As an example we work out the case S = R 2 and T = T so G is the Euclidean motion group. The idea to consider semi-direct products is not new, several articles have been written on this subject. See for example [FNP], [FM]. The final chapter concerns a transformation introduced by Sherman in [S]. Although this Sherman transform is a frame transform but not a wavelet transform, it has several resemblances with wavelet transforms. In [S], Sherman poses a key lemma concerning a singular integral. This singularity however, can be avoided. Moreover, we suggest an alternative transform which has the advantage that the unitarity relations appear in a more natural way. 4

7 2 Frame transforms and functional Hilbert spaces 2. Introduction Denote the space of all complex-valued functions on E by C E. We say that a Hilbert space H consisting of functions on a set E, i.e. a vector subspace of C E, is a functional Hilbert space, if point evaluation at every point is continuous, i.e. δ x : H C : f fx) 2.6) is a continuous linear functional on H for all x E. Then, by the Riesz-representation theorem, there exists a set { x x E} with x, f) H = fx), 2.7) for all x E and f H. We use the convention that the inner product is linear in the second entry. It follows that the span of the set { x x E} is dense in C E. Indeed, if f H is orthogonal to all x then f = 0 on E. Then define the function : E E C by x, x ) = x x) = x, x ) H, for all x, x E. The function is called the reproducing kernel. It is obvious that is a function of positive type on E, i.e., n n x i, x j )c i c j 0, 2.8) i= j= for all n N, c,..., c n C, x,..., x n E. So to every functional Hilbert space there belongs a reproducing kernel, which is a function of positive type. Conversely, as Aronszajn pointed out in his paper [Ar], a function of positive type on a set E, induces uniquely a functional Hilbert space consisting of functions on E with reproducing kernel. We will denote this space with C E. Without giving a detailed proof we mention that C E can be constructed as follows; start with : E E C, a function of positive type and define x =, x). Take the span { x x E} and define the inner product on this span as l α i xi, i= n ) β j xj j= C E = l i= n α i β j x i, x j ). 2.9) j= This is a pre-hilbert space. After taking the completion we arrive at the functional Hilbert space C E. There exists a useful characterization of the elements of C E. Lemma 2. Let be a function of positive type on E and F a complex-valued function on E. Then the function F belongs to C E if and only if there exists a constant γ > 0 such that l l α j F x j ) 2 γ α k α j x k, x j ), 2.0) j= k,j= 5

8 for all l N and α j C, x j E, j l. See [Ma, Lemma.7, pp.3] or [An, Th. II..]. This lemma enables us to give an expression for the norm of an arbitrary element in C E. Lemma 2.2 Let F C E. Then { l F 2 C = sup α E j F x j ) 2 j= l k,j= l N, α j C, x j E, ) α k α j x k, x j ) } l C α k xk 0. 2.) E For a detailed discussion of functional Hilbert spaces see [Ar], [An] or [Ma]. k= 2.2 Construction of a frame transform Starting with some labeled subset V of H, we will construct a functional Hilbert space by means of a function of positive type on the index set, using the construction as described in the introduction. Moreover, there exists a natural unitary map from V to this functional Hilbert space. Let H be a Hilbert space. Let E be an index set and V := {φ x x E}, 2.2) be a subset of H. We call the set V a frame. Define the function : E E C of positive type on E by x, x ) = φ x, φ x ) H, 2.3) for all x, x E. From this function of positive type the space C E can be constructed. The following theorem is the central observation of this thesis. Theorem 2.3 Frame Theorem) The map W : V C E : f W f, where W f : E C : x φ x, f) H, 2.4) is a unitary map. Here V inherits the inner product from H. First we show that W f C E for 6

9 any element f V and that W is bounded and therefore continuous). If f V then l ) α j W f xj ) 2 = j= l 2 α j φ xj, f) H = j= l 2 l α j φ xj f 2 H = j= H k,j= l j= ) α k α j x k, x j ) f 2 H, ) α j φ xj, f for all l N, α,..., α l C, and x,..., x n E. So W f C E by Lemma 2. and W f 2 f 2 C E H, by Lemma 2.2. Next we prove that W is an isometry. Because φ x ) x) = x, x ), W maps a linear combination i α iφ xi onto the linear combination i α i, x i ). So W V ) = {, x) x E}. Moreover, it maps V isometrically onto {, x) x E}, because W ) ) ) α i φ xi, W β j φ x j = α i, x i ), ) β j, x j) C E i j C E i j 2 H = i,j α i β j x i, x j) = i,j α i β j φ xi, φ x j ) H. Since V is dense in V and W is bounded on V it follows that W is an isometry. Furthermore, W [ V ] is dense in C E. So W is also surjective and therefore unitary. We will call the unitary map W a frame transform. In the sequel we will mostly introduce a frame transform by W : V C E defined by W f ) x) = φ x, f) H for all x E and f V, instead of writing W : V C E : f W f, where W f : E C : x φ x, f) H. The latter has the advantage that the structure of the objects is more tangible, but it has the disadvantage that it is a bit lengthy for simple calculations. Therefore the first phrase will be used when we deal with an explicit example of a frame transform and the second phrase otherwise. In most cases we are mainly interested in the case V = H, i.e. V is total in H. To get a feeling for what is happening we deal with two illustrating examples. Example: The special case E = N. Let H be a separable Hilbert space consisting of functions on the set E = N. Let V = {φ m m N} consist of an orthonormal basis, so V = H. Then, m, m ) = φ m, φ m ) H = δ mm, 2.5) for all m, m N. This means that we just get C N = l 2N). The unitary map W gives us the sequence of expansion coefficients c m of a vector f H with respect to the orthonormal basis. 7

10 Example: The special case E = H. Let E = H and V = {m m H} = H. The function of positive type is just the inner product m, m ) = m, m ) H. 2.6) This means that C E = CH, ) H. This is the functional Hilbert representation of an arbitrary Hilbert space. It is equal to the topological dual space H, the space of all continuous linear functions on H. The functional Hilbert space C E is an abstract construction. We are challenged to find alternative characterizations of these functional Hilbert spaces. In the literature two major classes of functional Hilbert spaces appear, functional Hilbert spaces of Bargmann-type and of Sobolev-type. The first type consists of a nullspace of unbounded operators on L 2 E, µ) and the second of the domain of unbounded operators on L 2 E, µ). For Bargmann-type spaces see [B]. For Sobolev-type, see [EG] and [EG2]. 8

