We give the best possible global bounds for a form of discrete

Jensen’s inequality. By some examples its fruitfulness is shown

Simic, Slavko

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Another converse of Jensen’s inequalitySlavko SimicAbstract. We give the best possible global bounds for a form of discreteJensen’s inequality. By some examples its fruitfulness is shown.1. IntroductionThroughout this paper x = {xi} represents a finite sequence of real numbersbelonging to a fixed closed interval I = [a, b], a < b and p = {pi},∑pi = 1 is apositive weight sequence associated with x.If f is a convex function on I, then the well known Jensen’s inequality ([1],[3])asserts that(1.1) 0 6∑pif(xi)− f(∑pixi).There are many important inequalities which are particular cases of Jensen’s in-equality such as the weighted A − G − H inequality, Cauchy’s inequality, the KyFan and Holder inequalities etc.One can see that the lower bound zero is of global nature since it does notdepend on p,x but only on f and the interval I whereupon f is convex.We give in [1] an upper global bound (that is, depending on f and I only)which happens to be better than already existing ones. Namely, we proved that(1.2) (0 6)∑pif(xi)− f(∑pixi) 6 Tf (a, b),withTf (a, b) := maxp[pf(a) + (1− p)f(b)− f(pa+ (1− p)b)].Note that, for a (strictly) positive convex function f , Jensen’s inequality canbe stated in the form(1.3) 1 6∑pif(xi)f(∑pixi).2000 Mathematics Subject Classification. Primary 26B25, 26D20.Key words and phrases. Jensen’s discrete inequality, Stolarsky means, best possible globalbounds.12 SIMICIt is not difficult to prove that 1 is the best possible global lower bound forJensen’s inequality written in the above form. Our aim here is to find the bestpossible global upper bound for this inequality. We shall show with examples thatin this way one may obtain at once converses of some important inequalities.2. ResultsOur main result is contained in the followingTheorem 2.1. Let f be a (strictly) positive, twice continuously differentiablefunction on I := [a, b], xi ∈ I and 0 6 p, q 6 1, p+ q = 1. We havei. if f is (strictly) convex function on I, then(2.1) 1 6∑pif(xi)f(∑pixi)6 maxp[pf(a) + qf(b)f(pa+ qb)] := Sf (a, b);ii. if f is (strictly) concave function on I, then(2.2) 1 6 f(∑pixi)∑pif(xi)6 maxp[f(pa+ qb)pf(a) + qf(b)] := S′f (a, b).Both estimations are independent of p.The next assertion shows that Sf (a, b)(S′f (a, b)) exists and is unique.Theorem 2.2. There is unique p0 ∈ (0, 1) such that(2.3) Sf (a, b) =p0f(a) + (1− p0)f(b)f(p0a+ (1− p0)b)Of particular importance is the followingTheorem 2.3. The expression Sf (a, b) represents the best possible global upperbound for Jensen’s inequality written in the form (1.3).3. ProofsWe shall give proofs of the above assertions related to the first part of Theorem2.1. Proofs concerning concave functions go along the same lines.Proof of Theorem 2.1. We apply the method already shown in [1]. Namely,since a 6 xi 6 b, there is a sequence ti ∈ [0, 1] such that xi = tia+ (1− ti)b.Hence,∑pif(xi)f(∑pixi)=∑pif(tia+ (1− ti)b)f(∑pi(tia+ (1− ti)b)) 6f(a)∑piti + f(b)(1−∑piti)f(a∑piti + b(1−∑piti).Denoting∑piti := p, 1−∑piti := q; p, q ∈ [0, 1], we get∑pif(xi)f(∑pixi)=pf(a) + qf(b)f(pa+ qb)6 maxp[pf(a) + qf(b)f(pa+ qb)] := Sf (a, b)ANOTHER CONVERSE 3Proof of Theorem 2.2. DenoteF (p) :=pf(a) + qf(b)f(pa+ qb).We get F ′(p) = g(p)/f2(pa+ qb) withg(p) := (f(a)− f(b))f(pa+ qb)− (pf(a) + qf(b))f ′(pa+ qb)(a− b).Also,g′(p) = −(a− b)2(pf(a) + qf(b))f ′′(pa+ qb),andg(0) = f(b)(f(a)− f(b)− f ′(b)(a− b)); g(1) = −f(a)(f(b)− f(a)− f ′(a)(b− a)).Since f is strictly convex on I and pa+qb ∈ I, we conclude that g(p) is monotonedecreasing on [0, 1] with g(0) > 0, g(1) < 0. Therefore there exists unique p0 ∈ (0, 1)such that g(p0) = F ′(p0) = 0. Also F ′′(p0) = g′(p0)/f2(p0a + q0b) < 0 and theproof is done. Proof of Theorem 2.3. Let Rf (a, b) be an arbitrary global upper bound.By definition, the inequality ∑pif(xi)f(∑pixi)6 Rf (a, b)holds for arbitrary p and xi ∈ [a, b].In particular, for #x = 2, x1 = a, x2 = b, p1 = p0 we obtain that Sf (a, b) 6Rf (a, b) as required. 