Dedicated to Prof. J. Kurzweil on the occasion of his 80th birthday

Transcription

1 131 (2006 MATHEMATICA BOHEMICA No. 3, THE HENSTOCK-KURZWEIL APPROACH TO YOUNG INTEGRALS WITH INTEGRATORS IN BV ϕ Boopogkrog Vryu, Tu Seg Chew, Sigpore (Received September 8, 2005 Dedicted to Prof. J. Kurzweil o the occsio of his 80th birthdy Abstrct. I 1938, L. C. Youg proved tht the Moore-Pollrd-Stieltjes itegrl f dg exists if f BV ϕ[, b], g BV ψ [, b] d ϕ 1 (1/ψ 1 (1/ <. I this ote we use the Hestock-Kurzweil pproch to hdle the bove itegrl defied by Youg. Keywords: Hestock itegrl, Stieltjes itegrl, Youg itegrl, ϕ-vritio MSC 2000 : 26A21, 28B15 1. Itroductio I 1936, L. C. Youg proved tht the Riem-Stieltjes itegrl f dg exists, if f BV p [, b], g BV q [, b], 1/p + 1/q > 1 d f, g do ot hve commo discotiuous poits, see [7], [11]. Two yers lter, he ws ble to drop the coditio o commo discotiuity for his ew itegrl (clled Youg itegrl, see [12]. The Youg itegrl is defied by the Moore-Pollrd pproch, see [2, pp , pp ] d [3], [8], [9]. I other words, the itegrl is defied by wy of refiemets of prtitios d the itegrl is the Moore-Smith limit of the Riem-Stieltjes sums usig the directed set of prtitios. However, modified Riem-Stieltjes sums ivolvig g(x+ d g(x re used i Youg itegrls. Furthermore, he geerlized his result d proved tht the Youg itegrl f dg exists if the followig Youg s coditio holds: f BV ϕ [, b], g BV ψ [, b] 233

2 d ϕ 1( 1 ψ 1( 1 <, where BV ϕ [, b] is the spce of fuctios of bouded ϕ-vritio o [,b]. The Youg itegrl with itegrtor i BV p usig the Hestock-Kurzweil pproch is give i [1]. I this ote we will gi use the Hestock-Kurzweil pproch to hdle the Youg itegrl with itegrtor i BV ϕ. Now we shll itroduce Hestock-Kurzweil itegrls, see [4]. Let P = {[u i, v i ]} be fiite collectio of o-overlppig subitervls of [, b], the P is sid to be prtil prtitio of [, b]. If, i dditio, [u i, v i ] = [, b], the P is sid to be prtitio of [, b]. Let δ be positive fuctio o [, b], [u, v] [, b] d ξ [, b]. The itervlpoit pir (ξ, [u, v] is sid to be δ-fie if ξ [u, v] (ξ δ(ξ, ξ + δ(ξ. Let D = {(ξ i, [u i, v i ]} be fiite collectio of itervl-poit pirs. The D is sid to be δ-fie prtil divisio of [, b] if {[u i, v i ]} is prtil prtitio of [, b] d for ech i, (ξ i, [u i, v i ] is δ-fie. I dditio, if {[u i, v i ]} is prtitio of [, b], the D is sid to be δ-fie divisio of [, b]. I this ote, deotes the set of rel umbers. Now, we shll defie itegrls of Stieltjes type by the Hestock-Kurzweil pproch. Defiitio 1.1. Let f, g : [, b]. The f is sid to be Hestock-Kurzweil itegrble (or HK-itegrble to rel umber A o [, b] with respect to g if for every ε > 0 there exists positive fuctio δ defied o [, b] such tht for every δ-fie divisio D = {(ξ i, [t i, t i+1 ]} of [, b], we hve S(f, δ, D A ε, where A is deoted by f dg. S(f, δ, D = f(ξ i (g(t i+1 g(t i. It is kow tht if f BV p [, b], g BV q [, b], 1/p + 1/q > 1, the f is HKitegrble with respect to g o [, b], see [1]. I this ote we follow ides of Youg to show tht if f BV ϕ [, b], g BV ψ [, b] d ϕ 1 (1/ψ 1 (1/ <, the f is HK-itegrble with respect to g o [, b]. 234

3 2. Youg s series The bove series is clled Youg s series. We shll preset some properties of Youg s series. Results d proofs re kow, see [5], [6], [12]. We give proofs here for esy referece. I this sectio, let λ, µ be strictly icresig cotiuous o-egtive fuctios o [0, with λ(0 = µ(0 = 0 d let ω, κ be icresig fuctios o [, β] with ω(β ω( A d κ(β κ( B. Lemm 2.1. For p = 0, 1, 2,..., there exists E p = {x 1, x 2,..., x p } [, β] such tht for y ξ, η (x i, x i+1, i = 1, 2,..., p 1, we hve ω(η ω(ξ A2 p d κ(η κ(ξ B2 p. Furthermore, E q E p if p q, #(E p 2 p+1 d #(E p+1 \ E p 2 p+1, where #(E p deotes the umber of elemets i the set E p.. Deote ω(ξ ω(η, κ(ξ κ(η by ω(ξ, η, κ(ξ, η respectively. Let E0 ω = {x (0 1, x(0 2 }, where x(0 1 =, x (0 2 = β. The for y ξ, η (x (0 1, x(0 2, we c see tht ω(ξ, η A. Let x (0 1 = sup{x [x(0 1, x(0 2 ]; ω(ξ, η A2 1 for y ξ, η (x (0 1, x} d let E ω 1 = {x(0 1, x(0 1, x(0 2 }.. We my ssume tht the bove isted of (x(0 1, x. We will ccordig to their order usig the ottio It is possible tht x (0 1 = x(0 2, i.e., Eω 1 = Eω 0 supremum is well-defied, otherwise we use (x, x (0 2 reme poits i E1 ω x (1 2 E ω 1 = {x (1 1, x(1 2, x(1 3 }. We clim tht for y ξ, η (x (1 2, x(1 3, ω(ξ, η A2 1. Suppose tht there exist ξ, η (x (1 2, x(1 3 such tht ω(ξ, η > A2 1. Sice ξ >, there exists poit β (x(1 1, x(1 2 such tht ω(β, ξ > A2 1. Sice ξ > x (1 2, ω(β, η = ω(β, ξ + ω(ξ, η > A2 1 + A2 1 = A. 235