11 3 The inverse frame transform 3. Inversion using projections Let H be a Hilbert space and V H a subset of H labeled with elements in a set E V = {φ x x E}. 3.) For the sake of simplicity assume that the span is dense in H.Otherwise, replace H by V. Consider the unitary frame transform W : H C E : f W f, where W f : E C : x φ x, f) H. We will analyze the inverse W of W. Suppose H = C I L is a functional Hilbert space itself, with reproducing kernel L. Set Ṽ = {W L ξ ξ I}. Then Ṽ is dense in CE, by unitarity of W. Moreover, W L ξ, W L ξ ) C E = L ξ, L ξ ) C I = Lξ, ξ ) for all ξ, ξ I. So if W : C E C I L : g W g, where W g : I E : ξ W L ξ, g) C E, 3.2) is the associated frame transform, then W g ) ξ) = W Lξ, g) C E = L ξ, W g) C I L = W g ) ξ), 3.3) for all g C E and ξ I. Hence, W = W is also a frame transform. Note that W L ξ )x) = φ x, L ξ ) C I L = φ x ξ). Although the second example at the end of section shows that every Hilbert space can be characterized as a functional Hilbert space, this characterization is not always very useful. In some cases, like L 2 R) or L 2 S ), the Hilbert space H can be regarded as limits of functional Hilbert spaces. The space L 2 S ) for example, admits a decomposition in spherical harmonics. This decomposition will be dealt with in section 8. Definition 3. We say that a sequence of functional Hilbert space {C I L n } n N converges to H if. each C I L n is a Hilbert subspace of H 2. the projections P n of H on C I L n satisfy a) P n P m = P min{n,m} b) lim n P n f f, for all f H Suppose the sequence of functional Hilbert space C I L n converges to H. Define V n = {P n ψ x x E}, 3.4) and the function n : E E C of positive type by n x, x ) = P n φ x, Pφ x ) H 3.5) 9

12 for all n N and x, x E. By Theorem 2.3 the map W n : C I L n C E n : f W n f, where W n f : E C : x P n ψ x, f) H is unitary for all n N. Note that for the frame transform W n the inverse frame transform is given by Wn : C E CI L : g Wn g where Wn g : I C : ξ W n L n;ξ, g) C E, for all n N. Theorem 3.2 The functional Hilbert spaces C E n converge to C E. Let f C E n. Then g = Wn f C I L n H. This means that C E W g = W n g = f. By the unitarity of W n and W we also find f C E n = g C I Ln = g H = f C E. Therefore, the spaces C E n are all closed subspaces of C E. Finally, to prove condition 2, we remark that the projection P n on the space C E n is given by P n = W P n W. The following theorem gives a formula for the inverse frame transform. Theorem 3.3 lim W n P nw f = f, 3.6) n for all f H. Since P nw f W f we also find by unitarity Wn P nw f = W P nw f f. 3.2 Inversion using Gelfand triples In this section we mention another method to obtain the inverse W. Assume V is a subset of a vector space labeled by a set E. Equip V with two inner products, ) and, ) 2 such that C 2, for some constant C > 0. Denote the completions of V under these two norms by H and H 2. Note that H 2 H. Then by 2.3) we obtain the functional Hilbert spaces C E and C E 2. The question arises how these two functional Hilbert spaces are related. The answer is straightforward. Since C 2, it is obvious that n i= n α i α j x i, x j ) = j= n i= C 2 n α i α j φ xi, φ xj ) = j= n α i φ xi 2 2 = C 2 i= n i= n α i φ xi 2 i= n α i α j 2 x i, x j ), for all n N, α... α n, x... x n and hence 2. By Lemma 2., it follows that C E C E 2 and C E 2 C C E on C E by Lemma j=

13 We return to the problem of inverting the frame transform. Assume V = {φ x x E} C I for some set I. Define φ ξ : E C by φ ξ x) = φ x ξ) for all x E and ξ I. Recall that if H is a functional Hilbert space C I L for some function L of positive type, then the inverse W equals the frame transform with respect to Ṽ = { φ ξ ξ I}. Then in addition Ṽ CE. In general L does not exist and Ṽ is not a subset of CE. Nevertheless, the functions φ ξ are still well-defined. In the sequel we use Gelfand triples to understand the role of these functions. Let R BH) such that R exists and is a self-adjoint operator on H. Hence R is in particular densely defined and closed. Consider the Gelfand triple H + H H, 3.7) constructed by this operator R. Recall that + = R H and = R H. Assume that V is a dense subspace of H +, i.e. {R φ x x E} = H. 3.8) Since R is bounded, this assumption implies that V is dense in all the space H +, H and H and therefore it makes the assumption V = H obsolete. The frame transform W maps H unitary onto C E. Define A BCE ) by A = W RW. Then A induces the Gelfand-triple C E ) + CE C E ). 3.9) Note ) that the frame transform W induces by restriction the unitary map W H+ : H + C E : f W f and after extension it induces in the same way a unitary map from H + onto ) C E. Next we introduce two other functions of positive type and frame transforms. Define : E E C by x, x ) = φ x, φ x ) +, 3.0) for all x, x E. Note the opposite signs. By Theorem 2.3, the transform W + : H + C E defined by W+ f ) x) = φ x, f) +, 3.) for all x E and f H +, is a unitary map from H + onto C E. Equivalently, define + : E E C by + x, x ) = φ x, φ x ), for all x, x E. By Theorem 2.3, the transform W : H C E + defined by W f ) x) = φ x, f), for all x E and f H, is a unitary map from H onto C E +. Lemma 3.4 C E + = ) C E and + CE = ) C E. First we show that { x x E} is dense in both ) C E and + CE +.

