Optimal Bell Functions for Biorthogonal Local ... - CiteSeerX

Optimal Bell Functions for Biorthogonal Local Trigonometric Bases KAI BITTNER Abstract

A general approach for biorthogononal local trigonometric bases in the twooverlapping setting was given by Chui and Shi. In this paper, we give error estimates for the approximation with such basis functions. In particular, it is shown that for a partition of the real axis into small intervals one obtains better approximation order if polynomials are reproduced locally. Furthermore, smooth trigonometric bases are constructed, which reproduce constants resp. linear functions by only one resp. a small number of basis functions for each interval.

Key Words. Local trigonometric bases, wavelets, biorthogonality AMS subject classi cations. 41A25,41A30, 42A10

1 Introduction Often it is useful to investigate the local properties of a signal. Therefore, many applications in signal and image processing use basis functions which are local in time and frequency. Because the most signals have both temporal and spectral correlation such basis functions have good approximation properties, i.e., one can obtain a reasonable approximation using only a few basis functions. One example of special interest are wavelets, where the basis functions are translates and dilates of one particular function. Another way to construct an orthogonal basis is to consider functions of the type k ik
?j ; k; j 2 Z; j = w(
? j )e where w is called the window or bell function. From a well-known theorem of Balian and Low it follows that if the family f j;k g is a frame for L (R) then either t w(t) 2= L (R) or  w^() 2= L (R). Thus, the functions j;k cannot be well localized in both time and frequency. In particular, from w^() 2= L (R) it follows that w0(t) 2= L (R), such that e.g. for bell functions with compact support we cannot obtain a frame or Riesz basis consisting of smooth functions. One can overcome this problem using certain sets of cosine or sine functions which form an orthonormal basis of L ([0; 1]) instead of the exponential monomials e ik
. In this 2

(

)

2

2

2

2

2

2

2

Fachbereich Mathematik, Universitat Rostock, Universitatsplatz 1, D-18051 Rostock, Germany, ([email protected]) 

1

way, smooth orthogonal local trigonometric bases were introduced by Malvar [11]. Here the so called \two-overlapping setting" is considered, where the window functions have compact support such that w(x ? r)w(x ? s)  0 if js ? rj > 1. The \Malvar bases" were independently discovered by Coifman and Meyer [7] in a generalized nonuniform setting, where the uniform spacing is replaced by an arbitrary partition

: : : < a? < a < a < : : : of R with aj ! 1 for j ! 1. An expository representation of these results can be found in [2]. As another example let us mention the Wilson bases described by Daubechies, Jaard and Journe [9] for window functions with arbitrary support. In [1], the connection of the both approaches is shown. Bivariate orthogonal local trigonometric bases are investigated in [14]. It is known that many applications only need the Riesz stability instead of the orthogonality of the basis functions. Furthermore, for non-orthogonal bases one obtains more freedom to choose the bell functions. Therefore, biorthogonal local trigonometric bases were introduced by Matviyenko [12] and by Jawerth and Sweldens [10], where one has orthogonality only between rk and sl for r 6= s. A more general approach is given by Chui and Shi [5, 6], where two arbitrary functions need not to be orthogonal. In [3] the results of Chui and Shi were applied to the bivariate case. In this paper, we consider local trigonometric functions of the form 1

k j (x) := wj (x)

s

0

1

?
x ? aj  aj ? aj cos k + aj ? aj  : 2

1 2

+1

+1

The following two questions arise, naturally. 1. What is the best choice for the splitting points aj ? 2. How one should choose the window functions wj ? The rst problem can be solved by an adaptive algorithm, where the splitting points are allowed to depend on the signal. Such algorithms where investigated for orthogonal bases by many authors, see e.g. Coifman and Wickerhauser [8, 13]. A disadvantage of this approach is that the determination of the best basis requires a higher amount of computation time. In particular for image processing in the bivariate situation it has to be investigated whether an adaptive algorithm is better than an explicit splitting in small uniform squares, like for JPEG or MPEG, with appropriate bell functions. This is a reason to discuss the second question. For that purpose one should investigate a large assortment of possible bell functions. Therefore, it is important to consider not only orthogonal bases, but more generally Riesz bases. In [10] and [12] window functions are introduced, for which the basis functions have good approximation properties for certain function classes. In particular, the approach of Jawerth and Sweldens [10] to consider smooth basis functions which reproduce constant functions is very powerful in many applications. Unfortunately, both papers [10, 12] are based on the assumption of orthogonality between rk and sl for r 6= s. This restriction leaves a lot of possible bell functions out of account. 2

The aim of this paper is to improve the approximation properties using the more

exible approach of Chui and Shi. To give a proper description of the approximation properties, we investigate the error for the approximation by local trigonometric bases. In particular, we are interested how the error behaves for a ner splitting of the real axis with a xed number of basis functions for each interval. We show that for such approximations arbitrary smooth bell functions do not imply a small error. On the other hand, applications in image processing like e.g. JPEG work with a xed number of basis functions for each interval. Therefore, in the second part of the paper we investigate local trigonometric bases which reproduce polynomials with a few basis functions. The paper is organized as follows. In x2, we recall the approach investigated by Chui and Shi. As a main tool for our purposes, a folding operator is introduced. With the aid of this folding operator, we establish error estimates for the approximation with smooth basis functions in x3. In Theorem 3, it turns out that the approximation becomes better for smooth bell and test functions if we have suciently many basis functions per interval. Furthermore, we investigate the error for the uniform splitting in more detail. In x4, we show that local trigonometric bases which reproduce polynomials have better approximation properties. Therefore, we investigate in the following sections how one can construct such bases. In x5, we present a method to construct bell functions of arbitrary smoothness such that the constant is reproduced by only one bell function per interval. The reproduction of linear functions is investigated in x6. Note, that the new results in x5 and x6 could be only established using the very general approach of Chui and Shi. Finally, in x7 we show that for the two-overlapping setting there does not exist a local trigonometric basis such that a polynomial of higher degree can be reproduced by a nite number of basis functions per interval.

