Preprint July 25, 2002 To pper in Illinois Journl of Mthemtics Henstock/Kurzweil Fourier trnsforms Erik Tlvil 1 Deprtment of Mthemticl nd Sttisticl Sciences University of Albert Edmonton AB Cnd T6G 2E2 etlvil@mth.ulbert.c Abstrct: The Fourier trnsform is considered s Henstock/Kurzweil integrl. Sufficient conditions re given for the existence of the Fourier trnsform nd necessry nd sufficient conditions re given for it to be continuous. The Riemnn-Lebesgue lemm fils: Henstock/Kurzweil Fourier trnsforms cn hve rbitrrily lrge point-wise growth. Convolution nd inversion theorems re estblished. An ppendix gives sufficient conditions for interchnging repeted Henstock/Kurzweil integrls nd gives n estimte on the integrl of product. AMS (MOS) subject clssifiction: 42A38, 26A39 1 Introduction If f : R R then its Fourier trnsform t s R is defined s f(s) e isx f(x) dx. The inverse trnsform is ˇf(s) (2π) 1 eisx f(x) dx. In this pper we consider Fourier trnsforms s Henstock/Kurzweil integrls. This is n integrl equivlent to the Denjoy nd Perron integrls but with definition in terms of Riemnn sums. We let HK A be the Henstock/Kurzweil integrble functions over set A R, dropping the subscript when A R. (The symbol llows set equlity.) Then HK properly contins the union of L 1 nd the Cuchy-Lebesgue integrble functions (i.e, improper Lebesgue integrls). The min points of HK integrtion tht we use cn be found in [1] nd [9]. Severl of our results depend on being ble to reverse the order of repeted integrls. In the Lebesgue theory this is usully justified with Fubini s Theorem. For HK integrls, necessry nd sufficient conditions were given in [12]. Lemm 25 in the Appendix gives sufficient conditions tht re redily pplicble to the cses t hnd. Also in the Appendix re some conditions for convergence of rpidly oscilltory integrls (Lemm 23) nd n estimte of the integrl of product (Lemm 24). There is substntil body of theory relting to Fourier trnsforms when they re considered s Lebesgue integrls. Necessry nd sufficient for existence of f on R is tht f L 1. This is becuse the multipliers for L 1 re the (essentilly) bounded mesurble functions nd e isx 1. The multipliers 1 Reserch prtilly supported by NSERC.
Henstock/Kurzweil Fourier trnsforms 2 for HK re the functions of (essentilly) bounded vrition. As x e isx is not of bounded vrition, except for s 0, we do not hve n elegnt existence theorem for HK Fourier integrls. Vrious existence conditions re given in Proposition 2. Exmple 3(f) gives function whose Fourier trnsform diverges on countble set. For L 1 convergence, f is uniformly continuous with limit 0 t infinity (the Riemnn-Lebesgue lemm). We show below (Exmple 3(e)) tht the Riemnn-Lebesgue lemm fils drmticlly in HK: f cn hve rbitrrily lrge point-wise growth. And, f need not be continuous. Continuity of f is equivlent to qusi-uniform convergence (Theorem 5). Some sufficient conditions for continuity of f pper in Proposition 6. Although f need not be continuous, when it exists t the endpoints of compct intervl, it exists lmost everywhere on tht intervl nd is integrble over tht intervl. See Proposition 7. As in the L 1 theory, we hve linerity, symmetry, conjugtion, trnsltion, modultion, diltion, etc. See formuls (2) (9) in [5, p. 117] nd [2, p. 9]. We drw ttention to the differentition of Fourier trnsforms (Proposition 8) nd trnsforms of derivtives (Proposition 9). One of the most importnt properties of Fourier trnsforms is their interction with convolutions. Propositions 10, 11, 13, 14 nd 15 contin vrious results on existence of convolutions; estimtes using the vrition, L 1 norm nd Alexiewicz norm; nd the trnsform nd inverse trnsform of convolutions. Proposition 16 gives Prsevl reltion. An inversion theorem is obtined using summbility kernel (Theorem 18). A uniqueness theorem follows s corollry. The pper concludes with n exmple of function f for which f exists on R but ˇˆf exists nowhere. As Henstock/Kurzweil integrls llow conditionl convergence, they mke n idel setting for the Fourier trnsform. We remrk tht mny of the Fourier integrls ppering in tbles such s [5] diverge s Lebesgue integrls but converge s improper Riemnn integrls. Thus, they exist s HK integrls. We use the following nottion. Let A R nd f rel-vlued function on A. The functions of bounded vrition over A re denoted BV A nd the vrition of function f over set A is V A f. All our results re stted for relvlued functions but the extension to complex-vlued functions is immedite. Note tht for complex-vlued functions, the vrition of the rel prt nd the vrition of the imginry prt re dded. The Alexiewicz norm of f HK A is f A sup I A f. The supremum is tken over ll intervls I A. I For ech of these definitions, the lbel A is omitted when A R or it is obvious which set is A. Wheres indefinite Lebesgue integrls re bsolutely continuous (AC), indefinite Henstock/Kurzweil integrls re ACG. See [1] for the definition of ACG nd the relted spce AC. Finlly, convergence theorem tht we use throughout is: Theorem 1 Let f nd g n (n N) be rel-vlued functions on [, b]. If f HK, V g n M nd g n g then fg n fg.