14 Since x = W φ x and φ x DR ), we also have x DA ) = ) C E. + Moreover, { x x E} is dense in ) C E by unitarity of W + H + and assumption 3.8). For the space C E + recall that φ x DR ) and hence R 2 φ x H for all x E. Moreover, {R 2 φ x x E} is dense in H. The operator W maps R 2 φ x onto x and therefore { x x E} is a subset of C E + for which the span is dense in C E +. Finally we show that the inner products are equal on this dense subspace. x, y ) C E + = R 2 φ x, R 2 φ y ) = R φ x, R φ y ) H = W R φ x, W R φ y ) C E = A x, A y ) C E for all x, y E. Hence in particular C E + = C E ) In the same way one can prove that C E = C) E which is dense in C E and ) C E. + on { x x E}. on { x x E} From now on we assume that R is such that H + is a functional Hilbert space with reproducing kernel L. Theorem 3.5 The set { φ ξ ξ I} is contained in C E. The inverse W H + is given by W H + : C E H + : F W H + F, where W H + F : E C : ξ φ ξ, F. 3.2) The first statement is trivial since H + is a functional Hilbert space and hence φ ξ = W L ξ where L is the reproducing kernel of H +. For the second statement we first show that A ;x = A x for all x E. ) A ;x y) = y, A ;x ) C E = A y, ;x ) C E for all x, y E. = A y ) x) = A y, x ) = y, A x ) = A x ) y) Secondly, we show that A 2 W L ξ = φ ξ for all ξ I. Denote the reproducing kernel of H by L. Note that W L ξ C E + hence A 2 W L ξ is well-defined. Then, A 2 W L ξ )x) = ;x, A 2 W L ξ ) C E = A ;x, A W L ξ ) C E = A x, A W L ξ ) C E = W R φ x, W R L ξ ) C E = R φ x, R L ξ ) H = φ x, L ξ ) H+ = W φ x, W L ξ ) C E = ;x, φ ξ ) C E = φ ξ x), 2

15 for all x E. Finally, we prove the inversion formula: fξ) = L ξ, f) H = W L ξ, W f) C E + = A W L ξ, A W f) C E = A 2 W L ξ, W f = φ ξ, W f, for all f H + and ξ I. Let t R t be a strongly continuous contraction semi-group such that the generator is self-adjoint. Then for all t 0, R t is self-adjoint, R t exists and is self-adjoint and hence densely defined). Moreover, assume that H t is a functional Hilbert space under the norm t = R t H, Rt V is dense in H for all t > 0. Then each of the operators R t defines a Gelfand triple, denoted by H t H H t, for all t > 0. Lemma 3.6 R t R s for all t > s. Let f H. Then, R t f H = R t s R s f H R t s R s f H R s f H. Hence R t R s. Lemma 3.7 The space H t is continuous embedded in H s for all t > s. Let f H t. There exist a g H such that f = R t g R t H) and hence f = R s R t s g H s. Moreover, f Hs = Rs f H Rt f H = f Ht. Hence the statement follows. Now we obtain the following the inversion formula 3.3) f = lim t 0 R t f = lim t 0 ξ φ ξ, W R t f t = lim t 0 ξ φ ξ, A t W f t. 3.4) This formula is the generalization of the formula Bargmann gives in his article to invert the Bargmann transform. We will return to this subject in the Section 4.2. In that section the assumption 3.8) is equivalent to the assumption that V = H since in that case R t V = V for all t > 0. 3

16 4 Frame transforms constructed from generating functions 4. An expansion theorem Since by Theorem 2.3) a frame transform is a unitary map, it maps an orthonormal basis onto an orthonormal basis. This leads to some nice consequences in case the set V in 2.2) is constructed from a generating function, since it provides a convenient basis for the image space. In this section, the image space typically is a functional Hilbert space consisting of analytic functions. An important result is the following Theorem 4. Let : E E C be a function of positive type and let {g n n N} be an orthonormal set in C E. Then n N g nx)g n y) is absolutely convergent for all x, y E. Moreover, the set {g n n N} is a basis for C E if and only if x, y) = n N g n x)g n y), 4.) for all x, y E. See [Ma, Lemma.]. Let H be a Hilbert space and V = {φ x x E} a subset of H labeled by a set E. Define the function : E E C of positive type by x, x ) = φ x, φ x ) H for all x, x E. By Theorem 2.3, the frame transform W : V C E : f W f, where W f : E C : x φ x, f) H, is a unitary map. Suppose {g n n N} is an orthonormal basis for H. Then φ x can be expanded in this basis as φ x = n N a n x)g n, 4.2) where a n x) = g n, φ x ), for all x E and for all n N. Theorem 4.2 Expansion Theorem) The reproducing kernel : E E C is given by x, x ) = n N a n x)a n x ), 4.3) for all x, x E, where the sum is absolutely convergent. Moreover, if V = H then {a n n N} is an orthonormal basis for C E. It is obvious that ) W g n x) = an x) for all x E and n N. By unitarity of W it follows that {a n n N} is an orthonormal basis for C E. The rest of the statement follows by Theorem 4.. 4

17 Suppose f V and write f = n N c ng n. Then 0 = ψ x, f) H = n N c n a n x), 4.4) for all x E. This implies that V = H if and only if for all {c n } n N l 2 N) ) x E c n a n x) = 0 n N c n = ) n N Theorem 4.2 can be used to construct special frame transforms based on generating functions. As an example we deal with the Bargmann-transform and two transforms based on generating functions for Laguerre Polynomials and Gegenbauer polynomials. In these examples E equals C or {z C z < } and the generating function is of the form φ z = g n z n. 4.6) Obviously, the functions z z n satisfy condition 4.5) and therefore V = H. 4.2 Bargmann-transform In this section the Bargmann-transform as an example to the previous sections. Some proofs, especially the unitarity, will become a lot easier with the aid of the theory developed so far. The following two results will be used. The proof is omitted. A result due to Cramer: H n x) < k n2 n/2 e x2 /2 for all x R and n N, where k is a constant. See [C]. Mehler s formula for Hermite polynomials [ ] H n x)h n y) w/2) n = w 2 ) 2xyw x 2 + y 2 )w 2 2 exp, 4.7) n! w 2 ) which is valid for all x, y R and w D = {z C z < }. See [MOS, 5.6, pp.252]. Let H = L 2 R). From the theory of special functions it is well-known that the generating function of the Hermite polynomials is given by e z2 +2zx = H n x) z n, 4.8) n! 5

