The Fourier transform and finite differences: an exposition and proof. of a result of Gary H. Meisters and Wolfgang M. Schmidt

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1 he Fourier transform and finite differences: an exposition and proof of a result of Gary H. Meisters and Wolfgang M. Schmidt Rodney Nillsen, University of Wollongong, New South Wales 1. Introduction. In 1972, Gary H. Meisters and Wolfgang M. Schmidt [4] proved an elegant and remarkable result concerning Fourier analysis on the circle group, which is the set of complex numbers of modulus 1. We denote the normalised Haar measure on by µ. hat is, for an integrable, complex-valued function f on, f dµ = 1 2π fe it ) dt. 2π 0 he space L 2 ) is the family of functions f on such that f 2 dµ <. Meisters and Schmidt proved that if a function f in L 2 ) has the property that f dµ = 0, then f is of the form 3 f j δ bj f j ), for suitable f 1, f 2, f 3 in L 2 ) and b 1, b 2, b 3 in, where δ b denotes the Dirac measure at b and denotes convolution. Note that δ b f is the function z fb 1 z) on and is called the translation of f by b. hey deduced that every linear form on L 2 ) such that δ a f) = f) for all a and all f in L 2 ) must be continuous. hese results of Meisters and Schmidt were generalized further on in [6], and to the non-compact case of R in [6, 7]. Meisters and Schmidt actually presented their results in the context of compact abelian groups and spaces of abstract distributions. he present purpose is to give a more accessible proof in the case of the circle group and L 2 ) only, with the intention as well of incorporating the more general results in [6]. he motivation for doing this was research work carried out recently with Susumu Okada [9], which makes central use of the main result presented here. he presentation is based on classical measure theory and the Fourier transform in the space L 2 ) of the circle group. he techniques are those of Meisters and Schmidt in all essentials, although they are used to obtain a more general result. As mentioned, consists of all numbers z C such that z = 1. We can identify with the interval [0, 2π) by means of the one-to-one and onto mapping x e ix from [0, 2π) to. Statements about may be expressed as equivalent statements about the interval [0, 2π). Suitable sources for background knowledge in integration theory and Fourier analysis may be found in [1, 2, 10], for example. he discussion also makes use of the integration of functions in multivariate analysis. he symbols N, Z, R and C, respectively denote the sets consisting of the positive integers, the integers, the real numbers and the complex numbers. 2. Notation and statement of the main theorem. Let M) denote the set of regular Borel measures on. hen, M) has an operation of convolution, which we denote by. If λ M), λ n denotes the measure λ λ λ, where λ appears n times. If b, 1

2 δ b M) denotes the Dirac measure at b. As above, the normalised Haar measure on is denoted by µ, and µ M). If λ M) and f L 2 ), then f λ L 2 ) and f λx) = fz 1 x)dλz), for almost all x. If s N and b put µ b,s = δ 1 δ b ) s. A function in L 2 ) is called a difference of order s if it is of the form µ b,s g, for some g L 2 ) and some b. he vector space of functions that are finite sums of differences of order s is denoted by D s L 2 )). Note that a difference of order 1 is a function of the form g δ b g for some b and g L 2 ). Also, a difference of order 2 is a function of the form g 2 1 δ b + δ b ) g for some b and g L 2 ). HE MAIN HEOREM. Let s N. hen, D s L 2 )) is independent of s and, in fact, { } D s L 2 )) = f : f L 2 ) and f du = 0. Also, if f D s L 2 )), f is equal to a sum of 2s + 1 differences of order s. he heorem was proved in the case of s = 1 by G. Meisters and W. Schmidt [1]. hus, they proved: a function in L 2 ) is a sum of three differences of order 1 if and only if f dµ = 0. An immediate consequence of this is: if is a linear form on L2 ) such that δ b f) = f) for all f L 2 ) and all b, then is a multiple of the Haar measure on and is therefore continuous. However, Meisters showed in [5] that the corresponding statement fails when the non-compact group R replaces the compact group. 3. he Fourier transform. he Fourier transform for the circle group is an essential concept in the proof of he Main heorem. If λ M), the Fourier transform of λ is the function λ on Z given by λn) = z n dλz), for n Z. Now L 2 ) M), and the Fourier transform is a Hilbert space isometry from L 2 ) onto l 2 Z). If λ, ν M), λ ν) c = λ ν. If b, δ b n) = b n. Note that the condition on f in he Main heorem above, namely that f dµ = 0, is equivalent to the condition f0) = Preliminary results. he main obstacle to proceeding is that we have no way of deciding whether a function is a finite sum of differences of order s. he way out is given by a result of Meisters and Schmidt [4], which changes the problem into a question of whether certain integrals are convergent or divergent. We then obtain estimates on these integrals to obtain our results. Our approach here is the same as in [4], the main difference being that we consider differences of order s, not only s = 1, and a corresponding larger class of integrals. In places like equation 1) below, and elsewhere, formal expressions of the form 0/0 and x/0 for some x > 0 may occur. We take 0/0 to be 0 and x/0 to be is such cases. 2