4 It cotrdicts the defiitio of E ω 0. Hece, for y ξ, η (x (1 2, x(1 3, ω(ξ, η A2 1. Tht is, for y ξ, η (x (1 i, x (1 i+1, i = 1, 2, we hve ω(ξ, η A2 1. Let x (1 1 = sup{x [x(1 1, x(1 2 ]; ω(ξ, η A2 2 for every ξ, η (x (1 1, x}, x(1 2 = sup{x [x (1 2, x(1 3 ]; ω(ξ, η A2 2 for every ξ, η (x (1 2, x} d E ω 2 = {x (1 1, x(1 1, x(1 2, x(1 2, x(1 3 }. It is still possible tht x (1 1 = x(1 2 or x (1 2 = x(1 3. We gi reme Eω 2 ccordig to their order by E2 ω = {x (2 1, x(2 2, x(2 3, x(2 4, x(2 5 }. Usig the sme rgumet s bove, we lso hve for y ξ, η (x (2 i, x (2 i+1 d for every i = 1, 2, 3, 4, ω(ξ, η A2 2. Usig this method, we c hve Ep ω = {x(p 1, x(p 2,..., x(p p }, p = 0, 1, 2,..., such tht for y ξ, η (x (p i, x (p i+1, i = 1, 2,..., p 1, we hve ω(ξ, η A2 p. We c lso see tht E ω q E ω p wheever q p, the umber of elemets i E ω p is t most 2 p + 1 d the umber of elemets i E ω p+1 \ Eω p is t most 2p. Usig the sme rgumet, we lso c defie Ep κ = {y(p 1, y(p 2,..., y(p m p } for p = 0, 1, 2,..., so tht for y ξ, η (y (p i, y (p i+1, i = 1, 2,..., m p 1, we hve κ(ξ, η B2 p. Furthermore, Eq κ Eκ p wheever q p, the umber of elemets i Eκ p is t most 2 p + 1 d the umber of elemet i Ep+1 κ \ Ep κ is t most 2 p. Now, let E p = Ep ω Ep κ = {z (p 1, z(p 2,..., z(p r p }. The for every ξ, η (z (p i, z (p i+1, i = 1, 2,..., r p 1, ω(ξ, η A2 p d κ(ξ, η B2 p. Furthermore, E q E p wheever q p, the umber of elemets i E p+1 \ E p is t most 2 2 p = 2 p+1 d umber of elemets i E p is t most 2(2 p = 2 p+1, sice, β Ep ω Eκ p. 236

5 Lemm 2.2. (i For y positive iteger v, the followig iequlities hold: ( B 2 +v λ(a2 (+v µ(b2 (+v 2 µ =0 =2 v 1 λ d (ii. (i 2 v ( B2 v λ µ =0 1 2 v 1 =2 v 1 λ ( B µ, ( B 2 λ(a2 µ(b2 3 λ µ. 2 +v λ(a2 (+v µ(b2 (+v = 2 k λ(a2 k µ(b2 k =0 Similrly, we hve k k=v =2 k 1 +1 =2 v 1 λ k=v ( B λ µ = 2 ( B µ. =2 v 1 +1 ( B λ µ 2 v ( B2 v v ( B2 v λ µ = λ µ k k =0 k=2 2 λ(a2 (v+ µ(b2 (v+ = 1 2 v 2 +v λ(a2 (+v µ(b2 (+v =0 1 2 v 1 =2 v 1 λ (ii As i the first prt of (i, =0 ( B µ. =0 2 λ(a2 µ(b2 = λ(aµ(b + 2 k λ(a2 k µ(b2 k λ(aµ(b + 2 ( B λ µ k=1 3 ( B λ µ. 237

6 Lemm 2.3. ( 1 ( 1 λ µ < if d oly if ( B λ µ <.. Suppose λ(1/µ(1/ <. Let m be positive iteger such tht A m d B m. The Coversely, suppose ( B m 1 ( B λ µ = λ µ + = m 1 m 1 ( B λ µ + The λ (1/µ (1/ <. Therefore, λ (1/(Aµ (1/(B <. Cosequetly, λ(1/µ(1/ <. =m ( B λ µ + m ( B λ µ (k+1m k=1 =km ( B λ µ ( 1 ( 1 λ µ <. k k k=1 λ(a/µ(b/ <. Let λ (x = λ(ax, µ (x = µ(bx. 3. Itegrls of step fuctios I this sectio we shll preset Youg s results o itegrls of step fuctios, see [12]. Let g be regulted fuctio o [, β] d s step fuctio o [, β] with s(x = q q+1 c i χ (ti,t i+1(x + d i χ {ti}(x, where χ G is the chrcteristic fuctio of G, d = t 1 < t 2 <... < t q+1 = β. It is kow, see [1], tht s dg = q q+1 c i (g(t i+1 g(t i + + d i (g(t i + g(t i. Furthermore, we lwys ssume tht the followig coditios hold: (1 { s(ξ s(η λ(ω(ξ ω(η, g(ξ g(η µ(κ(ξ κ(η 238

