The decay of the Walsh coefficients of smooth functions

Transcription

1 The decay of the Walsh coefficients of smooth functions Josef Dick Astract We give upper ounds on the Walsh coefficients of functions for which the derivative of order at least one has ounded variation of fractional order. Further, we also consider the Walsh coefficients of functions in periodic and non-periodic reproducing kernel Hilert spaces. A lower ound which shows that our results are est possile is also shown. Mathematical Suject Classification 2: Primary 42C1 Keywords: Wavelet, Walsh series, Walsh coefficient, Soolev space, smooth function 1 Introduction In this paper we analyze the decay of the Walsh coefficients of smooth functions. Walsh functions wal k : [, 1 {1, ω,..., ω 1 }, where k is a nonnegative integer and ω = e 2πi/, were first introduced in [11] and further early results were otained in [2, 6]. See for example [9] for an overview. It is well known that Walsh functions form a complete orthonormal system of L 2 [, 1, see [2, 7]. In analogy to our aim for Walsh functions, consider Fourier series for a moment: A classical result says that the kth Fourier coefficient of an r times School of Mathematics and Statistics, University of New South Wales, Sydney 252, Australia. josef.dick@unsw.edu.au 1

2 differentiale function decays with order k r. An analogous result for the Walsh coefficients of r times differentiale functions has een missing in the literature and will e provided here. The Walsh coefficients of functions which satisfy a Hölder condition were already considered in [6]. Here we consider the decay of the Walsh coefficients of functions which satisfy even stronger smoothness assumptions, i.e., have at least one smooth derivative. It has long een known from [6] that the only asolutely continuous functions for which all Walsh coefficients decay faster than 1/k are the constants. Here we refine this result e showing that for r times differentiale functions, the Walsh coefficients decay with order a 1 a minv,r, where k = κ1 a κ v av 1 with < κ 1,..., κ v < and a 1 > > a v >. I.e., only the coefficients k which have only one non-zero digit in their ase expansion decay with order 1/k, the others decay faster. We also prove a lower ound which shows that this result is est possile. The question of how the Walsh coefficients of smooth functions decay plays a central role in numerical integration of smooth functions. In [3, 4] this decay was implicitly used to give explicit constructions of quasi-monte Carlo rules which achieve the optimal rate of convergence for the numerical integration of functions with smoothness r > 1. Throughout the paper we use the following notation: We assume that 2 is a natural numer, and that k N where N denotes the set of natural numers has ase expansion k = κ 1 a κ v av 1, where v 1, < κ 1,..., κ v <, and a 1 > > a v >. For k = we will assume that v =. Let the real numer x [, 1 have ase representation x = x 1 + x 2 +, 2 with x i < and where infinitely many x i are different from 1. For k N we define the kth Walsh function y wal k x = ω κ 1x a1 + +κ vx av, where ω = e 2πi/. For k = we set wal x = 1. For a function f : [, 1] R we define the kth Walsh coefficient of f y ˆfk = and we can form the Walsh series fx k= fxwal k x dx ˆfkwal k x. 2

3 Among other things, it was shown in [4] that if a function f : [, 1] R has r 1 derivatives for which f r 1 satisfies a Lipschitz condition, then ˆfk C r a 1 a minr,v for some constant Cr > independent of f and k. An explicit constant was also given in [4]. In this paper we improve upon the results in [3] and [4] in several ways. We improve the constant C r mentioned aove and otain also a constant for r = in [4] we have C r for r. In the context of numerical integration this is interesting as we want to know how the integration error depends on the smoothness. If the function and all its derivatives are periodic then the result can e strengthened. This was already implicitly used in [3], ut will e explicitly shown here. We need the following lemma which was first shown in [6] and appeared in many other papers see for example [4] for a more general version. The following notation will e used throughout the paper: k = k κ 1 a 1 1 and hence k < a 1 1. Lemma 1 For k N let J k x = x wal kt dt. Then J k x = a ω κ 1 1 wal k x + 1/2 + ω κ wal k x 1 c ω ϑ 1 1 wal ϑ a 1 +c 1 +kx. For k =, i.e., J x = x J x = 1/2 + 1 dt = x, we have 1 c ω ϑ 1 1 wal ϑ c 1x. 1 We also need the following elementary lemma. Lemma 2 For any < κ < we have 1 ω κ sin π and 1/2 + ω κ sin π. We introduce some further notation which will e used throughout the paper: For v > 1 let k = k κ 2 a2 1, and hence k < a2 1. For l N let l = λ 1 d λ w dw 1, where w 1, < λ 1,..., λ w <, and d 1 > > d w >. Further let l = l λ 1 d1 1 and hence l < d1 1. For w > 1 let l = l λ 2 d2 1, and hence l < d2 1. 3

