Functional inequalities involving numerical differentiation formulas of order two

We write expressions connected with numerical differentiation formulas of order $2$ in the form of Stieltjes integral, then we use Ohlin lemma and Levin-Stechkin theorem to study inequalities connected with these expressions. In particular, we present a new proof of the inequality \begin{equation} \label{Dr} f\left(\frac{x+y}{2}\right)\leq\frac{1}{(y-x)^2}\int_x^y\hspace{-2mm}\int_x^yf\left(\frac{s+t}{2}\right)ds\:dt \leq\frac{1}{y-x}\int_x^yf(t)dt \end{equation} satisfied by every convex function $f:\R\to\R$ and we obtain extensions of \rf{Dr}. Then we deal with nonsymmetric inequalities of a similar form.


Introduction
Writing the celebrated Hermite-Hadamard inequality we can see that (1) is, in fact, an inequality involving two very simple quadrature operators and a very simple differentiation formula. In papers [10] and [11] the quadrature operators occurring in (1) were replaced by more general ones whereas in [8] the middle term from (1) was replaced by more general formulas used in numerical differentiation. Thus inequalities involving expressions of the form where n i=1 a i = 0, α i + β i = 1 and F = f were considered. In the current paper we deal with inequalities for expressions of the form (where = f ), which are used to approximate the second-order derivative of F, and surprisingly, we discover a connection between our approach and some inequality which was considered by Dragomir. First we make the following simple observation.
Remark 1 Let f, F, : [x, y] → R be such that = F, F = f, let n i , m i ∈ N ∪ {0}, i = 1, 2, 3; a i, j ∈ R, α i, j , β i, j ∈ [0, 1], α i, j + β i. j = 1, i = 1, 2, 3; j = 1, . . . , n i ; b i, j ∈ R, γ i, j , δ i, j ∈ [0, 1], γ i, j + δ i, j = 1, i = 1, 2, 3; j = 1, . . . , m i . If the inequality The simplest expression used to approximate the second-order derivative of f is of the form Remark 2 From numerical analysis it is known that This means that for convex g and for G such that G = g we have In this paper we shall obtain some inequalities for convex functions, which do not follow from numerical differentiation results. In order to get such results we shall use Stieltjes integral. In paper [9] it was observed that the classical Hermite-Hadamard inequality (1) easily follows from the following Ohlin lemma Lemma 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.
Ohlin lemma was used also in paper [10]. However, in the present approach (similarly as in [8] and [11]) we are going to use a more general result from [5], (see also [6] Theorem 4.2.7). In this theorem we use the notations from [6].
for all continuous and convex functions f : [a, b] → R it is necessary and sufficient that F 1 and F 2 verify the following three conditions: Proof Let F: [0, 1] → R be such that = F. Now, to prove this proposition it is enough to do the following calculations

Remark 4
Observe that if and f are such as in Proposition 1 then the following equality is satisfied After this observation it turns out that inequalities involving the expression (9) were considered in the paper of Dragomir [3] where (among others) the following inequalities were obtained As we already know (Remark 1) the first one of the above inequalities may be obtained using the numerical analysis results. Now the inequalities from the Dragomir's paper easily follow from Ohlin lemma, but there are many possibilities of generalizations and modifications of inequalities (11). These generalizations will be discussed in the following chapters.

The Symmetric Case
We start with the following remark.
It is impossible to obtain inequalities involving y x f dF * and any of the expressions: for all convex f : [x, y] → R. Without loss of generality we may assume that F * (x) = 0, then from Theorem 1 we have F * (y) = 1 Also from Theorem 1 we get This remark means that in order to get some new inequalities of the Hermite-Hadamard type we have to integrate with respect to functions constructed with use of (at least) two quadratic functions, as it was the case in Proposition 1. Now we may present the main result of this section.
Theorem 2 Let x, y be some real numbers such that x < y and let a ∈ R. Let f, F, : [x, y] → R be any functions such that F = f and = F and let T a f (x, y) Then the following inequalities hold for all convex functions f : if a ≥ 0, then if a ≤ 0, then if a ≤ 2, then if a ≥ 6, then if a ≥ −6, then are not comparable in the class of convex functions.
Proof In view of Remark 1 we may restrict ourselves to the case x = 0, y = 1. Take a ∈ R, let f : [0, 1] :→ R be any convex function and let F, First we shall prove that T a f (0, 1) 1]. Then the functions F 1 , F 2 have exactly one crossing point (at 1 2 ) and Moreover, if a > 0 then the function F 1 is convex on the interval (0, 1 2 ) and concave on ( 1 2 , 1). Therefore, it follows from Ohlin lemma that for a > 0 we have which,in view of Remark 1, yields (12), and for a < 0 the opposite inequality is satisfied which gives (13). Take It is easy to check that for a ≤ 2 we have 2 , 1 and this means that from Ohlin lemma we get (14). Suppose that a > 2. Then there are three crossing points of functions F 1 and F 3 : is increasing on intervals [0, x 0 ], [ 1 2 , x 1 ] and decreasing on [x 0 , 1 2 ] and on [x 1 , 1]. This means that ϕ takes its absolute minimum at 1 2 . As it is easy to calculate ϕ 1 2 ≥ 0 if a ≥ 6 which, in view of Theorem 1, gives us (15).
To see that for a ∈ (2, 6) expressions T a f (x, y) and f x+y 2 are not comparable in the class of convex functions it is enough to observe that in this case ϕ(x 0 ) > 0 and ϕ 1 2 < 0. Now let Similarly as before, if a ≥ −2 then we have F 1 (t) ≥ F 4 (t) for t ∈ 0, 1 2 , F 1 (t) ≤ F 4 (t) for t ∈ 1 2 , 1 , i.e., there is only one crossing point of these functions and (16) is obvious. However, for a ∈ (−2, −6] we have and therefore, in view of Theorem 1, we still have (16). In the case a < −6 inequality (21) is no longer true which means that expressions T a f (x, y) and f (x)+ f (y) 2 are not comparable in the class of convex functions.
This theorem provides us with a full description of inequalities which may be obtained using Stieltjes integral with respect to a function of the form (17). Some of the obtained inequalities are already known. For example from (12) and (13) we obtain the inequality whereas from (14) for a = 2 we get the inequality However, inequalities obtained for "critical" values of a, i.e., −6, 6. are here particularly interesting. In the following corollary we explicitly write these inequalities.