7 for y ξ, η [, β] with ξ > η, where λ, µ, ω, κ re give i Sectio 1. I this sectio we lwys ssume tht λ(a/µ(b/ <. Recll tht ω(β ω( A d κ(β κ( B. Defiitio 3.1. Let s be step fuctio defied o [, β] d E p the fiite set s defied i Lemm 2.1. Let E = {x i : i = 1, 2,..., m + 1} be y fixed fiite set cotiig E 0. We defie s E to be the step fuctio iduced by s d E s follows: s E (x = m m+1 s(x i +χ (xi,x i+1(x + s(x i χ {xi}(x. We hve, by the formul for the vlue of the itegrl of step fuctio with respect to g preseted bove, s E dg = m m+1 s(x i +(g(x i+1 g(x i + + s(x i (g(x i + g(x i. We remrk tht if E cotis ll poits of discotiuity of s, the s E = s d s E dg = s dg. Lemm 3.2. Let E E 0. The ( (s E Ep s Ep dg A ( B N pλ 2 p µ 2 p, where N p = #(E \ E p. Furthermore, lim (s E Ep s Ep dg = 0. p. Let N p deote #(E \ E p. Let s deote the step fuctio s E Ep s Ep. Suppose s is iduced by prtitio {[y i, y i+1 ]} m of [, β]. If y i E p, the s hs zero vlues over hlf-ope subitervl [y i, y i+1. Therefore, the umber of subitervls where s hs ozero vlue is t most N p. The (s E Ep s Ep dg N pλ(a2 p µ(b2 p N 0 λ(a2 p µ(b2 p. Hece, for y fixed fiite set E, lim (s E Ep s Ep dg p lim p N 0λ(A2 p µ(b2 p = 0. I the bove, we use the fct tht λ, µ re cotiuous t 0 d λ(0 = µ(0 =

8 Theorem 3.3. Let s be step fuctiod E 0 s bove. The (s s E0 dg 6 ( B λ µ.. From Lemm 3.2, E p+1 = E p+1 E p d #(E p+1 E p 2 p+1, we hve β (s Ep+1 s Ep dg = (s Ep+1 E p s Ep dg 2 p+1 λ(a2 p µ(a2 p = 2 2 p λ(a2 p µ(a2 p. Now, let E be fiite set cotiig E 0 d ll poits of discotiuity of s, the s dg = s E dg = s E E v dg for ll v = 0, 1, 2,.... Hece we hve (s s Ep dg β (s E s Ep+q dg + (s Ep+q s Ep+q 1 dg (s Ep+1 s Ep dg ( β β (s E E p+q s Ep+q dg + (s Ep+q s Ep+q 1 dg (s Ep+1 s Ep dg ( β lim q = lim q 0 + lim 4 q 1 q m=0 =2 p 1 λ 2 2 p+m λ(a2 (p+m µ(a2 (p+m ( B µ The lst iequlity holds by Lemm 2.2 (i. Whe p = 0, by Lemm 2.2 (ii we get for p = 1, 2,.... (s s E0 dg 6 ( B λ µ. 240

9 Corollry 3.4. Suppose tht s 1 (x = d i χ [ui,u i+1(x + d χ {u+1}(x, s 2 (x = m e i χ [vi,v i+1(x + e m χ {vm+1}(x re step fuctios defied o [, β]. Let (1 hold with s = s 1, s 2 d d 1 e 1 λ(a. The (2 m d i (g(u i+1 g(u i e i (g(v i+1 g(v i 13 ( B λ µ.. First we shll prove tht the followig iequlity holds: d i (g(u i+1 g(u i d 1 (g(β g( 6 ( B λ µ. Let g (u i = g(u i, g (t = g(u i for those t close to u i from the left, otherwise, let g (t = g(t. The g (u i = g (u i d g lso stisfies g (ξ g (η µ( κ(ξ κ(η for y ξ, η [, β]. The s 1 dg = = d i (g (u i+1 g (u i + d (g (u +1 g (u +1 d i (g (u i+1 g (u i = d i (g(u i+1 g(u i. Applyig Theorem 3.3 to s = s 1 d g = g, s E 0 dg = d 1 (g (β g ( + d (g (β g (β+ = d 1 (g(β g(, we get the iequlity (2. Thus m d i (g(u i+1 g(u i e i (g(v i+1 g(v i ( B λ µ + d 1 (g(β 1 g( 1 e 1 (g(β 1 g( 1 ( B λ µ + d 1 e 1 g(β 1 g( 1 ( B λ µ + λ(aµ(b = 13 ( B λ µ. 241

10 4. Itegrble fuctios Now we shll itroduce BV ϕ [, b], which is geerliztio of BV[, b], the spce of fuctios of bouded vritio o [, b], d prove existece theorem (Theorem 4.6 i the Hestock-Kurzweil settig. Defiitio 4.1. A fuctio ϕ: [0, is sid to be N-fuctio if 1. ϕ(0 = 0; 2. ϕ is cotiuous o [0, ; 3. ϕ is strictly icresig d 4. ϕ(u s u. Exmples of N-fuctios re ϕ 1 (u = u p, p 1 d ϕ 2 (u = e u 1. Defiitio 4.2. Let ϕ: [0, be N-fuctiod f : [, b]. We defie V ϕ (f; [, b] = sup ϕ( f(x i+1 f(x i, where supremum is tke over ll prtitios {[x i, x i+1 ]} of [, b]. The umber V ϕ (f; [, b] is clled the ϕ-vritio of f o [, b]. Let BV ϕ [, b] deote the collectio of ll fuctios f : [, b] stisfyig V ϕ (f; [, b] <, see [5], [6], [12]. Such fuctios re sid to be of bouded ϕ-vritio. Whe it is cler tht we re cosiderig the itervl [, b], we shll deote V ϕ (f; [, b] by V ϕ (f. For exmple, where ϕ(u = u p, p 1, BV ϕ [, b] is the spce of fuctios of bouded p-vritio o [, b]. The followig lemm d its proof re kow. Lemm 4.3. If f BV ϕ [, b], the f is bouded o [, b] d f is regulted fuctio.. Suppose f is ubouded. Let M be y positive rel umber. The there exists x [, b] such tht M f(x f(. Hece Therefore M f(x f( = ϕ 1 (ϕ( f(x f( ϕ 1 (ϕ( f(x f( + ϕ( f(b f(x ϕ 1 (V ϕ (f; [, b]. ϕ(m V ϕ (f; [, b] for ll M > 0. Sice ϕ(m s M, we hve V ϕ (f; [, b] =. This leds to cotrdictio. The proof tht f is regulted is stdrd. 242