4 2 On the Walsh coefficients of polynomials and power series In the following we otain ounds on the Walsh coefficients of monomials x r. Let χ r,v a 1,..., a v ; κ 1,..., κ v = For k = we define χ r,, which is given y χ r, = x r dx = 1 r + 1. x r wal k x dx. We know from [4, Lemma 3.7] that the Walsh coefficients of x r are if v > r, hence we have χ r,v = for v > r. The Walsh series for x is already known from Lemma 1, thus note that we need to take the complex conjugate of 1 to otain the Walsh series for x χ 1,1 a 1 ; κ 1 = a 1 1 ω κ It can e checked that χ 1,1 1. Indeed, we always have 2 χ r,v a 1,..., a v ; κ 1,..., κ v x r wal k x dx = x r dx = 1 r + 1 for all r, v. We otain a recursive formula for the χ r,v using integration y parts, namely x r wal k x dx = J k xx r 1 r x r 1 J k x dx = r x r 1 J k x dx. 3 Using Lemma 1 and 3 we otain for 1 v r and r > 1 that χ r,v a 1,..., a v ; κ 1,..., κ v 4 = r a 1 1 ω κ 1 1 χ r 1,v 1 a 2,..., a v ; κ 2,..., κ v +1/2 + ω κ χ r 1,v a 1,..., a v ; κ 1,..., κ v 1 + c ω ϑ 1 1 χ r 1,v+1 a 1 + c, a 1,..., a v ; ϑ, κ 1,..., κ v. 4

5 From 4 we can otain χ r,r a 1,..., a r ; κ 1,..., κ r = 1 r r! a 1 a r and, with a it more effort, χ r,r 1 a 1,..., a r 1 ; κ 1,..., κ r 1 r 1 = 1 r r! a 1 a r 1 1 ω κw w=1 r 1 ω κw w=1 1 r 1 1/2 + 1/2 + ω κw 1 1 aw, w=1 for all r 1. In principle we can otain all values of χ r,v recursively using 4. We calculated already χ r,v for v = r, r 1 and we could continue doing so for v = r 2,..., 1. But the formulae ecome increasingly complex, so we only prove a ound on them. For any r and a non-negative integer k we define for r =, k, for k =, r, µ r k = a a v for 1 v r, a a r for v > r. Lemma 3 For 1 r < v we have χ r,v = and for any 1 v r we have χ r,v a 1,..., a v ; κ 1,..., κ v r! min u v µuk r u + 1! 3 min1,u 2 sin π u max,u Proof. The first result was already shown in [4]. For the second result we use induction on r. We have already shown the result for r = v = 1. Now assume that χ r 1,v a 1,..., a v ; κ 1,..., κ v 5

6 r 1! 3 min1,u min µuk u v r u! 2 sin π u max,u We show that the result holds for r. We have already shown that χ r,v 1 r+1, which proves the result for u =. By taking the asolute value of 4 and using the triangular inequality we otain χ r,v a 1,..., a v ; κ 1,..., κ v 5 r a 1 1 ω κ 1 1 χ r 1,v 1 a 2,..., a v ; κ 2,..., κ v + 1/2 + ω κ χ r 1,v a 1,..., a v ; κ 1,..., κ v 1 + c ω ϑ 1 1 χ r 1,v+1 a 1 + c, a 1,..., a v ; ϑ, κ 1,..., κ v. Using Lemma 2, χ r 1,v 1 and r c=1 c = 1, we otain from 5 1 that χ r,v a 1,..., a v ; κ 1,..., κ v 3 a 1 2 sin π, which proves the ound for u = 1. To prove the ound for 1 < u v we proceed in the same manner. Using Lemma 2, and χ r 1,v a 1,..., a v ; κ 1,..., κ v we otain µ r 1! 3 u 1k min1,u 1 r u + 1! 2 sin π u 1 χ r,v a 1,..., a v ; κ 1,..., κ v r a 1 2 sin π χ r 1,v 1 a 2,..., a v ; κ 2,..., κ v max,u 2 + 1, χ r 1,v a 1,..., a v ; κ 1,..., κ v 1 + c χ r 1,v+1 a 1 + c, a 1,..., a v ; ϑ, κ 1,..., κ v 6

7 max,u 2 µuk r! 3 min1,u r u + 1! 2 sin π u a 2 a 1 + a 2 a µuk r! 3 min1,u r u + 1! 2 sin π max,u 1 u + 1, + 1 as 1 2c = 1 and a +1 1 > a 2. Thus the result follows. Let now fx = f + f 1 x + f 2 x 2 +. The kth Walsh coefficient of f is given y ˆfk = = = fxwal k x dx f r x r wal k x dx r= f r χ r,v a 1,..., a v ; κ 1,..., κ v. r=v We can estimate the kth Walsh coefficient y ˆfk = χ r,v a 1,..., a v ; κ 1,..., κ v f r r=v χ r,v a 1,..., a v ; κ 1,..., κ v f r r=v r=v r! 3 min1,u f r min u v µuk r u + 1! 2 sin π u min1,u min µuk u v 2 sin π u r=v Hence we have shown the following theorem r! f r r u + 1!. max,u 1 max,u 1