Henstock/Kurzweil Fourier trnsforms 3 The theorem holds for [, b] R, where R R {± } is the extended rel line. For proof see [13]. 2 Bsic properties We first tckle the problem of existence. If f : R R then f exists s Lebesgue integrl on R if nd only if f L 1. This is becuse function is Lebesgue integrble if nd only if its bsolute vlue is integrble. And, e isx 1 for ll s, x R. No such simple necessry nd sufficient conditions re known for existence of HK Fourier integrls. However, we do hve the following results. Proposition 2 Let f :R R. () In order for f to exist t some s R it is necessry tht f HK loc. (b) If f HK loc then f exists on R if f is integrble in neighbourhood of infinity or if f is of bounded vrition in neighbourhood of infinity with limit 0 t infinity. (c) Let f HK. Define F 1 (x) x f nd F 2(x) x f. Then f exists t s R if nd only if both the integrls e isx F 0 1 (x) dx nd 0 e isx F 2 (x) dx exist. Proof: () For ech s R, the function x e isx is of bounded vrition on ny compct intervl. (b) This follows from the Chrtier Dirichlet convergence test. See [1]. (c) Let T > 0. Integrte by prts to obtin T 0 e isx f(x) dx F 1 (0) F 1 (T )e ist is T 0 e isx F 1 (x) dx. Since F 1 is continuous with limit 0 t infinity, e isx f(x) dx exists if nd 0 only if e isx F 0 1 (x) dx exists. The other prt of the proof is similr. Although F 1 is continuous with limit 0 t infinity, it need not be of bounded vrition. So, f HK does not imply the existence of f. See Exmple 3(c) below. Notice tht prt (b) (with HK loc replced by L 1 loc ) nd prt (c) re flse for L 1 convergence of f. Titchmrsh [14] gives severl sufficient conditions for existence of conditionlly convergent Fourier integrls ( 1.10 1.12). However, these ll require tht f L 1 loc.
Henstock/Kurzweil Fourier trnsforms 4 When f L 1 nd s, h R, f(s + h) e i(s+h)x f(x) dx. By dominted convergence this tends to f(s) s s h. So, f is uniformly continuous on R. When f exists in HK in neighbourhood of s we hve x e isx f(x) in HK but the fctor e ihx is not of bounded vrition on R except for h 0. In generl we cnnot tke the limit h 0 under the integrl sign nd f need not be continuous. And, for f L 1 nd s 0, chnge of vribles x x + π/s gives f(s) (1/2) e isx [f(x) f(x + π/s)] dx. Writing f y (x) f(x + y) for x, y R, we hve f(s) (1/2) f f π/s 1. Continuity of f in the L 1 norm now yields the Riemnn-Lebesgue lemm: f(s) 0 s s. It is true tht if f HK then f is continuous in the Alexiewicz norm [10]. However, since the vrition of x e isx is not uniformly bounded s s, existence of f does not let us conclude tht f tends to 0 t infinity. The following exmples show some of the differences between L 1 nd HK Fourier trnsforms. Exmple 3 The trnsforms () (d) pper in [5]. Convergence in () is by Lemm 23, (b) is similr, fter integrting by prts, nd (c) is Frullni integrl. () If f(x) sgn(x) x 1/2 then f is not in HK or in ny L p spce (1 p ) nd yet f(s) 2π sgn(s) s 1/2 for s 0. Notice tht, even though f is odd, f does not exist t 0 since HK convergence does not llow principl vlue integrls. (b) Let g(x) e ix2. Then ĝ(s) π e i(π s2 )/4. In this exmple, ĝ is not of bounded vrition t infinity, nor does ĝ tend to 0 t infinity, nor is ĝ uniformly continuous on R. The sme cn of course be sid for g. (c) Let h(x) sin(x)/x. Then ĥ(s) i log (s )/(s + ) for s. (d) Let k(x) x/(x 2 + 1). Then k(s) iπ sgn(s)e s for s 0. Note tht f does not exist t 0, even though its principl vlue is 0. (e) Fourier trnsforms in HK cn hve rbitrrily lrge point-wise growth. Given ny sequence { n } of positive rel numbers, there is continuous function f on R such tht f exists on R nd f(n) n for ll n 1 [11]. (f) Let { n } nd {b n } be sequences in R. Define f(x) n1 n sin(b n x)/ x for x 0 nd f(0) 0. Assume tht n > 0, n < nd n b n <. Then f is continuous on R, except t the origin, where it hs finite jump