11 Lemm 4.4. Let g BV ψ [, b], E = {x 1, x 2,..., x } E 0, d ε > 0, where E 0 is give i Lemm 2.1. The there exists costt δ > 0 such tht for y fiite collectio of disjoit subitervls {[u i, v i ]} with [u i, v i ] (x i, x i + δ or [u i, v i ] (x i δ, x i for ech i, we hve g(v i g(u i ε.. Let ε > 0 be give. First, sice g is regulted fuctio, there exists costt δ > 0 such tht g(t g(x i ε 2 wheever 0 < x i t < δ d g(x i + g(t ε 2 wheever for ech i. Therefore, we get the required result. 0 < t x i < δ Next, we shll prove Lemm 4.5 usig Lemm 2.2 d Corollry 3.4. We eed the followig ottio. Let A V ϕ (f d B V ψ (g. Defie ω(x = V ϕ (f; [, x] d κ(x = V ψ (g; [, x]. Let λ = ϕ 1, µ = ψ 1. Hece, for y ξ, η [, b] with η > ξ, λ(ω(η ω(ξ = ϕ 1 (ω(η ω(ξ f(η f(ξ. Similrly, µ(κ(η κ(ξ g(η g(ξ. Let E v = {x 1, x 2,..., x v } be give s i Lemm 2.1 with v 1 d [, β] = [, b]. The #(E v 2 v+1. Furthermore, f(η f(ξ λ(ω(η ω(ξ λ(a2 v = ϕ 1 (A2 v d g(η g(ξ µ(κ(η κ(ξ µ(b2 v = ψ 1 (B2 v for y η, ξ (x k, x k+1 with η > ξ, k = 1, 2,..., v 1. The bove is equivlet to (1 metioed before Defiitio 3.1. From ow owrds, divisio D = {(ξ i, [u i, v i ]} is lwys deoted by D = {(ξ, [u, v]}. 243

12 Lemm 4.5. Let f BV ϕ [, b] d g BV ψ [, b] with ϕ 1 (1/ψ 1 (1/ <. Let v 1 be fixed d E v = {x 1, x 2,..., x v } give s bove. Suppose D = {(ξ, [u, v]} d D = {(ξ, [u, v ]} re two prtil divisios of [, b] such tht [u, v] = [u, v ] d [u, v] (x k, x k+1, [u, v ] (x k, x k+1. The for y ξ [u, v], ξ [u, v ], we hve (D f(ξ(g(v g(u (D f(ξ (g(v g(u 52 where v 1. =2 v 1 ϕ 1 ψ 1( B. Let D k = {(ξ, [u, v] D ; [u, v] (x k, x k+1 } d D k = {(ξ, [u, v ] D ; [u, v ] (x k, x k+1 }, k = 1, 2,..., v 1. It is cler tht f(ξ f(ξ < ϕ 1 (A2 v. Note tht {[u, v]; [u, v] (x k, x k+1 } = {[u, v ]: [u, v ] (x k, x k+1 } =: [, β]. Applyig Corollry 3.4, for y ξ, ξ we hve (D k f(ξ(g(v g(u (D k f(ξ (g(v g(u 13 ϕ 12 v ψ 1( B2 v for k = 1, 2,..., v 1. Note tht D = v 1 D k d v 2 v+1. Hece, by Lemm 2.2 (i, k=1 (D f(ξ(g(v g(u (D f(ξ (g(v g(u 13(2 v+1 1 ϕ 12 v ψ 1( B2 v 13(2 v v 1 ψ 1( B 52 =2 v 1 ϕ 1 =2 v 1 ϕ 1 ψ 1( B The followig existece theorem is proved i [12] by the Moore-Pollrd pproch. Now we will prove it by the Hestock-Kurzweil pproch.., 244

13 Theorem 4.6 (Existece Theorem. Let f BV ϕ [, b] d g BV ψ [, b]. Suppose tht ϕ 1 (1/ψ 1 (1/ <. The f dg exists.. First, let A V ϕ (f; [, b] d B V ψ (f; [, b]. The by Lemm 2.3, ϕ 1 (A/ψ 1 (B/ <. Let ε > 0, choose v such tht λ(a/µ(b/ ε/52. =2 v 1 Let E v = {x 1, x 2,..., x v } be give s i Lemm 2.1. Let δ be give s i Lemm 4.4 with E = E v. Let δ be positive fuctio defied o [, b] with δ(x < δ for ll x [, b] such tht if D = {(ξ, [u, v]} is δ-fie divisio of [, b], the [u, v] (ξ δ, ξ + δ d ξ E v, [u, v] (x k, x k+1 d ξ (x k, x k+1, k = 1, 2,..., v 1. Now let D = {(ξ, [u, v]} d D = {(ξ, [u, v ]} be two δ-fie divisios of [, b]. Let D = D 1 D 2, D = D 1 D 2 where D 1 = {(ξ, [u, v] D ; ξ E v }, D 1 = {(ξ, [u, v ] D ; ξ E v }, D 2 = D \D 1 d D 2 = D \D 1. Suppose (ξ, [u, v] D 2 d x i + δ(x i [u, v] (or x i δ(x i [u, v]. The we divide [u, v] ito two prts [u, x i + δ(x i ], [x i + δ(x i, v] ([u, x i δ(x i ], [x i δ(x i, v], respectively. Let D 1 be the uio of D 1 d (ξ, [u, x i + δ(x i ], (ξ, [x i δ(x i, v]. Let D 2 = D \ D 1. Similrly, we costruct D 1 d D 2 = D \ D 1. The, by Lemms 4.4 d 4.5, we get (D f(ξ(g(v g(u (D f(ξ (g(v g(u (D 1 f(ξ(g(v g(u ( D 1 f(ξ(g(v g(u + (D 2 f(ξ(g(v g(u (D 2 f(ξ (g(v g(u 8 f ε + ε, where f = sup{f(x; x [, b]}. Thus f dg exists. 5. Approximtio I this sectio we show tht f dg c be pproximted by s dg, where s is step fuctio. This pproximtio theorem c be foud i [12]. Theorem 5.1. Let f BV ϕ [, b], g BV ψ [, b] d ϕ 1 (1/ψ 1 (1/ <. The, givey ε > 0, there exists step fuctio s o [, b] such tht (f s dg ε. 245

