Henstock-Kurzweil Laplace Transform

Transkript

1 Chper 5 Hensock-Kurzweil Lplce Trnsform 5.1 Inroducion The Lplce rnsform of funcion f : [, ) R, s C, is defined s he inegrl [15] e s f() d. (5.1) Mny uhors, in some previous decdes, showed heir ineres in sudying he Lplce rnsform in clssicl sense; by clssicl mening Lebesgue inegrl nd/or Riemnn inegrl on he rel line. In his chper, we focus on n inegrl rnsform h generlizes he clssicl Lplce rnsform. Here he generlizion is given by he replcemen, in he definiion of Lplce rnsform, i.e., in (5.1), of Lebesgue inegrl wih Hensock-Kurzweil (HK) inegrl. The Hensock-Kurzweil inegrl is defined in erms of Riemnn sums bu wih sligh modificion, ye i includes he Riemnn, improper Riemnn, nd Lebesgue inegrls s specil cses. This inegrl is equivlen o he Denjoy nd Perron inegrls. 117

2 The firs quesion rises is h wheher (or in which condiions) he Lplce rnsform when reed s HK inegrl mke sense. The sufficien condiion for he exisence of Lplce rnsform is h he funcion f is loclly inegrble on [, ), becuse he produc of loclly inegrble funcion nd bounded mesurble funcion is loclly inegrble. Since Hensock- Kurzweil inegrl llows nonbsolue convergence, i mkes n idel seing for he Lplce rnsform. I ws proved [9] h, on compc inervl, he produc of nonbsolue inegrble funcion nd bounded vriion funcion is sill nonbsolue inegrble nd his remins vlid in cse of n unbounded inervl [1]. We observed in (5.1) h he kernel funcion e s, s C, is no bounded vriion funcion on [, ) hence we cnno clim (s in he clssicl sense) h he HK Lplce rnsform exiss for ny HK inegrble funcion. The convoluion plys n imporn role in pure nd pplied mhemics, pproximion heory, differenil nd inegrl equions nd mny oher res. Lplce rnsform hs he propery h i cn inerc wih convoluions so we obin vrious resuls on convoluion. We give necessry nd sufficien condiions so h he convoluion operor s HK inegrl is coninuous. Finlly we solve he problem of inversion by using generlized differeniion. The resuls in his chper re ppered in [, 14]. 5. Exisenil Condiions In his secion, we ckle he problem of exisence of Lplce rnsform s HK inegrl. According o Zyed [16], he clssicl sufficien condiion for he exisence 118

3 of Lplce rnsform is h he funcion f is loclly inegrble on [, ), i.e., f L loc [, ). This is becuse he muliplier for he Lebesgue inegrble funcions is bounded mesurble funcion nd he funcion e s, s C, is bounded mesurble funcion. We know h he mulipliers for HK inegrble funcions re precisely he funcions of bounded vriion [1]. Noe h in (5.1), he funcion e s, s C, is no of bounded vriion on [, ) bu i is of bounded vriion on ny compc inervl J R. Hence we cn no consider (5.1) s HK inegrl direcly. So we cnno ge ny specific condiion for he exisence of Lplce rnsform s HK inegrl. However we do hve he following differen condiions. Theorem Le f : [, ) R be ny coninuous funcion such h F (x) = x f, x <, is bounded on [, ). Then he Lplce rnsform L{f}(s), i.e., L{f}(s) = e s f() d, Re.s > exiss. Proof. Le < x <, s = σ + iω, σ, ω R. We hve (F () e s ) = s F () e s + f() e s. Since e s is of bounded vriion on ny compc inervl, sy [, b] nd F () is bounded on [, ). Then F () e s is HK inegrble on [, b]. Therefore s F () e s is inegrble over I = [, b]. Now (e s ) d = s e s d = = σ + ω e σ d σ + ω [e σ e σb ]. σ Wihou loss of generliy le us ke =. Therefore (e s ) σ + ω d = [1 e σb ] <. σ 119

4 Hence (e s ) is bsoluely inegrble over [, b]. Now by Hke s heorem [1], we cn wrie lim (e s ) σ + ω d = lim [1 e σb ] b b σ σ + ω =, σ > σ σ + ω =, Re.s >. σ This shows h (e s ) is bsoluely inegrble on [, ). Agin lim e s =, Re.s >. Hence he funcion e s : [, ) R is coninuous on [, ) wih lim e s = nd (e s ) is bsoluely inegrble over [, ) nd we hve ssumed h f : [, ) R is coninuous funcion such h F (x) = x f, x <, is bounded on [, ). Therefore by Dedekind s es [1], we cn sy h he inegrl e s f() d exiss for Re.s >. Remrk 5... The bove resul cn lso be proved by using Du-Bois Reymond s es [1] s follows: Tke φ() = e s, s = σ + iω, J = [, ). Then clerly φ() is differenible nd φ L 1 (J). Since F is bounded, we hve lim F () e s =, Re.S >. Hence F (x) = x f, x < is bounded on J, e s is differenible on J, (e s ) L 1 (J) nd F () e s s, Re.s >. Therefore by Du-Bois Reymond s es, f()e s HK(J). Th is, he Lplce rnsform of f(), e s f() d exiss, Re.s >. 1