Henstock/Kurzweil Fourier trnsforms 5 discontinuity. Suppose s is not in the closure of { b n, b n }. Then f(s) n n1 e isx sin(b n x) dx x i n (cos [(s + b n )x] cos [(s b n )x]) dx (2) n1 0 x i n log s b n s + b n. (3) n1 The reversl of summtion nd integrtion in (1) is justified using Corollry 7 in [12]. Hence, f exists on R, except perhps on the closure of { b n, b n }. We will now show f diverges t ech b k with k b k 0. Let T 1, T 2 > 0 nd consider T2 T 1 e ib kx n sin(b n x) dx x n1 T2 n e ibkx sin(b n x) dx n1 T 1 x T2 sin [(b k + b n )x] sin [(b k b n )x] n T 1 2 n1 (1) (4) i sin(b k x) sin(b n x) dx x.(5) In (4), convergence of n b n permits reversl of summtion nd integrtion. The rel prt of (5) converges for ll k 1, uniformly for T 1, T 2 0. Hence, the rel prt of f exists on R. The k th summnd of the imginry prt of (5) is T2 k sin 2 (b k x) dx T2 b k T 1 x k sin 2 x dx x. T 1 b k This diverges s T 1, T 2. Hence, f(b k ) does not exist. If {b n } hs no limit points then we hve n exmple of function whose Fourier trnsform exists everywhere except on countble set. It is not difficult to rrnge for {b n } to be subset of compct intervl nd { n } to be such tht f converges on R \ {b n }, i.e., f lso exists t the limit points of {b n }. Suppose s is limit point of {b n }. As noticed bove, the rel prt of f(s) exists. And, 1 1 sin(sx) sin(b nx) dx/ x 2 s. So, f(s) exists if nd only if lim T n1 T n (cos [(s + b n )x] cos [(s b n )x]) dx x 1
Henstock/Kurzweil Fourier trnsforms 6 exists. Suppose s 0 nd T > 1. If s b n T > 1 then T cos( s b n x) dx s bn T cos x dx 1 x s b n x 1 cos x dx s bn T x + cos x dx x s b n log (1/ s b n ) + c. The constnt c is equl to the supremum of t cos x dx/x over t > 1. When 1 s b n T 1, we hve T cos( s b n x) dx s bn T cos x dx x x 1 s b n log T log (1/ s b n ). The cse for s + b n T is similr. It follows tht the condition n log s b n s + b n < (6) n1 is sufficient for existence of f(s). If 1/T < s b n < 1 then, s bove, Therefore, T 1 1/T < s b n <1 cos( s b n x) dx x cos(1) log (1/ s b n ) c. T n cos( s b n x) dx x 1 1/T < s b n <1 n [cos(1) log (1/ s b n ) c]. 1 Let T, then condition (6) is lso necessry for existence of f(s). Exmples 3(), (c), (d) nd (f) show tht f need not be continuous. However, continuity of f is equivlent to qusi-uniform continuity. Definition 4 (Qusi-uniform continuity) Let f : R 2 R. If F (x) : f(x, y) dy exists in neighbourhood of x 0 R then F is qusi-uniformly continuous t x 0 if for ll ɛ > 0 nd for ll M > 0 there exist m x0,ɛ,m > M nd δ x0,ɛ,m > 0 such tht if x x 0 < δ then f(x, y) dy < ɛ. y >m
Henstock/Kurzweil Fourier trnsforms 7 This is modifiction of similr definition for series, originlly introduced by Dini. See [3], pge 140. Theorem 5 Let f :R R. Then, f is continuous t s 0 R if nd only if f is qusi-uniformly continuous t s 0. Proof: For m > 0, let F m (s) m m e isx f(x) dx. Let h R. Then, F m (s+h) F m (s) m m [e ihx 1]e isx f(x) dx. Note tht either ssumption implies x e isx f(x) is in HK loc for ech s R. And, V [ m,m] [x e ihx 1] 4m h. Tking the limit h 0 inside the bove integrl now shows F m is continuous on R for ech m > 0. Suppose f is qusi-uniformly continuous t s 0 R. Given ɛ > 0, tke M > 0 such tht x >t eis0x f(x) dx < ɛ for ll t > M. From qusi-uniform continuity, we hve m > M nd δ > 0. Then, for s s 0 < δ, [ e isx e is0x] f(x) dx e isx f(x) dx + e is0x f(x) dx x >m 2ɛ. x >m x >m It follows tht f is continuous t s 0. Suppose f is continuous t s 0 nd we re given ɛ > 0 nd M > 0. Since f exists t s 0, there is N s0,ɛ > 0 such tht x >m e is0x f(x) dx < ɛ whenever m > N s0,ɛ. Continuity of f t s0 implies the existence of ξ s0,ɛ > 0 such tht f(s) f(s 0 ) < ɛ when s s 0 < ξ s0,ɛ. And, F m is continuous on R. Hence, there exists η s0,ɛ,m > 0 such tht when s s 0 < η s0,ɛ,m we hve F m (s) F m (s 0 ) < ɛ. Let m mx(m, N s0,ɛ) nd δ min(ξ s0,ɛ, η s0,ɛ,m). Then for s s 0 < δ we hve e isx f(x) dx e is0x f(x) dx + f(s) f(s 0 ) + F m (s) F m (s 0 ) x >m < 3ɛ. x >m And, f is qusi-uniformly continuous t s 0. We now present two sufficient conditions for Fourier trnsform to be continuous. The first is in the spirit of the Chrtier Dirichlet convergence test nd the second is in the spirit of the Abel convergence test. For simplicity, the results re stted for functions on [0, ). The generl cse follows esily. Proposition 6 Let g nd h be rel-vlued functions on [0, ) where g BV. Define f gh.