18 and the sum converges for all z C and x R. Then rewrite this identity as π /4 e z2 /2+ 2zx x 2 /2 = H n x)e 2 x2 2n n! π for all z C and x R. Define the subset V = {φ z z C} where z n n!. 4.9) φ z x) = π /4 e z2 /2+ 2zx x 2 /2, 4.0) for almost all x R and all z C. Define the function : C C C of positive type by z n w z, w) = n n! = e zw, 4.) for all z, w C. By Theorem 4.2, the frame transform W : H C C defined by ) W f z) = π /4 e z2 /2+ 2zx x 2 /2 fx) dx, 4.2) R for all z C and f L 2 R) is a unitary map. Moreover the set {z zn n n N 0 } is an orthonormal basis in C C, since {g n : x Hnx)e 2 x2 2 n n! n N 0 } is an orthonormal basis in π L 2 R). The inner product of the functional Hilbert space is characterized by the integral Φ, Ψ) C E = Φz)Ψz)e z 2 dz, 4.3) π C for all Φ, Ψ C E zn, since this inner product makes the set {z n! n N 0 } orthonormal. The functional Hilbert space C C is called the Bargmann-space of dimension. It will be denoted by B. The generalization to higher dimensions is straightforward and therefore we only give the reference [B]. Define the harmonic oscillator operator H on L 2 R) by ) Hf x) = x 2 )fx) d2 2 dx fx)), 4.4) 2 d for almost all x R and all f DH) = {f L 2 R) 2 f L dx 2 2 R) Mf L 2 R)}, where Mf)y) = y 2 fy) for all y R.. The basis elements g n are eigenfunctions of H with eigenvalue n for all n N 0, i.e. Hg n = ng n for all n N 0. The operator H is a self-adjoint positive operator and therefore H is an infinitesimal generator of a strongly continuous contraction semi-group of self-adjoint operators denoted by t e th. Obviously, e th g n = e tn g n for all t > 0 and n N 0. Denote the inverse of e th by e th for all t > 0. With the operators e th and e th we want to construct a Gelfand triple as in 3.3). Therefore H t = e th L 2 R)) must be a functional Hilbert space. Strictly speaking, 6

19 H t L 2 R). As such, it consists of equivalence classes of measurable functions instead of functions and cannot be a functional Hilbert space. For this moment we will write [f] instead f for elements of L 2 R). It turns out that every class [f] e th L 2 R)) has a unique continuous representant f. Now H t is defined to the linear space of this continuous representants, equipped with the norm h Ht = e th [h] L2 R). The space H t is a Hilbert space of its own right. Moreover, it turns out that H t is a functional Hilbert space. Let [f] H t and write [f] = c n[g n ]. Since [f] H t, e th [f] 2 L 2 R) = c n 2 e 2tn. Define the sequence {f N } N N of functions on R by f N x) = N c n g n x) 4.5) for all x R and N N. Then f N x) f M x) N c n g n x) k n=m N c n = n=m k N c n 2 e 2tn N e 2tn n=m n=m N c n e tn e tn n=m k e 2t f H t 4.6) for all x R and N > M N. Hence f N converges pointwise to a function f on R. Moreover, it converges uniformly on R. Hence it follows that f is the continuous representant of the equivalence-class [f]. Naturally this representant must be unique. Moreover, by a same estimate as above fx) k e 2t eth [ f] L2 R) = k e 2t f Ht 4.7) for all x R. In the sequel we will not use the notation [f] for elements in L 2 R) anymore. Lemma 4.3 Let t > 0. The space H t = e th L 2 R)) with inner product, ) t = e th, e th ) L2 R) is the functional Hilbert space with reproducing kernel L t x, y) = [ H n x)h n y) e x2 /2 y 2xy 2 /2 e 2nt = e 4t ) 2 n 2 exp 2 x2 + y 2 ] ) cosh2t) n! sinh2t) for all x, y R. From 4.7) it follows that H t is a functional Hilbert space. The set {e tn g n n N} is an orthonormal basis for H t, hence the statement follows by Theorem 4. and Mehlers formula )

20 By applying the Cauchy-Schwartz inequality one easily obtains the following inequality, [ ] fx) 2 f 2 H t L x 2 H t = f 2 H t e 4t ) x 2 2 cosh2t)) 2 exp, 4.9) sinh2t) for all x R. For all f H t and x R. Lemma 4.4 Let t R, then V H t. Moreover, e th φ z = φ e t z, 4.20) for all z C. Let z C. Then e th φ z = e th z n g n n! = g n e tn zn = n! g n e t z) n n! = φ e t z This proves the statement. The lemma implies that e th V is dense in L 2 R) since e th V = V and V is dense in L 2 R). This proves the final assumption in the Gelfand triple 3.3). Analogous to section 3.2, we introduce the operator A t = W e th W. Corollary 4.5 Let t > 0. The operator A t = W e th W is given by At f ) z) = fe t z), 4.2) for all z C and f C C t. Let f C C t. Then At f ) z) = W e th W f) z) = e th W f, φ z ) L2 R) = W f, e th φ z ) L2 R) = W f, φ e t z) L2 R) = W W f ) e t z) = fe t z), for all z C. Since A t f is only a re-scaling of the functions f, it easily follows that f 2 B t = Bt f 2 B = fe t z) 2 e z 2 dz 4.22) π C for all f B t. Moreover, the reproducing kernel t : C C C is given by t z, w) = e e 2t zw, 4.23) 8