8 Theorem 1 Let fx = f + f 1 x + f 2 x 2 + and let k N. Then we have ˆfk 3min1,u min µuk u v 2 sin π max,u 1 u + 1 r! f r + 1 r u + 1!. r=v Remark 1 This result cannot e directly otained from [4], as there the constant for a power series would e infinite. r=v and The ound in the theorem makes of course only sense for u for which r! f r is finite. We give some examples: r u+1! For f C [, 1] we have f r = r!f r. If f r grows exponentially like for fx = e ax with a > 1, then f r r=v will e finite r v+1! for any v N. The theorem implies that the Walsh coefficients decay with order O µvk. r! Using Sterling s formula we otain that r v + 1 v 1 as r r v+1! tends to. For fx = 1 with < c < 1 we have f 1 cx r = c r. In this case we have r! f r r v + 1! r v + 1 v 1 c r = c v 1 r v 1 c r <, r=v r=v for all v N. The theorem implies that the Walsh coefficients decay with order O µvk. For f C [, 1] with fx = r= f rx r we define the semi-norm f r f = f r =. r! r=1 r=1 Then the v 1th derivative of f is given y f v 1 x = r= f v 1 = v 1 + r! f v 1+r x r = r! r=v r=v 1 r=1 r! r v + 1! f rx r v+1 r! f r r v + 1! = f r r v + 1!. r=v Hence we otain the following corollary from Theorem 1. 8

9 Corollary 1 Let f C [, 1] with f z < for all z N. Then for every k N we have ˆfk µvk 3 2 sin π v 1 v + 1 f v Let us consider another example: Let f r = r δ, with δ > 1. So, for example, we can choose u = minv, δ 2 in the theorem aove, which will guarantee that r! f r r=v <. On the other hand, this sum is not finite r u+1! for δ 2 < u v. The theorem implies that the Walsh coefficients decay with order O µ minv, δ 2k. Note that this function f is only δ 2 times continuously differentiale. We will consider this case in the next section. 3 On the Walsh coefficients of functions in C r [, 1] In this section we prove an explicit constant C r, which is etter than the constant which can e otained from [4]. Before the next lemma we introduce a variation of fractional order: For < λ 1 and f : [, 1] R let V λ f = sup =x <x 1 <...<x N 1 <x N =1 N n=1 x n x n 1 fx n fx n 1 x n x n 1 λ, where the supremum is taken over all partitions of the interval [, 1]. If f has a continuous first derivative on [, 1], then V 1 f = f x dx. If f satisfies a Hölder condition of order < λ 1, i.e., fx fy C f x y λ for all x, y [, 1], then V λ f C f. The following lemma appeared already in [6] aleit in a slightly different form, see also [4, 8]. Lemma 4 Let < λ 1 and let f L 2 [, 1] satisfy V λ f <. Then for any k N, the kth Walsh coefficient satisfies ˆfk 1 1+λ λa 1 V λ f. 9

10 Thus, the decay of the Walsh coefficients of functions with smoothness < r 1 has already een considered and we deal with r > 1 in the following. Let now f L 2 [, 1] with V λ f < and let F 1 x = x fy dy. Then using integration y parts as in the previous section, we otain for k N ˆF 1 k = F 1 xwal k x dx = fxj k x dx. Sustituting the Walsh series for J k from Lemma 1, we otain ˆF 1 k = a 1 1 ω κ 1 1 ˆfk + 1/2 + ω κ ˆfk + c ω ϑ 1 1 a ˆfϑ 1 +c 1 + k. Taking the asolute value on oth sides and using the same estimations as in the previous section, we otain ˆF 1 k a 1 2 sin π ˆfk + ˆfk + 1 c ˆfϑ a1+c 1 + k. 6 Thus, using Lemma 4 we otain for k N with v 2, that ˆF 1 k a 1 λa 2 V λ f For k = κ 1 a 1 1 we otain ˆF 1 k a 1 2 sin π c=1 11+λ 2 sin π λ. ˆf λ λa 1 V λ f. Defining F r x = x F r 1y dy for r 1, we can otain ounds on the Walsh coefficients of F r y using induction on r. Using similar arguments as in the proof of Lemma 3 we otain for v > r that ˆF r k µrk λa r+1 V λ f 11+λ λ 2 sin π r r 1 + 1, + 1 7