Henstock/Kurzweil Fourier trnsforms 8 () Suppose there re positive constnts M, δ nd K such tht, if s s 0 < δ nd M 1, M 2 > M then M 2 M 1 e isx h(x) dx < K. If g(x) 0 s x then f is continuous t s 0. (b) Let H s (x) x 0 e ist f(t) dt. If ĥ is continuous t s 0 nd there re δ, K > 0 such tht for ll x > 0 nd s s 0 < δ we hve H s (x) K then f is continuous t s 0. Proof: Write φ s (x) e isx h(x). With no loss of generlity, g( ) 0. For (), let s s 0 < δ nd M 1, M 2 > M. Using Lemm 24, M2 M2 e isx f(x) dx φ M 1 s (x) dx inf M 1 g + φ s [M1,M [M 2]V [M1,M 2]g 1,M 2] [ ] K inf g + V [M, ]g [M 1,M 2] 0 s M. Therefore, f exists t s 0. Whence, given ɛ > 0 nd N > 0 we cn tke m > M such tht whenever s s 0 < δ we hve m e isx f(x) dx M 2 m e isx f(x) dx < ɛ for ll M 2 > N. It follows tht f is qusi-uniformly continuous nd hence continuous. For (b), since g BV we hve lim x g(x) c R. Writing f h(g c) + ch we need only consider M φ s(g c) φ s [M, ) V [M, ) (g c). By our ssumption, φ s [M, ) 2K for s s 0 < δ. And, V [M, ) (g c) 0 s M. Although f need not be continuous, when it exists t the endpoints of compct intervl it is integrble over the intervl. Proposition 7 Let [, b] be compct intervl. If f exists t nd b then f exists lmost everywhere on (, b), f is integrble over (, b) nd f i f(x)[e ibx e ix ] dx/x. Proof: The integrl I : i f(x) [ e ibx e ix] dx x exists since x [ f(x)e ibx /x nd x f(x)e ix /x re integrble over R \ ( 1, 1) nd x e ibx e ix] /x is of bounded vrition on [ 1, 1]. And, I f(x)e ibx e i(s b)x ds dx f. f(x)e isx dx ds
Henstock/Kurzweil Fourier trnsforms 9 Hence, f exists lmost everywhere on (, b) nd is integrble over (, b). Lemm 25(b) justifies the reversl of x nd s integrtion. The usul lgebric properties of linerity, symmetry, conjugtion, trnsltion, modultion, diltion, etc., fmilir from the L 1 theory, continue to hold for HK Fourier trnsforms. See formuls (2) (9) in [5, p. 117] nd [2, p. 9]. The proofs re elementry. There re lso differentition results nlogous to the L 1 cse (pges 117 nd 17, respectively, of the previous references). Proposition 8 (Frequency differentition) Suppose f exists on the compct intervl [α, β]. Define g(x) xf(x) nd suppose g HK. Then ˆf iĝ lmost everywhere on (α, β). In prticulr, ˆf (s) iĝ(s) for d s ll s (α, β) such tht ds α ĝ ĝ(s). Proof: The necessry nd sufficient condition tht llows differentition under the integrl, ˆf (s) i e isx xf(x) dx, for lmost ll s (α, β) is tht e isx xf(x) ds dx e isx xf(x) dx ds (7) for ll [, b] [α, β]. See [12, Theorem 4]. We hve g HK, e isx 1 nd V [α,β] [x e isx ] 2(β α) s. The left member of (7) is i[ f(b) f()]. Hence, by Lemm 25(b), (7) holds, nd ˆf (s) iĝ(s) for lmost ll s (α, β). Exmining the proof of [12, Theorem 4], we see tht we get equlity ˆf (s) iĝ(s) when d s ds α ĝ ĝ(s). There re similr results for n-fold differentition when the function x x n f(x) is in HK for positive integer n. Proposition 9 (Time differentition) () If f ACG (R) nd f(x) 0 s x then for ech s 0, both f(s) nd f (s) fil to exist or f (s) is f(s). (b) Suppose f ACG (R) nd f, f HK. Then for ech s 0, either both f(s) nd f (s) fil to exist or f (s) is f(s). Proof: () Let M 1, M 2 > 0. Integrte by prts to get M2 M 1 e isx f (x) dx e ism2 f(m 2 ) e ism1 f(m 1 ) + is M2 M 1 e isx f(x) dx. Now tke the limits M 1, M 2. (b) Consider x f M f + f(m) f(x) for x, M R. Since f HK, the limits s x exist. Hence, f hs limit t infinity. But, f HK so this limit must be 0 nd we hve reduction to cse ().
