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In the present paper we will consider a multivariate generalization of theShannon theory which is based on a semi-orthogonal Wavelet Analysis using

polysplines, and which has been recently developed in [9], [10], [1], [12], [11]. Inthe case of what we call spherical sampling our approach provides a formula forthe recovery of a function f from a band-limited class P V0 by taking its valuesover the spheres centered at the origin ofRn and having radii ej for all j 2 Z:In the second case, of what we call parallel sampling we recover a function ffrom a band-limited class P V0 by taking its values on all hyperplanes in R

n

dened by fx1 = jg for j 2 Z:An interesting feature of our results is that they use essentially all advances in

the one-dimensional sampling theorems for Riesz basis as developed by GilbertWalter in his book [22], see also the more recent paper [23].

2 Shannon-Walter sampling with exponential splines(Lsplines)

In the present Section we will construct a formula of Shannon type which isbased on exponential splines (these splines are called sometimes Lsplines), cf.[9].

2.1 Shannon sampling for Riesz basis according to GilbertWalter

First, we will provide the construction of Gilbert Walter of sampling and thecorresponding Shannon-Walter functions for semi-orthogonal Wavelet Analysisgenerated by a scaling function from a Riesz basis; the proofs follow the proofs

outlined in [22] for the case of orthogonal scaling functions.Let us assume that the real-valued function (t) is continuous and hasshifts f (t j)gj2Z which represent a Riesz basis for the Hilbert subspace V0in L2 (R) ; i.e.

V0 := closL2(R) f (t j) : j 2 Zg ;

and there exist two constants A ; B > 0 such that for arbitrary constants cjholds

A1X

j=1

jcj j2

1X

j=1

cj (t j)

2

L2(R)

B1X

j=1

jcj j2 ;

following the tradition we call scaling function of V0.For every element f 2 V0 we have

f(t) =1X

j=1

fj (t j) : (1)

Assume that the space V0 has the property that for every f 2 V0 with

f(j) = 0 for j 2 Z

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follows f 0: Then there should exist a basis fj (t)gj2Z of V0 such that aformula of Shannon type holds, i.e. for every f 2 V0

f(t) =1X

j=1

f(j) j (t) for t 2 R:

The problem of nding the proper conditions on and the basis fj (t)gj2Z has

been resolved by Gilbert Walter, [22]. Below we provide in detail his construc-tion.

It is well-known that to the Riesz basis f (t j)gj2Z there exists unique

dual Riesz basisne (t j)o

j2Zwhere duality meansD

(x j) ;

e (x `)

E= j`; (2)

cf. [21], [22]. Let us assume that the functions and e satisfy the followingasymptotic conditions: (t) = O

jtj1"

; e (t) = O jtj1" (3)

as t ! 1; t 2 R;

or, which is equivalent, there exist positive constants C1 > 0 and C2 > 0 suchthat for all t 2 R holds

j (t)j C1

(1 + jtj)1+"e (t)

C2

(1 + jtj)1+":

Then a direct estimate shows that the functionq(x; y) =

Xj2Z

e (x j) (y j) (4)is uniformly convergent on every compact set in R2.

On the other hand for every xed x1 2 R (or uniformly for x1 from a compactsubset in R ) we have the estimate

jq(x1; y)j2

Xj;k2Z

j (y j) (y k)jC22

(1 + jx1 jj)1+" (1 + jx1 kj)

1+" :

We have also

ZR j (y j) (y k)j dy ZRC1

(1 + jy jj)1+"

C1

(1 + jy kj)1+" dy

C21

ZR

1

(1 + jy jj)1+"dy

= C21

ZR

1

(1 + jyj)1+"dy < 1:

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The last implies the convergence of the integral

ZR

jq(x1; y)j2 dy:

Thus for every element f 2 L2 (R) the integralsZR

q(x1; y) f(y) dy

make sense. In a similar way follows that q(x; y1) 2 L2 (R) uniformly for y1from a compact subset in R:

Thus, from (1) and the representation for the dual basis we obtain the fol-lowing reproduction property for every f 2 V0;

Zq(x; y) f(y) dy = f(x) ; (5)Zq(x; y) f(x) dx = f(y) : (6)

Let us note that by the denition it is not clear whether the function q issymmetric.

Let us remark that from (4) it is obvious that

q(t j; 0) = q(t; j) : (7)

We summarize the main results of Gilbert Walters approach from [22] inthe following Proposition.