4. ApplicationsIn the sequel we shall give some examples to demonstrate the fruitfulness ofthe assertions from Theorem 2.1. Since all bounds will be given as a combinationof means from the Stolarsky class, here is its definition.Stolarsky (or extended) two-parametric mean values are defined for positivevalues of x, y as(4.1) Er,s(x, y) :=(r(xs − ys)s(xr − yr))1/(s−r), rs(r − s)(x− y) 6= 0.E means can be continuously extended on the domain{(r, s;x, y)|r, s ∈ R;x, y ∈ R+}by the followingEr,s(x, y) =(r(xs−ys)s(xr−yr))1/(s−r), rs(r − s) 6= 0exp(− 1s + xs log x−ys log yxs−ys), r = s 6= 0(xs−yss(log x−log y))1/s, s 6= 0, r = 0√xy, r = s = 0,x, x = y > 0,and in this form are introduced by Keneth Stolarsky in [2].4 SIMICMost of the classical two variable means are special cases of the class E. Forexample, E1,2 = x+y2 is the arithmetic mean A(x, y), E0,0 =√xy is the geo-metric mean G(x, y), E0,1 = x−ylog x−log y is the logarithmic mean L(x, y), E1,1 =(xx/yy)1x−y /e is the identric mean I(x, y), etc. More generally, the r-th powermean(xr+yr2)1/ris equal to Er,2r.Example 1 Taking f(x) = 1/x, after an easy calculation it follows thatS1/x(a, b) = (A(a, b)/G(a, b))2. Therefore we obtain at onceProposition 1 If 0 < a 6 xi 6 b, then the inequality(4.2) 1 6 (∑pixi)(∑ pixi) 6 (a+ b)24ab,holds for an arbitrary weight sequence p.This is the extended form of Schweitzer inequality.Example 2 For f(x) = x2 we get that the maximum of F (p) is attained atthe point p0 = ba+b .HenceProposition 2 If 0 < a 6 xi 6 b, then the following means inequality(4.3) 1 6√∑pix2i∑pixi6 A(a, b)G(a, b)holds for an arbitrary weight sequence p.Specializing the above inequality, that is, putting pi = u2i /∑i u2i , xi = vi/uiand noting that 0 < u 6 ui 6 U, 0 < v 6 vi 6 V imply a = v/U 6 xi 6 V/u = b,we obtain a converse of the well-known Cauchy’s inequality.Proposition 3 If 0 < u 6 ui 6 U, 0 < v 6 vi 6 V , then(4.4) 1 6∑u2i∑v2i(∑uivi)26(√UV/uv +√uv/UV2)2.In this form the Cauchy’s inequality was stated in [3, p.80].Note that the same result can be obtained from Schweitzer’s inequality (4.2)taking pi = uivi/∑i uivi, xi = ui/vi.Example 3 Let f(x) = xα, 0 < α < 1. Since in this case f is a concavefunction, applying the second part of Theorem 2.1, we getProposition 4 If 0 < a 6 xi 6 b, then(4.5) 1 6 (∑pixi)α∑pixαi6(Eα,1(a, b)E1−α,1(a, b)G2(a, b))α(1−α),independently of p.In the limiting cases we obtain two important converses. Namely, writing (4.5)as(4.6) 1 6∑pixi(∑pixαi )1/α6(Eα,1(a, b)E1−α,1(a, b)G2(a, b))1−α,and, letting α→ 0+, the converse of generalized A−G inequality arises.ANOTHER CONVERSE 5Proposition 5 If 0 < a 6 xi 6 b, then(4.7) 1 6∑pixi∏xpii6 L(a, b)I(a, b)G2(a, b).Note that the right-hand side of (4.7) is exactly the Specht constant (cf [1]).Analogously, writing (4.5) in the form(4.8) 1 6( (∑ pixi)α∑pixαi) 11−α 6(Eα,1(a, b)E1−α,1(a, b)G2(a, b))α,and taking the limit α→ 1−, we getProposition 6 If 0 < a 6 xi 6 b, then(4.9) 0 6∑pixi log xi −∑pixi log(∑pixi)∑pixi6 log L(a, b)I(a, b)G2(a, b).Finally, putting in (4.5) pi = vi/∑i vi, xi = ui/vi, α = 1/p, 1 − α = 1/q, weobtain the converse of discrete Ho¨lder’s inequality.Proposition 7 If p, q > 1, 1p +1q = 1; 0 < a 6uivi6 b, then(4.10) 1 6 (∑ui)1/p(∑vi)1/q∑u1/pi v1/qi6(E1/p,1(a, b)E1/q,1(a, b)G2(a, b)) 1pqIt is interesting to compare (4.10) with the converse of Ho¨lder’s inequality forintegral forms (cf [4]).References[1] S. Simic, On an upper bound for Jensen’s inequality, J. Inequal. Pure Appl. Math. 10/2Article 60 (2009), 5pp.[2] K. B. Stolarsky, Generalizations of the logarithmic mean, Math. Mag. 48/2 (1975), pp. 87-92.[3] 3) G. Polya, G. Szego, Aufgaben und Lehrsatze aus der Analysis, Springer-Verlag (1964).[4] S. Saitoh, V. K. Tuan, M. Yamamoto, Reverse weighted Lp norm inequalities in convolutions,J. Inequal. Pure Appl. Math. 1/1, Article 7 (2000), 7pp.Mathematical Institute SANU, Kneza Mihaila 36, 11000 Belgrade, SerbiaE-mail address: ssimic@turing.mi.sanu.ac.rs

Another Converse of Jensen's Inequality

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