summary:In the paper, we show that the space of functions of bounded variation and the space of regulated functions are, in some sense, the dual space of each other, involving the Henstock-Kurzweil-Stieltjes integral

Aye, Khaing Khaing

Lee, Peng Yee

English

Czech digital mathematics library

The dual of the space of functions of bounded variation

Institute of Mathematics AS CR, v. v. i.

Mathematica BohemicaKhaing Khaing Aye; Peng Yee LeeThe dual of the space of functions of bounded variationMathematica Bohemica, Vol. 131 (2006), No. 1, 1–9Persistent URL: http://dml.cz/dmlcz/134078Terms of use:© Institute of Mathematics AS CR, 2006Institute of Mathematics of the Czech Academy of Sciences provides access to digitized documentsstrictly for personal use. Each copy of any part of this document must contain these Terms of use.This document has been digitized, optimized for electronic delivery andstamped with digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://dml.cz131 (2006) MATHEMATICA BOHEMICA No. 1, 1–9THE DUAL OF THE SPACE OF FUNCTIONSOF BOUNDED VARIATIONKhaing Khaing Aye, Peng Yee Lee, Singapore(Received May 31, 2005)Dedicated to Prof. J.Kurzweil on the occasion of his 80th birthdayAbstract. In the paper, we show that the space of functions of bounded variation andthe space of regulated functions are, in some sense, the dual space of each other, involvingthe Henstock-Kurzweil-Stieltjes integral.Keywords: bounded variation, two-norm space, dual space, linear functional, Henstockintegral, Stieltjes integral, regulated functionMSC 2000 : 26A39, 26A42, 26A45, 46B26, 46E991. IntroductionThe classical Riesz representation theorem is well known [5]. It states that acontinuous linear functional on the space C[a, b] of continuous functions on [a, b]can be represented in terms of the Riemann-Stieltjes integral∫ baf dg for f ∈ C[a, b]where g is a function of bounded variation on [a, b]. In other words, the Rieszrepresentation theorem showed that the dual of C[a, b] is the space BV of functionsof bounded variation or more precisely a subspace of BV. However we fail to say inwhatever sense that the space C[a, b] is the dual of BV .In 1966, Hildebrandt [9] has characterized continuous linear functionals on thespace BV regarding BV as a two-norm space. By a two-norm space we mean thatthe space is endorsed with two-norm topology or two-norm convergence. More pre-cisely, a sequence is convergent in the two-norm sense if the sequence is bounded inone given norm and convergent in another given norm. Further Hildebrandt used theleft Cauchy integral [8, p. 87] among others. It is not explicit how his representation1theorem is connected with the Riesz theorem. In this paper, we reformulate Hilde-brandt’s results in terms of the Henstock-Kurzweil-Stieltjes integral, and establish aduality in which a part of it is the Riesz representation theorem. More precisely, letBVN be the subspace of BV such that the condition f(x) = 12 [f(x+) + f(x−)] holdsfor x ∈ (a, b) with f (a) = f (a+) and f (b) = f (b−) whenever f ∈ BVN . A regulatedfunction [2] on [a, b] is a function whose one-sided limits exist at every point of [a, b].Let RF denote the space of regulated functions on [a, b] , and RFN the subspace ofRF satisfying the condition as in BVN . Regarding BVN as a two-norm space andRFN a normed space, then the dual of BVN is RFN and vice versa. Note that C[a, b]is a subspace of RFN and its dual is BVN . The fact that the dual of C [a, b] is BVNcan now embedded into the duality of RFN and