In this paper, we point out a Grüss type inequality and apply it for special means (logarithmic mean, identric mean, etc... ) and in Numerical Analysis in connection with the classical trapezoid formula

Dragomir, Sever S

McAndrew, Alasdair

English

Victoria University Eprints Repository

ON TRAPEZOID INEQUALITY VIA A GRU¨SS TYPE RESULTAND APPLICATIONSS.S. DRAGOMIR AND A. MCANDREWAbstract. In this paper, we point out a Gru¨ss type inequality and apply itfor special means (logarithmic mean, identric mean, etc... ) and in NumericalAnalysis in connection with the classical trapezoid formula.1. IntroductionIn 1935, G. Gru¨ss (see for example [1, p. 296]), proved the following integralinequality which gives an approximation for the integral of a product in terms ofthe product of integrals:Theorem 1. Let f, g : [a, b]→ R be two integrable mappings so that ϕ ≤ f (x) ≤ Φand γ ≤ g (x) ≤ Γ for all x ∈ [a, b], where ϕ,Φ, γ,Γ are real numbers. Then wehave: ∣∣∣∣∣∣ 1b− ab∫af (x) g (x) dx− 1b− ab∫af (x) dx · 1b− ab∫ag (x) dx∣∣∣∣∣∣≤ 14(Φ− ϕ) (Γ− γ)(1.1)and the inequality is sharp, in the sense that the constant 14 can not be replaced bya smaller one.For a simple proof of this fact as well as for extensions, generalizations, discretevariants etc... see the book [1, p. 296] by Mitrinovic´, Pec˘aric´ and Fink and thepapers [2]-[7] where further references are given.In this paper, we point out a different Gru¨ss type inequality and apply it forspecial means (logarithmic mean, identric mean, etc... ) and in Numerical Analysisin connection with the classical trapezoid formula.2. A Gru¨ss’ type inequalityWe start with the following result of Gru¨ss’ type.Theorem 2. Let f, g : [a, b] → R be two integrable mappings. Then we have thefollowing Gru¨ss’ type inequality:∣∣∣∣∣∣ 1b− ab∫af (x) g (x) dx− 1b− ab∫af (x) dx · 1b− ab∫ag (x) dx∣∣∣∣∣∣Date: March, 1999.1991 Mathematics Subject Classification. Primary 26D15, 26D20; Secondary 41A05.Key words and phrases. Gru¨ss Inequality, Trapezoid Inequality.12 S.S. DRAGOMIR AND A. MCANDREW≤ 1b− ab∫a∣∣∣∣∣∣f (x)− 1b− ab∫af (y) dy ·g (x)− 1b− ab∫ag (y) dy∣∣∣∣∣∣ dx.(2.1)The inequality (2.1) is sharp.Proof. First of all, let observe thatI :=1b− ab∫af (x)− 1b− ab∫af (y) dy ·g (x)− 1b− ab∫ag (y) dy dx=1b− ab∫af (x) g (x)− g (x) · 1b− ab∫af (y) dy − f (x) · 1b− ab∫ag (y) dy+1b− ab∫af (y) dy · 1b− ab∫ag (y) dy dx=1b− ab∫af (x) g (x) dx− 1b− ab∫ag (x) dx · 1b− ab∫af (y) dy− 1b− ab∫af (x) dx · 1b− ab∫ag (y) dy + (b− a) · 1b− ab∫af (y) dy · 1b− ab∫ag (y) dy=1b− ab∫af (x) g (x) dx− 1b− ab∫ag (x) dx · 1b− ab∫af (x) dx.On the other hand, by the use of modulus properties, we have|I| ≤ 1b− ab∫a∣∣∣∣∣∣f (x)− 1b− ab∫af (y) dy ·g (x)− 1b− ab∫ag (y) dy∣∣∣∣∣∣ dxand the inequality (2.1) is proved.Choosing f (x) = g (x) = sgn(x− a+b2), the equality is satisfied in (2.1).The following corollaries are interesting.Corollary 1. Let f : [a, b] → R be a differentiable mapping on (a, b) having thefirst derivative f ′ : (a, b)→ R bounded on (a, b). Then we have the inequality:∣∣∣∣∣∣f (a) + f (b)2 − 1b− ab∫af (x) dx∣∣∣∣∣∣≤ b− a4maxx∈(a,b)∣∣∣∣f ′ (x)− f (b)− f (a)b− a∣∣∣∣(2.2)TRAPEZOID INEQUALITY AND APPLICATIONS 3Proof. A simple integration by parts gives that:f (a) + f (b)2(b− a)−b∫af (x) dx =b∫a(x− a+ b2)f ′ (x) dx.(2.3)Applying the inequality (2.1) we get that:∣∣∣∣∣∣b∫a1b− a(x− a+ b2)f ′ (x) dx− 1b− ab∫a(x− a+ b2)dx · 1b− ab∫af ′ (x) dx∣∣∣∣∣∣≤ 1b− ab∫a∣∣∣∣∣∣x− a+ b2− 1b− ab∫a(y − a+ b2)dy·f ′ (x)− 1b− ab∫af ′ (y) dy∣∣∣∣∣∣ dx.Asb∫a(x− a+ b2)dx = 0we get∣∣∣∣∣∣b∫a(x− a+ b2)f ′ (x) dx∣∣∣∣∣∣ ≤b∫a∣∣∣∣(x− a+ b2)(f ′ (x)− f (b)− f (a)b− a)∣∣∣∣ dx≤ maxx∈(a,b)∣∣∣∣f ′ (x)− f (b)− f (a)b− a∣∣∣∣b∫a∣∣∣∣(x− a+ b2)∣∣∣∣ dx=(b− a)24maxx∈(a,b)∣∣∣∣f ′ (x)− f (b)− f (a)b− a∣∣∣∣ .(2.4)Now using the equality (2.3), the inequality(2.4) becomes the desired result(2.2) .Corollary 2. Let f : [a, b] → R be a differentiable mapping on (a, b) having thefirst derivative f ′ : (a, b)→ R, q-integrable on (a, b) where p, q > 1, 1p + 1q = 1.Thenwe have the inequality: ∣∣∣∣∣∣f (a) + f (b)2 − 1b− ab∫af (x) dx∣∣∣∣∣∣≤ 12(b− ap+ 1) 1p b∫a∣∣∣∣f ′ (x)− f (b)− f (a)b− a∣∣∣∣q dx1q.(2.5)4 S.S. DRAGOMIR AND A. MCANDREWProof. Using Ho¨lder’s inequality, we have that:b∫a∣∣∣∣(x− a+ b2)(f ′ (x)− f (b)− f (a)b− a)∣∣∣∣ dx≤ b∫a∣∣∣∣x− a+ b2∣∣∣∣p dx1p b∫a∣∣∣∣f ′ (x)− f (b)− f (a)b− a∣∣∣∣q dx1q=(b− a) 1p+12 (p+ 1)1p b∫a∣∣∣∣f ′ (x)− f (b)− f (a)b− a∣∣∣∣q dx1qas a simple computation shows thatb∫a∣∣∣∣x− a+ b2∣∣∣∣p dx =a+b2∫a(a+ b2− x)pdx+b∫a+b2(x− a+ b2)pdx= −(a+b2 − x)p+1p+ 1∣∣∣∣∣a+b2a+(x− a+b2)p+1p+ 1∣∣∣∣∣ba+b2=(b− a)p+1(p+ 1) 2p+1+(b− a)p+1(p+ 1) 2p+1=(b− a)p+1(p+ 1) 2p.Now, using the first part of (2.4) and the identity (2.3), we get the desired result(2.5) .The following result also holds.Corollary 3. Let f : [a, b] → R be a differentiable mapping on (a, b) and supposethat f ′ : (a, b)→ R is integrable on (a, b) .Then we have the inequality:∣∣∣∣∣∣f (a) + f (b)2 − 1b− ab∫af (x) dx∣∣∣∣∣∣ ≤ 12b∫a∣∣∣∣f ′ (x)− f (b)− f (a)b− a∣∣∣∣ dx(2.6)Proof. We haveb∫a∣∣∣∣(x− a+ b2)(f ′ (x)− f (b)− f (a)b− a)∣∣∣∣ dx ≤ maxx∈(a,b)∣∣∣∣x− a+ b2∣∣∣∣×b∫a∣∣∣∣f ′ (x)− f (b)− f (a)b− a∣∣∣∣ dx = b− a2b∫a∣∣∣∣f ′ (x)− f (b)− f (a)b− a∣∣∣∣ dx.Using the first part of (2.4) and the identity (2.3) ,we get the desired result(2.6) .TRAPEZOID INEQUALITY AND APPLICATIONS 53. Applications for some special meansLet us recall some special means we shall use in the sequel:(a) The arithmetic meanA = A (a, b) :=a+ b2, a, b ≥ 0;(b) The geometric meanG = G (a, b) :=√ab, a, b ≥ 0;(c) The harmonic meanH = H (a, b) :=21a +1b, a, b > 0;(d) The logarithmic meanL = L (a, b) :={a if b = ab−aln b−ln a if b 6= a, a, b > 0;(e) The identric meanI = I (a, b) := a if a = b1e(bbaa) 1b−aif a 6= b, a, b > 0;(f) The p−logarithmic meanLp = Lp (a, b) := a if b = a[ bp+1−ap+1(p+1)(b−a)] 1pif b 6= a, a, b > 0where p ∈ R\ {−1, 0} .It is well known thatH ≤ G ≤ L ≤ I ≤ A(3.1)and the mapping Lpis monotonically increasing in p ∈ R with L0 := I and L−1 :=L.I. Now, let consider the inequality∣∣∣∣∣∣f (a) + f (b)2 − 1b− ab∫af (x) dx∣∣∣∣∣∣≤ b− a4maxx∈(a,b)∣∣∣∣f ′ (x)− f (b)− f (a)b− a∣∣∣∣ ,(3.2)where f is as in Corollary1.1. Consider the mapping f : (0,∞)→ R, f (x) = xr, r ∈ R\ {0,−1} . Then for0 < a < b, we havef (a) + f (b)2= A (ar, br) ,1b− ab∫af (x) dx = Lrr (a, b) ,6 S.S. DRAGOMIR AND A. MCANDREWf ′ (x)− f (b)− f (a)b− a = rxr−1 − rLr−1r−1 = r(xr−1 − Lr−1r−1),and by the inequality3.2 we get:|A (ar, br)− Lrr (a, b)| ≤|r| (b− a)4maxx∈(a,b)∣∣xr−1 − Lr−1r−1∣∣(3.3)2. Consider the mapping f : (0,∞) → R, f (x) = 1x . Then for 0 < a < b, wehavef (a) + f (b)2=A (a, b)G2 (a, b),1b− ab∫af (x) dx =1L (a, b),f ′ (x)− f (b)− f (a)b− a = −1x2+1ab=x2 −G2G2x2maxx∈(a,b)∣∣∣∣f ′ (x)− f (b)− f (a)b− a∣∣∣∣ = maxx∈(a,b){∣∣b2 − ab∣∣ab · b2 ,∣∣a2 − ab∣∣ab · a2}=(b− a)abmaxx∈(a,b){1b,1a}=(b− a)a2band by the inequality (3.2)we get∣∣∣∣ AG2 − 1L∣∣∣∣ ≤ (b− a)24aG2which is equivalent to0 ≤ LA−G2 ≤ (b− a)24aL.(3.4)3. Consider the mapping f : (0,∞) → R, f (x) = lnx. Then for 0 < a < b, wehavef (a) + f (b)2= lnG,1b− ab∫af (x) dx = ln I,f ′ (x)− f (b)− f (a)b− a =1x− 1L,maxx∈(a,b)∣∣∣∣f ′ (x)− f (b)− f (a)b− a∣∣∣∣ = 1a − 1L = L− aaLand by the inequality (3.2) we get|lnG− ln I| ≤(L− aaL)TRAPEZOID INEQUALITY AND APPLICATIONS 7which is equivalent to:1 ≤ IG≤ exp(L− aaL)(3.5)II. Now, let consider the inequality:∣∣∣∣∣∣f (a) + f (b)2 − 1b− ab∫af (x) dx∣∣∣∣∣∣≤ 12b∫a∣∣∣∣f ′ (x)− f (b)− f (a)b− a∣∣∣∣ dx(3.6)1. Consider the mapping f : (0,∞) → R f (x) = x2, r ∈ R\ {0,−1} and0 < a < b. Thenb∫a∣∣∣∣f ′ (x)− f (b)− f (a)b− a∣∣∣∣ dx = |r|b∫a∣∣xr−1 − Lr−1r−1∣∣ dx.For simplicity, let