On the Integral Representation and the Raşa, Jensen and Hermite–Hadamard Inequalities for Box-Convex Functions

We give the integral representation of Popoviciu’s (Mathematica 8:1–85, 1934) box-(m, n)-convex functions. Based on this integral representation, we obtain a characterization of box-(m, n)-convex orders, which we then use in the proofs of the Raşa type, the Hermite–Hadamard type and the Jensen inequalities for box-(m, n)-convex functions as well as for strongly box-(m, n)-convex functions. We extend the results obtained by Gal and Niculescu (J Math Anal Appl 479:903–925, 2019), and by Gavrea and Gavrea (Mediterr J Math 19:18–33, 2022).


Introduction
Hopf [8] and Popoviciu [13,14] introduced the notion of higher order convex functions based on the so called divided differences. Let n ≥ 1. Given a function of one real variable on an interval I and a system x 0 , x 1 , . . . , x n+1 of pairwise distinct points of I the divided differences of order 0, 1, . . . , n + 1 are respectively defined by the formulas The function f (x) is called n-convex (n-concave), if [x 0 , x 1 , . . . , x n+1 ; f ] ≥ 0 (≤ 0) for any pairwise distinct points x 0 , x 1 , . . . , x n+1 . The function f is convex when all divided differences of order two are nonnegative for all systems of pairwise distinct points. Proposition 1. [13] A function f (x) is n-convex if and only if its derivative f (n−1) (x) exists and is convex (with the convention f (0) (x) = f (x)).
If f = f (x, y) is a function defined on a rectangle I ×J and x 0 , x 1 , . . . , x m are pairwise distinct points of I and y 0 , y 1 , . . . , y n are pairwise distinct points of J, one defines the divided double difference of order (m, n) by the formula This paper is inspired by the results of two recent papers: by Gal and Niculescu [3] and by Gavrea and Gavrea [5,6].
In Sect. 2, we give the integral representation of box-monotone functions, without any additional assumptions about their differentiability.
In Sect. 3, we give necessary and sufficient conditions for probability measures satisfying Raşa type inequality for box-monotone functions.
In Sect. 4, we give the integral representation and several characterizations of box-(m, n)-convex functions.
In Sect. 5, we introduce a definition of box-(m, n)-convex orders. Based on the integral representation, we obtain a characterization of box-(m, n)-convex orders, which will be used in Sects. 6 and 7.
In Sect. 6, we obtain necessary and sufficient conditions for probability measures satisfying Raşa type inequality for box-(m, n)-convex functions (Theorem 24). In this paper, we present a probabilistic version of the Raşa type inequality. Theorem 24 gives a generalization of an analytical version of the Raşa type inequality for binomial distributions, which has been recently proved in [5,6].
In Sect. 7, we obtain the Hermite-Hadamard type and the Jensen inequalities for box-(m, n)-convex functions. The Jensen inequalities presented in [3], have been proved based on the box analog of the subdifferential inequality (see [3,Sect. 6]). In this paper, we present a new approach to obtaining the Hermite-Hadamard type and the Jensen inequalities, based on box-(m, n)convex orders.
In Sect. 8, we introduce the notion of strongly box-(m, n)-convex functions. We obtain the Hermite-Hadamard type and the Jensen inequalities for strongly box-(m, n)-convex functions.
In the following theorem, we give an integral representation of boxmonotone functions without any assumptions about their differentiability. The integrals used in the theorem are the Lebesgue-Stieltjes integrals. where ii) g(x, y) = f (x, y) up to a pseudo-polynomial of order (0, 0), iii) in place of g, it can be taken the non-decreasing right-continuous box- Proof. (⇒) Assume that f : S → R is a box-monotone function. From our assumptions and Lemma 1, we obtain and g = f * .
(⇐) Let f be of the form (3) with right-continuous box-monotone function g. Let a ≤ a 1 < a 2 ≤ b and c ≤ c 1 < c 2 ≤ d. [c 1 , c 2 ]) ≥ 0, we obtain that Δ(f ; [a 1 , a 2 ] × [c 1 , c 2 ]) ≥ 0, which implies that f is box-monotone. We will prove that i), ii) and iii) are satisfied. Let f be of the form (3). Let a ≤ a 1 < a 2 ≤ b and c ≤ c 1 < c 2 ≤ d. It is not difficult to prove that By (4), f can be written in the form (3), with f * in place of g. Taking into ac- Condition ii) follows immediately from (4). It is not difficult to prove that the function g * (x, y) = g(x, y) − g(a, y) − g(x, c) + g(a, c) is non-decreasing. Since g = g * up to a pseudo-polynomial of order (0, 0), the condition iii) is proved.

