Ohlin’s lemma and some inequalities of the Hermite–Hadamard type

Using the Ohlin lemma on convex stochastic ordering we prove inequalities of the Hermite–Hadamard type. Namely, we determine all numbers a,α,β∈[0,1]\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${a,\alpha,\beta\in[0,1]}$$\end{document} such that for all convex functions f the inequality af(αx+(1-α)y)+(1-a)f(βx+(1-β)y)≤1y-x∫xyf(t)dt\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$af(\alpha x+(1-\alpha )y)+(1-a)f(\beta x+(1-\beta) y)\leq \frac{1}{y-x} \int\limits_{x}^yf(t)dt$$\end{document}is satisfied and all a,b,c,α∈(0,1)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${a,b,c,\alpha\in(0,1)}$$\end{document} with a + b + c = 1 for which we have af(x)+bf(αx+(1-α)y)+cf(y)≥1y-x∫xyf(t)dt\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$af(x)+bf(\alpha x+(1-\alpha)y)+cf(y)\geq\frac{1}{y-x} \int\limits_{x}^yf(t)dt$$\end{document}.


Introduction
Inspired by the paper [8] we use the Ohlin lemma to prove some new inequalities of the Hermite-Hadamard type. Namely we find all numbers a, α, β ∈ [0, 1] such that the inequality is satisfied for all convex functions f : [x, y] → R. In the second part we deal with the second inequality of the Hermite-Hadamard type We also show that some inequalities which may be found in literature (with quite long proofs) are particular cases of our results.
For the sake of completeness we cite the main tool which we will use throughout the whole paper. Lemma 1.1. (Ohlin [7]) Let X 1 , X 2 be two random variables such that EX 1 = EX 2 and let F 1 , F 2 be their distribution functions. If F 1 , F 2 satisfy for some x 0 the following inequalities for all continuous and convex functions f : R → R.
may easily be obtained with the use of the Ohlin lemma. Indeed, let F 1 be the distribution function of δ x+y 2 , F 2 the distribution function of the measure which is equally distributed in [x, y] and let F 3 be the distribution function of δx+δy 2 . Then it is easy to see that pairs (F 1 , F 2 ) and (F 2 , F 3 ) satisfy the assumptions of the Ohlin lemma and, using (1.2), we get (1.3).
As we can see, the Ohlin lemma is a strong tool, however, it is worth noticing that in the case of inequalities of the type which we consider the distribution functions involved cross more than once. Therefore a simple application of the Ohlin lemma is impossible and some additional idea is needed.