Proposition 1 Let (t) be a scaling function satisfying (3), i.e. (t) and itsdual dened by (2) satisfy (t) ; e (t) = O jtj1" as t ! 1 for t 2 R: Letus dene the function by

() :=Xj2Z

(j) eij : (8)

We assume that the function () satises the non-zero condition

() 6= 0 for 2 R: (9)

Then the following hold:1. The system of functions fq(t; j)gj2Z = fq(t j; 0)gj2Z represents a

Riesz basis of V0; where q is dened in (4)2. The uniquedual Riesz basis fSj (t)gj2Z corresponding to fq(t j; 0)gj2Z

satisesSj (t) = S0 (t j) :

Hence fS0 (t j)gj2Z is a Riesz basis as well.

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3. The following Shannon type formula holds for all f in the space V0:

f(t) = Xj2Z

f(j) S0 (t j) for t 2 R: (10)

4. The function S0 (t) has a Fourier transform which satises

cS0 () =b () ()

for 2 R: (11)

Remark 2 We will call the function S0 Shannon-Walterfunction. Formula(10) may be considered as the analog to the Shannon-Kotelnikov-Whittaker for-mula whereby the condition for band-limitedness of f (i.e. the compactness of

the support of the Fourier transform bf() ) is replaced by f 2 V0. In the contextof Wavelet Analysis the frequency variable of Fourier Analysis is replaced bythe index j.

Proof. 1. We will prove that the linear operator T which is dened by themap

The ( j)i (x) = q(x; j)

has a continuous extension to V0: For this purpose we will consider the space cV0of the Fourier transforms of the functions in V0:

Let us consider the operator T1 dened on the space cV0 as the operator ofmultiplication by (), i.e.

T1 [f] () := f() ():

Since by (3) the function has a fast decay it follows that () is a bounded,continuous and periodic function and satises

j ()j c0 > 0 for 2 R:

Hence, the linear operator T1 is bounded and its inverse dened by multiplication

with

()1

is also bounded.

Let us see thatT = F1T1F

where F denotes the Fourier transform and F1 its inverse. Indeed, it is enoughto check this on the elements e (x j) : We have

Fhe ( j)i () = eijbe ()

T1Fhe ( j)i () = ()eijbe ()

=X

`2Z

(`) ei`!

eijbe ()

=be () X

`2Z

(`) ei(`j);

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and taking F1 shows nally that F1T1F

he ( j)

i(t) = q(t j; 0) :

So we have proved that T is a bounded invertible operator, and by a basicresult in [21] (p. 30 ) it follows that the system fq(t; j)gj2Z is a Riesz basis ofV0:

2. By a theorem in [21] there is a unique Riesz basis Sj (t) which is biorthog-onal to fq(t j; 0)gj2Z ; i.e.

hq(t j; 0) ; Sk (t)i = jk :

On the other hand due to a change of variable = t k we have

hq(t j; 0) ; S0 (t k)i =

Zq(t j; 0) S0 (t k) dt

=

Zq( + k j; 0) S0 () d

= kj;0

which shows that the family of functions fS0 (t j)gj2Z is biorthogonal tofq(t j; 0)gj2Z : Hence, by the uniqueness it follows Sj (t) = S0 (t j) :

3. Equation (10) is obtained by the expansion off(t) in the basis fS0 (t j)gj2Z :Indeed, from

f(t) =Xj

fjS0 (t j)

and by the biorthogonality relation (6) and (7) follows

f(j) = hf; q(t j; 0)i = fj :

4. For proving (11) we need to apply the Shannon expansion (10) to the

function and to take the Fourier transform.

2.2 The exponential splines (Lsplines)

2.2.1 Preliminaries on exponential splines

At the beginning of Wavelet Analysis the splines played and important role (seesome history in [4] and [13]). In 1991 Chui and Wang have constructed thecompactly supported spline wavelets, cf. [2]. A characteristic feature of thisWavelet Analysis is that it is semi-orthogonal. Further Wavelet Analysis usingexponential splines has been initiated in [6] and has been discussed in detailin [9], [10]. We will refer to the monograph [9] for all details of the outline on

exponential splines following below.The exponential splines are dened by means of ordinary dierential opera-tors with constant coecients given by polynomials

L (z) :=NYj=1

(z j) ; (12)