2 Biorthogonal local trigonometric bases In the following, we will consider biorthogonal, local trigonometric bases in the so called two-overlapping setting, investigated by Chui and Shi [5, 6]. Here, we shortly recall the de nition of these bases. Let be given sequences (aj )j2Z, (aj )j2Z, (a?j )j2Zwith aj < aj  a?j < aj and aj ? a?j = aj ? aj for j 2 Z: We de ne now bell functions wj : R ! R with support [aj ; aj ]  supp(wj )  [a?j ; aj ]; (1) which yields a two-overlapping setting, i.e., supp wj \ supp wi has measure zero for jj ? ij > 1. Further, we consider the functions +

+

+

+1

+1

+ +1

+1

Cjk (x) :=

s

2

?
x ? aj  cos k +  1 2

aj ? aj aj ? a j which for each j 2 Zform an orthonormal basis of L ([aj ; aj ]). Now we introduce the cosine wavelets k k j (x) := wj (x) Cj (x); j 2 Z; k 2 N : +1

2

+1

+1

0

3

wj wj+1

wj-1

_

aj

+

aj

_

aj

aj+1

+

aj+1

aj+1

Figure 1: An example for bell functions in the two-overlapping setting. Note, that the functions Cjk can be replaced by

q

Sjk (x) := or

8q > aj > < q Djk (x) := > aj > :q

?

?aj sin k +

2

aj+1

1 2




x?aj aj+1 ?aj 



?aj ;

k = 0; j even, x?aj ?aj cos k aj ?aj ; k = 1; 2; : : : ; j even, x?aj aj ?aj sin k aj ?aj ; k = 1; 2; : : : ; j odd as well as suited mixtures of them (cf. [2, 5, 6]). Because we obtain similar assertions for all these bases we will consider here only the cosine bases fCjk g. 1

+1

2

+1

+1

2

+1

+1

1

0.5

-0.5

1

0.5

1.5

-0.5

-1

Figure 2: Bell function (dashed line) with corresponding basis functions lines). To investigate the basis properties of

M j (x) := M

w (x) := j

k j

we introduce the matrices

 w (x) j

wj (2aj ? x) ?wj? (x) wj? (2aj ? x) 1

4

1



0

j

and

6

j

(solid

(cf. [5, 6]), as well as the total folding Operator Tw (cf. [3, 10]) de ned by  T f (xT)w f (x) := wj (x) f (x); f (x)  for aj < x < a?j ; w T f (2a ? x) := M j (x) f (2a ? x) ; for aj < x < aj : +

w

j

+1

(2)

+

j

Note that this de nition determines Tw f uniquely a.e. on R. We want to emphasize that the introduction of Tw is motivated by the following observation. Since Cjk is even with respect to aj and odd with respect to aj we obtain that +1

Z

Z

aj+1

k j (x) f (x) dx =

R

aj

Cjk (x) Tw f (x) dx

(3)

if the integrals are well{de ned. Furthermore, if f and wj are smooth, then Tw f is smooth in [aj ; aj ] and there exists a smooth extension which is even with respect to aj and odd with respect to aj . The functions jk form a Riesz basis of L (R) i the folding operator Tw and its inverse operator are bounded, (cf. [3]). In [3, 5, 6] one nds also the following statements on the Riesz bounds. The best lower Riesz bound is given by minfAj ; Aj g A := kTw? k?L !L = jinf 2Z +1

+1

2

2

1

0

2

1

2

2

with

Aj := 1

ess inf? jwj (x)j ; 2

x2(aj ;aj+1 )

s

+


(x) Aj := ess?inf kM ?j (x)k? = ess?inf
j2(x) ? j4 ? j det M j (x)j ; x2 aj ;aj x2 aj ;aj 1

2

+

(

q

2

)

2

2

2

+

(

)

where kAk := (AH A) is the spectral norm of the matrix A and 2


j (x) := jwj (x)j + jwj (2aj ? x)j + jwj? (x)j + jwj? (2aj ? x)j : Analogously, the best upper Riesz bound is given by B := kTw kL !L = sup maxfBj ; Bj g 2

2

2

0

2

2

1

j 2Z

2

1

1

2

2

with

Bj := ess sup jwj (x)j ; ? 1

2

x2(a+j ;aj+1 )

s


(x) Bj := ess ?sup kM j (x)k = ess ?sup
j2(x) + j4 ? j det M j (x)j : x2 aj ;aj x2 aj ;aj 2

(

+

2 2

)

2

2

(

+

)

In particular, the operator Tw is bounded i all bell functions wj are contained in L1 (R). 5

Furthermore, dual basis functions are given by ~jk := w~j Cjk with the dual bell

8 ; if aj  x < a?j ; > w x j > < wj? aj?x ? w~j (x) = > wj Majj x ?x; if aj?  x < aj ; > : M j x ; if aj  x < aj ; +

1 ( ) ) 1 (2 det ( ) ) +1 (2 +1 det +1 ( )

+1

+

+ +1

+1

0; otherwise: We are interested in the question how the smoothness of the bell functions is connected with the smoothness of the dual bells. For this, we establish the following assertion.

Theorem 1 Let the functions f j;k g form a Riesz basis. The dual bell functions w~j , j 2 Z are contained in C m(R) i wj 2 C m(R) for all j 2 Z. Proof. Assume that wj 2 C m(R). From the Riesz basis property and the continuity of wj it follows that there exists values "j > 0 such that wj (x) = 6 0 for x 2 [aj ?"j ; a?j +"j ] and det M j (x) = 6 0 for x 2 [aj? ?"j? ; a?j +"j ]. Hence, w~j is piecewise smooth. Indeed +

+

we have

1

1

+1

+1

+1

+1

j ? (2aj ? x) m ([a ? " ; a? + " ]); f (x) := wdet 2 C j? j j j? M j (x) f (x) := w 1(x) 2 C m([aj ? "j ; a?j + "j ]); j j (2aj ? x) m ([a ? " ; a? + " ]): f (x) := wdet 2 C j j j j M (x) 1

1

+

+

2

+1

3

+1

1

1

+1

+1

+

+1

+2

j +1

+1

+2

Obviously, the functions f , f , f and 0 coincide piecewise, i.e., f (x) = 0 for x 2 [aj? ; a?j ]; f (x) = f (x) for x 2 [aj ; aj ]; f (x) = f (x) for x 2 [aj ; a?j ]; f (x) = 0 for x 2 [aj ; aj ]: 1

