Inequalities of the Hermite–Hadamard Type Involving Numerical Differentiation Formulas

We observe that the Hermite–Hadamard inequality written in the form fx+y2≤F(y)-F(x)y-x≤f(x)+f(y)2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f\left(\frac{x+y}{2}\right)\leq\frac{F(y)-F(x)}{y-x}\leq\frac{f(x)+f(y)}{2}$$\end{document}may be viewed as an inequality between two quadrature operators fx+y2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${f\left(\frac{x+y}{2}\right)}$$\end{document}f(x)+f(y)2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\frac{f(x)+f(y)}{2}}$$\end{document} and a differentiation formula F(y)-F(x)y-x\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\frac{F(y)-F(x)}{y-x}}$$\end{document}. We extend this inequality, replacing the middle term by more complicated ones. As it turns out in some cases it suffices to use Ohlin lemma as it was done in a recent paper (Rajba, Math Inequal Appl 17(2):557–571, 2014) however to get more interesting result some more general tool must be used. To this end we use Levin–Stečkin theorem which provides necessary and sufficient conditions under which inequalities of the type we consider are satisfied.


Introduction
We shall obtain some class of inequalities of the Hermite-Hadamard type. First we write the classical Hermite-Hadamard inequality [see [2] for many generalizations and applications of (1)]. Now, let us write (1) in the form This inequality is, clearly, satisfied by every convex function f and its primitive function F. However (2) may be viewed as an inequality involving two types of expressions used, respectively, in numerical integration and differentiation. Namely f x+y 2 and f (x)+f (y) 2 are the simplest quadrature formulas used to approximate the definite integral, whereas F (y)−F (x) y−x is the simplest expression used to approximate the derivative of F. Moreover, as it is known from numerical analysis, if F = f then the following equality is satisfied for some ξ ∈ (x − h, x + h). This means that (3) provides an alternate proof of (1) (for twice differentiable f ). This new formulation of the Hermite-Hadamard inequality inspires us to replace the middle term of Hermite-Hadamard inequality by more complicated expressions than those used in (1). In the paper [7] all numbers a, α, β ∈ [0, 1] such that for all convex functions f the inequality were replaced by longer ones while the integral mean remained unchanged. In the current paper we do the opposite job, namely, we are going to prove inequalities of the type α, β ∈ (0, 1) and a 1 + a 2 + a 3 + a 4 = 0.

Preliminaries
In recent papers [6] and [7] Ohlin lemma on convex stochastic ordering was used to obtain inequalities of the Hermite-Hadamard type. In the current paper we are going to work in a similar spirit. Thus now we cite this lemma. Vol. 67 (2015) Inequalities of the Hermite-Hadamard Type 405 Lemma 1. (Ohlin [5]) Let X 1 , X 2 be two random variables such that EX 1 = EX 2 and let F 1 , F 2 be their cumulative distribution functions. If F 1 , F 2 satisfy for some x 0 the following inequalities for all continuous and convex functions f : R → R.
However, in the present approach we are going to use a result from [3] (see also [4,Theorem 4.2.7]).
for all continuous and convex f , it is necessary and sufficient that F 1 and F 2 verify the following three conditions: Remark 1. Observe that if measures μ 1 , μ 2 corresponding to the random variables occurring in Ohlin's lemma are concentrated on the interval [x, y] then Ohlin's lemma is an easy consequence of Theorem 1. Indeed, μ 1 , μ 2 are probabilistic measures thus we have F 1 (x) = F 2 (x) = 0 and F 1 (y) = F 2 (y) = 1. Moreover EX 1 = EX 2 yields (8) and, from the inequalities (4), we get (7). Now we shall use Theorem 1 to make an observation which is more general than Ohlin lemma and concerns the situation when functions F 1 , F 2 have more crossing points than one. First we need the following definition.
We say that the pair (F 1 , F 2 ) crosses n−times (at points x 1 , . . . , x n ) if the inequalities where i is even, are satisfied and i= 0, 1, . . . , n.

Further, let numbers A i be defined by the following formulas
for odd i and for even i. Then the following assertions hold true.
is not satisfied by all continuous and convex f.

(ii) If n is odd then inequality (10) is satisfied for all continuous and convex f if and only if the following inequalities are satisfied
. . .
Proof. First we prove (i). For an indirect proof assume that n is even and that (10) is satisfied for all convex f. Then from Theorem 1 (and from the definition of A 0 ) we get which is a contradiction with (7).

Vol. 67 (2015)
Inequalities of the Hermite-Hadamard Type 407 Now we shall show the condition (ii) to this end assume that n is odd and that (10) is satisfied for all convex f. Then Assume, on the other hand that all the above inequalities hold true and take To finish the proof it is enough to apply the Levin-Stečkin theorem.