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where j are some real constants. For simplicity sake we will denote by thenon-ordered vector of the numbers j

:= [1; 2;:::;N] ; (13)

where some of the j s may repeat (the number of repetitions of j is themultiplicityj of this j ); this notation is a convenient way to avoid every timeexplicitly writing the multiplicities of the j s. We will have the operator

L

d

dt

:= L

d

dt

:=

NYj=1

d

dt j

(14)

and the space UN of analytic solutions dened by

UN = UN () := f(t) : L d

dt f(t) = 0 for t 2 R (15)which is well known to be of dimension N, cf. [15]. The space UN is generatedby the exponents

tset for s = 0; 1;:::; 1

where 2 and is its multiplicity, where obviously we have

N =X

;

and the sum is over all dierent values in :The classical polynomial case is obtained when j = 0 for all j = 1; 2;:::;N:The basic objects which we will consider are the so-called cardinal expo-

nential splines (Lsplines ) generated by the operator L

ddt

: Namely, we

dene the space S by puttingS :=

f : f(t) 2 CN2 (R) \ C1 (R n Z) ; fj(j;j+1) 2 UN

; (16)

i.e. they are piecewise analytic solutions to the equation Lddt

f(t) = 0:

We will consider the Wavelet Analysis generated by the Bsplines arisingfrom the operator L

ddt

which are called sometimes T Bsplines (cf. [17]

these are cardinal exponential splines having minimal support). As in the classi-cal polynomial spline theory there exists up to a factor only one such T Bspline,i.e. a cardinal exponential spline QN [](t) = QN (t) 2 S supported in the in-terval [0; N] and satisfying

QN (t) > 0 for t 2 (0; N) ;

cf. [5], [2], [17].It is convenient to dene QN (t) by means of its Fourier transform (cf. p.274 in [9]), namely we have

dQN () = QNj=1 ej eiQNj=1 (i j)

for 2 R: (17)

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Many important quantities which we will need are dened by means of theT Bspline QN (t) :

In the case of the classical polynomial splines Schoenberg introduced thesocalled Euler-Frobenius polynomial. In order to dene a generalization of theEuler-Frobenius polynomials introduced by Schoenberg we will considerthe function (see Corollary 13:24; p. 235 in [9], Micchelli [14])

AN1 (x; ) :=1

2i

Z

1

L (z)

exz

ez dz; (18)

where the closed contour surrounds all points j 2 but excludes all zerosof the function e

xz

ez . If we put

r () :=N

Yj=1 ej

; s () :=

N

Yj=1 ej

(19)

then we obtain the following representation (see Corollary 13:25 p. 235 in [9])

N1 (x; ) = r () AN1 (x; ) (20)

where N1 (x; ) is a polynomial of degree N 1 of: The so-called Euler-Frobenius polynomial

N1 () := N1 (0; ) (21)

has degree N 2; cf. Corollary 13:25 in [9].We have the following important Proposition (cf. [14], [16], or Theorem

13:31 on p. 237 and Corollary 13:53 on p. 253 in [9]).

Proposition 3 The polynomial N1 () has exactly N 2 negative zeros.By means of the T Bspline we may dene the Euler-Schoenberg expo-

nential spline, (see p. 254 in [9] ):

N1 (x; ) =1X

j=1

jQN (x j) : (22)

Note that the word exponential has nothing to do with the exponential meantin the present paper, but it is related to the following easy to check property

N1 (x + 1; ) = N1 (x; ) : (23)

We have the following important (cf. p. 255; Proposition 13:55 and Theorem

13:56 on p. 256; Corollary 13:57 p. 256 in [9]).

Proposition 4 1. The Euler-Schoenberg exponential spline in (22) satises thefollowing relation for all x with 0 x 1;

N1 (x; ) =(1)N1

N1eP

jjN1 (x; ) : (24)

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2. If the vector is symmetric, i.e. satises = ; then

N1N2

; 1z = N1N

2; z for all z 2 C; (25)

and the polynomial N1 (0; z) 6= 0 satises

N1 (0; z) 6= 0 for all z 2 C with jzj = 1: (26)

In particular, we see that from (23) and (24) the following representationholds:

N1

N

2;

=

N2 N1 (0; ) (27)

= (1)N1 N2+1 exp0@Xj j1AN1 () :