BVN .Let V (g; [a, b]) or V (g) denote the total variation of g on [a, b]. A function g is ofbounded variation on [a, b] or g ∈ BV if V (g) is finite. A sequence {gn} of functionson [a, b] is said to be two-norm convergent to g in BV if V (gk) 6 M for all k and gkis uniformly convergent to g on [a, b]. The space BV is now a two-norm space withthe two-norm structure provided by the two-norm convergence in BV . Further, RFis a normed space with norm given by ‖f‖ = sup {|f (x)| : x ∈ [a, b]}.2. The Henstock-Stieltjes integralWe define the Henstock-Kurzweil-Stieltjes integral [6], [7] and prove a convergencetheorem. A real-valued function f is said to be Henstock-Kurzweil-Stieltjes integrableto A with respect to g on [a, b] or (f, g) Henstock-Kurzweil-Stieltjes integrable on[a, b] , if for every ε > 0, there exists a function δ(ξ) > 0 for ξ ∈ [a, b] such that forany division D of [a, b] given by a = x0 < x1 < . . . < xn = b, with ξ1, ξ2, . . . , ξnsatisfying ξi ∈ [xi−1, xi] ⊂ (ξi − δ (ξi) , ξi + δ (ξi)) for i = 1, 2, . . . , n, we have∣∣∣∣n∑i=1f(ξi)[g(xi)− g(xi−1)]−A∣∣∣∣ < ε.For brevity, we write D = {(ξ, [u, v])} in which (ξ, [u, v]) denotes a typical point-interval pair (ξi, [xi−1, xi]) in D, and also we write the Riemann sum in the form(D)∑f(ξ)[g (v)− g (u)]. Here D is said to be δ-fine if the above condition holds [6],[7]. That is, (f, g) is Henstock-Kurzweil-Stieltjes integrable to A on [a, b] if for everyε > 0, there is a positive function δ such that for any δ-fine division D = {(ξ, [u, v])}of [a, b] , we have(∗) |s(f, g, D)−A| < ε2where s (f, g, D) = (D)∑f (ξ) [g (v)− g (u)]. For simplicity, we write A =∫ ba f dg.A division D1 is said to be finer than a division D2 if every interval of D1 is asubinterval of some interval of D2. If g (x) = x, then the integral reduces to theHenstock-Kurzweil integral. If δ is a constant, then it reduces to the Riemann-Stieltjes integral. Moreover, if g(x) = x and δ is a constant, then it reduces to theRiemann integral [10]. We remark that the Henstock-Kurzweil-Stieltjes integral of(f, g) does allow both functions f and g to be discontinuous on the same side at apoint.Let f be a real-valued function defined on [a, b] . A family F of Henstock-Kurzweil-Stieltjes integrable functions (f, gk) , k = 1, 2, . . ., is said to be equi-integrable on [a, b],if the above inequality (∗) holds with g and A replaced by gk and the integral Ak ofgk, and δ being independent of k [7, p. 104]. It is clear that if f ∈ RF and g ∈ BV,then∫ baf dg exists, for reference see [3].Lemma 1. Suppose f ∈ RF . If gk ∈ BV and V (gk) 6 M for every k, then{(f, gk) : k = 1, 2, . . .} is equi-integrable on [a, b]. . Since f ∈ RF, it is known [1] that for every ε > 0, there existsa division D0 : a = x0 < x1 < . . . < xn = b such that |f (ξ)− f (η)| < ε forξ, η ∈ (xi−1, xi) , i = 1, 2, . . . , n. Fix i and put g∗k (x) = gk (x) for every x ∈ (xi−1, xi)and g∗k (xi−1) = gk (xi−1+) , g∗k (xi) = gk (xi−). Then we have∫ xixi−1f dgk =∫ xixi−1f dg∗k + f (xi−1) [gk (xi−1+)− gk (xi−1)]+ f (xi) [gk (xi)− gk (xi−)].Define a function δ (ξ) > 0 for every ξ ∈ [a, b] such that every δ-fine division D ={(ξ, [u, v])} of [a, b] is finer than D0. Take any δ-fine division D. Then we can writeD = D1 ∪D2 ∪ . . . ∪Dn where Di is a δ-fine division of [xi−1, xi]. Thus we have∣∣∣∣s (f, gk, Di)−∫ xixi−1f dgk∣∣∣∣ =∣∣∣∣s (f, g∗k, Di)−∫ xixi−1f dg∗k∣∣∣∣ 6 εV (gk; [xi−1, xi]).Hence for every δ-fine division D = {(ξ, [u, v])} of [a, b] ,∣∣∣∣s (f, gk, D)−∫ baf dgk∣∣∣∣ 6n∑i=1∣∣∣∣s (f, g, Di)−∫ xixi−1f dgk∣∣∣∣ 6 εV (gk; [a, b]) 6 εMfor every k. That is, {(f, gk) : k = 1, 2, . . .