5 Theorem If he Lplce rnsform of f(), e s f() d exiss for Re.s >, hen he funcion f HK loc. Proof. Suppose he inegrl e s f() d exiss for Re.s >. Noe h he funcion e s, s C, s funcion of rel vrible, is no of bounded vriion on [, ) bu is of bounded vriion on ny compc inervl I = [, b]. By Hke s heorem [1], we cn wrie e s f() d = lim b e s f() d. Since he limis on R.H.S. exis, we cn sy h he HK inegrl e s f()d exiss, Re.s > for some compc inervl [, b]. Hence he funcion f() is HK inegrble necessrily on he compc inervl I = [, b], h is, f HK loc. Theorem Le f HK([, )). Define F (x) = f() d. Then x L{f}(s) exiss if nd only if e s F () d exiss s C. Proof. Le T >. The funcion e s, s C, is of bounded vriion on compc inervl [, T ] nd he funcion f is HK inegrble on [, ). Hence e s f(), s C, is HK inegrble on [, T ] nd by inegring by prs, we ge T T e s f() d = e st F (T ) F () + s e s F () d. Since he funcion f is HK inegrble on [, ), F () is finie, sy A nd by [9.1(), [5]], F is coninuous funcion such h lim F (T ) =. T Therefore by Hke s heorem [1], T { T } lim e s f() d = lim e st F (T ) F () + s e s F () d. T T 11

6 Hence e s f() d = s e s F () d A. This shows h L{f}(s) exiss if nd only if e s F () d exiss s C, Re.s >. Theorem (Uniqueness heorem) Le L{f}(s) = e s f() d nd L{g}(s) = e s g() d. If f() = g().e. on [, ), hen L{f}(s) = L{g}(s). 5.3 Bsic Properies The usul elemenry properies such s lineriy, dilion, modulion, rnslion re remins sme for he HK Lplce rnsform, he proofs of which re quie similr o h of clssicl ones. Also here re some oher resuls h re nlogous o he clssicl one wih some modificion in he hypohesis which we discuss below. Theorem (HK Lplce rnsform of derivive) Le f : [, ) f() R be funcion which is in ACG δ (R) such h lim =. Then for Re.s > e s boh L{f}(s) nd L{f }(s) fil o exis or L{f }(s) = s L{f}(s) f(). Proof. Le T >. Consider he inegrl J = T e s f () d. This inegrl is vlid since he funcion e s, s C, is bounded vriion on ny compc inervl [, T ] nd he funcion f is HK inegrble on R. By inegring by prs, we ge By Hke s heorem [1], T lim T J = e st f(t ) f() + s T e s f() d. T e s f () d = lim T e st f(t ) f() + lim s e s f() d T 1

7 Therefore = s e s f () d = s Th is, L{f }(s) = s L{f}(s) f(). e s f() d f(). e s f() d f(). Remrk The bove resul cn be generlize by imposing he condiions on f s: For k =, 1,,..., (n 1), f (k) f re in ACG δ (R) nd lim (k) =, Re.s >. e s Then L{f (n) }(s) nd L{f}(s) fils o exis or L{f (n) }(s) = s n L{f}(s) s n 1 f() s n f ()... f (n 1) (). Theorem (Differeniion of HK Lplce rnsform) Le f : [, ) R be funcion such h L{f}(s) exiss on compc inervl, sy J = [α, β], α, β R. Define g() = f() nd ssume h g HK(R + ). Then L{g}(s) exiss nd L{g}(s) = (L{f}(s)).e. Here s R. Proof. Suppose L{f}(s) = Define g() = f(). e s f() d exiss on [α, β], α, β R. We know he necessry nd sufficien condiion for differeniing under he HK inegrl is h for ll [, b] [α, β] [1]. Now e s f() ds d = = s= s= = e s f() ds d = = s= = e s f() d ds (5.) (e e b )f() d 13

8 = L{f}() L{f}(b) which exiss. Therefore he double HK inegrl = s= e s f() ds d exiss. Hence by [lemm 5(), [13]], he equliy in (5.) holds nd we cn differenie under he HK inegrl. By (5.), we hve s= = e s f() d ds = L{f}() L{f}(b) = (L{f}(s)) ds. Therefore e s f() d ds = s= = for ll [, b] [α, β]. Which implies h (L{f}(s)) ds e s f() d = (L{f}(s)).e. on [α, β]. Th is, L{ f()}(s) = (L{f}(s)).e. on [α, β]. Hence he derivive of he HK Lplce rnsform of f() exiss.e. on [α, β]. Remrk Le G(s) = L{ f()}(s) nd H(s) = L{f()}(s). Then (H(s)) = G(s).e. on [α, β]. And, we know h [5], A funcion f : [, b] R is HK inegrble on [, b] if nd only if here exiss n ACG δ funcion F on [, b] such h F = f.e. on [, b]. Agin, by fundmenl heorem of clculus [1], (H(s)) is HK inegrble, i.e., every derivive is HK inegrble. Hence G(s) is lso HK inegrble. Consequenly, he funcion H(s) is ACG δ on [α, β]. Coninuing in his wy n-imes, we cn sy h if n f() is HK inegrble, hen ny order of derivive of L{f}(s) exiss.e. So L{f}(s) is nlyic.e. 14