We will use further the following basic

Lemma 5 Let the vector be symmetric, i.e. satises = : Then the following holds

jN1 (1)j N1 ei jN1 (1)j for 2 R:

Proof. By Proposition 3 the polynomial N1 () has precisely N 2 zerosvj < 0; and by Proposition 4, 1) and 2) they satisfy

vjvN1j = 1 for j = 1; 2;:::;N 2:

Hence, we have the representation

N1 () = DN2Yj=1

( vj) ;

where D is a non-zero coecient. By the reality of the zeros vj < 0 it follows

N1 ei = jDj N2Yj=1

ei vj = jDj N22Yj=1

ei vj ei 1vj

= jDj

N22

Yj=1 ei vj

2

jvj j

= jDj

N22

Yj=11 2vj cos + v2j

jvj j

jDj

N22Y

j=1

1 2vj + v2jjvj j

= jN1 (1)j :

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In a similar way we obtain

N1 ei jDjN22Y

j=1

1 + 2vj + v2jjvj j

= jN1 (1)j :

2.3 Checking the conditions of Gilbert Walter

Now we put for the scaling function

(t) = QN (t) : (28)

A fundamental result in [9] (Theorem 14:6 ) says that for arbitrary non-orderedvector the system f (t j)gj2Z represents a Riesz basis for the space V0dened by means of the cardinal splines S as

V0 = S \ L2 (R) ; (29)

the explicit values of the Riesz constants is found in [9].According to (17) the Fourier transform of the basic T Bspline satises the

asymptotic

b () = dQN () = NYj=1

ej ei

i j= O

1

jjN

!for ! 1;

and it is clear that the derivatives of b () satisfy similar asymptotic.As we mentioned before formula (2), the basis f (t j)gj2Z has a dual Riesz

basis generated by a unique function e, i.e. the set of functions ne (t j)oj2Z

generates a Riesz basis of V0: For the dual function e one has the Fourier trans-form (see p. 356 in [9] )

be () = C2S b ()2N1

hei (N; ei) for 2 R (30)with CS = exp

0

@1

2

N

Xj=1j

1

A;

where e = [; ] , i.e. it is the symmetrized vector of :Proposition 6 For arbitrary integer N 1 and arbitrary non-ordered vector of order N; the functions and e dened by (28) and (30) satisfy condition(3).

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Proof. The asymptotic of the function is clear since QN (t) is a com-pactly supported function, in particular, for every integer m 0 the following

asymptotic holds:

(t) = O

1

jtjm

for t ! 1:

Now from Lemma 5, applied to the vector e and to the corresponding func-tion 2N1

hei N; ei (whereby we use essentially the symmetry of the vectore ), and from formula (27), it follows that the function 2N1 hei N; ei isbounded, namely

j2N1 (1)j

2N1

he

i N; ei

j2N1 (1)j for 2 R;

here the Euler polynomial 2N1 () corresponds to the vector e:On the other hand, we have the following equality proved by integration by

parts:

e (t) = 12

Z11

eitbe () d

=1

2

Z11

1

1 + t2

1

d2

d2

eit

be () d=

1

2

1

1 + t2

Z11

eit

1 d2

d2

be () d:Let us estimate the last integral. Since 2N1 hei (N; ) is a polynomial in

; it follows that the derivatives of2N1hei N; ei with respect to have nice

asymptotic for ! 1: Now by the above remarks about the asymptotic ofthe derivatives of the function b () and the representation formula (30) followsthe inequality e (t) 1 + t2 C for t 2 R:This justies the asymptotic condition for e in (3).

For proving the asymptotic of and e we needed no restrictions on thevector : However the non-zero condition (9) is more restrictive. In fact, G.Walter has shown that the classical splines (this corresponds to the case of all

j = 0 in ) generate a simple and natural Shannon type formula only inthe case of odd degree cardinal splines, [22]. Similar is the situation with theexponential splines. For that reason we will restrict our attention to the case

N = 2p: (31)

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Proposition 7 1. We have the equality

() := Xj2Z

(j) eij = N1 [] 0; ei for 2 R;hence the non-zero condition (9) is equivalent to the condition

N1 [] (0; ) 6= 0 for jj = 1: (32)

2. For arbitrary symmetric non-ordered vector of order N = 2p and for given by (28) condition (9) holds, i.e.

() 6= 0 for 2 R:

Proof. By the denition of N1 in (22) and by (24) follows

Xj2Z

(j) j

= Xj2Z

(j) j

= N1 [] (0; )

=(1)N1

N1exp

0@ NXj=1

j

1AN1 () :(Please note the dierence between N1 [] and 2N1

hei since they corre-spond to dierent vectors). Now Lemma 5 gives us the lower and upper boundsofN1 ei from which we obtain immediately (9), i.e.X

j2Z

(j) eij 6= 0 for 2 R:

Note that by means of Lemma 5 we have used essentially the symmetry ofthe vector e and the fact that N = 2p: We will apply the above Proposition 7to the case of non-ordered vectors = k = [1;:::;N] given by

k :=

j = k for j = 1;:::;pj = k for j = p + 1; :::; 2p

; (33)

where k is some integer.On the other hand there are other vectors which are important for our

multivariate theory, which are not symmetric but close to being symmetric,and for which the non-zero condition (9) holds; they have been considered inthe paper [12] and called nearly symmetric there. For us the following vectors = k = [

1;:::;N] will be important:

k :=

1+j = k + 2j for j = 0;:::;p 1;

1+p+j = n k + 2 + 2j for j = 0;:::;p 1;(34)

here k is some integer.We have the following analog to Proposition 7.

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Proposition 8 Let = k be dened by (34). Then satises the non-zerocondition (9).

Proof. In the proof of Proposition 13 in the paper [12], we proved that

AN1 (0; 1) 6= 0

for every k 0: On the other hand by Proposition 3 the polynomial N1 ()can have only real zeros with < 0. Since by (20) and Proposition 4 we have

N1 [k](0; ) = s

1

AN1 (0; )

it follows that the function N1 [k](0; ) has no zeros for jj = 1:

Hence, in the above cases we may apply the results of G. Walter as presentedin the previous section. By Proposition 1, 5) it follows that there exists a

Shannon-Walter exponential spline S0 2 V0 = S \ L2 (R) such that itsFourier transform satises

cS0 () =b () ()

=b ()

N1 [] (0; ei): (35)

From the above Propositions we obtain the following:

Theorem 9 LetN = 2p and be symmetric non-ordered vector of order N; or be a vector given by (34). Then for every cardinal exponential splinef 2 V0there holds a Shannon type formula:

f(t) =1

Xj=1S0 (t j) f(j) : (36)

3 The multivariate case

Our main purpose is to nd a multivariate generalization of the Shannon-typeformula (10). For that purpose we will use a Wavelet Analysis which is gen-uinely multivariate. We will work in the framework of the polyharmonic WaveletAnalysis introduced and studied in [9]. It uses piecewise polyharmonic functions(polysplines) which generalize the piecewise polynomial functions (and splines)in the one-dimensional spline theory.

We assume that the integer p 1 is xed.We will consider the solutions to the polyharmonic equation

p

u (x) = 0

in annular domains, i.e. in annuli Aa;b = fx 2 Rn : a < jxj < bg given by two

numbers a;b > 0: Here

=nX

j=1

@2

@x2j

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is the Laplace operator and p is its p-th iterate.For every integer k 0 we will assume that we are given an orthonormal

basis of the spherical harmonics of homogeneity degree k; which we will denoteas usually by fYk;` (x)g

ak`=1 ; cf. [20], [9]. Thus we haveZSn1

Yk;`1 () Yk;`2 () d = 1`2 ;

here Sn1 is the unit sphere in Rn, d is the spherical area measure, and j isthe Kronecker symbol taking on 1 for j = 0 and 0 elsewhere. We will work withthe spherical variables x = r where r = jxj ; 2 Sn1: It is known that forspecial functions of the form f(r) Yk;` () one has (see [9], p. 152 )

p (f(r) Yk;` ()) = Lp

(k)f(r) Yk;` ()

where the operator L(k) is given by

L(k) :=@2

@r2+

n 1

r

@

@r

k (n + k 2)

r2:

On the other hand by the change v = log r we obtain another operator Mk;psatisfying

Lp(k)

d

dr

= e2pvMk;p

d

dv

where

Mk;p (z) :=

p1Yj=0

(z k 2j) (z + n + k 2 2j) ;

cf. [9], Theorem 10:34: Here it is important that unlike Lp(k) ddr the opera-tor Mk;p

ddv

has constant coecients. Thus we are within the framework of

the exponential polynomials and we may consider exponential splines for theoperator L = Mk;p dened by the vector = k = [1;:::;N] given by (34).