Henstock/Kurzweil Fourier trnsforms 10 3 Convolution If f nd g re rel-vlued functions on R then their convolution is f g(x) f(x t)g(t) dt. The following proposition gives the bsic properties of convolution. Proposition 10 Let f nd g be rel-vlued functions on R. Define f x :R R by f x (y) f(x + y) for x, y R. For intervl I [α, β] R nd y R, define I y [α y, β y]. () If f g exists t x R then f g(x) g f(x). (b) If f HK, g BV nd h L 1 then (f g) h f (g h) on R. (c) Let f HK. Suppose tht for ech compct intervl I R there re constnts K I nd M I such tht g h (z) K I for ll z I nd the function y h(y)v I y g is in L 1. If f (g h) exists t x R then (f g) h(x) f (g h)(x). (d) (f g) x f x g f g x wherever ny one of these convolutions exists. (e) supp(f g) {x + y : x supp(f), y supp(g)}. Proof: For (), (d) nd (e), the L 1 proofs hold without chnge. See [6, Proposition 8.6]. To prove (b), write (f g) h(x) f g(x y)h(y) dy f (g h)(x). f(x y z)g(z) dz h(y) dy f(x z)g(z y)h(y) dy dz Lemm 25(b) llows us to chnge the order of y nd z integrtion. The proof of (c) is similr but now we use Lemm 25(). The next proposition gives some sufficient conditions for existence of the convolution nd some point-wise estimtes. Proposition 11 () Let f HK nd g BV. Then f g exists on R nd f g(x) f [inf g + V g]. (b) Let f HK loc nd g BV with the support of g in the compct intervl [, b]. Then f g exists on R nd f g(x) x x b f inf [,b] g + f [x,x b] V [,b] g.
Henstock/Kurzweil Fourier trnsforms 11 Proof: () Using Lemm 24, (b) Now, f g(x) f g(x) x x b f(x t)g(t) dt f inf g + f V g f [inf g + V g]. f(x t)g(t) dt f inf g + f [x b,x ]V [,b] g. [,b] These conditions re sufficient but not necessry for existence of the convolution. Exmple 12 Let f(x) log x sin(x) nd g(x) x α, where 0 < α < 1. Then f nd g do not hve compct support nd re not in HK, BV or L p (1 p ). And yet f g exists on R. When f HK nd g L 1 BV then f g exists on R nd we cn estimte it in the Alexiewicz norm. Proposition 13 Let f HK nd g L 1 BV. Then f g exists on R nd f g f g 1. Proof: Existence comes from Proposition 11. Let < b. Using Lemm 25(), we cn interchnge the repeted integrls, And, f g dx g(t) f g dx g 1 sup t R g 1 sup t R f(x t)g(t) dt dx (8) f(x t) dx dt. (9) t t f g 1. f(x t) dx f
Henstock/Kurzweil Fourier trnsforms 12 Under suitble conditions on f nd g, we hve the usul interctions between convolution nd Fourier trnsformtion nd inversion. Proposition 14 If f exists t s R nd g L 1 BV then f g(s) f(s) ĝ(s). Proof: We hve f g(s) f(s) ĝ(s). e isx [ e ist f(t) ] [ e ist g(x t) ] dt dx e ist f(t) e is(x t) g(x t) dx dt The interchnge of integrls is vlidted by Lemm 25(), since ] V [,b] [t e is(x t) [ g(x t) dx V [x b,x ] t e ist g(t) ] dx 2 s (b ) g 1 + 2(b )V g. Proposition 15 If f nd g re in HK loc such tht f exists lmost everywhere, ĝ L 1, s s ĝ(s) is in L 1 nd (ĝ)ˇ g lmost everywhere then f g ( f ĝ)ˇon R. Proof: Let x R. Then ĝ(x t) exists for lmost ll t R. And, f g(x) 1 2π 1 2π ( f ĝ)ˇ(x). f(t) e isx ĝ(s) e is(x t) ĝ(s) ds dt e ist f(t) dt ds Suppose f exists t s 0. Then V I [t e is(x t) e is0t ĝ(s)] 2 ĝ(s) s s 0 I nd the reversl of s nd t integrtion order is by Lemm 25(). 4 Inversion A well-known inversion theorem sttes tht if f nd ˇˆf re in L 1 then f ˇˆf lmost everywhere. These re rther restrictive conditions s both f nd f must be continuous (lmost everywhere) nd vnish t infinity. In Exmple 3() nd (b), f is multiple of f nd ĝ is multiple of g so we certinly hve f ˇˆf nd g ˇĝ lmost everywhere nd yet none of these integrls exists in L 1. However, they do exist in HK. And, we hve similr inversion theorem in HK. First we need the following Prsevl reltion.