2

3

1

2

1

3

2

+

1

+ +1

3

+1

+1

+2

Therefore, it holds that f k (a?j ) = 0, f k (aj ) = f k (aj ), f k (a?j ) = f k (a?j ) and f k (aj ) = 0 for k = 0; : : : ; m. Hence, we obtain w~j 2 C m(R). Note that this assertion holds even if aj = a?j , because none of the intervals above shrinks to one point in this case. Analogously, we can show that wj 2 C m(R) follows from w~j 2 C m(R), because the functions wj are the dual bells with respect to w~j .  ( ) 3

( ) 1

+ +1

+

( ) 1

+

( ) 2

+

( ) 2

+1

( ) 3

+1

+1

3 Error estimates

3.1 The non-uniform grid

Let us now investigate the approximation properties of local trigonometric bases. Here we will consider the error in the norm of L (R) for the approximation of a smooth function 2

6

by a nite number of smooth basis functions for each interval. Because the bell functions have compact support one can obtain analogous local estimates in the same way. In the sequel, we consider the Sobolev spaces Wpm(R) with the norm m X

kf kWpm R := ( )

 =0

kf

! =p 1

 kp Lp (R)

( )

:

For our further investigations we need the following assertion. Lemma 2 If f 2 W m(R) and wj 2 W1m(R) then

X

2

kDm (T

j 2Z

w f )kL2 ([aj ;aj+1 ])  2

m ?
X m

2



 =0

kTw  kL ( )

2

R)!L2(R)kf

2(

m? ) k2 2 < 1: L (R)

(

Proof. Because the matrices D Mjw (x) and Mjw  (x) only dier in the sign of the second column, we have the equality ( )

kD Mjw (x)k = kMjw  (x)k ( )

2

and it follows that

kTw kL ( )

2

R)!L2(R) = kTw( ) kL2 (R)!L2(R) ;

2(

where Tw is the operator we obtain by replacing in (2) wj and Mj by w  and D Mj , respectively. With the Leibniz formula we have ( )

X

( )

kDm (T

j 2Z

w f )kL2 ([aj ;aj+1 ]) 2

m ?
X = k m Tw f m? kL R  m ?
X m kT  k m? k  w L R !L R kf L R:  ( )

(

)

2

2(

)

=0

2

 =0

2

( )

2(

)

2(

)

(

)

2

2(

)

From wj 2 W1 (R) we conclude that Tw  is bounded and the lemma is shown. ( )



If f jk g is a Riesz basis then any function f 2 L (R) has the representation

f=

XX

j 2Zk2N0

with the coecients

fjk

= hf; ~jk i =

Z

aj+1

2

fjk

k j

Tw f (x)Cjk (x) dx : ~

aj

Now we consider the approximation XX k fj SNf := j 2Zk < wM?xx ; w~(x) = > M x? ; : 0;

?
? j
, where

(7)

2 2

h

if ? < x < ; (8) if < x < ; otherwise: Analogously to Theorem 3, we conclude the following error estimate. Corollary 4 For Nj = N 2 N, f 2 W m(R) and w 2 W1m(R) we can estimate the approximation error of the partial sum SN f by pB ?  ?m
m?
w  (x)k kf k m : (9) kf ? SNf kL R  m N + ?m max h ess sup k M W R  m  x2 ; = (1 ) det ( ) (1 ) det ( 1)

1 2

1 2

1 2

3 2

2

2(

)

0

~( )

1 2

[0 1 2]

2

2 (

)

?
Proof. With w~j() = h? w~() h
? j and Nj = N the assertion (9) follows immediately from Theorem 3.  This corollary does not imply a good approximation order for a xed N and h ! 0, because the right-hand side in (9) is a positive constant for suciently small h. In particular, for an arbitrary bell function we cannot expect that the approximation order depends on h, as stated in the following. 9

Lemma 5 Let N 2 N be xed. There exists a bell function w~ 2 W1m(R) such that sup lim sup kf ? SN f kL R  c > 0: 2(

kf kW2m =1 h!0

)

We choose a function f ? 2 W m(R) with kf ?kW m = 1 and f ?(x) = c for ? < x < . For h = l , l 2 N we estimate 1 2

Proof.

3 2

kf ? ? S

? 2 N f kL2 (R)

2

1

0

2

Z= l? X X k hc w~(x) Cj (x) dx =B  B j kN j kN ?= X Z= k w~ (x) Cj (x) dx = c : = Bc kN l? X X 1

0

3 2

1

jhf ? ; jkij2

0

0

=0

=0

3 2

2

1 2

2

2

0 0

?1=2

Obviously, one can nd a bell w~, such that for every N 2 N it exists a k  N with

Z=

3 2

?1=2

w~(x) Cjk (x) dx 6= 0

(10)

and thus c > 0. Because c is independent of l we have lim kf ? ? SN f ?kL R  c

l!1

and the assertion follows immediately.

2(

)



Therefore, we investigate how the bell function w resp. the dual bell w~ has to be chosen, such that our cosine wavelets have good approximation properties for a xed small N , too. Obviously, this can be achieved only if (10) does not hold for any k  N for a suciently large N .

4 The reproduction of polynomials An important property of a basis is how polynomials are represented. Let n be the class of (algebraic) polynomials of degree less or equal n. We say a basis reproduces the polynomials p 2 n, if on any nite domain these polynomials are represented with a nite number of basis functions. If a function f can be approximated well by a polynomial for each interval (a?j ; aj ) we can expect a good approximation by the functions f j;k : j 2 Z; k < N g. For example, suppose we have an image with a constant background. Surely, we do not want to spend many coecients in the representation of the background. In what follows, we show that for a basis which reproduces polynomials one obtains a good approximation even if N is small. + +1

10

Theorem 6 Let N 2 N and a bell function w be given such that each polynomial p 2 n 0

has the representation

p=

Then for N  N0, that

X NX? 0

1

hp; ~jk i jk :

j 2Z k=0 m m > n, f 2 W2 (R), w~ 2 W1m(R) and for suciently n+1 kf ? SN f kL2 (R)  C h 1 m kf (n+1)kW2m?n?1 (R);

(11) small h it holds

?N +


(12)

2

where C depends only on w~ , m and n.