21 for all z, w C. Hence, fz) f Bt e 2 e 2t z 2, 4.24) for all z C and f B t. This space also frequently occurs in [B], but is not used in the context of a Gelfand triple. We recall that the function φ x : C C is defined by φ x z) = φ z x) = for all z C and x R. g n x) zn n!, 4.25) Lemma 4.6 The functions φ x satisfy the following inequality φ x z) A z e z 2 /2, 4.26) for some A > 0 which is independent of x R. Let x R and z C. Then, φ x z) = φ z x) = g n x) zn gn x) zn z n k. n! n! n! The entire function F : R C defined by F r) = m=0 r m m! e r2 /2, for all r R, has the asymptotic expansion, F r) = 8π) /4 r{ 6r + O )}, r. 2 r4 See [O, Ch. 9, 8, pp ]. Hence the statement follows. Theorem 4.7 Let t > 0 and f H t. Then fx) = φ x, W f = φ x z) W f ) z)e z 2 dz, 4.27) for all x R. C 9

22 Let f H t and t > 0. Then W f B t. Let x R. By 4.23) and Lemma 4.6, fz) φ x z) A z e 2 +e 2t ) z 2, for all z C. Therefore the integral in 4.27) is well-defined. Moreover, C φ x z) W f ) z)e z 2 dz = = lim N = lim N lim N C N N g n x) g n x) zn n! W f ) z)e z 2 dz C z n n! W f ) z)e z 2 dz N g n x)g n, f) L2 R) = fx) Hence the statement follows. By 3.4) we obtain that f = lim e th f = lim x φ x z) t 0 t 0 C W f ) e t z)e z 2 dz ). 4.28) for all f L 2 R). 4.3 Laguerre polynomials In this section a transform introduced by Bargmann in [B] will be dealt with. The image space frequently occurs in the sequel. Consider the generalized Laguerre polynomials defined by L α) n x) = ) n d n! ex x α [e x x n+α ], 4.29) dx for all α >, x > 0 and n N 0. The generalized Laguerre polynomials have the following generating function exp[ x + z)/2 z)] z) α+ = e x/2 for, α >, x > 0 and z <. For fixed α, the set {g n : x N 0 is an orthonormal basis in L 2 0, ), x α dx). L α) n x)z n, 4.30) n! Γn+α+) ) 2 e x/2 L α) n n 20

23 Let α > be fixed and let D = {z C z < } be the unit disc. Set V = {φ α) z z D} where φ α) z is defined by φ α) z x) = exp[ x + z)/2 z)] z) α+ 4.3) for all x 0, ) and z D. Define the function α) : D D C of positive type by α) Γn + α + ) z, w) = zw) n Γα + ) =, 4.32) n! zw) α+ for all z, w D. Then by Theorem 2.3 and 4.2 the frame transform W α : L 2 0, ), x α dx) C D defined by α) Wα f ) z) = 0 exp[ x + z)/2 z)] z) α+ fx)x α dx, 4.33) for all f L 2 0, ), x α dx), is a unitary map. Moreover, the set {a n : z Γn+α+) n! ) 2 z n n N 0 } is an orthonormal basis for C D α). If α > 0 there exists a useful characterization of the functional Hilbert space. First we recall an elementary result. By definition of the beta-function, 0 r 2m+ r 2 ) α dr = for all α > 0 and m N 0. m!γα) 2Γα + m + ), 4.34) Theorem 4.8 The space C D consists of all analytic functions f : D C for which α) fz) 2 z 2 ) α dµz) <. 4.35) πγα) D Here dµ stands for the normal Lebesgue measure on C. Moreover, f, g) C D = fz)gz) z 2 ) α dµz), 4.36) α) πγα) for all f, g C D α). D First, we prove the orthonormality of the set {a n n N 0 }. Let n, m N 0. Then 2π a n z)a m z) z 2 ) α dµz) = r n+m+ e im n)φ r 2 ) α dφdr D 0 0 = 2πδ nm r n+m+ e im n)φ r 2 ) α dr = πδ nm Γα)m! Γm + α + ) 2 0

24 where 4.34) was used in the last step. This proves the orthonormality. Secondly, let f be an analytic function on D. Write fz) = c nz n for all z D and define f N by f N z) = N c mz n for all z D. Then f N z) 2 z 2 ) α dµz) = π D N Γα)m! Γm + α + ) c n 2 Hence the integral in 4.35) converges if and only if the sum Γα)m! c Γm+α+) n 2 converges. Moreover, in case of convergence fz) 2 z 2 ) α dµz) = π D Γα)m! Γm + α + ) c n 2 = f F0, and therefore the theorem follows. In accordance to [B] the space C D α) will be denoted by F α for α > 0. Finally, consider α = 0 which leads to the space F 0 = C D 0). Note that the assumption α > 0 is crucial in Theorem 4.8 and therefore the theorem is not applicable to F 0. Since {z z n z N 0 } is an orthonormal basis for F 0, the space consists of all analytic functions f that can be represented in power series of the form fz) = for all z D. c n z n, {c n } n N0 l 2 N 0 ). 4.37) Lemma 4.9 Let f F 0. Then f 2 F 0 = 2π lim R 2π 0 fre iφ ) 2 dφ. 4.38) Let {c n } n N0 l 2 N 0 ) such that fz) = c nz n for all z D. Since {z z n n N 0 } is an orthonormal basis for F 0, the norm of f is given by f 2 F 0 = c n 2. Moreover 2π fre iφ ) 2 dφ = 2π 0 2π = m=0 2π c n c m R n+m e im n)φ dφ a n 2 R 2n for all 0 < R <. Take the limit R on both sides and the statement follows. 22 0

25 Note that the above proof also shows that F 0 consist of all analytic functions on D such that the limit in 4.38) exists. The reproducing property of the reproducing kernel can be proved directly from Cauchy s integration theorem. By 4.38) the inner product of f, g F 0 is given by f, g) F0 = lim R 2π = lim R 2πi = lim R 2πi 2π 0 2π 0 fre iφ )gre iφ )dφ fre iφ )gre iφ ) ireiφ dφ Reiφ C R fw)gw) w dw, where C R = {z D z = R} for all 0 < R < with positive direction). Let z D and f = z 0) then z, g) F0 = lim R 2πi C R zw gw) w dw = lim gw) dw R 2πi C R w zr = lim grz) = gz). R There also exists a characterization of the norm in H 0 which is similar to the characterization of the norm of F α as formulated in Theorem 4.8. Define the Euler operator E on F 0 by ) df Ef z) = z z), 4.39) dz for all z D. It is obvious that the basis elements z z n are eigenvectors with eigenvalue n for all n N 0. The domain of E can therefore be defined by DE) = {x c nz n {c n n 2 } n N l 2 N 0 )}. Theorem 4.0 The space F 0 consists of all analytic functions f on D for which I + E)fz)fz) dµz) <. 4.40) π D Here dµz) stands for the normal Lebesgue measure on C. Moreover, f, g) F0 = I + E)fz)gz) dz, 4.4) π for all f, g F 0. D 23