} is equi-integrable on [a, b]. 3Theorem 1. Suppose gn ∈ BV for n = 1, 2, . . . , and f ∈ RF . If gn is two-normconvergent to g in BV, then∫ ba f dg exists and limn→∞∫ ba f dgn =∫ ba f dg. . It is clear that g ∈ BV and∫ baf dg exists. Further, by Lemma 1,{(f, gn) : n = 1, 2, . . .} is equi-integrable on [a, b]. Then for every ε > 0, there existsa function δ (ξ) for ξ ∈ [a, b] such that for every δ-fine divisionD = {(ξ, [u, v])} of [a, b]and for every n, we have |s (f, g, D)−∫ ba f dg| < ε and |s (f, gn, D)−∫ ba f dgn| < ε.Since gn is uniformly convergent to g on [a, b] , we can prove that there exists D0such that |s (f, gn, D0)− s (f, g, D0)| < ε for large n. Therefore we have∣∣∣∣∫ baf dgn −∫ baf dg∣∣∣∣ 6∣∣∣∣∫ baf dgn − s (f, gn, D0)∣∣∣∣ + |s (f, gn, D0)− s (f, g, D0)|+∣∣∣∣∫ baf dg − s (f, g, D0)∣∣∣∣ < 3εfor large n. 3. Two-norm continuous linear functionalsA functional F defined on BV or RF is linear if F (αf + βg) = αF (f) + βF (g)for f, g ∈ BV or RF, and real α, β. A functional F defined on BV is said to be two-norm continuous if F (gn) → F (g) as n → ∞ whenever gn is two-norm convergentto g in BV . Theorem 1 shows that if f ∈ RF, then F (g) =∫ ba f dg is two-normcontinuous in BV . A functional F defined on RF is continuous if F (fn) → F (f) asn →∞ whenever ‖fn − f‖ → 0 as n →∞. In this section, we characterize the dualor the two-norm dual of the space BV, that is, the space of all two-norm continuouslinear functionals on BV, in terms of the Henstock-Kurzweil-Stieltjes integral. Wedefine characteristic functions γt and δt for a fixed t ∈ [a, b]. When t ∈ (a, b) , defineγt (x) = 0 for every x ∈ [a, t), 12 when x = t and 1 for every x ∈ (t, b]. When t = a,define γa (x) = 0 when x = a and 1 for every x ∈ (a, b].When t = b, define γb (x) = 0for every x ∈ [a, b) and 1 when x = b. We also define δt(x) = 1 when x = t and0 otherwise. We now give a series of lemmas leading to the main theorem of thispaper.Lemma 2. If F is a two-norm continuous linear functional defined on BV, thenf ∈ RFN where f (t) = F (γt) for every t ∈ [a, b]. . Suppose f /∈ RF . Then there exists a point x such that f(x+) or f(x−)does not exist. Suppose f(x+) does not exist. Then there exist η > 0 and a sequenceof pairwise disjoint intervals [x′j , x′′j ] on the same right side of x and approaching4x such that |f(x′′j ) − f(x′j)| > η for every j. Define gn = 1nn∑j=1εj(γx′′j − γx′j ) whereεj = sgn(f(x′′j )− f(x′j)). It is clearly seen that gn is uniformly convergent to zero on[a, b] and V (gn) 6 2 butF (gn) =1nn∑j=1εj(F (γx′′j )− F (γx′j )) =1nn∑j=1|f(x′′j )− f(x′j)| > η,for every n. That is, F is not a two-norm continuous linear functional on BV . There-fore f(x+) and f(x−) must exist for every x. Hence f is a regulated function. Inview of the definition of γt, indeed f ∈ RFN . Lemma 3. Suppose F is a two-norm continuous linear functional defined on BV.If f(t) = F (γt) for every t ∈ [a, b] , then f ∈ RFN, the integral∫ baf dg exists andF (g) =∫ ba f dg for g ∈ BVN . . By Lemma 2, f ∈ RFN and consequently∫ ba f dg exists. In view ofTheorem 1, it is sufficient to prove Lemma 3 for the case when g is a step functionin BVN . Let ti, i = 1, . . . , n− 1, be the discontinuity points of g. Consider the casewhen a = t0 < t1 < . . . < tn = b. Then∫ baf dg =n−1∑i=1f(ti) [g (ti+)− g (ti−)] = F( n−1∑i=1γti · [g (ti+)− g (ti−)]).Here we note thatn−1∑i=1γti · [g (ti+)− g (ti−)] = gand hence∫ baf dg = F (g) for the step function g ∈ BVN and