11 and for v = r that ˆF r k 8 µrk 2 sin π r 1 r ˆf λ λar V λ f. For 1 v < r we have ˆF r k µrk 2 sin π v Note that we also have F r x = v ˆF r v λ λav V λ F r v x tr 1 + ft r 1!. 9 dt, where x t r 1 + = x t r 1 1 [,x t for x, t 1 and 1 [,x t is 1 for t [, x and otherwise. A function f C r [, 1] for which V λ f r < can e represented y a Taylor series fx = f + f 1 x + + f r 1 1! r 1! xr 1 + f r t x tr 1 + r 1! With this we can now otain a ound on the Walsh coefficients of f. For v r we know from [4] that f + f 1 x + + f r 1 1! r 1! xr 1 wal k x dx =. To ound the Walsh coefficient of f r t x tr 1 + dt for v > r we can use r 1! 7 to otain ˆfk µrk λa r+1 V λ f r 11+λ λ 2 sin π r dt r For v = r we can use 8 to otain ˆfk µrk 2 sin π r 1 r ˆf r λ λar V λ f r 11.

12 For 1 v < r we have f + f 1 x + + f r 1 1! µrk 3 2 sin π v r 1! xr 1 wal k x dx v 1 r s=v f s s v + 1! and therefore, using 9, we otain ˆfk µrk 2 sin π v 1 v [ r 1 f s 3 s v + 1! + ˆf v λ λav V λ f ] v, s=v where ˆf v denotes the th Walsh coefficient of f v. We have shown the following theorem. Theorem 2 Let f C r [, 1] with V λ f r <, and let k N. Then for v > r we have ˆfk µrk λa r+1 V λ f r 11+λ λ 2 sin π r 1 r + 1, + 1 for r = v we have ˆfk µrk 2 sin π r 1 r f r x dx λ λar V λ f r, and for v < r we have ˆfk µrk 2 sin π [ v r 1 3 s=v f s s v + 1! + v ] f v x dx λ λav V λ f v. We also prove ounds on the decay of the Walsh coefficients of functions from Soolev spaces. For this, we first need ounds on the Walsh coefficients of Bernoulli polynomials, which we consider in the next section. 12

13 4 On the Walsh coefficients of Bernoulli polynomials For r let B r denote the Bernoulli polynomial of degree r and r = B r r!. For example we have B x = 1, B 1 x = x 1/2, B 2 x = x 2 x+1/6 and so on. Those polynomials have the properties r x = r 1x and r x = for all r 1. We oviously have x = and x dx = 1. Further, B r 1 x = 1 r B r x and also r 1 x = 1 r r x. The numers B r = B r are the Bernoulli numers and B r = for all odd r 3. Further, for r 2, we have r x = 1 h r e 2πihx, for x πi r h Z\{} It is more convenient to calculate with r rather than the Bernoulli polynomials. For r 1 and k N let β r,v a 1,..., a v ; κ 1,..., κ v = r xwal k x dx. As for χ r,v, we also have β r,v = for v > r. Further, for k = let v = and we have β r, = for all r 1. The Walsh series for 1 can e otained from the Walsh series of J from Lemma 1 and is given y 1 x = x 1/2 = 1 c ω ϑ 1 1 wal ϑ c 1x. Thus β 1,1 a 1 ; κ 1 = a 1 1 ω κ 1 1. Using integration y parts and J k = J k 1 = we otain for all r > 1 that r xwal k x dx = r 1 xj k x dx

14 Using Lemma 1 and 11 we otain for 1 v r and r > 1 that β r,v a 1,..., a v ; κ 1,..., κ v 12 = a 1 1 ω κ 1 1 β r 1,v 1 a 2,..., a v ; κ 2,..., κ v +1/2 + ω κ β r 1,v a 1,..., a v ; κ 1,..., κ v 1 + c ω ϑ 1 1 β r 1,v+1 a 1 + c, a 1,..., a v ; ϑ, κ 1,..., κ v. From 12 we can otain β r,r a 1,..., a r ; κ 1,..., κ r = 1 r a 1 a r for all r 1. The first few values of β r,v are as follows: r 1 ω κs s=1 r = 1: β 1, =, β 1,1 a 1 ; κ 1 = a 1 1 ω κ 1 1 ; r = 2: β 2, =, β 2,1 a 1 ; κ 1 = 2a 1 1 ω κ 1 1 1/2 + ω κ 1 1 1, β 2,2 a 1, a 2 ; κ 1, κ 2 = a 1 a 2 1 ω κ ω κ 2 1 ; In principle we can otain all values of β r,v recursively using 12. We calculated already β r,v for v = r and we could continue doing so for v = r 1,..., 1. But the formulae ecome increasingly complex, so we only prove a ound on them. For any r and a non-negative integer k we introduce the function for r =, k, for k =, r, µ r,per k = a a v + r va v for 1 v < r, a a r for v r. Lemma 5 For any r 2 and 1 v r we have β r,v a 1,..., a v ; κ 1,..., κ v µr,perk 2 sin π r 2 r