26 Let f be analytic function on D. Write fz) = c nz n, for all z D. Define f N by f N z) = N c nz n. Then I + E)f N z)f N z) dµz) = N π D π N = N c n c m m=0 0 D 2n + 2)r 2n+ dr = n + )z n z m dµz) N c n 2. The theorem follows by a same argument as in Theorem Gegenbauer polynomials 4.4. Introduction The Gegenbauer polynomials of order λ are defined by the following generating function 2xz + z 2 ) λ = Cnx)z λ n, 4.42) which is valid for < x <, z < and λ 0. In this section we consider the Gegenbauer polynomials of order λ > 0. Apply the operator z d + 2λ on both sides to dz obtain 2λ xz) 2xz + z 2 ) = Cnx)n λ + 2λ)z n, 4.43) λ+ for all < x <, z < and λ 0. For fixed λ > 0, the set {g n : x in L 2, ), dµ) where, ) n!λ+n) 2 C λ Γn+2λ) nx) n N 0 } is an orthonormal basis dµx) = Γλ)2 2 2λ π x2 ) λ 2 dx. 4.44) Let λ > 0 be fixed. Set V = {φ λ) z φ λ) z x) = z D} where φ z is defined by 2λ xz) 2xz + z 2 ) λ+ 4.45) for all z D and almost all x, ). Obviously V L 2, ), dµ). Define the function λ) : D D C of positive type by λ) z, w) = Γn + 2λ)n + 2λ) 2 w n z n 4.46) n!n + λ) 24

27 for all z, w D. By Theorem 2.3 and 4.2 the frame transform W λ : L 2, ), µ) C D λ defined by Wλ f ) z) = Γλ)2 2 2λ π 2λ xz) 2xz + z 2 ) fx) λ+ x2 ) λ 2 dx, 4.47) for all f L 2, ), µ) and z D, is unitary. Moreover, the set {a n n N 0 } is an orthonormal basis for C D where a C D n is defined by λ) a n z) = ) ) Γn + 2λ) W λ g n z) = n!n + λ) 2 n+2λ)z n = Γn + 2λ + ) n! ) 2 n + 2λ n + λ zn, 4.48) for all z D and n N 0. The following theorem characterizes the space C D λ). Theorem 4. The space C D λ) equals F 2λ as a set. Moreover, f F2λ < f C D λ) 2 f F2λ, 4.49) for all f C D λ). ) Note that z Γn+2λ+) 2 z n was an orthonormal basis for F n! 2λ. Now the statement easily follows by the estimate < n + 2λ n + λ 2 for all n N The semi-group generated by the Euler operator Define the Euler operator E on C D λ) by ) df Ef z) = z z), 4.50) dz for all z D and that z z n are eigenvectors with eigenvalue n. The domain of E can therefore be defined as {f = c na n {c n n} n N0 l 2 N 0 )}. The following lemma follows by straightforward calculations. Lemma 4.2 f, g) C D λ) = E + λ)e + 2λ) f, g ) F 2λ 4.5) for all f, g C D λ). 25

28 Obviously E +2λ is a positive operator with the eigenvectors a n with eigenvalues n+2λ. Therefore E +2λ) is well-defined and has the same set of eigenvectors a n with eigenvalue n + 2λ). The operator E + λ)e + 2λ) eliminates the factor n+λ n+2λ in 4.48). The theorem follows by Theorem 4.8. It is easily seen that E is a closed dissipative operator and therefore generates a contraction semi-group t e Et. The bounded) operator e Et has eigenvectors a n with eigenvalues e nt for all n N 0 and t 0. Let f C D and write fz) = λ) c nz n for all z D. Then, e te f ) z) = c n e tn z n = fe t z), 4.52) for all z D. Hence e te is a scaling of the functions in C D. Note that it is crucial that λ) t > 0, since only then we obtain e t z D for all z D. From standard theory of evolution equations it follows that E si) f = 0 e st e te fdt, 4.53) for all f C D. Since C D is a functional Hilbert space, this simplifies to an ordinary λ) λ) integral E si) f ) z) = e st e te f ) z)dt = for all f C D λ) and z D. Theorem 4.3 f, g) C D = f, g) F2λ + 2λ α) for all f, g C D α). Let f, g C E λ). Then e te f, e te g ) F 2λ e st fe tz )dt, 4.54) e 4λt dt 4.55) f, g) C D λ) = E + λ)e + 2λ) f, g ) F 2λ = f, g ) F 2λ λ E + 2λ) f, g ) F 2λ = f, g ) + λ E 2λ) f, g ) F 2λ F 2λ = f, g ) + λ e 2λt e te fdt, g ) F 2λ F 2λ = f, g ) F 2λ + λ 0 0 e 2λt e te f, g ) F 2λ dt. 26

29 = t/2 and use the self- For the sake of symmetry use the substitution t adjointness of e te to obtain the statement. Using 4.53) one also obtains an alternative formula for the reproducing kernel. Theorem 4.4 The reproducing kernel is given by the following integral z, w) = Γ2λ + ) Let w D. By 4.46), e λt 0 E + 2λ) E + λ) w ) z) = for all z D. Therefore, 2λ + e tz w dt 4.56) e tz w) 2λ+2 E + λ)w ) z) = Γ2λ + )2λ + zw) zw) 2λ+2 Γn + 2λ + ) w n z n = n! Γ2λ + ) zw) 2λ+ for all z D. The statement now follows by applying 4.53). 27