14 . First, let A V ϕ (f; [, b] d B V ψ (f; [, b]. The, by Lemm 2.3, ϕ 1 (A/ψ 1 (B/ <. Let E v = {x 1, x 2,..., x v } be give s i Lemm 4.5 with v 1. Defie v v 1 s(x = f(x k χ {xk }(x + f(x k +χ (xk,x k+1 (x. k=1 The (f s(x k = 0 for ll x k E v. By Theorem 4.6, f dg s dg = b (f s dg exists. Let ε > 0; there exists positive fuctio δ o [, b] such tht wheever D = {(ξ, [u, v]} is δ-fie divisio of [, b], x i is tg for every i = 1, 2,..., v, d (f s dg (D (f s(ξ(g(v g(u ε 2. By Lemm 4.5, d ll x i beig tgs, we hve (D (f s(ξ(g(v g(u = (D (f s(ξ(g(v g(u ξ / E v 52 k=1 =2 v 1 λ ( B µ for v 1. Therefore (f s dg ε =2 v 1 λ ( B µ. Choosig v big eough, we get the required result. Corollry 5.2. Let f BV ϕ [, b], g BV ψ [, b] d. The f dg 6 ϕ 1( V ϕ (f ψ 1( V ψ (g ϕ 1 (1/ψ 1 (1/ < + f((g(+ g( + f(+(g(b g(+ + f(b(g(b g(b.. Let ε > 0. By Theorem 5.1 there exists step fuctio s o [, b] such tht (f s dg ε. Hece b f dg ε + s dg. 246

15 By Theorem 3.3, s dg 6 ( Vϕ (f ( Vψ (g λ µ + s E0 dg. Note tht s E 0 dg = f((g(+ g(+f(+(g(b g(++f(b(g(b g(b. Hece we get the required result. 6. Itegrtio by prts A geerl result for itegrtio by prts i the settig of Hestock-Kurzweil itegrls of Stieltjes type c be foud i [10]. I this sectio, we will prove this result i more cocrete forms. For y prtil divisio D = {(ξ, [u, v]} o [, b], defie S (f, g, D = (D (f(ξ f(u(g(ξ g(u, S + (f, g, D = (D (f(v f(ξ(g(v g(ξ d S(f, g, D = S (f, g, D S + (f, g, D. We sy tht S (f, g exists if there exists S (1 such tht for every ε > 0 there exists positive fuctio δ o [, b] such tht whe D is δ-fie divisio of [, b], we hve S (f, g, D S (1 ε. We the deote S (1 by S (f, g. Similrly, we c defie S + (f, g d S(f, g. Clerly, if two of S (f, g, S + (f, g d S(f, g exist, the the third exists d S(f, g = S (f, g S + (f, g. Lemm 6.1. Let f BV ϕ [, b] d g BV ψ [, b] with ϕ 1 (1/ψ 1 (1/ <. The d S + (f, g = S (f, g = (f(t i + f(t i (g(t i + g(t i (f(t i f(t i (g(t i g(t i, 247

16 where t i re the commo poits of discotiuity of f d g d the bove series coverge bsolutely.. Let ε > 0, let v be positive iteger such tht ϕ 1 (A/ψ 1 (B/ =2 v 1 ε/52. Let E v = {x 1, x 2,..., x v } be give s i Lemm 2.1. E v my coti some poits of {t i }. We my ssume tht there exists positive iteger N such tht t j / E v wheever j N. Now tke y two positive itegers m, N with m <. Let δ be positive umber such tht for every i = m, m + 1,...,, if t i (x j, x j+1 for some j, the (t i, t i + δ (x j, x j+1 d {(t i, t i + δ} i=m re o-overlppig itervls. Let η i (t i, t i + δ for ll i = m, m + 1,...,, d D = {t i, [t i, η i ]} i=m d D = {η i, [t i, η i ]} i=m. From the defiitio of D d D it is cler tht D d D re prtil divisios of [, b] d stisfy the coditio of Theorem 4.5. Hece, we hve The (f(η i f(t i (g(η i g(t i = (D f(η i (g(η i g(t i (D f(t i (g(η i g(t i 52 ψ 1( B 52 ε 52 = ε. i=m =2 v 1 ϕ 1 (f(t i + f(t i (g(t i + g(t i ε. i=m Observe tht D d D re prtil divisios. Therefore (f(t i + f(t i (g(t i + g(t i 2ε, i=m where m, N. Hece S + (f, g coverges bsolutely. Similrly, S (f, g coverges bsolutely. Lemm 6.2. Let f BV ϕ [, b] d g BV ψ [, b], E = {x 1, x 2,..., x } E 0, d ε > 0. The there exists costt δ > 0 such tht for y fiite collectio of disjoit subitervls {[u i, v i ]} with [u i, v i ] (x i, x i + δ for ech i or [u i, v i ] (x i δ, x i for ech i, we hve 248 f(v i f(u i g(v i g(u i ε, f(v i f(u i ε V ψ (B

17 d g(v i g(u i ε V ψ (A.. The proof is similr to tht of Lemm 4.4. Let ε > 0 be give. First, observe tht f d g re regulted fuctios. Therefore, there exists costt δ > 0 such tht {[ ε g(t g(x i mi 2 {[ ε g(x i + g(t mi 2 {[ ε f(t f(x i mi 2 ] 1 2, ] 1 2, ] 1 2, ε } 2V ϕ (A ε } 2V ϕ (A ε } 2V ψ (B wheever 0 < x i t < δ, wheever 0 < t x i < δ, wheever 0 < x i t < δ d {[ ε f(x i + f(t mi 2 ] 1 2, ε } 2V ψ (B wheever 0 < t x i < δ for ech i. Therefore, the required result follows. Lemm 6.3. Let f BV ϕ [, b] d g BV ψ [, b] with ϕ 1 (1/ψ 1 (1/ <. Let ε > 0. If E v = {x 1, x 2,..., x v } is the set give i Lemm 2.1 d {t j } m j=1 E v, where {t j } m j=1 re ll commo poits of discotiuity of f d g such tht ϕ 1 (A/ψ 1 (B/ < ε/312 d (f(t j f(t j (g(t j g(t j =2 v 1 j=m+1 ε/6, the there exists positive rel umber δ such tht for y δ -fie prtil divisio D = {(x i, [u i, v i ]} v of [, b] we hve S(f, g, D (S + (f, g S (f, g 2 3 ε.. Applyig Lemm 6.2 to ε/18 d E = E v, we get positive costt δ. Let D = {(x i, [u i, v i ]} v, the v S (f, g, D = (f(x i f(u i (g(x i g(u i v = [(f(x i f(x i + (f(x i f(u i ][(g(x i g(x i + (g(x i g(u i ] 249