15 Proof. We prove the ound y induction on r. Using Lemma 2 it can easily e seen that the result holds for r = 2. Hence assume now that r > 2 and the result holds for r 1. By taking the asolute value of 12 and using the triangular inequality together with Lemma 2 we otain β r,v a 1,..., a v ; κ 1,..., κ v a 1 2 sin π β r 1,v 1 a 2,..., a v ; κ 2,..., κ v + β r 1,v a 1,..., a v ; κ 1,..., κ v + 1 c β r 1,v+1 a 1 + c, a 1,..., a v ; ϑ, κ 1,..., κ v. We can now use the induction assumption for β r 1,v 1, β r 1,v, β r 1,v+1. Hence, for v > 1, we otain β r,v a 1,..., a v ; κ 1,..., κ v µr,perk 2 sin π r 3 r a 2 a 1 + 2c a 2 a 1. By noting that 1 2c = 1, and a +1 1 > a 2 we otain the result. For v = 1 note that β r, =. In this case we have β r,1 a 1 ; κ 1 µr,perk 2 sin π r 3 r c + 1 µr,perk 2 sin π r 2 r + 1, + 1 which implies the result. The r are polynomials, ut using 1 we can extend r periodically so that it is defined on R. We denote those functions y r. Then for r 1 we have 2r x = 2 1r+1 h 2r cos 2πhx for x R, 2π 2r and 2r+1 x = 2 1r+1 2π 2r+1 h=1 h 2r 1 sin 2πhx for x R. h=1 15

16 From this it can e seen that r x = 1 r r x for all r 2. Note that for x, y [, 1] we have 2r x y = 2r x y and 2r+1 x y = 1 1x<y 2r+1 x y, where 1 x<y is 1 for x < y and otherwise. We also extend B r periodically to R, which we denote y B r. In the next section we will also need a ound on the Walsh coefficients of r x y. For k, l let where γ r k, l = = 1 2πi r τ h,k = r x ywal k xwal l y dx dy 13 h Z\{} h r τ h,k τ h,l, e 2πihx wal k x dx. We have γ r k, = γ r, l = for all k, l, as +z r x dx = for z any z R. Further we have γ r l, k = 1 r γ r k, l and therefore also γ r k, l = γ r l, k. We otain ounds on γ r y induction. In the next lemma we calculate the values of γ 2. Lemma 6 For all k, l we have γ 2 k, = γ 2, l =. For k, l > we have 2a 1 if k = l, γ 2 k, l = 1 2 sin 2 κ 1 π/ 1 3 a 1 d 1 ω κ ω λ if k = l >, and k l, a 1 d 1 1/2 + ω λ ω κ a 1 1/2 + ω κ ω κ 1 1 if k = l, a 1 d 1 1/2 + ω κ ω λ d 1 1/2 + ω λ ω λ 1 1 if k = l, a 1 a 2 1 ω κ 2 1 ω κ if k = l, d 1 d 2 1 ω λ 2 1 ω λ if k = l, otherwise. Proof. Note that γ 2 k, = γ 2, l = for all k, l, as +z 2 x dx = z for any z R. 16

17 Now assume that k, l >. The value of γ 2 k, k has een otained in [5, Appendix A] ut can also e otained from the following. The Walsh series for 2 x y = 2 x y = x y2 x y + 1 can e calculated in the following way: We have x = J x and y = J y and so Further x y 2 2 = J x J y 2. 2 x y = x + y 2 minx, y = x + y 2 1 [,x t1 [,y t dt, where 1 [,x t is 1 for t [, x and otherwise. Note that J k x = x wal kt dt = 1 [,xtwal k t dt, which implies Thus 1 [,x t = J k xwal k t. k= minx, y = = = m,n= 1 [,x t1 [,y t dt J m xj n y wal m twal n t dt J m xj m y. m= The Walsh series for 2 x y is therefore given y 2 x y = J x 2 + J y 2 J x J y 2 We have γ 2 k, l = 2 x ywal k xwal l y dx dy 17 + J m xj m y m=1