Henstock/Kurzweil Fourier trnsforms 13 Proposition 16 Let ψ nd φ be rel-vlued functions on R. Suppose ψ exists t some s 0 R. Suppose φ L 1 nd the function s sφ(s) is lso in L 1. If ψ φ exists, then ψ exists lmost everywhere nd ψ φ ψ φ. Proof: Let f(x) ψ(x)e is0x nd g(x, y) e i(s0 y)x φ(y). A simple computtion shows V [,b] g(, y) O((b )yφ(y)) s y. The conditions of Lemm 25 () re stisfied. Now we hve the inversion theorem. The proof uses the method of summbility kernels. Using Proposition 16, one inserts summbility kernel in the inversion integrl. There is prmeter z x + iy tht is sent to x 0, yielding inversion t x 0. We cn ctully let z x 0 in the upper complex plne, provided the pproch in non-tngentil. This is nlogous to the Ftou theorem for boundry vlues of hrmonic functions. Define the upper hlf plne by Π + {z x + iy : x R, y > 0}. We identify Π + with R. For x 0 Π +, we sy z x 0 non-tngentilly in Π + if z Π + nd z x 0 such tht x x 0 /y C for some C > 0. Definition 17 (Summbility kernel) A summbility kernel is function Θ : R R such tht Θ L 1 AC, Θ 2π, s s Θ(s) is in L1, Θ L 1 BV, s s Θ (s) is in L 1, nd x V [x, ) Θ nd x V(, x] Θ re O(1/x) s x. Theorem 18 (Inversion) Let f : R R such tht f exists lmost everywhere. Define F (x) x x 0 f for x 0 R. If F (x 0 ) f(x 0 ) nd ˇˆf exists t x 0 then f(x 0 ) ˇˆf(x0 ). If ˇˆf exists lmost everywhere then f ˇˆf lmost everywhere. Proof: Let z x + iy for x R nd y > 0. Define φ z : R R by φ z (s) Θ(ys)e isx, where Θ is summbility kernel. Then φ z (t) Θ((t x)/y)/y. And, 1 φ z (s) 2π f(s) ds 1 φ z (t)f(t) dt (10) 2π 1 Θ((t x)/y)f(t) dt. (11) 2πy The inversion theorem now follows, provided we cn prove the following. I. The conditions of Proposition 16 re stisfied so tht (10) is vlid. II. As z x 0 non-tngentilly in Π +, the left side of (10) becomes ˇˆf(x0 ). III. As z x 0 non-tngentilly in Π +, (11) becomes f(x 0 ).
Henstock/Kurzweil Fourier trnsforms 14 I. In Proposition 16, let ψ f nd φ φ z. We hve existence of f t some s 0 R. And, φ z nd s sφ z (s) re in L 1 if nd only if Θ nd s s Θ(s) re in L 1. Since Θ L 1, φ z is continuous with limit 0 t infinity. So, if φ z is of bounded vrition t infinity, the integrl f φ z will exist. It suffices to hve Θ of bounded vrition t infinity. Proposition 16 now pplies. II. Write the left side of (10) s (2π) 1 [ ] [ Θ(ys)e is(x x 0) isx0 e f(s)] ds. The function s e isx0 f(s) is in HK. And, we hve V [s Θ(ys)e is(x x0) ] 2V Θ + 2 Θ 1 x x 0 /y. So, for non-tngentil pproch, this function is of bounded vrition, uniformly s z x 0. This llows us to tke the limit inside the integrl on the left side of (10), yielding ˇˆf(x0 ). III. Let δ > 0. Write 1 ( ) t x Θ f(t) dt 1 ( ) t x Θ f(t) dt+ 1 ( ) t x Θ f(t) dt. y y y y y y t x <δ t x >δ (12) Consider the lst integrl in (12). There is s 0 R such tht t e is0t f(t) is in HK. Now, V [x+δ, ) [t 1 ( ) ] y Θ t x e is0t 2 y y V Θ [δ/y, ) + 2 s 0 Θ 1. With our ssumptions on Θ, this lst expression is bounded s z x 0. And, when Θ L 1, Θ AC loc nd Θ L 1 then Θ(t) o(1/t) s t [2, pge 20]. The sme pplies on the intervl (, x δ]. Hence, tking the limit z x 0 inside the integrl yields 0 for ech fixed δ > 0. Tret the first integrl on the right side of (12) s follows. Becuse 1 x+δ Θ((t 2πy x)/y) dt 1 δ/y Θ(t) x δ 2π dt 1 s y 0 +, we cn ssume δ/y f(x 0 ) 0 (otherwise replce f( ) with f( ) f(x 0 )). Let F (t) t x 0 f. We hve F (x 0 ) f(x 0 ) 0. And, Θ(s) o(1/s) s s. Given ɛ > 0, we cn tke 0 < δ < 1 smll enough such tht F (x 0 + t) ɛ t nd Θ(1/t) ɛ t for ll 0 < t 2δ. Without loss of generlity, ssume x x 0. Tke z x 0 δ with x x 0 /y C for some constnt C > 0. Integrte by prts, 1 y x+δ x δ ( t x Θ y ) f(t) dt 1 [ Θ y ( ) ( δ F (x + δ) y Θ δ ) ] F (x δ) y +J 1 + J 2 + J 3, (13) where J 1 y 2 x 0 Θ ((t x)/y)f (t) dt, J x δ 2 y 2 x x Θ 0 ((t x)/y)f (t) dt, J 3 y 2 x+δ Θ ((t x)/y)f (t) dt. Note tht if 0 < y δ 2 then y/δ δ nd x x ± δ x 0 < 2δ so Θ(±δ/y)F (x ± δ) /y 2ɛ 2.