Results
First we prove the following simple lemma.
for all continuous and convex functions f : R → R then EX 1 = EX 2 .
Proof. If Ef (X 1 ) ≤ Ef (X 2 ) is satisfied for all continuous and convex functions then it is satisfied for f (x) = x which yields EX 1 ≤ EX 2 and for g(x) = −x gives us EX 1 ≥ EX 2 .
Vol. 89 (2015) Ohlin's lemma and some inequalities 917 Now, observe that if we want inequality (1.1) to be satisfied then we cannot use the endpoints of the interval at the left-hand side of (1.1).
Remark 2.2. If inequality (1.1) with some a ∈ (0, 1), α, β ∈ [0, 1], α > β is satisfied for all x, y ∈ R and all continuous and convex functions f : [x, y] → R then α, β ∈ (0, 1). Indeed, suppose that α = 1, take x = 0, y = 1 and Then f is convex and This function is convex and using (1.1) for f 1 , we obtain this inequality for f. and one of the following conditions holds true: Proof. In view of Remark 2.3, we may assume that x = 0, y = 1. Let X 1 be the random variable such that and let F 1 be its distribution function. Further let X 2 be such that Assume that (1.1) is satisfied for all continuous and convex functions f : which means that we have proved (2.1). We consider three cases. If a + α ≤ 1 then together with (2.1) this means that our assertion follows from the Ohlin lemma. Similarly, if a + β ≥ 1 then and, also in this case, we get the inequality (1.1) using the Ohlin lemma. Now we pass to the most interesting case. If a + α > 1 and a + β < 1 then functions F 1 , F 2 cross three times and it is no longer possible to use the Ohlin lemma.
Assume first that a + 2α = 2. Then we define functions G 1 and H 1 by the following formulas Then pairs G 1 , F 2 and H 1 , F 2 satisfy the assumptions of the Ohlin lemma thus we obtain for every convex function f Using the above inequalities we arrive at which means that the proof in this case is finished.
On the other hand, since a < 1 − β We have proved that g(1 − β) < (1 − β)(1 − a) < g (1), which means that there exists t 1 ∈ (1 − β, 1) such that (2.4) is satisfied. Now, using this t 1 , we shall define functions G 1 , H 1 and H 2 by the following formulas: It is easy to see that pairs (G 1 , F 2 ) and (H 1 , H 2 ) satisfy the assumptions of the Ohlin lemma. Indeed, using (2.4), we may write This means that for both pairs we may use the Ohlin lemma, getting for every convex f the following inequalities Now from (2.5) and (2.6) we get Vol. 89 (2015) Ohlin's lemma and some inequalities 921 From the above inequalities we get To finish the proof it remains to show that, in the case a+2α > 2, inequality (1.1) is not satisfied for some convex function h. To this end it suffices to take We get Remark 2.5. If we take a = 1 2 in Theorem 2.4 then we get the inequality proved in [6] which may be found also in [5] [inequality (2.73)].
On the other hand, considering the particular case α = a 2 , we obtain an inequality proved in [4]. Now we turn our attention to the second of the Hermite-Hadamard inequalities. It is easy to see that expressions of the form for all convex f (if α, β ∈ (0, 1)). Therefore we shall consider expressions of the form where a, b, c, α ∈ (0, 1), a + b + c = 1. One of the examples is the inequality (satisfied for all continuous and convex f ) which was proved in [2], see also [3]. Therefore now we formulate the following result.

922
T. Szostok AEM Theorem 2.6. Numbers a, b, c, α ∈ (0, 1), such that a + b + c = 1 satisfy the inequality for all x, y ∈ R and all continuous and convex functions f : and one of the following conditions holds true: Proof. So as in the proof of Theorem 2.4 we assume that x = 0, y = 1. Let X 1 be the random variable such that and let F 1 be its distribution function. Further let X 2 be such that and let F 2 be its distribution function. Equality (2.9) is an easy consequence of Lemma 2.1. The proof in cases (i) and (ii) is very similar to the respective part of the proof of Theorem 2.4. Thus we assume that (iii) is satisfied. Moreover we assume that the third inequality occurring in (iii) is strict (the equality case is similar to that of Theorem 1). We shall show that there exists t 1 ∈ (a + b, 1) such that To this end we define Then, since 2a + α > 1, On the other hand, we have Vol. 89 (2015) Ohlin's lemma and some inequalities 923 further from (2.9) we obtain (2.12) From (2.11) and (2.12) we get which means that, in particular, and, consequently, On the other hand we have The above two equalities together with (2.13) give us i.e.
Thus the existence of t 1 such that (2.10) is satisfied has been proved.

924
T. Szostok AEM Using this t 1 , we define functions: and Then pairs (G 2 , G 1 ) and (F 2 , H 1 ) satisfy the assumptions of the Ohlin lemma. Indeed, we have and, in view of (2.10) Using (2.10) once more, we obtain Vol. 89 (2015) Ohlin's lemma and some inequalities 925 This means that we may use the Ohlin lemma, getting for every convex function f : [0, 1] → R the following inequalities: (1) and The above inequalities give us the inequality as claimed.
It is easy to show that in the case 2a + α < 1 there exists a convex function f such that inequality (2.8) is not satisfied. This observation finishes the proof.
Note that the original Hermite-Hadamard inequality consists of two parts. We treated these cases separately. However it is possible to formulate a result containing both inequalities. To prove this corollary it suffices to check that the constants occurring at the right-hand side satisfy the assumptions of Theorem 2.6.

926
T. Szostok AEM Remark 2.8. As it is known from the paper [1], if a continuous function satisfies inequalities of the type which we have considered then such a function must be convex. Therefore inequalities obtained in this paper characterize convex functions (in the class of continuous functions).
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