18 v v = (f(x i f(x i (g(x i g(x i + (f(x i f(x i (g(x i g(u i v v + (f(x i f(u i (g(x i g(x i + (f(x i f(u i (g(x i g(u i. Let F = E v \ {t j } m j=1, the F is the set of poits i E v which re ot commo poits of discotiuity of f d g. Hece (f(x i f(x i (g(x i g(x i = 0. Cosider v x i F (f(x i f(x i (g(x i g(x i = (f(x i f(x i (g(x i g(x i + (f(x i f(x i (g(x i g(x i x i F = 0 + (f(x i f(x i (g(x i g(x i = The x i / F m (f(t j f(t j (g(t j g(t j. j=1 v S (f, g, D = (f(x i f(u i (g(x i g(u i = x i / F m v (f(t j f(t j (g(t j g(t j + (f(x i f(x i (g(x i g(u i j=1 v v + (f(x i f(u i (g(x i g(x i + (f(x i f(u i (g(x i g(u i. Therefore S (f, g, D S (f, g v = (f(x i f(u i (g(x i g(u i (f(t j f(t j (g(t j g(t j v m (f(x i f(u i (g(x i g(u i (f(t j f(t j (g(t j g(t j j=m+1 j=1 j=1 (f(t j f(t j (g(t j g(t j

19 v v = (f(x i f(x i (g(x i g(u i + (f(x i f(u i (g(x i g(x i v + (f(x i f(u i (g(x i g(u i + ε 6. By Lemm 6.2 we hve d Thus Similrly, Hece v (f(x i f(u i (g(x i g(u i ε 18, v (f(x i f(x i (g(x i g(x i ε 18V ϕ (A V ϕ(a = ε 18 v (f(x i f(x i (g(x i g(x i ε 18V ψ (B V ψ(b = ε 18. S (f, g, D S (f, g ε 18 + ε 18 + ε 18 + ε 6 ε 3. S(f, g, D (S + (f, g S (f, g S + (f, g, D S + (f, g ε 3. S (f, g, D S (f, g + S + (f, g, D S + (f, g ε 3 + ε 3 = 2 3 ε. Lemm 6.4. Let f BV ϕ [, b] d g BV ψ [, b] with ϕ 1 (1/ψ 1 (1/ <. The for y give ε > 0 there exists positive fuctio δ such tht for y δ-fie divisio D of [, b] we hve S(f, g, D (S + (f, g S (f, g ε.. Let ε > 0, choose v such tht ϕ 1 (A/ψ 1 (B/ ε/312. =2 v 1 Let E v = {x 1, x 2,..., x v } be give s i Lemm 2.1. Applyig Lemm 6.3 to E = E v, we get positive costt δ. Let δ be positive fuctio defied o [, b] with δ(x < δ for ll x [, b] such tht if D = {(ξ, [u, v]} is δ-fie divisio 251

20 of [, b], the [u, v] (ξ δ, ξ + δ whe ξ E v d [u, v] (x k, x k+1 whe ξ (x k, x k+1, k = 1, 2,..., v 1. Now let D = {(ξ, [u, v]} be δ-fie divisio of [, b]. Let D = D 1 D 2, where D 1 = {(ξ, [u, v] D : ξ E v }, D 2 = D \ D 1. Hece, by Lemm 6.3, S(f, g, D (S + (f, g S (f, g S (f, g, D 2 + S + (f, g, D 2 + S(f, g, D 1 (S + (f, g S (f, g S (f, g, D 2 + S + (f, g, D ε. By Lemm 4.5, we hve S (f, g, D 2 = (D 2 (f(ξ f(u(g(ξ g(u = (D 2 f(ξ(g(ξ g(u (D 2 f(u(g(ξ g(u 52 ψ 1( B ε 6 =2 v 1 ϕ 1 d S + (f, g, D 2 = (D 2 (f(v f(ξ(g(v g(ξ 52 ψ 1( B ε 6. =2 v 1 ϕ 1 Hece, S(f, g, D (S + (f, g S (f, g S (f, g, D 2 + S + (f, g, D ε 1 6 ε ε ε = ε. We c verify tht S (f, g, S + (f, g exist d S (f, g = (f(t i + f(t i (g(t i + g(t i, S + (f, g = (f(t i f(t i (g(t i g(t i. Theorem 6.5. Let f BV ϕ [, b] d g BV ψ [, b] with <. The 252 f dg + m=1 g df = f(bg(b f(g( + S(f, g, ϕ 1 (1/mψ 1 (1/m

21 where S(f, g = (f(t i + f(t i (g(t i + g(t i (f(t i f(t i (g(t i g(t i d {t i } re ll commo poits of discotiuity of f d g.. Sice S(f, g = S + (f, g S (f, g, by Lemm 6.1, S(f, g exists. Let ε > 0 be gived let f, g : [, b]. The there exists positive fuctio δ 1 o [, b] such tht for y δ 1 -fie prtil divisio D = {([u i, v i ], ξ i } of [, b], S(f, g, D S(f, g ε 2. Sice f is itegrble with respect to g, there exists positive fuctio δ 2 o [, b] such for y δ 2 -fie divisio D = {([t i, t i+1 ], ξ i } of [, b], we hve (D f(ξ i (g(t i+1 g(t i f dg ε 2. Choose δ(ξ = mi{δ 1 (ξ, δ 2 (ξ}. Let D = {([t i, t i+1 ], ξ i } be δ-fie prtil divisio of [, b]. We c see tht ( (D ( g(ξ i (f(t i+1 f(t i f(bg(b f(g( + S(f, g f dg = (D ( g(ξ i (f(t i+1 f(t i f(t i+1 g(t i+1 + f(t i g(t i ti+1 + f dg S(f, g t i = (D f(ξ i (g(t i+1 g(t i + (f(ξ i f(t i (g(ξ i g(t i ti+1 (f(t i+1 f(ξ i (g(t i+1 g(ξ i + f dg S(f, g t i (D ti+1 f(ξ i (g(t i+1 g(t i f dg + S(f, g, D S(f, g ε 2 + ε 2 = ε. t i Thus, we c coclude tht g is itegrble to f(bg(b f(g(+s(f, g f dg o [, b] with respect to f. Hece, we hve f dg + g df = f(bg(b f(g( + S(f, g. 253