9 Theorem (Muliplicion by 1 ) Le f : [, ) R be funcion { } such h L{f}(s) exiss nd ssume h f() HK(R + ). Then L f() (s) exiss nd Here s R. { } f() L (s) = Proof. Assume h f() HK(R + ). s L{f}(u) du. Since he funcion e s is of bounded vriion on ny compc inervl I = [, b], we hve e s f() HK([, b]). Th is, e s f() d exiss. Wihou loss of generliy, le us ke = so h e s f() d exiss. By Hke s heorem [1], lim } Hence L (s) exiss. { f() Now consider he inegrls I 1 = s b e s f() d exiss. e u f() du d, I = I 1 = = Therefore I 1 exiss on R R. [e s ] f() d s f() e d which exiss. s e u f() d du. Also, noe h he funcion e s is of bounded vriion on ny compc inervl J R. Therefore R V J(e s ) d < M J, for some consn M J >. Hence by [lemm 5(), [13]], I exiss on R R nd I 1 = I on R R. So Th is, L s f() e d = = s { } f() (s) = L{f}(u) du. s s e u f() d du L{f}(u) du. 15

10 Theorem (HK Lplce rnsform of inegrl) Le f : [, ) R { } be HK inegrble funcion such h L{f}(s) exiss. Then L f(u) du (s) { } exiss nd L f(u) du (s) = 1 L{f}(s). s Proof. Define F () = f(u) du, R+. Then F () = f().e. on R + [9.1(b), [5]]. Since f HK(R + ), he inegrl Th is, lim Hence lim F () is bounded on R +. Now for T >, consider T f(u) du exiss on R +. f(u) du exiss on R +. e s f() d = T e s F () d. This inegrl exiss since e s is of bounded vriion on [, T ] nd he funcion f() is HK inegrble on R +. By inegring by prs, we ge T T e s f() d = e st F (T ) + s e s F () d. Since lim F (T ) is bounded on R, se lim F (T ) = A. T T By Hke s heorem [1], we cn wrie T e s f() d = lim e s f() d T = lim T e st F (T ) + s lim = s T T e s F () d Re.s >. e s F () d 16

11 Therefore Th is, e s F () d = 1 s e s f() d, Re.s >. e s f(u) du d = 1 L{f}(s), Re.s >. s { } Hence L f(u) du = 1 L{f}(s), Re.s >. s 5.4 Convoluion In his secion we shll discuss some clssicl resuls of Lplce convoluion in he HK inegrl seing, clled HK-convoluion. Firs we give wo resuls bou he exisence of HK-convoluion. Theorem Le f HK(R + ) nd g BV(R + ). Then we hve f g exiss on R +. Proof. By definiion [4], f g() = f( τ) g(τ) dτ which exiss s HK-inegrl by muliplier heorem [1]. By inegring by prs, we ge Hence f g exiss on R +. f g() A f { } inf g + V R + [,] g(τ). Theorem Le f HK loc nd g BV such h he suppor of g is in he compc inervl I = [, b]. Then } f g() A f [ b, ] {inf g + V [,b] g(τ). [,b] 17

12 Proof. Suppose supp(g) [, b]. Then we hve f g() = f( τ)g(τ) dτ. Inegre by prs, we ge f g() g(τ) b f( τ) dτ + f( τ) dg(τ) inf g(τ) b b τ [,b] f( τ) dτ + f( τ) dτ V [,b] g(τ) { } inf g(τ) + V [,b] g(τ) sup τ [,b] f(u) du. [ b, ] b Therefore } f g() A f [ b, ] {inf g + V [,b] g(τ). [,b] The following resul shows h he HK-convoluion is bounded. Theorem Le f HK(R + ) nd g L 1 (R + ) BV(R + ). Then we hve A f g A f g 1. Proof. For < b, we hve Le f g() d = f( τ)g(τ) dτ d. I 1 = Consider f( τ)g(τ) dτ d nd I = I = f( τ)g(τ) d dτ. Since f HK(R + ), we hve f( τ) d exiss nd finie. Le b f(τ) dτ = M. 18 f( τ)g(τ) d dτ.