P

Proof. First we will estimate the terms kN jfjk j2 separately. Without loss of generality, we restrict ourselves to j = 0. Because m > n the function f is in W2n+1(R) and has the Taylor expansion

Zx (x ? t)n n f  ?h
X
? k+ n ?t + h
dt : x f f x+ h = k! n! | {z } | {z } ( )

( +1)

2

2

2

=0

0

u0 (x+h=2)

u1 (x+h=2)

Since u is a polynomial of degree n we obtain for k  N the coecients f k = hu ; ~ ;ki. Analogously to Theorem 3, one shows 0

X

kN

jf k j2 0

0

m ?
 ?
? m X m   N+ 1 2

2

 =0



2

w m?





(x) h u L ess sup M x2 ; = ~(

1

0

2

)

2

2

(0 1 2)

( ) 1

0

2

?h=2;3h=2])

2 ([

:

(13)

The higher order derivatives of u are given by 1

8 Zx > (x ? t)n? f n ?t + h
dt; <
 ? u x + h = > (n ?  )! : f  ?x + h
; ( ) 1

( +1)

2

2

0

( )

2

for   n; for  > n:

Applying the Cauchy-Schwarz inequality, we obtain for   n that

Zh Zx (x ? t)n? ?
n h dt dx = f t + (n ?  )! ?h Zx h Zx Z ?
 ((n ?1  )!) (x ? t) n? dt f n t + h dt dx: ?h 2

ku 

kL

( ) 2 2 ([ 1

?h=2;3h=2])

( +1)

2

0

2(

( +1)

)

2

2

0

0

11

Using Holder's inequality it follows that

ku  kL

( ) 2 2 ([ 1

?h=2;3h=2])

h Zx Z  ((n ?1  )!) (x ? t) ?h Zh x n? 1

Zh n? dt dx f n ?t + h
dt

2(

2

0

)

( +1)

2

?h

= ((n ?  )!) 2 2(n ?  ) + 1 dx f n L n?

= ((n ?  )!) (2nh? 2 + 1)(n ?  + 1) f n Inserting this into (13), we obtain 2(

)+1

( +1)

2

2

?h=2;3h=2])

2 ([

0

2(

+1)

( +1)

2

kf ? SN f kL 2

R B

2(

)

0

XX j 2ZkN

jfjk j

1 2

2

 =0

+

?h=2;3h=2]) :

2

2 ([

2

n ?
 ?
? m X m  2B  N + 0

L



2

n

ess sup kM w m? (x)k ((n ?h  + 1)!) f n x2 ; = ~(

 =n+1

2



2( +1)

2 2

( +1)

2

!

 ess sup kM w m? (x)k h f L R : x2 ; =

(0 1 2)

m ?
X m

)

~(

(0 1 2)

)

For suciently small h the assertion follows immediately.



2 2

2

( )

L 2

R)

2(

2

2(

)



5 Reproduction of constants We study now the question how a constant function is represented. In particular, we demand that the constant is already reproduced by the functions j; (cf. [10]), i.e., 1=

X

j 2Z

0

By (3), this is equivalent to

k; = h ~jk ; p i = 0

x 2 R:

j;0(x);

Zj

(14)

+1

Tw p (x) Cjk (x) dx ~ 0

0

j

with p (x)  1. Because the functions Cjk = C ;k(
? j ), k 2 N , form an orthonormal basis of L ([j; j + 1]), the equality (14) is also equivalent to p Twp (x) = C ;k (x) = 2 cos  x; x 2 (0; 1): (15) In the following, we want to determine smooth dual bell functions, which satisfy (15). To describe such a bell, we consider the left part 0

0

2

~ 0

0

0

2

l := w~ ? [

12

;

1 1 ] 2 2

r := w~(
+ 1) ?

and the right part

[

of w~ separately. From (15), we obtain

;

1 1 ] 2 2

p

p

l(x) + l(?x) = 2 cos  x and r(x) ? r(?x) = ? 2 sin  x : 2

(16)

2

If we demand a symmetric bell function, i.e., l(x) = r(?x), then we obtain only the function w~ (x) = sin(  x +  ) for x 2 (? ; ) (cf. [10, 12]). This bell function generates an orthonormal basis. Unfortunately, w~ is continuous but not continuously dierentiable. To obtain a smoother bell we have to cut out the symmetry. It turns out, that with a nonsymmetric bell we can construct cosine wavelets of arbitrary smoothness which reproduce a constant. 0

2

1 2

4

3 2

0

2.5 2 1.5 1 0.5

-0.5

1

0.5

1.5

Figure 3: A bell function w (dashed line) and it's dual bell w~ (solid line) for which the basis functions j reproduce a constant. We have chosen l (x) = sin 2x and r (x) = ? cos x such that w; w~ 2 C (R). The dashed-dotted line is the graph of w~ (x) = sin(  x +  ). 0

1 2

1

0

2

2

1 4

1

4

Theorem 7 Assume the functions l ; r 2 C m([? ; ]) satisfy l (x) = ?l (?x) and r (x) = r (?x) as well as 1

1

1

1 2

1

1 2

1

1

k d   = dxk sin( x + ) x ? ; dk sin(  x +  ) = dx x k for k = 0; : : : ; m. Then the dual bell w~ given by 8 sin(  x +  ) + l (x) for ? < x < ; < w~(x) = : sin(  x +  ) + r (x ? 1) for < x < ; 0 otherwise ?
l ? ?
rk k

( ) 1

( ) 1

1 2

2

4

1 2

2

4

2

4

1

2

4

1

satis es (14) and is contained in C m (R).

13

1 2

=

3

=2

1 2

1 2

3 2

1 2

(17) (18)

(19)

Proof. Because l1 is odd and r1 is even it holds that l1(x) + l1(?x) = 0 and r1(x) ? r1(?x) = 0. Hence, p Tw~ p0 (x) = Tw~ p0(x) = 2 cos 2 x; i.e., (14) holds. Obviously, w~?is piecewise smooth. Using that l1 is odd and r1 is even,

? (k ) (k ) 1 1 one shows that l1 2 = r1 ? 2 follows from (17) and (18) and therefore, it is shown that w~ 2 C m(R).  0

6 Reproduction of linear functions

Now we want additionally, that our basis reproduces linear functions, i.e., for all a; b 2 R there exist coecients j;k 2 R such that

X NX?