30 5 The classical Fourier and Laplace transforms 5. The Fourier transform as a frame transform on a Gelfand triple In his section the classical Fourier transform on L 2 R) is interpreted as a special kind of frame transform. Of course the function x e iωx is not an element of L 2 R) for all ω R. Nevertheless, by making use of a Gelfand triple H I L 2 R) H I for which the space H I does contain this functions, a frame transform can be defined which leads to the Fourier transform. In this section the following two result will be used, which we will not prove. A result due to Cramer: H n x) < k n2 n/2 e x2 /2 for all x R and n N, where k is a constant. See [C]. Mehler s formula for Hermite polynomials [ ] H n x)h n y) w/2) n = w 2 ) 2xyw x 2 + y 2 )w 2 2 exp, 5.) n! w 2 ) which is valid for all x, y R and w D = {z C z < }. See [MOS, 5.5, pp. 252]. Define the operator R on L 2 R) by Rf ) x) = x 2 d2 dx 2 )f) x), 5.2) for all f DR) = {f L 2 R) d2 dx 2 f L 2 R) Mf L 2 R)}, where Mf)y) = y 2 fy) for all y R. It is well-know that this operator has a complete set of eigenvectors {g n n N 0 } with eigenvalues λ n = 2n +. Moreover, R is a positive unbounded and self-adjoint unbounded) operator, with bounded inverse. Next, we construct the Gelfand triple H I R) L 2 R) H I R). 5.3) Note that the sets { gn 2n+ n N 0 }, {g n n N 0 } and {2n + )g n n N 0 } are orthonormal bases for respectively H I R), L 2 R) and H I R). The following theorem is very easily proved by using the Fourier transform on L 2 R). But since our goal is to construct the Fourier transform, it would be sloppy to use it in the proof. Therefore the proof is somewhat lengthy and uses the result by Cramer. 28

31 Theorem 5. The space H I R) is a functional Hilbert space with reproducing kernel I given by I x, y) = 2 n n!2n + ) 2 π H nx)h n y)e x2 /2 e y2 /2, 5.4) for all x, y R. Moreover, the sum converges absolutely. Since H I R) as a set is equal to DR) and hence a subset of L 2 R). Strictly spoken, its elements are not functions but classes of functions. Therefore it would be better but tiresome) to write [h] instead of h for its elements. The main purpose of this proof is to prove that making those classes is obsolete. It turns out that every class [h] in H I R) has a unique analytic representant. Instead of considering classes, we only consider the analytic representant. Let f H I R). The set { g 2n+ n n N 0 } is an orthonormal basis for H I R). Hence f = a ng n H I R) if and only if a n 2 2n + ) 2 <. Define the function f N by f N x) = N a ng n x), for all N N and x R. Since f N x) f M x) N a n g n x) N a n 2 2n + ) 2 N n=m n=m N a n 2 2n + ) 2 N n=m f N f M I kπ 2 2 n=m k 2 2n + ) 2 n=m g n x) 2 2n + ) 2 for all N > M N and x R, the sequence of analytic functions {f N } N N uniformly on R to an analytic function f. By uniform convergence, it is also true that Rf N converges uniformly to R f. In addition R f L 2 R) and [ f] = [f]. By a same estimate as above we obtain kπ fx) f I 2 2, for all x R. Hence H I is a functional Hilbert space. The rest of the statement follows by Theorem 4. and Mehler s formula. 29

32 Define for α R the vector φ α H I φ α := i n g n α)g n = i n g n α) 2n + 2n + )g n. 5.5) This vector is well-defined because {2n+)g n n N 0 } is an orthonormal basis in H I R) and {g n α)/2n + )} n N0 l 2 N 0 ) for all α R. Set V = {φ α α R}. Define the function : R R C of positive type by α, β) = φ α, φ β ) E = n= g n α)g n β) 2n + ) 2, 5.6) for all α, β R. Obviously = I and therefore C R = H IR). By Theorem 2.3, the frame transform W : V H I defined by W f ) α) = φα, f) I, 5.7) for all f H I R) and x R is unitary. With a simple argument it follows that V = H I. To this end let f V, then 0 = φ α, f) I = i n g n α) 2n + 2n + )g n, f) I, 5.8) for all α. It is obvious that α φ α, f) I belongs to H I R), by the fact that {2n + )g n n N 0 } is an orthonormal basis for H I R) and that { gn n N} is an orthonormal basis for H I R). Moreover, 5.8) implies that 2n + )g n, f) I = 0 for all n N and 2n+ therefore f = 0. Since W is a unitary map from H I R) onto H I R) and R a unitary map from H I onto L 2 R) and from L 2 R) onto H I R) it follows that RW R is a unitary map from L 2 R) onto L 2 R). Note that for f = a ng n we have RW Rf = i n a n g n. 5.9) Hence RW Rf equals α φ α, f for all f H I R). Finally we connect the transform RW R to the classical way of introducing the Fourier transform. Suppose f H I R), then fx) = fx) + x 2 ) for almost all x R. +x 2 By definition Mf L 2 R) and therefore f L R) since f is a product of two L 2 R) functions. Let α R. Then e iαx fx) dx, 5.0) R 30

33 is well-defined for all f H I R). Write f = a ng n. Then by Mehler s formula e iαx fx) dx = lim e itαx t 2 x 2 +α 2 ) +t 2 fx) dx 5.) 2π R t + t2 )π R = lim g n x)g n α)it) t R n fx) dx 5.2) = lim g n α) g n x)it) n fx) dx 5.3) t = lim t R a n g n α)it) n = a n g n α)i n = φ α, f. 5.4) Hence we can conclude that φ α equals x e iαx in distributional sense. In addition, ) RW Rf α) = e iαx fx) dx, 5.5) 2π for all α R and f H I R). 5.2 The Laplace transform R In this section we interpret the Laplace transform as a frame transform. Let H = L 2 0, )) and C + = {z C Rez) > 0}. Moreover define the set V = {φ z L 2 0, )) z C + } where φ z t) = e zt, 5.6) for almost all t 0, ) and all z C +. Lemma 5.2 The set V has a dense span in L 2 0, )). As a consequence of the approximation theorem of Weierstrass, the set of polynomials is dense in L 2 0, ), dx). Suppose f V for some f L 2 0, )). Then in particular 0 = e tn ft) dt = x n f logx)) x dx = x n f logx)) dx, for all n N. Hence f = 0. Define the function : C + C + C of positive type by z, w) = φ z, φ w ) L2 0, )) = 0 e zt e wt dt = z + w, 5.7) 3