3 heorem 1. Let f L 2 ), let r N and let µ 1, µ 2,..., µ r M). hen the following statements hold. If fn) 2 r µ <, 1) 2 jn) n= then f = r µ j f j, for some f 1, f 2,..., f r L 2 ). he converse statement is also true. Proof. Let f L 2 ), let r N and let µ 1, µ 2,..., µ r M) be such that 1) holds. hen define { } hn) = max µ 1 n)), µ 2 n),..., µ r n), for each n Z. hen define subsets B 0, B 1,..., B r of Z, as follows: we put { } B 0 = n : n Z and hn) = 0, and then { } B j = n : n Z, n j 1 k=0 B k and hn) = µ j n) for j = 1, 2,..., r. he sets B 0, B 1,..., B r are disjoint and their union is Z. Note that if n B j for some j {1, 2,..., r}, then µ j n) = hn) > 0. hus, for j {1, 2,..., r}, a complex-valued function h j may be defined on Z by fn) h j n) = µ j n), if n B j; 0, if x B j. hen, for j {1, 2,..., r}, n Z h j n) 2 = n B j fn) 2 µ j n) 2, = r n Z r n Z <. 1 r fn) 2 [max{ µ 1 n) 2,..., µ r n) 2 }], fn) 2 r µ jn) 2, Hence h j l 2 Z). By the Riesz-Fischer heorem, there is f j L 2 R) such that f j = h j, for all j = 1, 2,..., r. he definition of h j gives fn) = µ j n)h j n) = µ j n) f j n), for all n B j. Now, note that B 0, B 1,..., B r are disjoint and have union equal to Z. Also, f j = 0 on B k if j k, and f = 0 on B 0. It follows that, on Z,, f = r µ j fj = r r ) c µ j f j ) c = µ j f j. 3

4 As functions that have equal Fourier transforms must be equal, we have the required identity f = r µ j f j. As mentioned above, the converse statement in heorem 1 is true, but it is not needed here and is not proved for further details see [4, pages ] and [6, pages 77-78]). Lemma 1. If x [0, π], x π sin x 2. Proof. Observe that x sin x/x is decreasing on 0, π/2]. So, we see that for all 0 x π, Lemma 2. Let s, m N with m > 2s. hen we have: m < and [0,1) m x2s j sin x 2 x/2 π/2 = x π. 0,2π) m m sin2s x j /2) <. Proof. Let m > 2s, let S m 1 denote the surface of the unit sphere in R m, and let σ m 1 denote the usual surface measure on S m 1. Observe that if x 1, x 2,..., x m ) [0, 1) m, then Also, there is c > 0 such that s xj) 2 c 1/2 xj) 2 m. m x 2s j, for all x 1, x 2,..., x m ) R m. hus, we have m [0,1) m x2s j c m m ) s [0,1) m c x2 j m Sm 1 r m 1 0 = c σ m 1 S m 1 ) r 2s m 0 dσ m 1 dr r m 2s 1 dr <, 2) because m > 2s. So we that the first-mentioned integral is finite. 4