assume that r > 1. Thenb∫a∣∣xr−1 − Lr−1r−1∣∣ dx =Lr−1∫a(Lr−1r−1 − xr−1)dx+b∫Lr−1(xr−1 − Lr−1r−1)dx= Lr−1r−1 (Lr−1 − a)−xrr∣∣∣∣Lr−1a+xrr∣∣∣∣bLr−1− (b− Lr−1)Lr−1r−1= Lrr−1 − aLr−1r−1 −Lrr−1 − arr+br − Lrr−1r− (b− Lr−1)Lr−1r−1=br + arr− Lr−1r−1 (a+ b) +2Lrr−1r=2r[A (ar, br)− rLr−1r−1A+ Lrr−1]and by the inequality (3.6) we get0 ≤ A (ar, br)− Lrr (a, b) ≤[A (ar, br)− rLr−1r−1A+ Lrr−1](3.7)orrLr−1r−1A ≤ Lrr (a, b) + Lrr−1 (a, b) .(3.8)Similar results can be obtained for r ≤ 1, r 6= 0,−1.We shall omit the details.2. Consider the mapping f : (a, b)→ R, f (x) = 1x . Then for 0 < a < b we have:b∫a∣∣∣∣x2 −G2G2x2∣∣∣∣ dx = 1G2b∫a∣∣∣∣x2 −G2x2∣∣∣∣ dx=1G2 G∫aG2 − x2x2dx+b∫Gx2 −G2x2dx8 S.S. DRAGOMIR AND A. MCANDREW=1G2[G2x−1−1∣∣∣∣Ga− (G− a) + (b−G)−G2 x−1−1∣∣∣∣bG]=1a2[−G2G+G2a+ b+ a− 2G+ G2b− G2G]=1G2[b+ a− 2G− 2G+G2(a+ bab)]=4G2(A−G) = 4 (A−G)G2and by inequality (3.6) we get:∣∣∣∣ AG2 − 1L∣∣∣∣ ≤ 2 (A−G)G2i.e.,0 ≤ AL−G2 ≤ 2L (A−G)(3.9)or equivalently:2LG ≤ G2 +AL(3.10)which is a very interesting inequality amongst A,L and G.3. Consider the mapping f : (a, b) → R, f (x) = lnx. Then for 0 < a < b, wehave:b∫a∣∣∣∣ 1x − 1L∣∣∣∣ dx =b∫a|x− L|xLdx =L∫a(L− x)xLdx+b∫Lx− LxLdx=1L[L lnx|La − (L− a) + (b− L)− L lnx|bL]=1L[L lnL− L ln a− L+ a+ b− L− L ln b+ L lnL]=1L[2L lnL− L (ln a+ ln b) + a+ b− 2L]and then by the inequality (3.6) we get|lnG− ln I| ≤ 12L[2L lnL− L (ln a+ ln b) + a+ b− 2L]= lnL− ln a+ ln b2+AL− 1 = lnL− lnG+ A− LL= ln[(LG)exp(A− LL)]i.e.,1 ≤ IG≤ LGexp(A− LL)(3.11)which implies1 ≤ IL≤ exp(A− LL).(3.12)TRAPEZOID INEQUALITY AND APPLICATIONS 94. Applications for the trapezoid formulaIn this section we shall assume that f : I ⊆ R → R is a differentiable mappingwhose derivative is satisfying the following condition:|f (b)− f (a)− (b− a) f ′ (x)| ≤ Ω (b− a)2 ,Ω > 0(4.1)for all a, b ∈ I and x between a and b.If f ′ is M−lipschitzian, i.e.,|f ′ (u)− f ′ (v)| ≤M |u− v| ,M > 0then|f (b)− f (a)− (b− a) f ′ (x)| = |f ′ (c)− f ′ (x)| |b− a|≤M |b− a| |c− x| ≤M (b− a)2where c is between a and b, too. Consequently, the mappings having the firstderivative lipschitzian satisfy the condition (4.1).The following trapezoid formula holds.Theorem 3. Let f : [a, b]→ R be a differentiable mapping on (a, b) whose deriva-tive f ′ : (a, b)→ R is satisfying the above condition (4.1)on (a, b) . If Ih : a = x0 <x1 < ... < xn−1 < xn = b is a division of [a, b] and hi = xi+1 − xi, i = 0, ..., n− 1,then we have:b∫af (t) dt = AT,Ih (f) +RT,Ih (f)(4.2)whereAT,Ih (f) =n−1∑i=0f (xi) + f (xi+1)2hi(4.3)and the remainder RT,Ih (f) satisfies the estimation:|RT,Ih (f)| ≤Ω4n−1∑i=0h3i .