13 Therefore I M g(τ) dτ. Since g L 1 (R + ), we hve for some finie K, g(τ) dτ = K. So I MK nd hence I exiss on R + R +. Now s g L 1 (R + ), we hve g 1 K 1 nd lso V R + [,] g M 1. Therefore by [lemm 5, [13]], I 1 = I on R + R +. Hence f g() d sup b f(u) du g(τ) dτ [, b] A f [, b] g 1. Therefore sup [,b] Th is, A f g A f g 1. f g() d A f [, b] g 1. Now we give some bsic elemenry properies of HK-convoluion. Theorem Suppose f HK(R + ) nd g BV(R + ). Then he following holds: I) f g() = g f() for ll R +. II) (λf) g = f (λg) = λ(f g) for some λ C. III) (f g) x () = f x g() = f g x (). IV) h (f + g)() = h f() + h g() for some h HK(R + ). Proof. The proof is similr o h of he clssicl cse. 19

14 The ssociive propery of HK-convoluion is given in he nex heorem. Theorem If f HK(R + ), g BV(R + ) nd h L 1 (R + ), hen (f g) h() = f (g h)(), R +. Proof. Consider (f g) h() = τ By chnging he order of inegrion, we ge Now Consider nd Le (f g) h() = I 1 = I = = ξ= ξ= f(ξ) f(ξ) g(τ ξ) h( τ) dξ dτ. ξ f(ξ) g h( ξ) dξ = f (g h)(). τ=ξ ξ= J = ξ= g(u) h( u ξ) du dξ f(ξ) g(τ ξ) h( τ) dξ dτ f(ξ) g(τ ξ) h( τ) dτ dξ. τ=ξ τ=ξ g(τ ξ) h( τ) dτ. Inegre by prs, we ge J g( ξ) u h( τ) dτ + h( τ) dτ dg(τ ξ) τ=ξ τ=ξ τ=ξ inf g [ξ,] h( τ) dτ + h( τ) dτ V [ξ,] g(τ ξ) inf [ξ,] g τ=ξ τ=ξ h( τ) dτ + τ=ξ τ=ξ inf [ξ,] g h 1 + h 1 V [ξ,] g(τ ξ) 13 h( τ) dτ V [ξ,] g(τ ξ)

15 Therefore } h 1 {inf g + V [ξ,] g(τ ξ). [ξ,] I sup R + f(ξ) dξ { h 1 { inf A f h 1 {inf [ξ,] g + V [ξ,] g(τ ξ) }} g + V [ξ,] g(τ ξ) [ξ,] }. Hence I exiss on R + R +. And, by [lemm 5, [13]] we re hrough. Now we show h he HK-Lplce rnsform of convoluion of wo funcions is he produc of heir HK-Lplce rnsforms. Theorem Le f 1 HK(R + ), f L 1 (R + ) BV(R + ). Then f g exiss nd if L {f 1 ()} nd L {f ()} exis s C, hen L {f 1 f ()} exiss s C nd L {f 1 f ()} (s) = L {f 1 ()} (s) L {f ()} (s). Proof. Consider L {f 1 f ()} (s) = = e s f 1 f () d By inerchnging he iered inegrls, we hve L {f 1 f ()} (s) = τ= Bu by hypohesis boh he inegrls = e sτ f 1 (τ)e s( τ) f ( τ) dτ d. e sτ f 1 (τ) e sτ f 1 (τ) dτ =τ e s( τ) f ( τ) d dτ e su f (u) du. e sτ f 1 (τ) dτ nd e sτ f (τ) dτ exis s C. Therefore L {f 1 f ()} exiss s C nd 131

16 L {f 1 f ()} (s) = L {f 1 ()} (s) L {f ()} (s). Now i remins o show he jusificion for inerchnging he iered inegrls. Since he funcion f L 1 (R + ), K I such h f 1 < K I, where I = [, b] R + is ny compc inervl. Clerly, V [,b] e s( τ) f ( τ) e σu V [ b, ] f (τ) + f (τ) e σr for some rels u nd r nd so V [ b, ] e sτ f (τ) dτ e σu V [ b, ] f (τ) dτ +e σr f (τ) dτ. Therefore V [ b, ] e sτ f (τ) dτ e σu V [ b, ] f (τ) dτ + K I e σr. Since f BV(R + ), here exiss consn M I such h Then V [ b, ] f (τ) dτ < M I. V [ b, ] e sτ f (τ) dτ e σu M I + K I e σr. Thus for ech compc inervl I = [, b] R +, he inegrl is finie nd f 1 K I. Hence he inegrl V [ b, ] e sτ f (τ) dτ τ= =τ e s f 1 (τ) f ( τ) d dτ exiss on R + R +. And, by [lemm 5, [13]], we cn inerchnge he iered inegrls. 13

17 Theorem In ddiion o he hypohesis of he bove heorem if f 1 f () is in HK(R + ) nd f is coninuous, hen f 1 f () is coninuous wih respec o he Alexiewicz norm. Proof. Choose ny δ > such h < δ 1 (The cse for negive δ is nlogous). Define D(, δ) = f 1 f ( + δ) f 1 f () = I 1 + I. Where I 1 = f 1 (τ) [f ( + δ τ) f ( τ)] dτ, I = Since f BV(R + ), f is bounded on R +. +δ So here exiss M such h f (x) M, for ll x R +. Then +δ I M f 1 (τ) dτ. Bu f 1 HK(R + ) implies h lim δ Therefore I s δ. Now consider I 1 = +δ f 1 (τ) dτ. f 1 (τ) [f ( + δ τ) f ( τ)] dτ. f 1 (τ)f (+δ τ) dτ. By inegring by prs, we ge I 1 [f ( + δ τ) f ( τ)] f 1 (τ) dτ u + f 1 (τ) dτ d(f (u + δ τ) f (u τ)) f ( + δ τ) f ( τ) sup f 1 (τ) dτ R + u + sup f 1 (τ) dτ d(f (u + δ τ) f (u τ)) u R + 133