The following representation for polyharmonic functions in annulus is a fun-damental result (cf. Theorem 10:39 in [9]):

Proposition 10 Every function u 2 C1 (Aa;b) satisfying pu (x) = 0 in theannulus Aa;b permits the representation

u (x) =Xk;`

uk;` (r) Yk;` () ; for x = r (37)

where uk;` (r) is a solution to

Lp(k)

d

dr

uk;` (r) = 0 for a < r < b:

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Let us assume that the integer k 0 is xed.Further we consider the space of exponential splines generated by the op-

erator L = Mk;p ddv : We dene the space of cardinal exponential splines andthe Wavelet Analysis corresponding to the operator L = Mk;p

ddv

as in (29),

namely we put

eVk0 = ef 2 L2 (R) : Mk;p ddv ef(v)j(j;j+1) = 0; and ef 2 Cp2 (R) : (38)

Let us recall that in the case of odd dimension n (since there are no repeatingfrequencies in of (34)) the analytic solutions of Mk;p

ddv

f = 0 are linear

combinations of the exponentialsne(k+2j)v; e(nk+2+2j)v

op1j=0

:

Respectively, by the change of the variable v = log r we obtain the space

Vk0 :=n

f : f(r) = ef(log r) ; for ef 2 eV0o : (39)Finally, we may apply Theorem 9, and we will obtain the Shannon-Walter

exponential spline given by (35) which we denote further by eS(k)0 (v) : By (36)eS(k)0 (v) satises a Shannon-type formula, i.e. for every element f 2 eV(k)0 wehave the equality ef(v) = 1X

j=1

ef(j) eS(k)0 (v j) : (40)By means of the change r = ev we have f(ev) :=

ef(v) and we put

S(k)0 (r) :=

eS(k)0 (log r) :Hence, for every f 2 Vk0 we have the Shannon-type formula

f(r) =1X

j=1

f

ej

S(k)0

rej

: (41)

3.1 Asymptotic ofS(k)0 for k ! 1

It is essential to see what is the asymptotic of eS(k)0 (v) and ofdeS(k)0 () as k ! 1:This will be important for the existence and the regularity of the multivariateShannon type function.

We assume as above that

(t) = QN [](t) (42)

for the nonordered vector = k = [1;:::;N] given by (34) (or (33) below),and that the non-zero condition (9) holds.

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By formula (35) we have

deS(k)0 () = cN ()N1 [k](0; ei) : (43)On the other hand by formula (24) we have the representation

N1 [k](0; ) =(1)N1

N1exp

0@Xj

j

1AN1 () : (44)Lemma 11 The Fourier transform of the Shannon-Walter function S(

k)0 and

its Fourier transform satisfy the following asymptotic:1. uniformly for 2 R

deS(k)0 () = O1k for k ! 1;2. uniformly for all t 2 R

eS(k)0 (t) = O (1) for k ! 1:Proof. By (42) it follows directly that for appropriate constants k0 0 and

C > 0 and for all 2 R and k k0 we havecN ()

=

QNj=1

ej ei

QN

j=1 (i j)

Cepk

(2 + k2)p:

Now we will take care of the denominator in (43). We will apply a subtleestimate of Theorem 11 in the paper [12] (please note the conventions in thepaper, where the size of the vector is numbered through N + 1 and here weuse N ). It says that there exist two constants C1; C2 > 0 and an integer k0 1such that for k k0 holds

C1kN1

jAN1 (0; )j C2

kN1for jj = 1:

From relation (20) we obtain

N1 () = r () AN1 (0; )

and from (44) we obtain

N1 [k](0; ) = s

1

AN1 (0; ) :

From (19) it follows that the asymptotic of the polynomial s () for k ! 1with = ei is

s () ekp:

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Hence, for k ! 1 and for appropriate constant C3 > 0 we have

deS(k)0 () C3 epk

(2+k2)p

epk

k2p1

= 1k

:

2. By denition we have

eS(k)0 (t) = Z11

eitcSk0 () d;hence by a direct estimate and a change of variable 0 = =k we obtain eS(k)0 (t) C3k2p1 Z1

1

1

(2 + k2)pd

= C03:

Obviously the results of Lemma 11 hold for the Shannon function S(k)0 (r) :=eS(k)0 (log r) in the variable r:

3.2 The Shannon polyspline

We have seen that for the vector k dened by (34) the non-zero condition (32)

is satised and we may dene the Shannon-Walter exponential splines S(k)0 :

We will call the following function (actually it is understood in a distribu-tional sense) Shannon polyspline kernel:

S0 (r;;) :=1

Xk=0ak

X`=1S(

k)0 (r) Yk;` () Yk;` () (45)

=1Xk=0

S(k)0 (r) Zk (; )

=1Xk=0

eS(k)0 (log r) Zk (; ) ;where Zk (; ) is the zonal harmonic of degree k, cf. [20]. Let us note that forevery xed the function S0 (r;;) is a polyspline ofr, and for every xed the function S0 (r;;) is a polyspline of r:

We will characterize the smoothness-convergence properties of this series interms of Sobolev spaces. Let us recall that the function h dened by the series

h (x) := Xk;` hk;`Yk;` ()belongs to the Sobolev space on the sphere Hs

Sn1

for a real number s if

and only if Xk;`

1 + k2

sh2k;` < 1;

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cf. section 7 and section 22 in [19].

Lemma 12 As a function of (or ) the Shannon polyspline S0 satises

S(r; ; ) 2 HsSn1

; S(r;; ) 2 Hs

Sn1

where

s = n +3

2 "

for every number" > 0:

Proof. We know that for some constant C > 0 and for all k 0; and` = 1; 2;:::;ak holds

jYk;` ()j Ckn21;

see e.g. [18], [8]. Also, and one has the estimate

ak C1kn2;

cf. e.g. [20].

Since the coecients of the series (45) are hk;` = S(k)0 (r) Yk;` () we have

the following inequalities

Xk;`

1 + k2

sh2k;`

Xk;`

1 + k2

sC2kn2 C2

1Xk=0

1 + k2

sk2n4

which proves the theorem.

3.3 The polyharmonic Wavelet Analysis

In [9] we have introduced a polyharmonic Wavelet Analysis by consideringthe following basic space

P V0 := closL2(Rn)

f(x) : pf(x) = 0 on ej < r < ej+1; x = r; f 2 C2p2 (Rn)

:

Let us note that the radii ej appear in a natural way since by the represen-tation (37) and Proposition 10 every function u 2 P V0 has coecients uk;` (r)satisfying

uk;` 2 Vk0 ;

where the spaces Vk0 are those dened in (39). Hence, we have the following

decomposition in the sense of L2 metric (cf. [9])

P V0 =1Mk=0

akM`=1

Vk0

which is generated by the expansion (37).

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As we have seen above, the Shannon polyspline kernel (45) is a function insome Sobolev space Hs and might be a distribution in the case of a negative

exponent. Hence, the formulas which we will write will be understood in adistributional sense. For two functions (distributions) f 2 Hs

Sn1

and g 2

HsSn1

with expansions

f() =Xk;`

fk;`Yk;` () ; g () =Xk;`

gk;`Yk;` () for 2 Sn1

we have the scalar product on Sn1 dened by an integral understood in adistributional sense

hf; gi :=

ZSn1

f() g () d =Xk;`

fk;`gk;`:

Note that the innite sum which gives sense to the above integral is absolutely

convergent for f 2 Hs Sn1 and g 2 Hs Sn1 :In our main Theorem below we will use the spaces P V0 to mimic the one-

dimensional cardinal spline space V0; and to formulate an analog to the Shan-non type formula.

Theorem 13 For every function f 2 P V0 \ C(Rn) the following Shannon

type formula holds:

f(r) =1X

j=1

ZSn1

S0

rej

; ;

f

ej

d;

where S0 (r;;) is the Shannon polyspline function dened in (45), andthe integral is understood in distributional sense.

Proof. Since by denition f 2 C(Rn) it follows that for all j 2 Z we havef

ej

2 L2; hence the following L2convergent expansion for every xed r > 0holds true

f(x) =1Xk=0

akX`=1

fk;` (r) Yk;` () =1Xk=0

akX`=1

efk;` (log r) Yk;` () ;where x = r:

We obtain in distributional sense the following equalities, for v = log r;

1Xj=1

ZSn1

S0

rej

; ;

f

ej

d =1X

j=1

1Xk=0

akX`=1

S(k)0

rej

fk`

ej

Yk;` ()

!