1

c x+c = 1

0

j 2Zk=0

j;k jk (x); x 2 R:

This implies in particular, that our basis has to reproduce a constant. Therefore, let us assume that w~ satis es (15). Because Tw is linear, we need further that

Z

1

h ~jk ; p (
? j )i = Tw p (x) C k (x) dx = 0; for k  N 1

~ 1

0

0

with p (x) = x. Hence, we have to look for a bell w~ which satis es (15) and 1

Tw p (x) = ~

1

X

N ?1 k=0

for certain N 2 N and ak 2 R. j;k =  ;k (c + jc ) + ak c . The equality (20) leads to 0

0

1

X

N ?1 k=0

1 2




(20)

Then one obtains the coecients j;k by

1

x(l(x) ? l(?x)) = and

?

ak cos k + x;

?

X

N ?1 k=0

?




ak cos k + x 1 2




(?1)k ak sin k + x = r(x) ? r(?x) + x(r(x) + r(?x)) +1

1 2

p

= ? 2 sin  x + x(r(x) + r(?x)) : 2

Together with (16), we conclude 2x l(x) = 2x r(x) =

p

2 x cos  x + 2

p

X

N ?1 k=0

ak cos(k + )x;

2(1 ? x) sin  x ? 2

14

(21)

1 2

X

N ?1 k=0

(?1)k ak sin(k + )x: 1 2

(22)

To obtain good approximation order we have to impose a certain smoothness on the bell. On the other hand, linear functions should be reproduced by a small number of basis functions. Therefore, we want to construct bell functions of highest smoothness for a xed N .

Theorem 8 For N  2 and m = N ? 1, there exists one and only one bell function w~ 2 W1m(R) with supp w~  [? ; ] such that hp ; ~jk i = k; hp ; ~jk i = 0; k  N : 1 2

3 2

0

0

1

This function is given by

8p N? >  x + P a (k )x > cos k x > k p > > >

xx?? cos  x + P ak (kx? )x  p NP? k > >  k > 2 ? (?1) (2k + 1)ak > k > :0 1

2 2

2

1

2 1

2

4

+1 2

cos

2(

=0

1

for x 2 [? 12 ; 21 )nf0g; for x = 0; for x 2 [ 21 ; 32 )nf1g;

+1 2

2

=0

2 2 2 2

cos

1)

(23)

for x = 1; otherwise;

=0

where the coecients ak are the solution of the linear system of equations

X

N ?1

X

N ?1 k=0

k=0

ak = 0

(24)

s(k ? ) (2k + 1) ak =  s(?) ?p4 s(1 ? ) ;  = 0; : : :; m ? 1 2

with

s(k) =

? 2 cos k +

p

1 2

8 1 >
 = < ?1 > : ?11

for k  0 for k  1 for k  2 for k  3

2

(25)

mod 4; mod 4; mod 4; mod 4:

Proof. 1. First we show w~ 2 W1m for each solution of (24), (25). From (21) with x = 0 it follows that w~ is bounded i (24) holds. In this case,

8p < l(x) = : p

2 2 2 2

and

P cos  x + a N ?1

2

k

k=0

cos





k+ 12 x 2x

for x = 0; otherwise

 8  p PN ? k > 2 ? ( ? 1) (2 k + 1) a < k k   NP ? r(x) = > p ?
k : x ? 1 sin  x ? k (?1)k ak x x 1 =0

4

2 2

1

1

2

=0

15

sin

1 + 2 2

for x = 0; otherwise

(26)

(27)

are both from C 1(R). Therefore, w~ 2 C m? (R) i ?
?
= r  ??
; r  ?
= 0;  = 0; : : : ; m ? 1 : (28) l  ? = 0; l  In the following we denote c(x) := dxd cos(x). By induction one deduces from (26) and (27) 1

( )

l

( )

1 2



( )

1 2

1 2



p

N X? (2k + 1)  ??

   2    1 (x) = a c k + x + 2 2 c 2 x 2x k k 2 ! p      ? 2    ? x l ? (x) ? 2 2 c? 2 x 1

( )

( )

1 2

1 2

=0

(

and

r

( )

1

1)

(29)

1





N X? k (2k + 1)  ?? 1 (x) = 2x (?1) ak c k + 2 k    p2  1     + 2 2 1 ? x c 2 x p  !    ?  2c x  ? x r ? (x) ? 2 2  2 1

+1

1 2


x


=0

+1

(

1

1)

(30)

for  2 N , x 6= 0 and l ? (x) := 0 resp. r ? (x) := 0. Now we consider the derivatives of l(x) and r(x) for x = ? and x = . For this purpose we use several equalities for s(k) in the following. Obviously, the correctness of these equalities for all integer numbers follows from the correctness for 0, 1, 2 and 3. From (29) and with c(?(k + )=2) = s(k ? ) it follows that ?
p ?
?   2 l ? N X? ?
p ?
= s(k ? ) (2k + 1)ak ?  s(?) ?p4 s(1 ? ) ? 2   2 l ? ? ; 2 k ( 1)

0

( 1)

1 2

1 2

1 2

( )

2

1 2

1

(

2

1)

1 2

=0

such that (25) is equivalent to l  (? ) = 2 l ? (? ),  = 0; : : : ; m ? 1, and therefore also equivalent to l  (? ) = 0,  = 0; : : : ; m ? 1. Further, with c((k + )=2) = s(k + ) we obtain from (29) and (30) ( )

( )

1 2

1 2

(

1 2

1)

1 2

?
 p2 l  ?
? r  ??
  NX ?   ( )

( ) 1 2

2

1 2

   (2k + 1) ak s(k + ) + (?1)k s(k ?  ? 1) + p1 s() ? 3s(? ? 1) 2 k    ?
+ r ? ??
: ?
p + p2m s( ? 1) + s(?) +   2 2 l ? 2 In the same way, one shows with s(k + ) = (?1)k s(k ?  ? 1) that for  > 0 ?
? p2 l ? ?
+ r ? ??
 = p4 s( ? 1)  2 =

1

=0

(

2

2

1

(

1) 1 2

(

16

1)

1 2

1) 1 2

(

1)