18 = f ( + δ τ) f ( τ) sup f 1 (τ) dτ R + u + sup f 1 (τ) dτ f (u + δ τ) f (u τ). u R + Therefore I 1 f ( + δ τ) f ( τ) sup f 1 (τ) dτ R + u + sup f 1 (τ) dτ f (u + δ τ) f (u τ). u R + Since f is coninuous on R +, for given ɛ > here exiss η > such h I 1 ɛ, δ < η. Therefore D(, δ) I 1 + I < ɛ, δ < η. Th is, for given ɛ > here exiss η > such h D(, δ) < ɛ, δ < η. Hence f 1 f is coninuous R +. Now, le α, β R +. Consider β α [f 1 f ( + δ) f 1 f ()] d = α+δ α f 1 f () d β+δ β f 1 f () d. Wrie F 1, (x) = x f 1 f () d. Then β sup [f 1 f ( + δ) f 1 f ()] d sup F 1, (α + δ) F 1, (α) α, β R + α R + α + sup β R + F 1, (β + δ) F 1, (β). Since F 1, is n indefinie HK inegrl of f 1 f on R +, i is coninuous on R + [5]. Therefore for given ɛ > here exiss η > such h Hence we hve F 1, (ξ + δ) F 1, (ξ) < ɛ, δ < η nd for ll ξ R+. sup α, β R + β α [f 1 f ( + δ) f 1 f ()] d < ɛ, δ < η. 134

19 Thus for given ɛ > here exiss η > such h A f 1 f (+δ) f 1 f () < ɛ, δ < η. This shows h f 1 f () is coninuous R + wih respec o he Alexiewicz norm. The following resul concerns wih necessry nd sufficien condiions on f n nd g n : [, ] R so h he HK-convoluion s n operor is coninuous. I involves eiher uniform boundedness or uniform convergence of he indefinie HK inegrl of f n. Theorem Le {f n } be sequence of HK-inegrble funcions, f n : [, ] R, n such h f n = f s n for some HK-inegrble funcion f : [, ] R. Define F n (x) = x f n, F (x) = x f. Le {g n } be sequence of uniform bounded vriion funcions, g n : [, ] R, n, such h g n g poinwise on [, ], where g : [, ] R. Then f n g n () f g() s n if nd only if i) F n F uniformly on [, ] s n. ii) F n F poinwise on [, ], {F n } is uniformly bounded nd V (g n g). Proof. i) Suppose F n F uniformly on [, ] s n. By ssumpion nd muliplier heorem [1], f n g n () exiss R + nd f n g n () = f n (u) g n ( u) du nd f g() = Consider f n g n () f g() = I 1 + I, where I 1 = [f n (u) f(u)] g n ( u) du, I = f(u) g( u) du. [g n ( u) g( u)] f(u) du. 135

20 By inegring by prs, we ge Then I 1 = g n ( u) [f n (u) f(u)] du = g n ( u)[f n () F ()] I 1 Mx u F n(u) F (u) s n Hence I 1 s n. And by inegring by prs, we ge y [F n (u) F (u)] dg n. [f n (u) f(u)] dg n dg n + M F n () F () (Since F n F uniformly). I = {g n ( u) g( u)} f F dg n + F dg. By ssumpion he firs erm ends o nd since F is coninuous nd ech g n is of bounded vriion wih g n g s n, so we hve F dg n F dg. Hence I s n. Therefore f n g n () f g() s n. Now for he converse pr, suppose h eiher F n F on [, ] or F n F no uniformly on [, ]. Then here is sequence {x i } i Λ in [, ] on which F n F uniformly. So, by Bolzno-Weiersrss heorem [7], we cn find subsequence {y i } i Γ of {x i } i Λ, Γ Λ, such h F n (y i ) F (y i ) for ll i nd y i y s i, i.e., F n (y i ) F (y i ). Le us ssume wihou loss of generliy h < y n y. Le H be Heviside sep funcion. Define H(u y n ), g n ( u) = H(u y), n Γ oherwise 136