Henstock/Kurzweil Fourier trnsforms 15 Estimte J 1 by writing J 1 1 x0 ( ) y 2 (x 0 t) t x Θ F (t) x δ y x 0 t dt ɛ (x0 x)/y (x 0 x yt) Θ (t) dt y δ/y 0 ɛc V Θ + ɛ t Θ (t) dt. (14) Similrly, For J 2 we hve J 3 ɛc V Θ + ɛ J 2 1 y 0 0 sup F (t) x 0 t x t Θ (t) dt. (15) (x 0 x)/y Θ (t) dt ɛc V Θ. (16) Putting (14), (16) nd (15) into (13) now shows tht the first integrl on the right side of (12) goes to 0 s z x 0 non-tngentilly. This completes the proof of prt III. Since, F f lmost everywhere, the proof of the theorem is now complete. Remrk 19 In plce of the condition t t Θ (t) is in L 1 we cn demnd tht Θ is incresing on (, 0) nd decresing on (0, ). The proof of III. then follows with minor chnges. The condition tht Θ AC cn lso be wekened. Corollry 20 Let f : R R. Then f 0 lmost everywhere if nd only if f 0 lmost everywhere. Proof: If f 0 lmost everywhere then f 0 on R. If f 0 lmost everywhere then f exists lmost everywhere nd ˇˆf exists lmost everywhere. Therefore, by the Theorem, ˇˆf f 0, lmost everywhere. Note tht the inversion theorem pplies to Exmple 3()-(d). The condition tht ˇˆf exists lmost everywhere cnnot be dropped. The following exmple shows tht existence of f on R does not gurntee existence of ˇˆf t ny point in R. Remrk 21 The most commonly used summbility kernels re ] 2 Θ 1 (x) 2π(1 x )χ [ 1,1] (x) Θ1 (s) 2π Cesàro/Fejér [ sin(s/2) s/2 Θ 2 (x) πe x Θ2 (s) 2π 1+s 2 Abel/Poisson Θ 3 (x) 2 π e x2 Θ3 (s) 2πe (s/2)2 Guss/Weierstrss.
Henstock/Kurzweil Fourier trnsforms 16 The Abel nd Guss kernels re summbility kernels ccording to Definition 17, while the Cesàro kernel does not stisfy this definition. Exmple 22 Let f(x) x α e ixν for x 0 nd f(x) 0 for x < 0. Using the method of Lemm 23 we see tht f exists on R for 1 < α < ν 1. And, f(s) 0 s α+1 ν 1 x α e i[xν sx] dx 0 x α e ip[xν x] dx ( p s ν/(ν 1)). (17) Write φ(x) x ν x. If ν > 1 then φ hs minimum t x 0 : ν 1/(ν 1). The method of sttionry phse [7] shows tht 2π f(s) ν(ν 1) eiπ/4 x α (ν 2)/2 0 e iφ(x0)sν/(ν 1) s 2α+2 ν 2(ν 1) s s. Let ν > 2. It now follows from Lemm 23 tht when ν/2 α < ν 1, f exists on R nd ˇˆf diverges t ech point of R. Note tht neither f nor f is in ny L p spce (1 p ). 5 Appendix Lemm 23 If γ > 0 nd δ R then () 1 0 eix γ x δ dx exists in HK if nd only if γ + δ + 1 > 0. The integrl exists in L 1 if nd only if δ > 1. (b) 1 e ixγ x δ dx exists in HK if nd only if γ > δ + 1. The integrl exists in L 1 if nd only if δ < 1. Proof: In (), integrte by prts to get 1 0 e ix γ x δ dx i γ [ ] e i γ lim x 0 eix x γ+δ+1 i(γ + δ + 1) + γ 1 0 e ix γ x γ+δ dx. The limit exists if nd only if γ + δ + 1 > 0, the lst integrl then converging bsolutely. Cse (b) is similr. For L 1 convergence, we simply tke the bsolute vlue of ech integrnd. Lemm 24 Let [, b] R nd let f HK [,b] nd g BV [,b]. Then fg f inf g + f [,b]v [,b] g. [,b]
Henstock/Kurzweil Fourier trnsforms 17 Proof: Given ɛ > 0, tke c [, b] such tht g(c) ɛ + inf [,b] g. Integrte by prts: fg c g(c) fg + c f fg c ( x ) f dg(x) + c ( ) b f dg(x). x And, fg [ ɛ + inf [,b] g [ ɛ + inf [,b] g ] ] f + sup x f V [,c]g + sup x c c x b f + f [,b]v [,b] g. x f V [c,b]g This lemm is n extension of inequlities proved in [8] nd [4] (Theorem 45, pge 36). Chnging g on set of mesure 0, such s singleton, does not ffect the integrl of fg but cn mke the infimum of g equl to zero. However, this reduction in inf g is reflected by corresponding increse in V g. This redundncy cn be eliminted by replcing g with its normlised version, i.e., for ech x [, b) replce g(x) with lim t x+ g(t) nd redefine g(b) 0. Then the inequlity becomes b fg f [,b] V [,b] g. The following lemm on interchnge of iterted integrls is n extension of Theorem 57 on pge 58 of [4]. Lemm 25 Let f HK nd let g : R 2 R. Let M denote the mesurble subsets of R. For ech (A, B) BV M, define the iterted integrls I 1 (A, B) f(x)g(x, y) dy dx x A y B I 2 (A, B) f(x)g(x, y) dx dy. y B x A () Assume tht for ech compct intervl I R there re constnts M I > 0 nd K I > 0 such tht R V Ig(, y) dy M I nd, for ll x I, g(x, ) 1 K I. If I 1 exists on R R then I 2 exists on BV M nd I 1 I 2 on BV M. (b) Assume there exist M, G L 1 such tht, for lmost ll y R, V g(, y) M(y) nd, for ll x R, g(x, y) G(y). Then I 1 I 2 on BV M.
