Correction: DOI:10.1017/S0004972713000713We study the question, whether a Riemann-Stieltjes integral of a positive continuous function with respect to a non-negative function of bounded variation is positive.Peer reviewe

Lukkarinen, Jani

S. Pakkanen, Mikko

arXiv

We study the question, whether a Riemann-Stieltjes integral of a positive continuous function with respect to a non-negative function of bounded variation is positive.Peer reviewe

Helsingin yliopiston digitaalinen arkisto

ON THE POSITIVITY OF RIEMANN–STIELTJES INTEGRALSJANI LUKKARINEN AND MIKKO S. PAKKANENAbstract. We study the question, whether a Riemann–Stieltjes integral of a positivecontinuous function with respect to a non-negative function of bounded variation ispositive.1. IntroductionLet f : [a, b]→ R be a continuous function and g : [a, b]→ R a function of boundedvariation. It is a classical result that for such f and g, and for any y ∈ (a, b], theRiemann–Stieltjes integral ∫ yaf(x)dg(x), (1.1)exists (see, e.g., [3, pp. 316–317]). The purpose of this note is to address the followingsimple question that seems to be less well-known.Question 1.2. If f is positive, g is non-negative, non-vanishing, and satisfies g(a) = 0,can we then select the upper limit of integration y so that the integral (1.1) is positive?The only reference known to us is a note by Satyanarayana [4] where an affir-mative answer is proven under a slightly different set of assumptions, most impor-tantly, assuming that g is non-decreasing. Let us stress that positive will here alwaysmean strictly positive. In particular, since f is continuous, under our assumptions0 < min f 6 max f <∞.Recall that, since g is of bounded variation, there exist non-decreasing functions g+and g− such that g = g+ − g−. Therefore, finite limx→y+ g(x) and limx→y− g(x) existfor any y ∈ [a, b] (apart from limx→a− g(x) and limx→b+ g(x), obviously). Moreover, theintegral (1.1) is equal to ∫ yaf(x)dg+(x)−∫ yaf(x)dg−(x) .Since g(a) = 0, we may assume that g+ and g− are non-negative and satisfy g±(a) = 0.We may now distinguish two special cases, where it is evident that the answer toQuestion 1.2 is yes.Date: March 26, 2012.2000 Mathematics Subject Classification. Primary 26A42; Secondary 26A45.Key words and phrases. Riemann–Stieltjes integral, positivity, function of bounded variation,Gro¨nwall’s inequality.1arXiv:1203.5276v1  [math.CA]  23 Mar 20122 J. LUKKARINEN AND M. S. PAKKANEN(1) If limx→x+L g(x) > 0, where xL := inf{x : g(x) > 0}, then there exists ε > 0 suchthat (1.1) is positive for y = xL + ε.(2) If g+(y) > 0 and g−(y) = 0, then (1.1) is at least g+(y) minx∈[a,y] f(x) > 0.The first item follows from the elementary lower bound, valid for 0 < ε ≤ b− xL,∫ xL+εaf(x)dg(x) > g+(xL + ε) mina,xL−ε≤x≤xL+εf(x)− g−(xL + ε) maxa,xL−ε≤x≤xL+εf(x),whereas the second item is a straightforward consequence of g±(a) = 0.In general, the integral (1.1) is positive if and only if∫ yaf(x)dg+(x) >∫ yaf(x)dg−(x). (1.3)Obviously, the condition g > 0 is equivalent to g+ > g−, and we have g(x) > 0 ifand only if g+(x) > g−(x). It would be tempting to conjecture that, as a continuousfunction, f is “nearly constant” in some neighborhood of xL and, hence, that theinequality (1.3) ought to hold for y = xL + ε with some “small” ε > 0, suggesting anaffirmative answer to Question 1.2 in general.Remark 1.4. The proviso g(a) = 0 may seem superfluous as the value of the integral(1.1) does not depend on g(a). However, together with the condition g > 0 it constrainsthe behavior of g near xL, which is a key part of the formulation of Question 1.2.It should also be stressed that the continuity of f is equally important—aside fromthe possible non-existence of the integral, there is no reason to expect the answer toQuestion 1.2 to be yes when f fails to be continuous.2. Negative answer to Question 1.2Unfortunately, the heuristic above is too simple-minded since mere continuity doesnot restrict the fine properties of the integrand f and leaves it with enough room tovary “too much” for our