1 2

and with s(k + ) = (?1) s(k ? ) it follows that

?
 p2 l  ?
? r  ??
  NX ?   ( )

( ) 1 2

2

= 2(?1)

1

k=0

1 2

(2k + 1) ak s(k ? ) ? p1 s(?) + p2m s(1 ? ) 2 2

!

such that (25) is equivalent to l  (1=2) = r  (?1=2),  = 0; : : :; m ? 1. Analogously, we show the equivalence of (25) to r  (1=2) = 0,  = 0; : : : ; m ? 1. Hence, (28) is equivalent to (25). Therefore, we obtain that w~ 2 C m? (R) i (24) and (28) are ful lled. Because l; r 2 C 1(R) the derivatives l m and r m are bounded. Hence, w~ m 2 L1 (R) and thus w~ 2 W1m(R). 2. We still have to show that the system of equations in Theorem 8 has a unique N ? with solution, i.e., the determinant of the matrix C = (cij )i;j ( )

( )

( )

1

(

)

(

)

(

)

1 =0

1

i = N ?1 cij = s(j ? i) (2j + 1)i for for i = 0; : : : ; N ? 2 does not vanish. To achieve this goal we show that the system of equations

X

N ?1 i=0

i cij = 0; j = 1; : : :; N ? 1;

has only the trivial solution. If we substitute i = ~i s(i ? 1) and multiply each equation with s(j ) we obtain the equivalent system of equations

X

N ?2 i=0

~i s(j ) s(i ? 1) s(j ? i) (2j + 1)i = ? N ? s(j ); j = 1; : : :; N ? 1: 1

With s(j ) s(i ? 1) s(j ? i) = (?1)ij , we have

X ? j


~i (?1) (2j + 1) i = ? N ? s(j ); j = 0; : : :; N ? 1:

N ?2

1

i=0

(31)

Note that the coecient matrix of this system is not a Vandermonde matrix, because the (N ? 1)-th row has the entries s(j ). However, the remaining entries of this matrix are elements of a Vandermonde matrix. Using this property we can formulate the following interpolation problem which is equivalent to (31). For the knot set x := 4j + 1 : j = ?  N  ; : : : ;  N ?  = (?1)j (2j + 1) : j = 0; : : : ; N ? 1 j the polynomial NX ? q(x) := ~i xi 1

2

2

2

i=0

should satisfy the interpolation conditions     q(xj ) = (?1)j N ? ; j = ? N ; : : : ; N ? : +1

1

1

2

17

2

Let us assume that

N ? 6=0. Then we have q(xj )q(xj ) < 0 and therefore q has a zero in   N N ? (xj ; xj ), j = ? ; : : : ; ? 1. This means we obtain N ? 1 zeros for a polynomial with degree N ? 2 and thus we conclude N ? = 0. Then it follows that q(x)  0, i.e., 1

+1

2

+1

1

2

1

i = ~i s(i ? 1) = 0; i = 0; : : : ; N ? 2 and therefore det C 6= 0. Substituting (26) and (27) in

8 l(x) < w~(x) = : r(x ? 1) 0

we obtain nally (23).

for ?  x < for  x < otherwise 1 2

1 2

3 2

1 2



1.5

1.5

1.25 1.25 1 0.75

1

0.5

0.75

0.25 -0.5

0.5 1

0.5

1.5

0.25

-0.25 -0.5

-0.5

1

0.5

1.5

Figure 4: The bell function w (dashed line) and the dual bell w~ (solid line) for N = 2 and N = 3. Note, that for N = 2 the bell w has poles in 0 and 1. The bell function w can be determined from the dual bell w~ by (8). We have still to investigate the question whether the bell function w resp. w~ generates a Riesz basis. It is well known that the functions f jk g form a Riesz basis with the bounds A and B i the dual system f ~jk g forms a Riesz basis with the bounds B ? and A? . Therefore, we consider here only the dual functions. For odd N we obtain a Riesz basis as stated in the following theorem. Theorem 9 Let N 2 N be odd. Then a Riesz basis is formed by the functions f ~jkg which are generated by the bell w~ from Theorem 8. 1

1

To prove this theorem we need the following assertion. Lemma 10 Let N 2 N be odd. If the coecients ak satisfy (25) then it holds that

a k ? a k? > 0; k = 1; : : : ; N ? : 2

2

1

1

2

18

Proof. From the proof of Theorem 8, namely from the regularity of the matrix C , it follows that for m = 1; : : :; N2?1 there exist uniquely determined coecients m; such that N X?2

m;?1 + m; s( ? 1)s(k ? )(2k + 1) = k;2m ? k;2m?1 (32) =0

for k = 0; : : : ; N ? 1 and n 2 N. We multiply each equation with s(k) and the equations read as follows

X

N ?2

s(k) m;? + 1

=0

?




m; (?1)k (2k + 1)  = (?1)m(k; m ? k; m? ): 2

2

1

Analogously to the proof of Theorem 8 we replace now (?1)k (2k + 1) by 4n + 1 with a suited n 2 Zand obtain (?1)n

m;?1

+

X

N ?2 =0

m; (4n + 1) = (?1)n(n;m ? n;?m )

for n = ? N ? ; : : :; N ? . Now we de ne 1

1

2

2

qm(x) :=

X

N ?2 =0

m; (4x + 1) :

Then it holds that (?1)n m;? + qm(n) = (?1)n (n;m ? n;?m ) Thus, we obtain for the even part of qm that ?(?1)n 2 m;? = qm(n) + qm(?n). Analogously to the proof of Theorem 8 we conclude now that ? = 0. Furthermore, it follows that qm is an odd function. With (25) and (32) we obtain 1

1

1

a m ? a m? = 2

2

1

=

X X

N ?1 N ?2 k=0 N ?2

X =0

=0

m;

s( ? 1)s(k ? )(2k + 1)

m; s( ? 1)

From (32) with k = 0 it follows that

 s(?)