21 nd g( u) = H(u y). Then we hve, for n Γ, f n g n () = F n () F n (y n ) nd f g() = F () F (y). Since F n (y n ) F (y n ) s n, y n y s n nd F is coninuous, we hve F n (y n ) F (y). Bu F n () F () s n. Therefore f n g n () f g() s n which is conrdicion. Hence we mus hve F n F uniformly on [, ]. ii) Suppose F n F poinwise on [, ], F n is uniformly bounded nd V (g n g). Consider f n g n () f g() = I 1 + I, where I 1 = f n (u) [g n ( u) g( u)] du, I = By inegring by prs, we ge I 1 = [g n ( u) g( u)] f n = [g n ( u) g( u)] F n () Therefore y [f n (u) f(u)] g( u) du. f n d(g n g) F n (y) d(g n g). I 1 g n ( u) g( u) F n () + F n (y) V (g n g). Since {F n } is uniformly bounded, we hve F n M, n nd for some consn M. Therefore I 1 g n ( u) g( u) M + M V (g n g). By ssumpion g n g s n nd lso V (g n g). Hence I 1 s n. Agin by inegring by prs, we wrie I = g( u) {F n () F ()} 137 (F n F ) dg. (5.3)

22 By ssumpion F n () F (), so he firs erm ends o. And, since F n M, F n F nd g is of bounded vriion, by he domined convergence heorem for Riemnn-Sieljes inegrl [6], we hve F n dg F dg s n. Therefore by (5.3), I s n. Hence f n g n () f g() s n. Now for he converse pr suppose here is c (, b) such h F n (c) F (c) s n. Le g n ( x) = g( x) = H(x c). Then f n (u) g n ( u) du = F n () F n (c), f(u) g( u) du = F () F (c). Since F n () F () nd F n (c) F (c), we hve F n () F n (c) F () F (c) s n. Th is, f n g n () f g() s n which is conrdicion. Hence we mus hve F n F poinwise on [, ]. Now if {F n } is no uniformly bounded, hen here is sequence {x i } i Λ in [, ] on which F n. Therefore by Bolzno-Weiersrss heorem [7], we cn find subsequence {y i } i Γ of {x i } i Λ, Γ Λ, such h F n (y i ) 1 for ll n nd F n (y i ) nd y i y. Wihou loss of generliy le us ke y i y. Define g n ( u) = H(u y i). Fn(yi ) Then V (g n ) 1, g = nd V (g n g) = nd f n (u) g n ( u) du = 138 f n (u) H(u y i) Fn (y i ) du

23 f n (u) = y i Fn (y i ) du = F n() Fn (y i ) F n (y i ). Therefore nd f n (u)g n ( u) du s n f(u)g( u) du =. Hence f n g n f g s n which is conrdicion nd so we mus hve {F n } is uniformly bounded sequence. 5.5 Inversion Since in cse of Hensock-Kurzweil inegrl, he inegrion process nd differeniion process re inverse of ech oher. Here we shll esblish he inverse of Hensock-Kurzweil Lplce rnsform by using Pos s generlized differeniion mehod [8]. For his we shll need following lemms: Lemm Le η be such h < η < b, nd h(x) C ( x + η), h () =, h () <, h(x) is nonincresing on (, b]. Then ( π e k h(x) dx e k h() k h () k. Proof. Since h(x) C ( x + η), h (x) is coninuous. Le ɛ > be such h < ɛ < h (). Choose δ >, δ < η such h h (x) h () < ɛ, x + δ. Th is, h () ɛ < h (x) < h () + ɛ, x + δ. (5.4) 139

24 Consider he inegrl We wrie I k = I k + I k, where I k = +δ I k = e k [h(x) h()] dx. e k [h(x) h()] dx, I k = e k [h(x) h()] dx. +δ Since h is nonincresing on (, b], we hve h(x) h( + δ) for ll x > + δ. Therefore nd Hence I k Now consider s k. I k e k [h(+δ) h()] dx +δ I k e k [h(+δ) h()] (b δ). I k = +δ By Tylor s series wih reminder, we hve Therefore h(x) h() h (ξ) So by (5.4), we cn wrie +δ Th is, I k = e k [h () ɛ] (x ) dx +δ +δ e k [h(x) h()] dx. (x ), ξ + δ. e k h (ξ) (x ) dx. +δ e k h (ξ) (x ) dx e k [h ()+ɛ] (x ) dx. Consider +δ +δ e k [h () ɛ] (x ) dx I k e k [h ()+ɛ] (x ) dx. (5.5) J = +δ e k [h () ɛ] (x ) dx. 14

25 Afer subsiuing x = u, we ge Therefore +δ Similrly, +δ J = = = δ e k [h () ɛ] u du 1 k (h () ɛ) k 1 (h () ɛ) 1 π k (h () ɛ) ( e k [h () ɛ] (x ) dx = ( e k [h ()+ɛ] (x ) dx = δ k (h () ɛ) Therefore (5.5) becomes ( ( π I k (h k () ɛ) Since ɛ > is rbirry, we hve ( π I k h k () Hence So Th is, ( I k π k h () ( e k [h(x) h()] dx e u du e u du s k s k. π k (h () ɛ) π k (h () + ɛ) ( π k h () ( e k h(x) dx e k h() π k h () π k (h () + ɛ) π. k h () s k. s k. s k.. s k. s k. 141