(4.4)Proof. Applying Corollary 1 on the interval [xi, xi+1] we can write:∣∣∣∣∣∣(xi+1 − xi) f (xi) + f (xi+1)2 −xi+1∫xif (t) dt∣∣∣∣∣∣≤ xi+1 − xi4maxx∈(xi,xi+1)∣∣∣∣f ′ (x)− f (xi+1)− f (xi)xi+1 − xi∣∣∣∣≤ Ω (xi+1 − xi)34i.e., ∣∣∣∣∣∣f (xi) + f (xi+1)2 hi −xi+1∫xif (t) dt∣∣∣∣∣∣ ≤ Ωh3i4for all i = 0, ..., n− 1.10 S.S. DRAGOMIR AND A. MCANDREWSumming the above inequality and using the generalized triangle inequality, weget the approximation (4.2) and the remainder satisfies the estimation (4.4).Remark 1. We have got in this way a trapezoid formula for a class larger thanthe class C2 [a, b] for which the usual trapezoid formula works with the remainderterm satisfying the estimation|RT,Ih (f)| ≤‖f ′′‖∞12n−1∑i=0h3iwhere ‖f ′′‖∞ = supt∈(a,b)|f ′′ (t)| <∞.References[1] MITRINOVIC´, D. S; PEC˘ARIC´, J. E and FINK, A. M.: Classical and New Inequalities inAnalysis, Kluwer Academic Publishers, Dordrecht, 1993.[2] DRAGOMIR, S.S., Some integral inequalities of Gru¨ss type, submitted[3] DRAGOMIR, S.S., Gru¨ss inequality in inner product spaces, submitted[4] DRAGOMIR, S.S., New estimation of the remainder in Taylor’s formula using Gru¨ss’ typeinequalities and applications, Mathematical Inequalities and Applications, (in press).[5] DRAGOMIR, S.S. and FEDOTOV, I., An inequality of Gru¨ss type for Riemann-Stieltjesintegral and applications for special means, Tamkang J. of Maths.,(4)29(1998), 287-292.[6] FEDOTOV, I. and DRAGOMIR, S.S., Another approach to quadrature methods, IndianJournal of Pure and Applied Mathematics, (in press).[7] DRAGOMIR, S.S., A Gru¨ss type integral inequality for mappings of r-Holder’s type andapplications for trapezoid formula, submittedSchool of Communications and Informatics, Victoria University of Technology, POBox 14428, MC Melbourne City, Victoria 8001, AustraliaE-mail address: {sever, amca}@matilda.vu.edu.auURL: http://matilda.vu.edu.au/~rgmia/dragomirweb.html

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On Trapezoid Inequality Via a Grüss Type Result and Applications

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In this paper, we point out a Grüss type inequality and apply it for special means (logarithmic mean, identric mean, etc... ) and in Numerical Analysis in connection with the classical trapezoid formula.\u

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On Trapezoid Inequality Via a Grüss Type Result and Applications

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