22 7. Covergece theorem I this sectio we will use Youg s ide, see [11], [12], to prove some covergece theorems for our settig. Defiitio 7.1 (Two-orm covergece. A sequece {f ( } of fuctios i BV ϕ [, b] is sid to be two-orm coverget to f if (i f ( is uiformly coverget to f o [, b], d (ii V ϕ (f ( A for every = 1, 2,.... I symbols, we deote the two-orm covergece by f ( f. It is cler tht BV ϕ [, b] is complete uder two-orm covergece, i.e., if f ( BV ϕ [, b], = 1, 2,..., d f ( f, the f BV ϕ [, b]. We eed the followig two lemms. Lemm 7.2. Let ϑ be strictly decresig cotiuous fuctio o (0, with lim ϑ(x = 0 d let x 1 ϑ(x dx exist. The there exists strictly icresig cotiuous fuctio ϱ o [0, with lim ϱ(x = such tht x ϱ(x lim x x = d 1 ϑ(x dϱ(x exists.. Sice ϑ(x dx exists, there exists positive fuctio o [0, 1 ι(x 0 with lim ι(x = d ι(x = 0 for x 1, such tht x 1 ϑ(xι(x dx d x ι(t dt exist for every x (0,. Let 0 ϱ(x = x + x 0 ι(t dt. The ϱ is strictly icresig fuctio with lim ϱ(x =. Therefore, x 1 ϑ(x dϱ(x = 1 ϑ(x[1 + ι(x] dx <. Now we shll prove tht lim x ϱ(x/x =. Let x > 2. The (x /x > 1 2. By Me-Vlue Theorem for itegrl, there exists y (, x such tht x x ι(x dx = ι(y.

23 Hece ϱ(x x = x x Sice ι( s, we hve 0 ι(x dx x [ 1 x x x ] ι(x dx 1 2 ι(y 1 2 ι(. ϱ(x lim x x =. Corollry 7.3. Let ϑ be strictly decresig cotiuous fuctio o (0, with lim ϑ(x = 0 d let ϑ(x dx exist. The there exists strictly icresig x 1 cotiuous fuctio ς o (0, with lim ς(x =, such tht x ς(x lim x x = 0 d 1 ϑ(ς(x dx exists.. Let ς = ϱ 1, where ϱ is give i Lemm 7.2. Thus we get the required result. Lemm 7.4. Suppose ϕ 1 (1/ψ 1 (1/ <. The there exist two N- fuctios ϕ, ψ such tht ϕ (u π(uϕ(u d ψ (u γ(uψ(u, where π, γ re icresig d lim π(x = lim γ(x = 0, with x 0 x 0 (ϕ 1( 1 (ψ 1( 1 <.. Give ϕ, ψ d ϕ 1 (1/ψ 1 (1/ <, we wt to costruct ϕ, ψ such tht ϕ (u π(uϕ(u, ψ (u γ(uψ(u, where π, γ re icresig fuctios with lim π(x = lim γ(x = 0 d x 0 x 0 (ϕ 1( 1 (ψ 1( 1 <. Let ϑ(u = ϕ 1 (1/uψ 1 (1/u for u (0,. The ϑ stisfies the coditios of Corollry 7.3. Hece there exists strictly icresig cotiuous fuctio ς o [0, with lim ς(x =, such tht x ς(x lim x x = 0 d 1 ϑ(ς(x dx exists. 255

24 Let θ(u = uς(u 1 for u (0, d θ(0 = 0. The lim θ(u = lim ς(u 1 /u 1 = u 0 u 0 0 d u/θ(u = 1/ς(u 1 is strictly icresig cotiuous fuctio o (0,. Let Φ(u = ϕ 1 (u/θ(u, Ψ(u = ψ 1 (u/θ(u, Φ(0 = 0 d Ψ(0 = 0. The Φ d Ψ re strictly icresig cotiuous fuctios o [0,. Furthermore, let ϕ = (Φ 1 d ψ = (Ψ 1. The (ϕ 1( 1 (ψ 1( 1 = = = ϕ 1( 1 ψ 1( 1 θ( 1 θ( 1 ϕ 1( 1 ψ 1( 1 ς( ς( ϑ(ς( ϑ(ς(x dx <, sice ϑ(ς(x is o-egtive. If t = (ϕ 1 (u = ϕ 1 (u /θ(u, the ϕ (t = u. t = ϕ 1 (u, the ϕ(t = u. Hece u = u /θ(u d ϕ (t ϕ(t = u u = θ(u = θ(ϕ (t =: π(t; clerly lim t 0 π(t = lim t 0 θ(ϕ (t = 0. Similrly, we hve ψ (t ψ(t = u u = θ(u = θ(ψ (t =: γ(t, 1 O the other hd, if d lim π(t = lim θ(ϕ (t = 0. Deotig by π(t, γ(t the upper bouds of t 0 t 0 π(u, γ(u for 0 < u t we see tht π, γ re icresig fuctios. The ϕ (t π(tϕ(t d ψ (t γ(tψ(t. Let D = {[u, v]} be prtitio of itervl [, β]. By Lemm 7.4, we hve (D ϕ ( f(v f(u = (D π( f(v f(u ϕ( f(v f(u π(2 f (D ϕ( f(v f(u. Hece, if A d A re the ϕ-vritiod ϕ -vritio of f, respectively, o [, β], we hve A Aπ(2 f Aπ(ϕ 1 (A. 256