5 When we come to the second integral, observe that for any function f on [0, 2π) m, fx 1, x 2,..., x m ) m = fx 1, x 2,..., x m )dx 1 dx 2..., dx m, 3) [0,2π) m J 1 J m J 1,J 2,...,J m where the sum is taken over all 2 m combinations of intervals J 1, J 2,..., J m where for each j, J j = [0, π) or J j = [π, 2π). Now if p + q = m we have [0,π) p [π,2π) q m sin2s x j /2) = = [0,π) p [0,π) q [0,π) p+q dx 1... dx p du 1... du q p sin2s x j /2) + q dx 1... dx m sin2s x j /2) sin2s 2π u j )/2) π 2s dx 1... dx m, using Lemma 1, [0,π) m x2s j = π m+2s dx 1... dx m, on putting x j /π in placed of x j, [0,1) m x2s j <, 4) upon using 2) and as m > 2s. We now see from 3) and 4) that for m > 2s, m sin2s x j /2) 2m π m+2s dx 1... dx m <. [0,1) m x2s [0,2π) m 5. Proof of he Main heorem. We have to show that a function f in L 2 ) is a finite sum of differences of order s if and only if f0) = 0. When this holds we have to show we need at most 2s + 1 differences of order s. First: the easy part. Let f D s L 2 )); we prove f0) = 0. For n Z, s N and b, s µ b,s n) = [δ 1 δ b ) s ] c n) = δ 1 δ b ) n)) c = 1 b n ) s. 5) However, if f is a sum of q differences of order s, there are b 1, b 2,..., b q and f 1, f 2,..., f q L 2 ) such that f = q µ b j,s f j. hen, by 5), fn) = q µ bj,sn) f j n) = q j 1 b n j ) s fj n). 6) Now, putting n = 0 means that 1 b n j = 1 1 = 0, so we deduce from 6) that f0) = 0. hus, if f is a finite sum of differences of order s, f0) = 0. Alternatively we can say that f D s L 2 R)) implies that f0) = 0, as required. Second: we prove the hard part. We show that if f L 2 ) and f0) = 0, then f is a sum of 2s + 1 differences of order s. Let n Z and n 0. Observe that if m N, and if µ m is the normalised Haar measure on m then for all n Z with n 0, and for f L 1 m ), fz1 n, z2 n,..., zm) n dµ m z 1, z 2,..., z m ) = fz 1, z 2,..., z m ) dµ m z 1, z 2,..., z m ). m m 5

6 he reason is that both sides define an invariant integral on m and so they must be equal by the uniqueness of the Haar integral. So, for all n Z with n 0, and for f L 1 m ), fz1 n, z2 n,..., zm) n dµ m z 1, z 2,..., z m ) = 1 2π) m fe ix 1, e ix 2,..., e ixm ) m. 7) m 0,2π) m We will use this result with fz 1, z 2,..., z m ) = 1 1 z 1 j. 2s NOW, let n Z with n 0 be given. We have dµ m y 1, y 2,..., y m ) dµ y 1, y 2,..., y m ) m m µ = y j,sn) 2 m m 1 y n j 2s = fz1 n, z2 n,..., zm) n dµ m z 1, z 2..., z m ) m = 1 2π) m fe ix 1, e ix 2,..., e ixm ) m, using 7), 0,2π) m = 1 m 2π) m 1 e ix j 2s = 1 4 <, 1 2π) m 0,2π) m 0,2π) m m sin2s x j /2) using Lemma 2 and the assumption that m > 2s. aking m = 2s + 1 we get that for all n Z with n 0, dµ 2s+1y 1, y 2,..., y 2s+1) 2s+1 2s+1 µ = M <. 8) y j,sn) 2 Note that M does not depend upon n. Now, fn) ) 2 2s+1 2s+1 n=, n 0 µ dµ 2s+1 y 1, y 2,..., y 2s+1 ) y j,sn) 2 ) = fn) 2 dµ 2s+1 y 1, y 2,..., y 2s+1 ) 2s+1 n=, n 0 2s+1 µ y j,sn) 2 M fn) 2, by 8),, <. n=, n 0 We deduce from this that n=, n 0 fn) 2 2s+1 µ y j,sn) 2 <, 6

