Several bounds in approximating the Riemann–Stieltjes integral in terms of s-convex integrands or integrator are given

Alomari, Mohammad Wajeeh

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Journals from University of Tartu

ACTA ET COMMENTATIONES UNIVERSITATIS TARTUENSIS DE MATHEMATICAVolume 16, Number 2, 2012Available online at www.math.ut.ee/acta/Bounds for the Riemann–Stieltjes integral vias-convex integrand or integratorMohammad Wajeeh AlomariAbstract. Several bounds in approximating the Riemann–Stieltjes in-tegral in terms of s-convex integrands or integrator are given.1. IntroductionA function f : R+ → R, where R+ = [0,∞), is said to be s-convex in thesecond sense iff (αx+ βy) ≤ αsf (x) + βsf (y)for all x, y ∈ [0,∞), α, β ≥ 0 with α + β = 1 and for some fixed s ∈ (0, 1].This class of functions is denoted by K2s . It can be easily seen that for s = 1,s-convexity reduces to the ordinary convexity of functions defined on [0,∞)(see [6]).In [2], Cerone and Dragomir have proved some error bounds in approx-imating the Riemann–Stieltjes integral in terms of some moments of theintegrand. Among others, they proved the following result.Theorem 1. Let u be p-convex with p > 0, f be monotonically increasingon [a, b] and such that the Riemann–Stieltjes integral∫ ba f (t) du (t) and theRiemann integrals∫ ba (t− a)p−1 f (t) dt ,∫ ba (b− t)p−1 f (t) dt exist. Then∫ baf (t) du (t) ≥ p(b− a)p[u (b)∫ ba(t− a)p−1 f (t) dt−u (a)∫ ba(b− t)p−1 f (t) dt].Received February 27, 2012.2010 Mathematics Subject Classification. 26A42, 26A51, 26D15.Key words and phrases. Riemann–Stieltjes integral, s-convex function.181182 MOHAMMAD WAJEEH ALOMARIFor other results concerning different bounds for the Riemann–Stieltjesintegral under various assumptions on f and u, see the recent papers [1]–[5]and the references therein.In this paper, several inequalities for the Riemann–Stieltjes integral∫ ba f (x) dg (x) are proved. Namely, the integrand f is assumed to bes-convex (s-concave) and the integrator g is monotonically increasing, boun-ded and s-convex (s-concave).2. Inequalities for s-convex integrands or integratorsWe may start with the following result.Theorem 2. Let f, g : [a, b] ⊆ R+ → R be such that f is s-convex on [a, b],g is monotonically increasing on [a, b] and the Riemann–Stieltjes integral∫ ba f (x) dg (x) and the Riemann integrals∫ ba (x− a)s−1 g (x) dx,∫ ba (b− x)s−1 g (x) dx exist. Then we have the inequalities∫ baf (x) dg (x) ≤ f (b)(b− a)s[(b− a)s g (b)− s∫ bag (x) (x− a)s−1 dx]+f (a)(b− a)s[− (b− a)s g (a) + s∫ bag (x) (b− x)s−1 dx]≤ [g (b)− g (a)] [f (a) + f (b)] . (2.1)Proof. Since f is s-convex on [a, b] and by using the integration by partsformula for Riemann–Stieltjes integral, we have∫ baf (x) dg (x) ≤∫ ba[(x− ab− a)sf (b) +(b− xb− a)sf (a)]dg (x)= f (b)∫ ba(x− ab− a)sdg (x) + f (a)∫ ba(b− xb− a)sdg (x)=f (b)(b− a)s∫ ba(x− a)s dg (x) + f (a)(b− a)s∫ ba(b− x)s dg (x)=f (b)(b− a)s[(b− a)s g (b)− s∫ bag (x) (x− a)s−1 dx](2.2)+f (a)(b− a)s[− (b− a)s g (a) + s∫ bag (x) (b− x)s−1 dx],which proves the first inequality in (2.1). To prove the second inequality in(2.1), using the monotonicity of g on [a, b], we get∫ ba(x− a)s−1 g (x) dx ≥ g (a)∫ ba(x− a)s−1 dx = 1sg (a) (b− a)sBOUNDS FOR THE RIEMANN–STIELTJES INTEGRAL 183and ∫ ba(b− x)s−1 g (x) dx ≤ g (b)∫ ba(b− x)s−1 dx = 1sg (b) (b− a)s .Therefore by (2.2), we get∫ baf (x) dg (x) ≤ f (b)(b− a)s[(b− a)s g (b)− s∫ bag (x) (x− a)s−1 dx]+f (a)(b− a)s[− (b− a)s g (a) + s∫ bag (x) (b− x)s−1 dx]≤ f (b)(b− a)s[(b− a)s g (b)− g (a) (b− a)s]+f (a)(b− a)s[− (b− a)s g (a) + g (b) (b− a)s]= [g (b)− g (a)] [f (a) + f (b)] ,which proves the second inequality in (2.1). The following result holds.Theorem 3. Let g : [a, b] ⊆ R+ → R be a monotonically increasingfunction on [a, b].(1) If f : [a, b]→ R+ is convex on [a, b], then we have the inequality∫ baf (x) dg (x)≤min{12[f (a)+f (b)+|f (a)−f (b)|] [g (b)−g (a)] ,[g (b)− g (a)2+∣∣∣∣g (a) + g (b)2− 1b− a∫ bag (x) dx∣∣∣∣] [f (a) + f (b)]} .(2.3)(2) If f is concave, then we have the inequality∫ baf (x) dg (x)≥max{12[f (a)+f (b)−|f (a)−f (b)|] [g (b)−g (a)] ,[g (b)− g (a)2−∣∣∣∣g (a) + g (b)2− 1b− a∫ bag (x) dx∣∣∣∣] [f (a) + f (b)]}(2.4)provided that the Riemann–Stieltjes integral∫ ba f (x) dg (x) exists.184 MOHAMMAD WAJEEH ALOMARIProof. (1) In (2.2), set s = 1, then we get∫ baf (x) dg (x) ≤∫ ba[x− ab− af (b) +b− xb− af (a)]dg (x)=f (b)b− a[(b− a) g (b)−∫ bag (x) dx]+f (a)b− a[− (b− a) g (a) +∫ bag (x) dx]= f (b)[g (b)− 1b− a∫ bag (x) dx]+ f (a)[1b− a∫ bag (x) dx− g (a)].Thus ∫ baf (x) dg (x) ≤ max {f (a) , f (b)} [g (b)− g (a)]=12[f (a) + f (b) + |f (a)− f (b)|] [g (b)− g (a)]and also∫ baf (x) dg (x) ≤ max{[g (b)− 1b− a∫ bag (x) dx],[1b− a∫ bag (x) dx− g (a)]}[f (a) + f (b)]=[g (b)− g (a)2+∣∣∣∣g (a) + g (b)2 − 1b− a∫ bag (x) dx∣∣∣∣] [f (a) + f (b)]which proves (2.3).(2) If f is concave, then we have∫ baf (x) dg (x) ≥∫ ba[x− ab− af (b) +b− xb− af (a)]dg (x).So, similarly to the proof of (1),∫ baf (x) dg (x) ≥ min {f (a) , f (b)} [g (b)− g (a)]and ∫ baf (x) dg (x) ≥ min{[g (b)− 1b− a∫ bag (x) dx],[1b− a∫ bag (x) dx− g (a)]}[f (a) + f (b)]which proves (2.4). BOUNDS FOR THE RIEMANN–STIELTJES INTEGRAL 185Theorem 4. Let f, g : [a, b] ⊆ R+ → R be, respectively, s1-, s2-convexfunctions on [a, b], s1, s2 ∈ (0, 1]. Then we have the inequality∫ baf (x) dg (x) ≤ s2s1 + s2[f (b) g (b)− f (a) g (a)]+ s1β (s2 + 1, s1) [f (a) g (b)− f (b) g (a)](2.5)provided that the Riemann–Stieltjes integral∫ ba f (x) dg (x) exists. If f, g ares1-, s2-concave, then the inequality (2.5) is reversed.Proof. Since f is s1-convex on [a, b], by using the integration