18 = = m=1 [ J x 2 + J y 2 J x J y + 2 wal k xwal l y dx dy J m xwal k x dx J m ywal l y dy. ] J m xj m y It remains to consider the integral J mxwal k x dx. Let m = η e 1 + m, with < η <, e >, and m < e 1. Then we have J m xwal k x dx = e 1 ω η 1 wal m xwal k x dx +1/2 + ω η c ω ϑ 1 1 m=1 wal m xwal k x dx wal ϑ e+c 1 +mxwal k x dx. This integral is not only if either m = k, m = k or m + ϑ e+c 1 = k for some ϑ, c. Analogously the same applies to the integral J mywal l y dy. Hence we only need to consider a few cases for which γ 2 k, l is non-zero, and y going through each of them we otain the result. Note that many values for γ 2 k, l are, in particular, if k and l are sufficiently different from each other. This property is inherited y r for r > 2 via the recursion γ r k, l = a 1 1 ω κ 1 1 γ r 1 k, l + 1/2 + ω κ 1 1 γ r 1 k, l + 1 c ω ϑ 1 1 γ r 1 ϑ c+a1 1 + k, l. 14 This recursion is otained from γ r k, l = r 1 x yj k xwal l y dx dy, 15 which in turn can e otained using integration y parts. In the following lemma we show that γ r k, l = for many choices of k and l. 18

19 Lemma 7 1. For any k, l we have γ r k, = γ r, l =. 2. For k, l > with v w > r we have γ r k, l =. 3. Let k, l > such that v w r. i If v = 1, ut κ 1, a 1 λ w, d w, then γ r k, l =. ii If w = 1, ut λ 1, d 1 κ v, a v, then γ r k, l =. iii If r 1 v w r, ut a v minv,w+1,..., a v, κ v minv,w+1,..., κ v then γ r k, l =. d w minv,w+1,..., d w, λ w minv,w+1,..., λ w, iv If v, w > 1 and v w r 2, ut a v minv,w+2,..., a v, κ v minv,w+2,..., κ v then γ r k, l =. d w minv,w+2,..., d w, λ w minv,w+2,..., λ w, Proof. a. This follows from z+1 r x dx = for all z R. z. We have γ 2 k, l = for v w > 2, which follows from Lemma 6. Let r > 2. Then y repeatedly using 14 we can write γ r k, l as a sum of γ 2 m i, n j for some values m i, n j, i.e., γ r k, l = i,j a i,jγ 2 m i, n j. But if v w > 2, then the difference etween the numer of digits of m i and n j will e igger than 2 and hence γ r k, l = y Lemma 6. c. For r = 2 the proof follows again from Lemma 6: If v = 1 w = 1, then k = l = resp. and we only have the cases k = l, k = l l = k resp., and k = l l = k resp. for which the result follows. The case 1 v w 2 comprises the cases k = l, k = l, k = l, and k = l. The case v = w can e otained y considering k = l, and k = l with k l. For r > 2, we can again use 14 repeatedly to otain a sum of γ 2 m i, n j. The result then follows y using Lemma 6. 19

20 In the following we prove a ound on γ r k.l for aritrary r 2. We set µ r,per k, l = max s r µ s,perk + µ r s,per l. Lemma 8 For r 2 and k, l > we have γ r k, l 2 µr,perk,l 2 sin π r r Proof. For r = 2 we use Lemma 6, and 1/2 + ω κ 2 sin π 1 to otain the result. 1 1, ω κ 1 1 Let now r > 2. By taking the asolute value of 14 and using the triangular inequality together with 1/2 + ω κ 1 1, ω κ sin π 1 we otain γ r k, l a 1 2 sin π γ r 1 k, l + γ r 1 k, l + 1 c γ r 1 ϑ a1+c 1 + k, l. 16 By using integration y parts with respect to the variale y in 13 we otain a similar formula to 15. Hence there is also an analogue to 16. W.l.o.g. assume that k l otherwise use the analogue to 16 and assume that the result holds for r 1. Then γ r k, l 2 a 1 2 sin π r r µ r 1,perk,l + µ r 1,perk,l + 1 c=1 c µ r 1,per a 1 +c 1 +k,l We have a 1 + µ r 1,per k, l = µ r,per k, l, a 1 + µ r 1,per k, l > µ r,per k, l, and a 1 +µ r 1,per a1+c 1 +k, l = 2a 1 +c+µ r 2,per k, l > c+µ r,per k, l. Therefore we otain γ r k, l 2 µr,perk,l 2 sin π r r c. As c=1 2c = 2 1 1, the result follows. 2 c=1