consequently for anyg ∈ BVN . The linear space c0 [a, b] [5] is the space of all functions f defined on [a, b] suchthat the set {t ∈ [a, b] : |f(t)| > ε} is finite for every ε > 0.Lemma 4. If F is a two-norm continuous linear functional defined on BV, thenf ∈ c0 [a, b] where f (t) = F (δt) for every t ∈ [a, b]. . Suppose f /∈ c0 [a, b]. Then there exists ε > 0 such that |f(ti)| > ε fora sequence {ti}i>1 ⊂ [a, b]. Define gn = 1nn∑i=1εiδti , where εi = sgn f (ti). Then weobtain V (gn) 6 2 and gn is uniformly convergent to 0 on [a, b] butF (gn) =1nn∑i=1εiF (δti) =1nn∑i=1|f(ti)| > ε for every n.5That is, F is not a two-norm continuous linear functional on BV . Therefore f ∈c0 [a, b]. We define the normalized function g∗ of g on [a, b] to be g∗(x) = 12 [g(x+)+g(x−)]for x ∈ (a, b) , and g∗ (a) = g (a+) , g∗ (b) = g (b−). If g ∈ BV, then g∗ ∈ BVN .The main result of this paper is the following theorem.Theorem 2. A functional F defined on BV is linear and two-norm continuous ifand only if there exist functions f1 ∈ RFN and f2 ∈ c0 [a, b] such thatF (g) =∫ baf1 dg∗ +∞∑i=1[g (ti)− g∗ (ti)] f2(ti)for every g ∈ BV, where ti, i = 1, 2, . . . , are the discontinuity points of g, and g∗ thenormalized function of g. . ⇒: Put f1(t) = F (γt) for every t ∈ [a, b]. By Lemma 2, f1 ∈ RFN . Letg ∈ BV and g∗ the normalized function of g. ThenF (g) = F (g∗) + F (g − g∗).Next, put f2(t) = F (δt) for every t ∈ [a, b]. By Lemma 4, f2 ∈ c0 [a, b]. Let ti,i = 1, 2, . . . , be the discontinuity points of g. Then for any nF (g − g∗) = F( ∞∑i=1[g (ti)− g∗ (ti)] δti)=n∑i=1[g (ti)− g∗ (ti)] f2(ti) + F( ∞∑i=n+1[g (ti)− g∗ (ti)] δti).Since {f2 (ti)} is a null sequence, that is, a sequence converging to 0, and∞∑i=1[g(ti)−g∗(ti)] absolutely convergent, the series∞∑i=1[g (ti)− g∗ (ti)] f2(ti) converges.Further,∞∑i=n+1[g (ti)− g∗ (ti)] δti is two-norm convergent to zero in BV as n →∞.Then, by the two-norm continuity of F, we havelimn→∞F( ∞∑i=n+1[g (ti)− g∗ (ti)]δti)= 0.Together with Lemma 3, we obtain the representation of F.6⇐: LetF1 (g) =∫ baf1 dg∗.Take gn ∈ BV such that gn is two-norm convergent to g in BV . So does g∗n. Since f1 ∈RFN, by applying Lemma 1 and Theorem 1, {(f, gk) : k = 1, 2, . . .} is equi-integrableon [a, b] ,∫ baf1 dg∗ exists and limn→∞∫ baf1 dg∗n =∫ baf1 dg∗. Hence limn→∞F1(gn) = F1(g).That is, F1(g) defines a two-norm continuous linear functional on BV .LetF2 (g) =∞∑i=1[g (ti)− g∗ (ti)]f2(ti).Since f2 ∈ c0 [a, b] , for every ε > 0, the set X = {x ∈ [a, b] : |f2(x)| > ε} is finite.Consequently, we have |f2(x)| < ε for every x /∈ X. Since gn is two-norm convergentto g in BV, for every ε > 0, there exists N such that for every n > N and for everyξ ∈ [a, b] , we have|gn(ξ)− g(ξ)| <εn (X)where n (X) denotes the number of the points in X. We may assume that |f2(x)| 6M1 for every x ∈ [a, b] and also we have V (gn) 6 M2 for every n and V (g) 6 M2.Therefore for every n > N, we have∑t∈X(|gn(t)− g(t)|+ |g∗n(t)− g∗(t)|)|f2(t)| 6∑t∈X2εn(X)M1and∑t/∈X(|gn(t)− g∗n(t)|+ |g(t)− g∗(t)|)|f2(t)| 6 ε[V (gn; [a, b]) + V (g; [a, b])].Therefore∣∣∣∣∑t∈(a,b)([gn(t)− g∗n(t)] − [g(t)− g∗(t)])f2(t)∣∣∣∣ 6 2εM1 + ε(2M2)and hence limn→∞F2(gn) = F2(g). That is, F2(g) defines a two-norm continuous linearfunctional on BV . Put F (g) = F1(g) + F2(g). Hence the proof is complete. As a special case, we obtain the following corollary.7Corollary 1. A functional F defined on BVN is linear and two-norm continuousif and only if there exists a function f ∈ RFN such thatF (g) =∫ baf dg for g ∈ BVN .That is, the dual of the space BVN