!



ak

p ? 4 sp(1 ? ) : 2 2

PN ? s( ? 1) s(?) = 0 such that m;  ? 1 NX  2 =0

2

(?1) 4 a m ? a m? = ? p 2  m; ?
= ? p1 qm0 ? : 2 Thus, to prove the lemma we have to show that qm0 ( ) < 0. To achieve this goal we use that the polynomial qm 2 m? is uniquely determined by the interpolation conditions qm(n) = (?1)n(n;m ? n;?m ); n = ? N ? ; : : : ; N ? : 2

2

1

=0

1 2

1 2

1

1

1

2

19

2

Let polynomials `j;m be given by ` ;m(x) = ? nx and

8 > < ?`j? ;m (x) mx ??jj `j;m (x) = > `j? ;m(x) : `j? ;m(x) mx ??jj 0

for j < m; for j = m; for j > m: Obviously, it holds that qm = ` N ? = ;m. By induction on j one shows for an arbitrary
? 0 m 2 N that `j;m ? < 0 for j 2 N . Hence, ?
a m ? a m? = ?`0 N ? = ;m ? > 0 which proves the lemma.  1

1

(

2

2

2

2

2

2

2

1

1 2

2

1) 2 0

2

2

1

(

1 2

1) 2

We use now this result to verify Theorem 9. Proof. 1. In the rst part of the proof we deduce a condition for the coecients ak , which ensures Riesz stability. Because the bell functions are bounded we know that the upper Riesz bound B is nite. Thus, we have still to show that the lower Riesz bound A does not vanish. For r
(x) ? (det M (x)) > 0; 0 < x <
( x ) ? A = xess inf 2 ;= 2 4 to be positive, it is sucient that
(x) > 0 and det M (x) 6= 0. Because
(x)  4(det M (x)) we have only to show that det M (x) 6= 0 for 0 < x < . For x 6= 0 the even function det M (x) is given by det M (x) = l(x)r(?x) + l(?x)r(x) ?
? cos  x sinp k + x cos  x sin  x NX k = ? (?1) ak x 2x k ?
N X? sin  x cos k + x p : + ak 2 x k By addition formulas we deduce 0

0

2

0

1 2

2

(0 1 2)

2

1 2

2

1

2

2

2

=0

1

1 2

2

=0

det M (x) = sin2xx + Thus, det M (x) > 0 is equivalent to sin x > 2x

X

X

N ?1)=2

(

k=1

(a k? ? a k) sinp2kx : 2x 2

1

2

N ?1)=2

(

k=1

(a k ? a k? ) sinp2kx 2x 2

2

1

and hence for 0 < x < equivalent to 1 2

1>

X

N ?1)=2 p

(

k=1

2kx : 2(a k ? a k? ) sin sin x 2

20

2

1

1 2

One easily shows that sin 2kx = 2k: = lim sup sinsin2kx x x! sin x 0 i

p

2> 4

X

N ?1)=2

(

k ja k ? a k? j: 2

k=1

2

1

By Lemma 10 this is equivalent to

p

2> 4

X

N ?1)=2

(

k (a k ? a k? ): 2

k=1

2

1

2. In the second part of the proof we derive from this inequality regarding the coef cients ak a condition which does not explicitly depend on these coecients. To achieve this aim we have to consider the system of equations (25). From the proof of Lemma 10 it follows that

X

N ?1)=2

(

k=1

k (a k ? a k? ) = ? p1 2 2

2

1

with

q :=

X

N ?1)=2

(

k=1

X

N ?1)=2

(

k=1

?
? k qk0 ? = ? p1 q0 ? 2 1 2

1 2




k qk :

Thus, the polynomial q 2 N ? is uniquely determined by the interpolation conditions 1

q(n) =

X

N ?1)=2

(

k=1

qm(n) = (?1)n n; n = ? N ? ; : : :; N ? : 2

3. To prove the theorem we have still to show, that ?
?q0 ? < 2 :

1

2

1 2

To achieve this goal we show rst that the polynomial q has the form

q(x) =

X

N ?3)=2

(

l=0

clrl(x);

with r (x) := x and rl(x) := rl? (x) (x ? l)(x + l) as well as 0

1

l l cl := ((2?l1)+ 1)!4 : +1

21

1

(33)

Because q and rl are odd functions, we have only to show that the polynomial above satis es (33) for n > 0. For n  N ? we obtain 1

2

X

N ?3)=2

(

l=0

n? X (?1)l

4l (n + l)! l (2l + 1)! (n ? l ? 1)! n+l n? X l l (?1) 4 2l + 1 : = l

clrl(n) =

1

+1

=0

1

+1

=0

By induction we show now together that n? X 1

l=0

and

(?1)l+14l

n X

(?1)l+1 4l

n +l

n 2l + 1 = (?1) n = q(n)

n + l

= (?1)n (2n + 1): +1

2l Obviously, the assertions are true for n = 0 and n = 1. Then we conclude l=0

n X l=0

(?1)l+14l

and

n X

(?1)l+14l

n + l 

n X

(?1)l+14l

n + l ? 1 X n? 1

(?1)l+14l

n + l ? 1

2l 2l ? 1 + l = ?4 (?1)n n + (?1)n (2n ? 1) = (?1)n (2n + 1)

=

2l

n + l + 1 

l=1

=0

+1

n X

(?1)l+14l

n + l X n? 1

(?1)l+14l

n+l

2l + 1 2l + l = (?1)n n ? (?1)n (2n + 1) = (?1)n (n + 1) = q(n + 1): Finally, we consider the value of q0(x) for x = ? . Again, by induction one shows that

l=0

=

2l + 1

l=0

=0

+1

1 2

rl0

??
= (?1)l  (2l)!  : 4ll ! 2

1 2

With the Taylor expansion of arcsin we obtain

?q0

??
1 2

=

< and the assertion is proved.

X

N ?3)=2

(

l=0 1

X l=0

(2l)!