purposes. Indeed, the general answer to Question 1.2 is no.We show that for any f that exhibits “enough” variation, there exists a suitable g suchthat the integral (1.1) is less than zero for all y ∈ (a, b].Theorem 2.1. Let f : [a, b] → (0,∞) be a continuous function. Suppose that thereexist two sequences (xn) and (xn) witha < · · · < xn < xn < · · · < x2 < x2 < x1 < x1 6 bsuch that for some α > 0 and γ ∈ (0, 1),f(xn)− f(xn) > αn−γ for all n ∈ N.Then, there exists a non-vanishing function g : [a, b] → [0,∞) of bounded variationsuch that g(a) = 0 and ∫ yaf(x)dg(x) < 0 for all y ∈ (a, b] .POSITIVITY OF RIEMANN–STIELTJES INTEGRALS 3Proof. Let β > 1 and consider h : [a, b]→ [0,∞) defined byh(x) :=∑n∈Nn−βχ[xn,xn)(x),where χE denotes the characteristic function of a set E. (Figure 1 illustrates thedefinition for Example 2.2 below.) This is, by construction, a function of boundedvariation such that h(a) = 0. We have for any n ∈ N,∫ xnaf(x)dh(x) = n−βf(xn)−∞∑k=n+1k−β(f(xk)− f(xk)),where∞∑k=n+1k−β(f(xk)− f(xk))> α∞∑k=n+1k−(β+γ)> α∫ ∞n+2x−(β+γ)dx=αβ + γ − 1(n+ 2)−(β+γ−1).Since γ < 1 and supn∈N f(xn) <∞, there exists n0 ∈ N such that∫ xnaf(x)dh(x) < 0 for all n > n0.We may also note that for all n > n0, with n > 2, and y ∈ (xn, xn−1),∫ yaf(x)dh(x) 6∫ xnaf(x)dh(x) < 0.Thus, defining g := hχ[0,xn0 ) yields a function with all the properties stated in thetheorem. Example 2.2. Function f : [0, 1]→ (0,∞) given byf(x) :={xγ sin(1/x) + 2, if x > 0,2, if x = 0,where γ ∈ (0, 1), satisfies the condition of Theorem 2.1 with xn = 14n−3 2pi , xn = 14n−1 2pi ,n ∈ N, using α = 2(2pi)−γ. Figure 1 illustrates the behavior of f and of the correspond-ing g and Riemann–Stieltjes integral, as defined in the proof of Theorem 2.1.4 J. LUKKARINEN AND M. S. PAKKANEN1.92.02.12.21.80.01 0.02 0.03 0.04xf(x)00.010.020.030.040.050 0.01 0.02 0.03 0.04xg(x)0−0.02−0.04−0.06−0.08−0.10−0.120.01 0.02 0.03 0.04xJ(x)Figure 1. Plot of f , as defined in Example 2.2 for γ = 12, and thecorresponding functions g and J , J(x) :=∫ x0f(x′)dg(x′), as defined inthe proof of Theorem 2.1 using β = 32and n0 = 7. The values have beencomputed numerically by considering only contributions with n 6 1 000.For clarity, only the region with x ∈ [0, 0.04] is shown.3. Integrands of bounded variationAny integrand f that satisfies the condition of Theorem 2.1 is clearly of unboundedvariation. This prompts us to ask, could we actually obtain an affirmative answer toQuestion 1.2 if f varied “less.” In fact, we are able to show that, if f is of boundedvariation, the answer to Question 1.2 is yes. Bounds for Riemann–Stieltjes integralsunder these assumptions have been derived by Beesak [1], but instead of building ourargument on them, we give a direct proof which relies on some elementary measuretheory and the measure-theoretic version of Gro¨nwall’s inequality.Theorem 3.1. Let f : [a, b] → (0,∞) be a continuous function and g : [a, b] → [0,∞)a non-vanishing function of bounded variation such that g(a) = 0. If f is of boundedvariation, then ∫ yaf(x)dg(x) > 0 for some y ∈ (a, b].POSITIVITY OF RIEMANN–STIELTJES INTEGRALS 5Proof. Let us denote by V dc f the total variation of f on [c, d] ⊂ [a, b]. Recall that totalvariation is additive in the sense that V da f = Vca f + Vdc f for a 6 c 6 d [3, Theorem12.1]. Moreover, since f is continuous, the mapping x 7→ V xa f from [a, b] to [0,∞) iscontinuous [3, Theorem 12.2]. Thus, there exists a finite, positive Borel measure ν on[a, b] such that ν([c, d)) = V dc f for any c and d such that a 6 c < d 6 b. Let us defineanother finite, positive Borel measure by µ(dx) := f(x)−1ν(dx). By construction, wehave then|f(d)− f(c)| 6∫[c,d)f(x)µ(dx) .By an approximation with suitable Riemann–Stieltjes sums, where the values of gare chosen to be sufficiently close to their respective infima on all subintervals of thepartitions, we can prove that∣∣∣∫ yag(x)df(x)∣∣∣ 6 ∫[a,y)f(x)g(x)µ(dx) . (3.2)The Riemann–Stieltjes integral on the left hand side is well-defined by the integrationby parts formula [3, Theorem 12.14]∫ yag(x)df(x) = f(y)g(y)− f(a)g(a)−∫ yaf(x)dg(x) , (3.3)whenever∫ yaf(x)dg(x) exists and this is always true under the present assumptions.Now suppose that, contrary to our assertion, we have∫ yaf(x)dg(x) 6 0 for all y ∈ (a, b]. (3.4)Rearranging the integration by parts formula (3.3) and using the assumption g(a) = 0,(3.2), and (3.4), we obtain for any y ∈ (a, b]f(y)g(y) 6∫ yag(x)df(x) 6∫[a,y)f(x)g(x)µ(dx) .But the measure-theoretic version of Gro¨nwall’s inequality [2, Theorem A.5.1] impliesthat then f(x)g(x) 6 0 for all x ∈ [a, b], whence g = 0, a contradiction. Remark 3.5. There are two straightforward refinements to Theorem 3.1. Firstly, itclearly suffices that f is of bounded variation on [xL, xL + ε] for some ε > 0. Secondly,if g is right-continuous, then also the mappingy 7→∫ yaf(x)dg(x)is right-continuous and, under the assumptions of Theorem 3.1, there exists an interval[c, d] ⊂ [a, b] such that ∫ yaf(x)dg(x) > 0 for all y ∈ [c, d].6 J. LUKKARINEN AND M. S. PAKKANENRemark 3.6. Theorem 3.1 explains why the answer to Question 1.2 is perhaps elusive.Experimentation with nicely behaving integrands will not suffice, since a “pathological”f is required in order to discover the general answer. However, such integrands need notbe mere curiosities. In fact, our study of Question 1.2 was originally motivated by anapplication in financial mathematics involving as the integrand a path of a continuous-time stochastic process, which is typically of unbounded variation.AcknowledgementsJ. Lukkarinen was supported by the Academy of Finland. M. S. Pakkanen acknowl-edges support from the Finnish Cultural Foundation, from CREATES, funded by theDanish National Research Foundation, and from the Aarhus University Research Foun-dation regarding the project “Stochastic and Econometric Analysis of Commodity Mar-kets”.References[1] P. R. Beesack, Bounds for Riemann–Stieltjes integrals, Rocky Mountain J. Math. 5 (1975) 75–78.Available at http://dx.doi.org/10.1216/RMJ-1975-5-1-75.[2] S. N. Ethier and T. G. Kurtz, Markov Processes: Characterization and Convergence, Wiley, NewYork, 1986.[3] M. H. Protter and C. B. Morrey, A First Course in Real Analysis, Springer, New York, 1977.[4] U. V. Satyanarayana, A note on Riemann–Stieltjes integrals, Amer. Math. Monthly 87 (1980)477–478. Available at http://www.jstor.org/stable/2320259.Jani Lukkarinen, Department of Mathematics and Statistics, University of Helsinki,P.O. Box 68, FI-00014 Helsingin yliopisto, Finland.E-mail address: jani.lukkarinen@helsinki.fiMikko S. Pakkanen, CREATES and Department of Economics and Business, AarhusUniversity, Bartholins Alle´ 10, DK-8000 Aarhus C, Denmark.E-mail address: msp@iki.fiURL: http://www.mikkopakkanen.fi/

On the positivity of Riemann-Stieltjes integrals

We correct two minor errors in Theorem 2.1 of Lukkarinen and Pakkanen [‘On the positivity of Riemann–Stieltjes integrals’, Bull. Aust. Math. Soc. 87 (2013), 400–405].</jats:p

JANI LUKKARINEN

MIKKO S. PAKKANEN

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Bulletin of the Australian Mathematical Society

‘ON THE POSITIVITY OF RIEMANN–STIELTJES INTEGRALS’

We study the question, whether a Riemann–Stieltjes integral of a positive continuous function with respect to a non-negative function of bounded variation is positive

Jani Lukkarinen

Mikko S. Pakkanen

CiteSeerX

ON THE POSITIVITY OF RIEMANN–STIELTJES INTEGRALS

We study the question, whether a Riemann-Stieltjes integral of a positive continuous function with respect to a non-negative function of bounded variation is positive.

10.1017/S000497271200063

Pakkanen, Mikko S.

Australian Mathematical Society (AustMS): E-Journals

AbstractWe study the question whether a Riemann–Stieltjes integral of a positive continuous function with respect to a nonnegative function of bounded variation is positive.</jats:p

On the positivity of Riemann-Stieltjes integrals

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Helsingin yliopiston digitaalinen arkisto

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