26 Lemm Le η be such h < η < b, nd h(x) C (b η x b), h (b) =, h (b) <, h(x) is nonincresing on [, b). Then ( e k h(x) dx e k h(b) π s k. k h (b) Proof. The proof is similr o h of previous lemm. Lemm Le η be such h < η < b, nd h(x) C ( x + η), h () =, h () <, h(x) is nonincresing on (, b]. Suppose φ(x) HK([, b]). Then ( φ(x) e k h(x) dx φ() e k h() π k h () s k. Proof. Since h(x) C ( x + η), we hve h (x) is coninuous. Le ɛ > be such h < ɛ < h (). Choose δ >, δ < η such h h (x) h () < ɛ, x + δ. Th is, Consider he inegrl h () ɛ < h (x) < h () + ɛ, x + δ. I k = We show h I k s k. Le We clim h g k s k. [φ(x) φ()] e k [h(x) h()] dx. e k [h(x) h()], x [, b] g k (x) =, x =. Since h is nonincresing on (, b], we hve h(x) h() < for ll x (, b]. Therefore g k (x) = e k [h(x) h()] s k. Se g(x) = for ll x [, b]. Then g k (x) = e k [h(x) h()] = g(x) for ll x [, b], s k. 14

27 Since φ(x) HK([, b]), he inegrl [φ(x) φ()] dx exiss. Observe h he sequence {g k (x)} is of uniform bounded vriion on [, b] wih g k (x) s k. Then by [Cor.3., [11]], we hve [φ(x) φ()] e k [h(x) h()] dx [φ(x) φ()] dx s k s k. Therefore Th is, φ(x) e k [h(x) h()] dx = φ() Bu by lemm (5.5.1), we hve Therefore φ(x) e k h(x) dx = φ() ( π e k h(x) dx e k h() k h () e k [h(x) h()] dx. e k h(x) dx. ( π φ(x) e k h(x) dx φ() e k h() k h () s k. s k. Lemm Le η be such h < η < b, nd h(x) C (b η x b), h (b) =, h (b) <, h(x) is nonincresing on [, b) nd suppose φ(x) HK([, b]). Then ( π φ(x) e k h(x) dx φ(b) e k h(b) k h (b) 143 s k.

28 Proof. The proof is similr o h of he previous lemm. Now we re redy o obin he inversion heorem for Hensock-Kurzweil Lplce rnsform. Theorem (Inversion Theorem) If f() HK(R + ) nd if he HK- Lplce rnsform of f(), L{f}(s) exiss, hen lim k 1 k! ( ) k+1 k e ku u k f(u) du = f(). Proof. The inversion heorem follows immediely, if we cn prove he following: I) II) lim k lim k I) We hve he relion 1 k! 1 k! ( ) k+1 k e ku u k f(u) du = f() (5.6) ( ) k+1 k e ku u k f(u) du = f(). (5.7) ( π f(x) e k φ(x) dx f() e k φ() k φ () Tke = nd φ(x) = ln(x) x Therefore Th is, f(x) e k [ln(x) x ] dx f() e k [ln() 1] ( s k. for ll x [, ). Then we hve π k ( 1 ) ( f(x) x k e kx π dx f() k e k k s k. s k. f(x) x k e kx dx f() k+1 e k ( π k s k. 144

29 Therefore s k, we hve 1 k! ( ) k+1 k b f(x) x k e kx dx = 1 k! = ( k e ( ) k+1 k ( π f() k+1 e k k ) k k ( π k! f(). k The Sirling s formul is given by ( ) n n e πn n!, using his we cn wrie 1 k! ( ) k+1 k b f(x) x k e kx dx = k f() πk = f() ( π k s k. s k This is rue for every b R, b >. Therefore by Hke s heorem [1], we hve lim k 1 k! ( ) k+1 k f(x) x k e kx dx = f(). II) Agin we hve he relion ( π f(x) e k φ(x) dx f(b) e k φ(b) k φ (b) s k. Tke b = nd φ(x) = ln(x) x, for ll x (, ]. Then we ge lim k 1 k! By Hke s heorem [1], we hve lim k 1 k! ( ) k+1 k f(x) x k e kx dx = f(). ( ) k+1 k f(x) x k e kx dx = f(). Therefore from (5.6) nd (5.7), we cn wrie lim k 1 k! ( ) k+1 k f(x) x k e kx dx = f(). 145

30 5.6 Exmples Here we shll give some exmples of HK-Lplce rnsformble funcions whose clssicl Lplce rnsform do no exis. However we could no ble o find he exc form of he HK-Lplce rnsform of hese funcions. Exmple Consider he funcion g() = e s sin ln(+), s C. Le f() = e s sin nd φ() = 1 ln(+). Observe h he funcion φ() is differenible on R + nd φ L 1 (R + ) nd if F () = e su sin u du, Re.s >, hen F () M for ll (, ), Re.s > for some consn M >. Now Then { e s (s sin cos ) F () φ() = s } s ln( + ). { e s (s sin cos ) lim F () φ() = lim + 1 } 1 s + 1 s + 1 ln( + ). Therefore lim F () φ() exiss. Hence by Du-Bois-Reymond s es [1], we hve e s sin ln(+) HK(R + ), Re.s >. Th is, e s sin ln( + ) d exiss, Re.s >. Hence he HK-Lplce rnsform of he funcion g() =. Exmple Consider he funcion f() = cos. We hve F () = cos u u du = cos u du. sin ln(+) exiss, Re.s > 146