21 5 On the Walsh coefficients of functions in Soolev spaces In this section we consider functions in reproducing kernel Hilert spaces. We consider the Soolev space H r of real valued functions f : [, 1] R, for which r > 1, and where the inner product is given y r 1 f, g r = s= f s x dx g s x dx + f r xg r x dx, where f s denotes the sth derivative of f and where f = f. Let f r = f, f r. The reproducing kernel see [1] for more information aout reproducing kernels for this space is given y K r x, y = = r B s xb s y 1 B r 2r x y s! 2 2r! r s x s y 1 r 2r x y, s= s= see for example [1, Section 1.2]. It can e checked that fy = f, K r, y r r = f s x dx s y 1 r f r x r x y dx. s= A ound on the Walsh coefficients of y,..., r y can e otained from Lemma 5. For the remaining term we use Lemma 8. We have r x y = γ r k, lwal k xwal l y k,l=1 and therefore the mth Walsh coefficient for the last term is given y 1 r = 1 r f r x r x y dx wal m y dy k,l=1 γ r k, l f r xwal k x dx wal l ywal m y dy 21

22 = k=1 γ r m, k f r xwal k x dx. We can estimate the asolute value of the last expression y k=1 γ r m, k f r x dx f r 2 x dx 2 sin π r r µr,perm,k. + 1 k=1 γ rm,k It remains to prove a ound on the rightmost sum, which we do in the following lemma. Lemma 9 For any r > 1 and m N we have k=1 γ rm,k µr,perm,k µr,perm Proof. Let m = η 1 e η z ez 1, where < η 1,..., η z < and e 1 > > e z >. We consider now all natural numers k for which γ r m, k. From Lemma 7 we know that γ r m, k = for v z > r. Hence we only need to consider the cases where v z r: v = maxz r, : If z r, then this case does not occur; otherwise there is only one k for which γ r m, k, and we otain the summand µr,perm. v = maxz r +1, : Again if z r +1, then this case does not occur; otherwise we can ound this summand from aove y µr,perm 1. maxz r + 1, < v z: First, let v = 1. Then κ 1 = η z and a 1 = e z. Therefore k is fixed, µ r,per m, k = µ r,per m, and µr,perm,k = µr,perm. Let now v > 1, which implies z > 1 as z v and z v + 2 r. In this case a 2,..., a v, κ 2,..., κ v = e z v+2,..., e z, η z v+2,..., η z. 22

23 Thus µ r,per m, k = µ z v+1,per m + a 1 + µ r z v+2,per k, k µ r,per m + a 1 a v z+r. Note that v z + r > 1. Let a v = a 1 a v z+r > v z + r 2. Then the sum over all k for which 1 < v z is ounded y µr,perm 1 z v=2 a =v 1 a µr,perm v=2 v+2 µr,perm 1. z + 1 v z + r 2: If z = 1 then 2 v r 1, and, y Lemma 7, we have η 1 = κ v and e 1 = a v. In this case µ r,per m, k = µ r,per k and µ r,per k µ r,per m = a 1 a v + + a v a v + r va v a v = a a v 1, where a i = a i a v and a 1 > > a v 1 >. The sum over all k for which 2 v r 1, and γ r m, k, is then ounded y r 1 1 v 1 v=2 µr,perm r 1 a 1 > >a v 1 >a v=e 1 > µr,perk 1 v 1 v=2 r 1 µr,perm v 2 v=2 µr,perm 1. For z > 1 and z + 1 v z + r 2 we have a 1 a v 1 a 1 > >a v 1 > a v z+2,..., a v, κ v z+2,..., κ v = e 2,..., e z, η 2,..., η z and v z + 2 r. Thus µ r,per m, k = a a v z+1 + e 1 + µ r v z+2,per m, m µ r,per m µ r 1,per m + a a v z+1 +µ r v z+2,per m, m 23

24 µ r,per m + a a v z+1, where a i = a i e 2 = a i a v z+2 and a 1 > > a v z+1 >. Thus the sum over all k for which z + 1 v z + r 2 and γ r m, k is ounded y µr,perm z+r 2 1 v z+1 v=z+1 a 1 a v z+1 a 1 > >a v z+1 > z+r 2 µr,perm v=z+1 1 v z µr,perm 1. v = z +r: In this case µ r,per m, k = a 1 + +a r µ r,per m+µ r,per m, where µ r,per m ra r+1. Thus a a r µ r,per m a 1 a r a r a r+1 and a 1 > > a r > a r+1. Hence, the sum over all k for which v = z + r is ounded y 1 r µr,perm a 1 a r µr,perm rr 1/2. a 1 > >a r> v = z + r 1: In this case µ r,per m, k = a a r µ r,per m + µ r,per m, where now a r = e 1 and κ r = η 1 are fixed. Hence, the sum over all k for which v = z + r 1 is ounded y 1 r 1 µr,perm a 1 a r 1 µr,perm r 1r 2/2. a 1 > >a r 1 > By summing up the ounds otained for each case, we otain the result. This implies the following theorem. Theorem 3 Let r > 1. Then for any k N we have ˆfk r s=v f s x dx µs,perk 2 sin π s max,s