25 Theorem 7.5. BV ϕ [, b] with m=1 If g BV ψ [, b] d {f ( } is two-orm coverget to f i ϕ 1 (1/mψ 1 (1/m <, the f dg exists d lim f ( dg = f dg.. Let ε > 0 be give. Let {f ( } be two-orm coverget to f i BV ϕ [, b] d g BV ψ [, b]. By the covexity of BV ϕ [, b], 1 2 (f ( f BV ϕ [, b]. Hece, (f ( f dg exists. Thus, there is positive fuctio δ such tht for every δ -fie divisio D = {([t i, t i+1 ], ξ i } of [, b], Let ( (f ( f dg (D (f ( (ξ i f(ξ i (g(t i+1 g(t i ε. V ϕ ( 1 2 (f ( f V ϕ (f ( + V ϕ (f A for every d V ψ (g = B. By Lemm 7.4, there exist two N-fuctio ϕ d ψ such tht ϕ (u π(uϕ(u d ψ (u γ(uψ(u, where π, γ re icresig d lim π(x = lim γ(x = 0, with x 0 x 0 (ϕ 1( 1 (ψ 1( 1 <. By Lemm 2.3, there exists positive iteger v such tht =v+1 (ϕ 1π(ϕ 1 (A (ψ 1( Bγ(ψ 1 (B < ε. For this v, choose τ > 0 such tht Hece for = 1, 2,..., v, (ϕ 1 (Aπ(τ (ϕ 1π(τ v(ψ 1 ( Bπ(ψ 1 (B v ε v(ψ 1 ( Bπ(ψ 1 (B ε.. 257

26 Sice f ( coverge to f uiformly o [, b], there is positive iteger N such tht for every N, we hve 1 sup 2 ( f ( (t f(t = 1 2 (f ( f < mi{ε, 1 2 τ}. t [,b] We my ssume tht whe N, the (f ( f((g(+ g( + (f ( f(+(g(b g(+ + (f ( f(b(g(b g(b ε. Hece for N, pplyig Corollry 5.2 to 1 2 (f ( f, we get f ( dg b f dg = 2 f ( f dg (ϕ 1( V ϕ ( 1 2 (f ( f v 12 (ϕ 1( V ϕ ( 1 2 (f ( f + 12 (ϕ 1( V ϕ ( 1 2 (f ( f =v+1 (ψ 1( V ψ (g (ψ 1( V ψ (g + ε (ψ 1( V ψ (g + ε v 12 (ϕ 1π(2 1 2 (f ( f (ψ 1( Bγ(ψ 1 (B + 12 (ϕ 1π(ϕ 1 (A (ψ 1( Bγ(ψ 1 (B + ε =v+1 v (ϕ 1π(2 1 2 (f ( f v (ϕ 1π(τ (ψ 1( Bγ(ψ 1 (B 12ε + 13ε = 25ε. (ψ 1( Bγ(ψ 1 (B + 13ε + 13ε Hece, lim f ( dg = f dg. Theorem 7.6. BV ϕ [, b] with m=1 If f BV ϕ [, b] d {g ( } is two-orm coverget to g i ϕ 1 (1/mψ 1 (1/m <, the f dg exists d 258 lim f dg ( = f dg.

27 . Sice g ( coverge to g uiformly, there exists positive iteger N such tht for every > N 1 we hve S(f, g ( S(f, g 1 4 ε, (f ( (b f(bg(b 1 4 ε d (f ( ( f(g( 1 4 ε. By Theorem 7.5 there exists positive iteger N > N 1 such tht for y > N we hve b g ( df g df ε 4. Hece f dg ( b f dg g ( df g df + (f ( (b f(bg(b + (f ( ( f(g( + S(f, g ( S(f, g ε. Hece, lim f dg( = f dg. Hece, we lso hve the followig theorem. Theorem 7.7. If {f ( } d {g ( } re two-orm coverget to f d g i BV ϕ [, b] d BV ψ [, b], respectively, with f dg exists d lim f ( dg ( = m=1 ϕ 1 (1/mψ 1 (1/m <, the f dg. Refereces [1] Boopogkrog Vryu, Tu Seg Chew: O itegrls with itegrtors i BV p. Rel Al. Exch. 30 (2004/2005, Zbl [2] R. M. Dudley, R. Norvis: Differetibility of Six Opertors o Nosmooth Fuctios d p-vritio. Spriger, Berli, Zbl [3] I. J. L. Grces, Peg-Yee Lee, Dogsheg Zho: Moore-Smith limits d the Hestock itegrl. Rel Al. Exchge 24 (1998/1999, Zbl [4] P. Y. Lee, R. Výborý: The Itegrl. A Esy Approch fter Kurzweil d Hestock. Cmbridge Uiversity Press, Zbl [5] R. Lesiewicz, W. Orlicz: O geerlized vritios (II. Stud. Mth. 45 (1973, Zbl [6] J. Musielk, W. Orlicz: O geerlized vritios (I. Stud. Mth. 18 (1959, Zbl

28 [7] E. R. Love, L. C. Youg: O frctiol itegrtio by prts. Proc. Lodo Mth. Soc., II. Ser., 44 (1938, Zbl [8] E. R. Love: Itegrtio by prts d other theorems for R 3 S-itegrls. Rel Al. Exch. 24 (1998/1999, Zbl [9] R. Norvis: Qudrtic Vritio, p-vritiod Itegrtio with Applictios to Strock Price Modellig. Preprit, [10] Š. Schwbik: A ote o itegrtio by prts for bstrct Perro-Stieltjes itegrls. Mth. Bohem. 126 (2001, Zbl [11] L. C. Youg: A iequlity of the Hölder type, coected with Stieltjes itegrtio. Act Mth. 67 (1936, Zbl [12] L. C. Youg: Geerl iequlities for Stieltjes itegrls d the covergece of Fourier series. Mth. A. 115 (1938, Zbl JFM Author s ddress: Boopogkrog Vryu, Tu Seg Chew, Deprtmet of Mthemtics, Ntiol Uiversity of Sigpore, 2, Sciece Drive 2, Sigpore , Republic of Sigpore, e-mil: mtcts@us.edu.sg. 260