is the space RFN . Conversely, the dual of RFNis BVN . This has been proved in [4] and is stated below. The proof is similar to thatof the Riesz theorem. A functional F on RFN is continuous if it is continuous withrespect to the norm in RFN .Theorem 3 [4]. A functional F defined on RFN is linear and continuous if andonly if there exist a function g ∈ BV and a real number d such thatF (f) = df(a) +∫ bag df for f ∈ RFN .Therefore the spaces BVN and RFN are the duals of each other. A representationtheorem for continuous linear functionals on RF of the form like Theorem 2 is alsoavailable [3], [12]. We shall not elaborate here.Finally, the following Riesz representation theorem is a consequence of Theorem 3,integration by parts, and the Hahn-Banach theorem, since C [a, b] is a closed subspaceof RFN .Corollary 2 (Riesz representation theorem). A functional F defined on C [a, b]is linear and continuous if and only if there exists a function g ∈ BVN such thatF (f) =∫ baf dg for f ∈ C [a, b] .Representation theorem for continuous linear functionals on RF was first given bySchwabik [13]. For another relevant reference, see [14].8References[1] D.Franková: Regulated functions. Math. Bohem. 116 (1991), 20–59. Zbl 0724.26009[2] J.Dieudonné: Foundations of Modern Analysis. New-York, 1960. Zbl 0176.00502[3] K.K.Aye: The duals of some Banach spaces. Ph.D Thesis, Nanyang Technological Uni-versity, 2002.[4] M.Tvrdý: Linear bounded functionals on the space of regular regulated functions. TatraMt. Math. Publ. 8 (1996), 203–210. Zbl 0920.46031[5] P.Habala, P.Hájek, V.Zizler: Introduction to Banach Spaces. 1996.[6] P.Y. Lee: Lanzhou Lectures on Henstock Integration. World Scientific, 1989.Zbl 0699.26004[7] P.Y. Lee, R.Výborný: The Integral: An Easy Approach after Kurzweil and Henstock.Cambridge University Press, 2000. Zbl 0941.26003[8] T.H.Hildebrandt: Introduction of the Theory of Integration. Academic Press, 1963.Zbl 0112.28302[9] T.H.Hildebrandt: Linear continuous functionals on the space (BV) with weak topologies.Proc. Amer. Math. Soc. 17 (1966), 658–664. Zbl 0152.13604[10] Tom M.Apostol: Mathematical Analysis. 1957. Zbl 0077.05501[11] W.Orlicz: Linear Functional Analysis. World Scientific, 1992. Zbl 0799.46002[12] K.K.Aye, P.Y. Lee: Orthogonally additive functionals on BV. Math. Bohem. 129(2004), 411–419.[13] Š. Schwabik: A survey of some new results for regulated functions. Seminario Brasileirode analise 28 (1988).[14] M.Brokate, P.Krejčí: Duality in the space of regulated functions and the play operator.Math. Z. 245 (2003), 667–668. Zbl 1055.46023Authors’ addresses: Khaing Khaing Aye, (1) Mathematics and Mathematics Educa-tion, National Institute of Education, Nanyang Technological University, 1, Nanyang Walk,Singapore 637616, (2) Department of Engineering Mathematics, Yangon Technological Uni-versity, Yangon, Myanmar, e-mail: wk.ygn@mptmail.net.mm; Peng Yee Lee, Mathematicsand Mathematics Education, National Institute of Education, Nanyang Technological Uni-versity, 1, Nanyang Walk, Singapore 637616, e-mail: pylee@nie.edu.sg.9

summary:This paper deals with regulated functions having values in a Banach space. In particular, families of equiregulated functions are considered and criteria for relative compactness in the space of regulated functions are given

Fraňková, Dana

Regulated functions with values in Banach space

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The dual of the space of functions of bounded variation

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