7 for almost all y 1, y 2,..., y 2s+1 ) in 2s+1. So, by heorem 1, for almost all y 1, y 2,..., y 2s+1 ) in 2s, there are g 1, g 2,..., g 2s+1 L 2 ) such that 2s+1 f = g j µ yj,s g j ). 9) hat is, we have shown that f is a sum of 2s+1 differences of order s. Note that the statement in 9), in that it holds for almost all b 1, b 2,..., b 2s+1 ) 2s+1, has proved more than the original statement in he Main heorem. Summarizing, we have the following. Let s N and f L 2 ). hen, f0) = 0 if and only if f is a sum of 2s + 1 differences of order s. In this case, for almost all b 1, b 2,..., b 2s+1 ) 2s+1, there are f 1, f 2,..., f 2s+1 L 2 ) such that f = 2s+1 δ 1 δ bj ) s f j. 6. Comments and conclusion. he main result mentioned here has analogues on other compact abelian groups apart from see [3, 4, 6]). here are also analogous results for non-compact groups see [6, 7]). here is also the the question as to the sharpness of the result in he Main heorem: if f0) = 0 do we need 2s + 1 differences in general in order to express f as sum of differences of order s, or will fewer suffice? For the space D 1 ), Meisters and Schmidt showed that 3 differences are needed, in general. For the corresponding spaces on the real line, it is shown in [8] that 2s + 1 differences are needed, in general. Acknowledgements. I wish to thank Susumu Okada for his suggestion to write an exposition of this result, and for his assistance in its preparation. Also, I wish to thank my former multivariate analysis student, Michael Van Der Kolff, for kindly pointing out some errors in a draft of this proof. References [1] R. Edwards, Fourier Series, volumes I and II, Holt, Rinehart and Winston, [2] E. Hewitt and K. Ross, Abstract Harmonic Analysis volume 1, Springer Verlag, Berlin, [3] B. Johnson, A proof of the translation invariant form conjecture for L 2 G), Bull. Sci. Math. de France, ), [4] G. Meisters and W. Schmidt, ranslation-invariant linear forms on L 2 G) for compact abelian groups G, Journ. Func. Anal., ), [5] G. Meisters, Some discontinuous translation-invariant linear forms, Journal of Functional Analysis, ), [6] R. Nillsen, Banach spaces of functions and distributions characterized by singular integrals involving the Fourier transform, Journal of Functional Analysis, ), [7] R. Nillsen, Difference Spaces and Invariant Linear Forms, Lecture Notes in Mathematics 1586, Springer Verlag, Berlin, [8] R. Nillsen and S. Okada, Sharp results concerning the expression of functions as sums of finite differences, Proc. Lond. Math. Soc ), [9] R. Nillsen and S. Okada, Aspects of the Fourier transform and finite differences on the circle group, preprint. [10] W. Rudin, Fourier Analysis on Groups, second edition, Interscience McGraw-Hill, New York, c Rodney Nillsen, Wollongong, May,

8 Go to Rod s home page at nillsen/differences on circle Download a pdf version of this paper at differences on circle.pdf Read an intuitive, more informal description of the results on this page at nillsen/differences on circle intuitive Rod s address: nillsen@uow.edu.au his pdf is taken from a LaeX version of the same material on Rod s website. here are no differences in content. 8