by partsformula for Riemann–Stieltjes integral, we have∫ baf (x) dg (x) ≤∫ ba[(x− ab− a)s1f (b) +(b− xb− a)s1f (a)]dg (x)= f (b)∫ ba(x− ab− a)s1dg (x) + f (a)∫ ba(b− xb− a)s1dg (x)=f (b)(b− a)s1∫ ba(x− a)s1 dg (x) + f (a)(b− a)s1∫ ba(b− x)s1 dg (x)=f (b)(b− a)s1[(b− a)s1 g (b)−s1∫ bag (x) (x− a)s1−1 dx](2.6)+f (a)(b− a)s1[− (b− a)s1 g (a)+s1∫ bag (x) (b− x)s1−1 dx].Since g(x) is s2-convex on [a, b], we haveg (x) ≤[(x− ab− a)s2g (b) +(b− xb− a)s2g (a)],which, by (2.6), gives∫ baf (x) dg (x)≤ f (b)(b− a)s1[(b− a)s1 g (b)−s1∫ ba((x− ab− a)s2g (b) +(b− xb− a)s2g (a))(x− a)s1−1 dx]+f (a)(b− a)s1[− (b− a)s1 g (a)+s1∫ ba((x− ab− a)s2g (b) +(b− xb− a)s2g (a))(b− x)s1−1 dx]186 MOHAMMAD WAJEEH ALOMARI=f (b)(b− a)s1[(b− a)s1 g (b)− s1g (b)(b− a)s2∫ ba(x− a)s1+s2−1 dx−s1g (a)(b− a)s2∫ ba(b− x)s2 (x− a)s1−1 dx]+f (a)(b− a)s1[− (b− a)s1 g (a) + s1g (b)(b− a)s2∫ ba(x− a)s2 (b− x)s1−1 dx+s1g (a)(b− a)s2∫ ba(b− x)s1+s2−1 dx].Simple calculations yield that∫ ba(b− x)s2 (x− a)s1−1 dx = (b− a)s1+s2 β (s1, s2 + 1) ,and ∫ ba(x− a)s2 (b− x)s1−1 dx = (b− a)s1+s2 β (s2 + 1, s1) ,where, ∫ ba(x− a)p (b− x)q dx = (b− a)p+q+1∫ 10(1− t)p tqdt= (b− a)p+q+1 β (p+ 1, q + 1)and β (·, ·) is the Euler Beta function.It follows that∫ baf (x) dg (x)≤ f (b)(b− a)s1[(b− a)s1 g (b)− s1g (b)(b− a)s2(b− a)s1+s2s1 + s2−s1(b− a)s1+s2 g (a)(b− a)s2β (s1, s2 + 1)]+f (a)(b− a)s1[− (b− a)s1 g (a) + s1(b− a)s1+s2 g (b)(b− a)s2β (s2 + 1, s1)+s1g (a)(b− a)s2(b− a)s1+s2s1 + s2]= f (b) g (b)− s1s1 + s2f (b) g (b)− s1β (s1, s2 + 1) f (b) g (a)− f (a) g (a) + s1β (s2 + 1, s1) f (a) g (b) +s1s1 + s2f (a) g (a)=s2s1 + s2[f (b) g (b)−f (a) g (a)]+s1β (s2+1, s1) [f (a) g (b)−f (b) g (a)] ,BOUNDS FOR THE RIEMANN–STIELTJES INTEGRAL 187since β (s1, s2 + 1) = β (s2 + 1, s1), which proves (2.5). Corollary 1. In Theorem 4, if s1 = s2 = 1, i.e., f, g are two convexfunctions on [a, b], then we have∫ baf (x) dg (x) ≤ f (a) + f (b)2[g (b)− g (a)] .Corollary 2. In Theorem 4, if f is convex and g is s-convex, then wehave∫ baf (x) dg (x) ≤ f (b)[sg (b)− g (a)s+ 1]+ f (a)[g (b)− sg (a)s+ 1](2.7)provided that the Riemann–Stieltjes integral∫ ba f (x) dg (x) exists. If f isconcave and g is s-concave, then the inequality (2.7) is reversed.Theorem 5. Let f, g : [a, b] ⊆ R+ → R be such that g satisfies φ ≤ g (t) ≤Φ for all t ∈ [a, b].(1) If f is s-convex on [a, b], then we have the inequality∫ baf (x) dg (x) ≤ f (b) [g (b)− φ] + f (a) [Φ− g (a)] . (2.8)(2) If f is s-concave on [a, b], then we have the inequality∫ baf (x) dg (x) ≥ f (b) [g (b)− Φ] + f (a) [φ− g (a)] (2.9)provided that the Riemann–Stieltjes integral∫ ba f (x) dg (x) exists.Proof. (1) From (2.2), we get∫ baf (x) dg (x) ≤ f (b)(b− a)s[(b− a)s g (b)− sφ∫ ba(x− a)s−1 dx]+f (a)(b− a)s[− (b− a)s g (a) + sΦ∫ ba(b− x)s−1 dx]≤ f (b)(b− a)s[(b− a)s g (b)− φ (b− a)s]+f (a)(b− a)s[− (b− a)s g (a) + Φ (b− a)s]= f (b) [g (b)− φ] + f (a) [Φ− g (a)]which proves (2.8).188 MOHAMMAD WAJEEH ALOMARI(2) If f is s-concave, then, similarly,∫ baf (x) dg (x) ≥ f (b)(b− a)s[(b− a)s g (b)− sΦ∫ ba(x− a)s−1 dx]+f (a)(b− a)s[− (b− a)s g (a) + sφ∫ ba(b− x)s−1 dx]≥ f (b)(b− a)s[(b− a)s g (b)− Φ (b− a)s]+f (a)(b− a)s[− (b− a)s g (a) + φ (b− a)s]= f (b) [g (b)− Φ] + f (a) [φ− g (a)]which proves (2.9). Remark 1. Define the function g : [a, b]→ R+, g(t) =∫ ta u(s)ds. Then gis differentiable on (a, b) and g′(t) = u(t). And we have∫ baf (x) dg (x) =∫ baf (x)u (x) dx.Therefore, we can point out some results for the Riemann integral of aproduct.(1) Under the assumptions of Theorem 2, we have∫ baf (x)u (x) dx ≤ [f (a) + f (b)]∫ bau(x)dx.(2) Under the assumptions of Theorem 3, we have the following.(a) If f : [a, b]→ R+ is convex on [a, b], then∫ baf (x)u(x)dx ≤ min{12[f (a) + f (b) + |f (a)− f (b)|]∫ bau(x)dx,12[∫ bau(x)dx+∣∣∣∣∫ bau(x)dx− 2b− a∫ ba∫ xau(t)dtdx∣∣∣∣] [f (a) + f (b)]} .BOUNDS FOR THE RIEMANN–STIELTJES INTEGRAL 189(b) If f is concave, then∫ baf (x)u(x)dx ≥ max{12[f (a) + f (b)− |f (a)− f (b)|]∫ bau(x)dx,12[∫ bau(x)dx−∣∣∣∣∫ bau(x)dx− 2b− a∫ ba∫ xau(t)dtdx∣∣∣∣] [f (a) + f (b)]} .(3) Under the assumptions of Theorem 4, we have∫ baf (x)u (x) dx ≤[s2s1 + s2f (b) + s1β (s2 + 1, s1) f (a)] ∫ bau (x) dx.(4) Under the assumptions of Theorem 5, we have the following.(a) If f is s-convex on [a, b], then∫ baf (x)u (x) dx ≤ f (b)[∫ bau (x) dx− φ]+ Φf (a) .(b) If f is s-concave, then∫ baf (x)u (x) dx ≥ f (b)[∫ bau (x) dx− Φ]+ φf (a) .References[1] N. S. Barnett and S. S. Dragomir, The Beesack–Darst–Pollard inequalities and approx-imations of the Riemann–Stieltjes integral, Appl. Math. Lett. 22 (2009), 58–63.[2] P. Cerone and S. S. Dragomir, Approximating the Riemann–Stieltjes integral via somemoments of the integrand, Math. Comput. Modelling 49 (2009), 242–248.[3] P. Cerone and S. S. Dragomir, Approximation of the Stieltjes integral and applicationsin numerical integration, Appl. Math. 51(1) (2006), 37–47.[4] S. S. Dragomir, Inequalities for Stieltjes integrals with convex integrators and applica-tions, Appl. Math. Lett. 20(1) (2007), 123–130.[5] S. S. Dragomir, Inequalities of Grüss type for the Stieltjes integral and applications,Kragujevac J. Math. 26 (2004), 89–122.[6] H. Hudzik and M. Maligranda, Some remarks on s-convex functions, AequationesMath. 48(1) (1994), 100–111.Department of Mathematics, Faculty of Science, Jerash University, 26150Jerash, JordanE-mail address: mwomath@gmail.com

Bounds for the Riemann–Stieltjes integral via s-convex integrand or integrator

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Bounds for the Riemann–Stieltjes integral via s-convex integrand or integrator

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