Corollary 1 For every convex function f : [x, y] → R the following inequalities are satisfied
Remark 6 In the paper [4] S.S. Dragomir and I. Gomm obtained the following inequality Inequality (23) from Corollary 1 is stronger than (24). Moreover, as it was observed in Theorem 2 inequalities (22) and (23) cannot be improved, i.e., the inequality for λ > 3 4 is not satisfied by every convex function f : [x, y] → R and the inequality In Corollary 1 we obtained inequalities for the triples: In the next remark we present an analogous result for the expressions Remark 7 Using the functions: F 1 defined by (10) and F 5 given by we can see that for all convex functions f : [x, y] → R. Moreover, it is easy to see that the above inequality cannot be strengthened which means that the inequality where a, b ≥ 0, 2a + b = 1, is not satisfied by all convex functions f if a < 1 6 .

The Non-symmetric Case
In this part of the paper we shall obtain inequalities for f (αx + (1 − α)y) and for α f (x) + (1 − α) f (y), when α is not necessarily equal to 1 2 .
Now, in contrast to the symmetric case (Remark 5), it is possible to prove inequalities using just one quadratic function but before we do this we shall present a non-symmetric version of Hermite-Hadamard inequality involving only the primitive function of f. Proposition 2 Let x, y be some real numbers such that x < y and let α ∈ [0, 1].
then the following inequality is satisfied: Proof As usually, the proof will be done on the interval [0, 1]. Define the functions F 6 , F 7 , F 8 : [0, 1] → R by the following formulas: and We have: then the following conditions hold true: and if α ∈ 1 3 , 1 2 ∪ 1 2 , 2 3 then S 1 α f (x, y) and S 2 α f (x, y) are incomparable in the class of convex functions.
Proof Take and let F 6 , F 7 , F 8 be defined so as in Proposition 2. Then we have It is easy to see that the functions F 9 , F 7 have exactly one crossing point; thus, we have which gives us (30). Now, assume that α ∈ 1 3 , 2 3 , then the function F 9 is increasing and, consequently, Thus for every convex function f we have which yields (31). Let now α < 1 3 .Then the function F 9 is decreasing on some interval [0, d] and increasing on [d, 1]. Observe that from the equality,  are incomparable in the class of convex functions (as claimed). The reasoning in the case α > 2 3 is similar. Now we shall prove the inequality (32). If α ∈ 0, 1 3 ∪ [ 2 3 , 1), then F 9 (0) ≤ α 1−α and F 9 (1) ≤ 1−α α , this means that the functions F 9 and F 8 have only one crossing point, and therefore, we have (32).

Remark 8 In order to obtain inequalities involving expressions of the form
, the functions of the form , must be used. Since the description of all possible cases in Theorem 2 was already quite complicated, we shall not present these inequalities in details here.

Remark 9
It is possible to use methods developed in this paper to get inequalities involving longer expressions of the form (3). In order to do that it is necessary to use more than two quadratic functions. For example considering the function and using Levin-Stechkin theorem, we get the following inequality we have 1 0 t 2 dF 9 (t) = 1 0 t 2 dF 6 (t). Moreover, as it was mentioned in the proof of Theorem 3, the functions F 9 , F 6 have in this case two crossing points. This implies (see [2,9]) that the inequalities: x + 3 + √ 3 6 y and S 2 3+ √ 3 6 (x, y) ≤ f 3 + √ 3 6 x + 3 − √ 3 6 y are satisfied by all 2-convex functions f : [x, y] → R

Remark 11
It is easy to see that all inequalities obtained in this paper in fact characterize convex functions (or 2-convex functions). This is a consequence of results contained in paper [1].
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