31 Le φ() = e s, s C. Then φ () = s e s L 1 (R + ), Re.s >. Now Then F () φ() = e s cos u du. lim F () φ() = lim e s = lim e s lim =, Re.s >. cos u du cos u du Therefore by Du-Bois-Reymond s es [1], we hve f() φ() HK(R + ), Re.s >. Hence he HK-Lplce rnsform of he funcion f() = cos exiss for Re.s >. Exmple Le φ() = e σ, R +, σ >. This funcion is coninuous on R + nd φ( + 1) = 1 ( + 1) e < 1 = φ() for ll σ(+1) R+ eσ 1 nd lso lim φ() = lim =, σ >. e σ Therefore φ() is decresing o zero s. Observe h he inegrl diverges. e σ d, σ > Now le n N be such h < (1 + 4n ) π 4ω. For R, we hve sin ω 1 if nd only if ω n=n [(1 + 4n) π 4, (3 + 4n) π 4 ]. 147

32 If n N, hen (3 + 4n) π < (1 + n) π. 4 (1+n)π e σ sin d = n (3+4j) π 4 e σ j=n (1+4j) π 4 n (3+4j) π 4 e σ j=n 1 = 1 (1+4j) π 4 n (3+4j) π 4 j=n n (1+4j) π 4 j=n 4 sin ω d 1 d 1 (3 + 4j) π 4 eσ(3+4j) π 4 (3 + 4j) π e s(3+4j) π 4 π n 1 (1 + j) π e. σ(1+j)π j=n π d On he oher hnd, we hve (1+n)π e σ Bu since e σ d = = = n π e σ n π e σ n π e σ n π e σ HK(R + ), we hve Therefore j=n e σjπ jπ =. So e σ sin ω / L 1 (R + ), σ >. Similrly, e σ cos ω / L 1 (R + ), σ >. Now le f() = sin ω, φ() = e σ, R. d + d + d + (1+n)π e σ d n π n (1+j)π e σ j=n jπ n (1+j)π e σjπ j=n jπ n d + e σjπ jπ π. j=n e σ d =. Observe h φ() is monoone funcion nd φ() s. And sin ωu du, ω for ll ω R+. Th is, F is bounded. Therefore by Chrier-Dirichle s es [1], we hve e σ sin ω HK(R + ). 148 jπ d d

33 Similrly, e σ cos ω HK(R + ). Hence e σ {cos ω i sin ω} HK(R + ). Th is, Thus e s e σ e iω d exiss, σ >. d exiss, Re.s > s HK inegrl, where s = σ + iω. 149

34 References [1] Brle R. G., A Modern Theory of Inegrion, Grd. Sud. Mh. 3, Amer. Mh. Soc. Providence, 1. [] Chudhry M. S., Tikre S. A., On Guge Lplce Trnsform, In. Journl of Mh. Anlysis, Vol. 5, No. 35 (11), [3] Erdelyi A., Tbles of Inegrl rnsforms, Vol. 1, McGrw Hill Book Compny, Inc [4] Gsque C., P. Wiomski P., Fourier nlysis nd pplicions, Springer- Verlg, New York, Inc [5] Gordon R. A., The Inegrls of Lebesgue, Denjoy, Perron, nd Hensock, Grd. Sud. Mh. 4, Amer. Mh. Soc. Providence, [6] McLeod R. M., The Generlized Riemnn Inegrl, Crus Mh. Monogrphs, No., The Mhemicl Associion of Americ, Wshingon DC, 198. [7] Nnson I. P., Theory of funcions of rel vrible, Vol. 1, Frederick Unger Publishing Co. New York, [8] Pos E. L., Generlized differeniion, Trns. Amer. Mh. Soc. Vol. 3 (193),

35 [9] Sks S., Theory of he Inegrl, Dover Publicions, New York, [1] Swrz C., Inroducion o Guge Inegrls, World Scienific Publishing Co. Singpore, 1. [11] Tlvil E., Limis nd Hensock inegrls of producs, Rel Anl. Exchnge 5 (1999-), [1] Tlvil E., Necessry nd sufficien condiions for differeniing under he inegrl sign, Amer. Mh. Monhly, 18 (1), [13] Tlvil E., Hensock-Kurzweil Fourier rnsforms, Ill. Jour. Mh. Vol. 46, No. 4(), [14] Tikre S. A., Chudhry M. S., On some resuls of HK-convoluion, In. Elecron. J. Pure Appl. Mh. Vol., No. (1), [15] Widder D. V., The Lplce rnsform, Princeon Universiy Press, Princeon, [16] Zyed A. I., The Hndbook of Funcion nd Generlized Funcion rnsformions, CRC Press, Inc