34 for all z, w C +. Since V = H, the frame transform L : L 2 0, )) C C+ defined by ) Lf z) = φz, f) L2 0, )) = e zt ft) dt, 5.8) for all f L 2 0, )) and z C +, is a unitary map. Note that C C+ functions. Fix x > 0 and define f x by 0 consists of analytic f x t) = { e tx ft) t > 0 0 t 0, 5.9) for all x 0 and f L 2 0, )). Then it is obvious that for all x > 0 we have f x L R) and F f ) ) x y) = 2π Lf x+iy) for almost all y R. Next we prove that l.i.m.x 0 fx = f 0. Let ε > 0. Then take a R > 0 such that f 0 t) 2 dt < ε R/B 0,R ) ε Take x > 0 such that e xr < 2. Then we see that f0 2 f x f 0 2 = f 0 f x )t) 2 dt + f 0 f x )t) 2 dt R/B 0,R B 0,R = f 0 t) e xt ) 2 dt + f 0 t) e xt ) 2 dt R/B 0,R B 0,R f 0 t) 2 dt + e xr ) 2 f 0 t) 2 dt < ε R/B 0,R B 0,R 2 + ε 2, 5.2) and the statement is achieved. Since l.i.m. x 0 fx ) ) = f 0 we also obtain l.i.m. x 0 F f x = F f 0 and l.i.m. x 0 y Lf x + iy) = F f0. Note that F f 0 L2 R) = f L2 0, )). As a result we obtain F, G) C C + for all F, G C C+ F w) = lim x 0 2π = lim F x + iy)gx + iy) dy, 5.22) x 0 2π R. In particular R F x + iy) dy, 5.23) x iy + w for all w C + and F C C+. We now give an alternative proof of 5.23) using the theory of complex functions. Let F C C+, w C+ and 0 < x < Rew). Let R > 0 be sufficiently large, then F w + x) = 2πi F z)dz 5.24) S I S II z w x by the residu Theorem and the fact that F is analytic. The contours are given by S = {z C Rez) = x Imz) < R} and S II = {z C z x = R Rez x) > 0}. 32

35 ir S II w O x S I x R First we prove that the integral over S II vanishes if R tends to infinity for all F C C+ and all x > 0. To this end, we make use of the inequality F z) = φ z, f) Rez) f 2 = Rez) F, 5.25) C C+ for all F = Lf and z C +, which follows by the Cauchy-Schwartz inequality. Let F C C+. Parameterizing the contour S II gives 2πi S II z w x F z)dz = 2π π/2 π/2 for all F C E. The integrand can be estimated by Re iφ w F x + Reiφ )Re iφ Re iφ w F x + Reiφ )Re iφ dφ F C C + Re iφ w x + R cos φ), 5.26) for all x > 0, F, w C + R > 0 and φ [0, 2π). Let ε > 0.There exists η > 0, R > 0 such that w R e iφ > R /2 for all φ [ π/2, π/2] and 2η < 2πε and < ε x 3 F C C + x+r. For R > R cos η 3 F the same inequalities are satisfied C C + and we find π/2 2π Re iφ w F x + Reiφ )Re iφ dφ π/2 2π Re iφ w F x + Reiφ )Re iφ dφ π/2 33 = 2π π/2 π/2 π/2 π/2+η π/2 = F C C+ 2π < 2ε 3 + ε 3 F C C + Re iφ w x + R cos φ) dφ + 4η x π/2 η π/2+η + ) 2π 2η) + x + R cos η π/2 π/2 η = ε. 5.27)

36 Hence we proved that for all F C C+ F w + x) = lim R we have F z) dz. 5.28) 2πi S I z x w Parameterizing S one finds F w + x) = F x + iy) dy. 5.29) 2π iy + x + w R Now let x 0 and by continuity of F we proved the reproducing property. Note that the structure of the space C C+ resembles structure of the space F 0. This resemblance will be clarified by a conformal mapping that relates the two spaces. First we construct an orthonormal basis in C C+. The set {e n : x e x/2 L n x) n N} is an orthonormal base in L 2 0, ). Lemma 5.3 The image of e n is given by ) z W en z) = 2 )n z +, 5.30) )n+ 2 for all z C + and n N 0. Hence, the set {a n : z z 2 )n z+ 2 )n+ } is an orthonormal basis for C C+. Let n N 0. Then W en ) z) = n! 0 e zx e +x/2 d ) n[e x x n ] dx dx e zx e +x/2 d ) n [e x x n ] dx dx = 0 + n! z 2 ) 0 =... = n! z 2 )n e zx e +x/2 e x x n dx = n! z 2 )n ) n d ) n e z+ 2 )x dx dz 0 = n! z 2 )n ) n d ) n dz z + 2 = z 2 )n z + 2 )n+ 0 for all z C +. 34

37 Corollary 5.4 For e zx the following decomposition holds e zx = for all z C + and x > 0. L n x)e x/2 z 2 )n z +, 5.3) )n+ 2 One readily verifies that the Möbius transform defined by z z 2 z+ 2 D = {z C z < }. Moreover, the inverse is given by z +z 2 2z T : C C+ CD by T f ) z) = z f for all f C C+ and z C+. maps C + onto. Define the operator ) + z, 5.32) 2 2z Lemma 5.5 The image of the basis element a n is given by T an ) z) = z n 5.33) for all z D and n N 0. Let n N 0. Then +z 2 ) + z ) T an z) = z a n = z 2 2 2z z + ) +z n+ 2 2 z +z 2 = ) n z + z + z) n ) z n+ = 2 ) n+ = z n z + + z + +z z ) n for all z D. Corollary 5.6 The operator T defines a unitary map from C C+ onto F 0. Note the resemblance between 5.22) for C C+ and 4.38). 35