25 + f r x dx 2 µr,perk 2 sin π r for all f H r, where for v > r the empty sum r s=v r 2, is defined as. Remark 2 This theorem can easily e generalized to tensor product spaces, for which the reproducing kernel is just the product of the one-dimensional kernels. 6 On the Walsh coefficients of smooth periodic functions We consider a suset of the previous reproducing kernel Hilert space, namely, let H r,per e the space of all functions f H r which satisfy the condition f s x dx = for s < r. This space also has a reproducing kernel, which is given y K r,per x, y = 1 r+1 B 2r x y 2r! The inner product is given y f, g r,per = We also have the representation = 1 r+1 2r x y. f r xg r x dx. fy = 1 r+1 f r x r x y dx. For this space we can otain an analogue to Theorem 3. Theorem 4 Let r > 1. Then for any k N we have ˆfk for all f H r,per. f r x dx 2 µr,perk 2 sin π r r , 1 25

26 Remark 3 This theorem can easily e generalized to tensor product spaces, for which the reproducing kernel is just the product of the one-dimensional kernel. Remark 4 For 2 4 we have 2 sin π < 1, and so, +1 for these cases, the constants in the theorems aove decrease as v, r increase. Remark 5 Because the Walsh coefficients considered in this paper converge fast, the Walsh series for functions f with smoothness r > 1 converges asolutely we have k= µr,perk k= µrk < for r > 1 and k= a 1 λa 2 < for λ > and we have see [4] fx = k= ˆfkwal k x for x < 1. 7 Lower ounds Fine [6, Theorem VIII] proved that the only asolutely continuous functions whose Walsh coefficients decay faster than 1/k are the constants. This result can e extended in the following way. Let f have smoothness r > 1 i.e., the r th derivative has at least ounded variation of order r r. Let k 1 = k, k 2 = k and k s = k s 1 κ s as 1 = κ s+1 as κ v av 1 for 1 s < v and let k v =. Using fxwal kx dx = f xj k x dx and Lemma 1 we have ˆfk = a 1 1 ω κ 1 1 ˆf k a 1 1/2 + ω κ 1 1 ˆf k 1 c ω ϑ 1 1 ˆf ϑ a1+c 1 + k, a 1 where ˆf k denotes the kth Walsh coefficient of f. Using Lemma 4 we otain for fixed k 1 and κ 1 that lim a 1 a 1 ˆfk = 1 ω κ 1 1 ˆf k 1. Applying this result inductively we otain for s min r, v and fixed k s and κ 1,..., κ s that lim lim a as s a 2 a 2 lim a 1 a 1 ˆfk 26

27 = lim lim a as s a 2 a 2 1 ω κ 1 1 ˆf k 1 s = 1 s 1 ω κ s 1 f ˆs k s. s =1 This implies that if ˆfk decays faster than a 1 a s for all k = κ 1 a κ v av 1 with < κ 1,..., κ v < and v s, then f ˆs k = for all k and therefore f s =, i.e., f is a polynomial of degree at most s 1. Theorem 5 Let f have smoothness r > 1. Then if for some 1 s r, ˆfk decays faster than a 1 a s for all k = κ 1 a κ v av 1 with < κ 1,..., κ v < and v s, i.e., lim lim a as s a 2 a 2 lim a 1 a 1 ˆfk = for all k with v s, then f is a polynomial of degree at most s 1. References [1] N. Aronszajn, Theory of reproducing kernels, Trans. Amer. Math. Soc., , [2] H.E. Chrestenson, A class of generalised Walsh functions, Pacific J. Math., , [3] J. Dick, Explicit constructions of quasi-monte Carlo rules for the numerical integration of high dimensional periodic functions, SIAM J. Numer. Anal., 45 27, [4] J. Dick, Walsh spaces containing smooth functions and quasi-monte Carlo rules of aritrary high order, SIAM J. Numer. Anal., to appear. [5] J. Dick, F. Pillichshammer, Multivariate integration in weighted Hilert spaces ased on Walsh functions and weighted Soolev spaces, J. Complexity, 21 25, [6] N.J. Fine, On the Walsh functions, Trans. Amer. Math. Soc., ,

28 [7] K. Niederdrenk, Die endliche Fourier- und Walshtransformation mit einer Einführung in die Bildverareitung, Vieweg, Braunschweig, [8] G. Pirsic, Emedding theorems and numerical integration of Walsh series over groups, PhD thesis, University of Salzurg, [9] W.R. Wade, Recent developments in the theory of Walsh series, Internat. J. Math. Math. Sci., , [1] G. Waha, Spline models for oservational data, CBMS-NSF Regional Conference Series in Applied Mathematics, Vol. 59, Society for Industrial and Applied Mathematics SIAM, Philadelphia, PA, 199. [11] J.L. Walsh, A closed set of normal orthogonal functions, Amer. J. Math., ,