Henstock/Kurzweil Fourier trnsforms 18 Proof: () Let I be the open intervls in R. First prove I 1 I 2 on I I. Fix (, b) nd (α, β) in I. For < < t <, define H (t) I 2 ((, t), (α, β)) β t α f(x)g(x, y) dx dy. (18) We will estblish the equlity of I 1 nd I 2 by ppeling to the necessry nd sufficient conditions for interchnging repeted integrls [12, Corollry 6]. For this, we need to show tht H is in ACG nd tht we cn differentite under the integrl sign in (18). Let F (x) x H (t) [F (t) F ()] β α β g(t, y) dy α f. Integrte by prts, t [F (x) F ()] d 1 g(x, y)dy. (19) The integrtor of the Riemnn-Stieltjes integrl over x [, t] is denoted d 1 g(x, y). Now, H (t) f [K [,b] + M [,b] ], nd I 2 (A, B) exists for ll A, B I with A bounded. We hve F ACG (R). So, there re E n R such tht R E n nd F is AC on ech E n, i.e., for ech n 1, given ɛ > 0, there is δ > 0 such tht if (s i, t i ) re disjoint with s i, t i E n nd s i t i < δ then f (si,t i) < ɛ. Fix n 1 with E n, ɛ nd δ s bove. Suppose (σ i, τ i ) re disjoint with σ i, τ i E n nd σ i τ i < δ. With no loss of generlity, we my ssume E n is subset of compct intervl [c, d]. Then [ ] sup H (p) H (q) f [σi,τ i] K[c,d] + M [c,d]. [p,q] [σ i,τ i] It follows tht H ACG (R). Now show tht we cn differentite under the integrl sign to compute H (t). Let 0 < h < 1 nd t R such tht F (t) f(t). Then 1 t+h f(x)g(x, y) dx sup F (t + h) F (t) h t 0< h <1 h g(t, y) + sup 1 h f [t h,t+ h ]V [t 1,t+1] g(, y). 0< h <1 It now follows from dominted convergence tht H (t) f(t) β g(t, y) dy α for lmost ll t R. And, by [12, Corollry 6], I 1 (A, B) I 2 (A, B) for ll A, B I with A bounded. By ssumption, I 1 (R, R) exists. For R, β β f(x) g(x, y) dy dx lim f(x) g(x, y) dy dx α t t t β lim t α lim H (t). t α f(x)g(x, y) dx dy
Henstock/Kurzweil Fourier trnsforms 19 Similrly, lim H (t) exists. Therefore, H is continuous on R nd hence t in ACG (R). It follows from [12, Corollry 6] tht I 1 (A, B) I 2 (A, B) for ll A, B I. We hve equlity of I 1 nd I 2 on BV M upon replcing f with fχ A nd g(x, ) with g(x, )χ B where A BV nd B M. (b) This is similr to prt (), but now the conditions on g ensure the existence of I 2 on R R. As in (), H ACG (R). To show H is continuous on R, note tht β t α f(x)g(x, y) dxdy f ( G 1 + M 1 ) nd t lim t f(x)g(x, y) dx exists for lmost ll y R. Whence, lim t H (t) exists nd H is continuous on R. Using [12, Corollry 6], we now hve equlity of I 1 nd I 2 on R R nd hence on BV M. References [1] R.G. Brtle, A modern theory of integrtion, Providence, Americn Mthemticl Society, 2001. [2] J.J. Benedetto, Hrmonic nlysis nd pplictions, Boc Rton, CRC Press, 1997. [3] T.J.I A. Bromwich, An introduction to the theory of infinite series, London, McMilln, 1926. [4] V.G. Čelidze nd A.G. Džvršeǐšvili, The theory of the Denjoy integrl nd some pplictions (trns. P.S. Bullen), Singpore, World Scientific, 1989. [5] A. Erdélyi, Tbles of integrl trnsforms, vol. I, New York, McGrw-Hill, 1954. [6] G.B. Follnd, Rel nlysis, New York, Wiley, 1999. [7] F.W.J. Olver, Asymptotics nd specil functions, Sn Diego, Acdemic Press, 1974. [8] M. Riesz nd A.E. Livingston, A short proof of clssicl theorem in the theory of Fourier integrls, Amer. Mth. Monthly, 62 (1955) 434 437. [9] C. Swrtz, Introduction to guge integrls, Singpore, World Scientific, 2001. [10] E. Tlvil, Continuity in the Alexiewicz norm, (to pper). [11] E. Tlvil, Rpidly growing Fourier integrls, Amer. Mth. Monthly, 108 (2001) 636 641.
Henstock/Kurzweil Fourier trnsforms 20 [12] E. Tlvil, Necessry nd sufficient conditions for differentiting under the integrl sign, Amer. Mth. Monthly, 108 (2001) 544 548. [13] E. Tlvil, Limits nd Henstock integrls of products, Rel Anl. Exchnge 25 (1999/00) 907 918. [14] E.C. Titchmrsh, Introduction to the theory of Fourier integrls, New York, Chelse, 1986.