Results and Applications
In this part of the paper we shall work with expressions of the type Therefore, now we make the following observation, l 1 stands here for the one-dimensional Lebesgue measure. where Proof. Let x j := α j x + (1 − α j )y, j = 1, . . . , n, and denote by Then, by the definition of μ, we obtain Next proposition will show that, in order to get some inequalities of the Hermite-Hadamard type, we have to use sums containing more than three summands. Proof. Using Proposition 1, we can see that Now, let then F 1 lies strictly above or below F 2 (on [x, y]). This means that But, on the other hand, if and This, together with (13), shows that neither Remark 3. Observe that the assumptions α 1 = 1 and α 3 = 0 are essential. For example, it follows from Ohlin lemma that inequality ).

Proposition 3. Let f : [x, y] → R be a continuous and convex function and let F be its antiderivative. Then f and F satisfy the following inequality
Proof. For the sake of simplicity we shall assume that x = 0 and y = 1. Then, according to Proposition 1 Then (F 1 , F 2 ) crosses only once (at 1 3 ) and we have

Vol. 67 (2015)
Inequalities of the Hermite-Hadamard Type 411 As we can see, it is possible to prove some results for three point formulas but to get more interesting results we shall use longer expressions. For example, from numerical analysis, it is known that for F = f we have It is natural to ask if the expression occurring at the right-hand side satisfies inequalities of the Hermite-Hadamard type. First we need some auxiliary results.

Lemma 3. If any of the inequalities
is satisfied for all continuous and convex functions f : and Proof. Taking x = 0, y = 1 and, using Proposition 1, we can see that Now we define F 1 , F 3 and F 4 by the formulas (12), (14) and (15), respectively. Then inequalities (18) and (19) may be written in the form.
This means that, if for example inequality (18) is satisfied then we must have     Let α i , i = 1, . . . , 4 satisfy 1 such that a 1 + a 2 + a 3 + a 4 = 0 and let equalities (20) and (21 and F 2 is the distribution function of a measure which is uniformly distributed in the interval [x, y] then (F 1 , F 2 ) crosses exactly once.
Proof. From (20) we can see that F 1 (x) = F 2 (x) = 0 and F 1 (y) = F 2 (y) = 1. Note that, in view of Proposition 1 the graph of the restriction of F 1 to the interval [x, y] consists of three segments. Therefore F 1 and F 2 cannot have more than one crossing point. On the other hand if graphs F 1 and F 2 do not cross then Vol. 67 (2015) Inequalities of the Hermite-Hadamard Type 413 Proof. We shall prove the first assertion. Other proofs are similar and will be omitted. It is easy to see that if inequalities which we consider are satisfied by every continuous and convex function defined on the interval [0, 1] then they are true for every continuous and convex function on a given interval [x, y]. Therefore we assume that x = 0 and y = 1. Let F 1 be such that (11) is satisfied and let F 2 be the distribution function of a measure which is uniformly distributed in the interval [0, 1]. From Proposition 1 and Remark 2 we can see that the graph of F 1 consists of three segments and, since a 1 > −1, the slope of the first segment is smaller than 1, i.e. F 1 lies below F 2 on some righthand neighborhood of x. In view of the Proposition 4, this means that the assumptions of Ohlin lemma are satisfied and we get our result from this lemma.
Now we shall present examples of inequalities which may be obtained from this theorem. Example 1. Using (i), we can see that the inequality is satisfied for every continuous and convex f and its antiderivative F.
Example 2. Using (ii), we can see that the inequality is satisfied by every continuous and convex function f and its antiderivative F.
Example 3. Using (iii), we can see that the inequality is satisfied by every continuous and convex function f and its antiderivative F.
Example 4. Using (iv), we can see that the inequality is satisfied by every continuous and convex function f and its antiderivative F. In all cases considered in the above theorem we used only Ohlin lemma. Using Lemma 2, it is possible to obtain more subtle inequalities. However (for the sake of simplicity) in the next result we shall restrict our considerations to expressions of the simplified form. Note that the inequality between f x+y 2 and expressions which we consider is a bit unexpected. Theorem 3. Let α ∈ 0, 1 2 let a, b ∈ R and let inequalities (20) and (21) be satisfied.
(i) if a > 0 then inequality is satisfied by every continuous and convex f and its antiderivative F if and only if (ii) if a < −1 and a 1 + a 2 > 0 then inequality is satisfied by every continuous and convex f and its antiderivative F if and only if Proof. We shall prove the assertion (i). The proof of (ii) is similar and will be omitted. Similarly as before we assume without loss of generality that x = 0, y = 1 and let F 1 be such that further let F 4 be given by (15). Then it is easy to see that (F 1 , F 4 ) crosses three times: at (1−α)b a+b , at 1 2 and at a+αb a+b , We are going to use Lemma 10. Since from (21) we know that it suffices to check that A 0 ≥ A 1 if and only if inequality (22) is satisfied. Since F 4 (x) = 0, for x ∈ 0, 1 2 , we get

+ 2F (y)
y − x ≤ f (x) + f (y) 2 is satisfied for every continuous and convex f and its antiderivative F. Remark 5. As it is known from the paper [1], if a continuous function satisfies inequalities of the type which we have considered then such function must be convex.
Therefore inequalities obtained in this paper characterize convex functions (in the class of continuous functions).
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