4l(l !)2 (2l + 1)

 (2l)! = arcsin(1) = 4l (l !) (2l + 1) 2 2



Unfortunately, for even N we have the following negative result. 22

Theorem 11 Let N 2 N be even and the functions ~jk be generated by the bell w~ from Theorem 8. Then the family f ~jk g is not a Riesz basis for L (R). Proof. Analogously, to the proof of Lemma 10 we conclude that for m = 0; : : : ; N ? 1 2

2

there exist uniquely determined coecients m; such that

X

N ?2

m;? + 1

=0

m; s( ? 1)s(k ? )(2k + 1) = k; m ? k; m 2

2

+1

(34)

for k = 0; : : :; N ? 1 and n 2 N. We multiply each equation with s(k) and obtain analogously to the proof of Lemma 10 (?1)n m;? + 1

X

N ?2 =0

m; (4n + 1) = (?1)n(n;m + n;?m? ) 1

for n = ? N ; : : :; N ? 1. With 2

2

qm(x) :=

X

N ?2 =0

m; (4x + 1)

it holds that (?1)n m;? + qm(n) = (?1)n(n;m + n;?m? ) and thus, we have qm(n) ? qm(?n ? 1) = ?(?1)n 2 m;? . Analogously to the proof of Theorem 8 we conclude now that ? = 0. Furthermore, it follows that qm(
? ) is an even function. With (25) and (34) we obtain 1

1

1

1 2

1

a m?a m 2

2

=

+1

X

N ?2 =0

m; s( ? 1)

 s(?)

p ? 4 sp(1 ? ) 2 2



?
= p1 m; ? p1 qm0 ? : 2 2 ?
Because qm(
? ) is even, it follows that qm0 ? = 0, and we deduce that for the solution of (24), (25) it holds a = p + a and a k = a k , k = 1; : : :; N ? 1. Using this assertion, we show that w~ vanishes for x = 1. By addition formulas we conclude 1 2

0

1 2

0

1 2

1 2

1

p

2

r(0) = ? 22 sin  x + 21x 2

2 +1

2

0p @ 2 sin  x 2

2

1  ? 
?
a k sin 2k + 1 + x ? sin 2k + 1 ? x A + k 0 p N= ? 1 p  X 2a k cos(2k + 1)xA : ? 2 sin  x + sin x @ 2 + X

N=2?1

1 2

2 +1

1 2

=0

=

2 1

2

2

2

2x

2

23

k=0

2 +1

Because the coecients ak satisfy (24) we obtain nally for x = 0 

w(1) = r(0) =

4

X

N ?1 k=0

ak = 0:

Thus, the matrix M w (x) is singular for x = 0. Because w~ is continuous, the assertion follows immediately from (7).  The table below shows the best possible Riesz bounds for small N which one can determine by (7) suciently precise. Apparently, for small odd N we obtain better Riesz constants. This and the smaller number of basis functions which are needed to reproduce a linear function are reasons to choose N not too large. ~

N a k , k = 0; : : : ; N ? 1 2 p ,? p 3 p , ? p  , p? 4 p? , ? p  , p? , p? 5 p? , ? p  , p?  , ?p  , ?p  6 ? p  , ? p  , ?p  , ?p  , ?p  , 7 ? p  , ? p  , ?p  , ?p  , 1

2

2

2

2

2+ 4 2

5 8

2

5 8

2

1

2 4 2

2 8 2

2+3 8 2

2

13+6 24 2

19+33 48 2

9

2

2

19+33 48 2

8

0

1

2

2 8 2

13 6 24 2

19+15 48 2

151+60 256 2

5 3 48 2

31 15 96 2

7 3 48 2

31 15 96 2

397 180 768 2

7 3 96 2

161 90 960 2

m A B 1 0 1 2 0:326::: 1:347::: 3 0 1:477::: 4 0:163::: 1:541::: 5 0 1:590::: ?p  , 6 0:108::: 1:637::: 

?p   73?p 30 , 320 2 7 3 96 2

0

131 60 3840 2

?p   1260 ?p 1260 ?p420 , ?757+1116 p  , ? 757+420 p  , 2629?p , 2629 , 1007 1536 2 1536 2 7680 2 7680 2 7680 2 1007?p 420 149?p 149?p 60 , 7680 602 , 7680 7680 2 2 ?757+1116  9731+3360  1120 ?1680 336 p ,? p , 2473?p p  , 835?p , 3097 , 1536 2 15360 2 5120 2 15360 2 3072 2 1523?p 672 8569?3360 2011?840 p  , 2161?840 p , 107520p2 , 215040 21504 2 2 215040 2

149 60 3840 2

7

0 1:671:::

8 0:081::: 1:696:::

Table 1: Coecients ak and corresponding Riesz bounds for small N .

7 Reproduction of polynomials with higher degree Now the question is, whether we can reproduce polynomials from n with n > 1. For the two-overlapping setting it turns out that we cannot construct a bell function such that polynomials of degree two are reproduced.

Theorem 12 There exists no basis f jk g such that for a certain N 2 N each polynomial p 2 n with n  2 has the representation p=

X NX?

1

j 2Zk=0

hp; ~jk i jk:

24

Proof. It is sucient to nd two polynomials from 2 such that for any basis jk only one of this polynomials has a representation of the above form. Indeed, the functions p0 (x) = 1 and p2 (x) = x2 are such polynomials as we will show in the following. Let us assume there exists a bell w~ such that

p = 0

X NX?

1

j 2Zk=0

ak jk

and p = 2

X NX?

1

j 2Zk=0

cj;k jk:

If we set ck = c ;k it follows for x 2 [? ; ] that 1 2

0

1 2

Tw p = l(x) + l(?x) = ~ 0

Tw p = x (l(x) + l(?x)) = 2

~ 2

From this equalities we deduce

x

2

X

N ?1 k=0

?




ak cos k + x = 1 2

X

N ?1 k=0 N ?1

X k=0

X

N ?1 k=0

?




ak cos k + x 1 2

?




ck cos k + x:

?

1 2




ck cos k + x 1 2

(35)

for x 2 [? ; ]. By analytic continuation we conclude, that the equation holds for all x 2 R. Since the right-hand side of (35) is 4-periodic for arbitrary coecients ck and the left-hand side is only if ak = 0, k = 0; : : : ; N ? 1, it follows that all coecients ak vanish. Hence, we do not have a basis. This proves the theorem.  1 2

1 2

The last result is a reason to consider not only the two-overlapping setting. If for three or more bells the intersection of the supports is not only a single point then one should have more freedom for the choice of a basis which reproduces linear functions. Using this freedom we can possibly nd a basis which reproduces polynomials of higher degree. In this way it could also be possible to construct smooth bases which reconstruct linear functions with fewer coecients than in the two-overlapping setting.

Acknowledgments The research of K. Bittner was supported by the Deutsche Forschungsgemeinschaft. The author thanks C. K. Chui for his support and the useful discussions.

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