=1Xk=0

akX=1

0@ 1Xj=1

S(k)0 rej fk` ejYk;` ()1A=

1Xk=0

akX`=1

0@ 1Xj=1

eS(k)0 (v j) fk` ejYk;` ()1A :

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On the other hand since for every k 0 we have fk` (ev) 2 Vk0 , by theShannon formula in (40) we obtain

1Xk=0

akX`=1

0@ 1Xj=1

eS(k)0 (v j) fk` ej1AYk;` () = 1X

k=0

akX`=1

fk` (ev) Yk;` ()

= f(ev)

= f(r)

This ends the proof.

Let us remark that the sampling

ff() : 2 Zng

is the one used in the usual multivariate generalizations of the Shannon typeformula. In Theorem 13 we use though the sampling

fk;`

ej

: k 0; ` = 1; 2;:::;ak; j 2 Z

which denes a new multivariate sampling paradigm.

4 The polysplines on parallel strips

We consider very briey for every integer p 1 the Wavelet Analysis generatedby the polysplines on strips (with periodic conditions), and the correspondingShannon polyspline function and Shannon formula.

We will introduce some convenient notations:

x = (t; y) 2 Rn

t 2 R and y 2 Tn1;

Tn1 := [0; 2]n1 :

The polyspline Wavelet Analysis on strips is dened by means of the followingspace, cf. [9],

P V0 := closL2(Rn)

f(t; y) : pf(t; y) = 0 for t =2 Z; f 2 C2p2R Tn1

;

these are functions which are piecewise polyharmonic on every strip (j;j + 1) Tn1y ; for all j 2 Z , and which are 2periodic in the variables y: The functions

f 2 P V0 permit the representation

f(t; y) =X

2Zn1

f (t) eih;yi;

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where for every 2 Zn1 the function f (t) is a solution to the equation

d2dt2

jj2p f (t) = 0 for t =2 Z;jj2 =

Xj

2j :

In the present case we have the vector = k = [1;:::;N] given by (33)for k = jj. For this = k and for every integer p 1 we obtain the operator

Lp(k) :=

d2

dt2 k2

p: (46)

Now as in (29) we have similar denition of the cardinal exponential splinespaces, namely we put

Vk0 :=

f 2 L2 (R) : L

p

(k)

d

dt

f(t)j(j;j+1) = 0; and f 2 C

2p2 (R)

:

For the vector k dened in (33), by Proposition 7 condition (32) alwaysholds. Hence, for every k 0 there always exists the Shannon-Walter expo-

nential spline S(k)0 dened in (35).

Remark 14 The case k = 0 corresponds to the classical cardinal splines of odddegree considered by G. Walter in [22].

Since all basic lemmata have been proved above also for this type of vectors,we obtain the following Shannon type formula:

Theorem 15 For every f 2 P V0 \ C(Rn) holds the Shannon type formula

f(t; y0) =1X

j=1

ZTn1

S0 (t j;y;y0) f(j;y) dy;

where S0 (t ; y; y0) is the Shannon polyspline distribution dened by

S0 (t ; y; y0) =

X2Zn1

S()0 (t) e

ih;yieih;y0i:

Again we may compare the usual multivariate sampling ff() : 2 Zng withour sampling paradigm

f (j) : 2 Zn1; j 2 Z

:

Finally, an idea for future research might be the error analysis of the Shannontype formulas of Theorem 13 and Theorem 15.

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References

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[17] Schumaker, L. L., Spline Functions: Basic Theory, J. Wiley, New York,1981.

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[20] Stein, E. M. and G. Weiss, Introduction to Fourier Analysis on EuclideanSpaces, Princeton University Press, Princeton, 1971.

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[22] Walter, G., Wavelets and Other Orthogonal Systems with Applications,CRC Press, Boca Raton, FL, 1994.

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Corresponding author:1. Ognyan Kounchev: Institute of Mathematics and Informatics, Bulgarian

Academy of Sciences, 8 Acad. G. Bonchev St., 1113 Soa, Bulgariakounchev@math.bas.bg; kounchev@gmx.de2. Hermann Render: Departamento de Matemticas y Computacin, Uni-

versidad de La Rioja, Edicio Vives, Luis de Ulloa s/n., 26004 Logroo, Spainrender@gmx.de

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