The Raşa Type Inequalities for Box-Monotone Functions
The Raşa inequality [19] can be written in terms of binomially distributed random variables [11] as follows where X, X 1 , X 2 , Y, Y 1 , Y 2 are random variables such that X, X 1 , X 2 ∼ B(n, x), Y, Y 1 , Y 2 ∼ B(n, y), X, Y are independent, X 1 , X 2 are independent and Y 1 , Y 2 are independent, and f : [0, 1] → R is a continuous convex function. Gavrea [4] presented the problem of generalization of the Raşa inequality, which can be written in terms of the expectation of binomially distributed random variables as follows. Problem 1. [4] Give a characterization of the class of convex functions g : [0, 1] 2 → R, satisfying Note, that condition (7) means that the function g is box monotone. Thus, condition (8) is a sufficient condition for the box monotone function to satisfy (9). In the next theorem, we prove, that (8) is also a necessary condition.
for all functions g : R 2 → R satisfying condition (7), then Proof. Let us assume that the assumptions of the theorem are satisfied. Let F , G, μ, ν be the tail distribution functions and distributions of the random variables X, Y , respectively, i.e. (11) is satisfied for functions g : R 2 → R satisfying condition (7), it is satisfied for all functions g(x, y) of the form There are three possible cases: The theorem is proved.

Integral Representation of Box-(m, n)-Convex Functions
Now we are going to develop an integral representation of box-(m, n)-convex functions for m, n ≥ 2. Proposition 2 seems to be a convenient tool to study (m + n)-differentiable functions. In this section, however, we are interested in all (even not necessarily continuous) box-(m, n)-convex functions defined on I × J, where I, J ⊂ R are open intervals (bounded or unbounded).
In the subsequent lemmas we need the following notion. Let m, n ≥ 1. We say that f : I × J → R is (m, n)-regular if for every y ∈ J the function f (·, y) is a linear combination of (m − 1)-convex functions, and for every x ∈ I the function f (x, ·) is a linear combination of (n − 1)-convex functions. Here, by a 0-convex function we mean a non-decreasing, right-continuous function.
Observe, that a pseudo-polynomial . . , A m are linear combinations of (n − 1)-convex functions, and B 0 , . . . , B n are linear combinations of (m − 1)convex functions. In that case, for any α ∈ I the function x α W (t, y)dt is an (m + 1, n)-regular pseudo-polynomial of order (m + 1, n). If, in addition, m > 1, then the right-derivative We also need several lemmas. Two of them concern a regularization of a box-(m, n)-convex function by subtracting an appropriate pseudo-polynomial. Proof. It is enough to prove (i). We note that for every y ∈ J and x ∈ I\{u 1 , . . . , u m }, we have The proof of part (i) is finished. The proof of (ii) is analogous.
Proof. For every y ∈ J, let A 0 (y), . . . , A m−1 (y) ∈ R be such that Clearly, if f is (m, n)-regular, then the functions A 0 , . . . , A m−1 are linear combinations of (n − 1)-convex functions (as linear combinations of the functions f (u k , ·), k = 1, . . . , m) and B 0 , . . . , B n−1 are linear combinations of (m − 1)-convex functions, hence V is (m, n)-regular. If In the next two lemmas, we describe how differentiation affects a box-(m, n)-convex function. Moreover, Proof. Due to symmetry, it is enough to show that (13) holds for k = 1. We skip an easy proof by induction on n. Using (13), we obtain Proof. Since f (·, y) is a linear combination of (m − 1)-convex functions and m−1 ≥ 1, we obtain that F is well defined and f (x, y) = f (α, y)+ which has the same sign as m i=2 (x−u i ). If follows that F (x, ·) is (n−1)-convex or (n − 1)-concave. Therefore, F (x, ·) is a linear combination of (n − 1)-convex functions. This finishes the proof of (m − 1, n)-regularity of F . Obviously, the same holds, when the roles of x and y are exchanged.
The final two lemmas describe how integration affects a box-(m, n)convex function. Proof. The proof of the lemma is by induction on n. For n = 1 we have In the induction step, we use the definition of the divided difference and the fact that it does not depend on the permutation of the points x 0 , . . . , x n+1 . Proof. Let f : I × J → R be a box-(m, n)-convex function, α ∈ I and let x 0 , x 1 , . . . , x m+1 be pairwise distinct points of I and y 0 , y 1 , . . . , y n be pairwise distinct points of J. Since the function ϕ is well defined, we may use Lemma 10, and we obtain Assume that f is (m, n)-regular. For every y ∈ J, the function f (·, y) is a linear combination of (m − 1)-convex, hence ϕ(·, y) is a linear combination of m-convex functions. Let x ∈ I. We fix pairwise distinct u 1 , . . . , u m ∈ I smaller than min(α, x), and pairwise distinct v 1 , . . . , v n ∈ J. Let V be an (m, n)-regular pseudo-polynomial of order (m − 1, n − 1) given by Lemma 7. We denote f = f − V and ϕ(x, y) = x α f (t, y)dy. Then ϕ − ϕ is an (m + 1, n)regular pseudo-polynomial of order (m, n−1). By Lemma 6, for every t between α and x the function f (t, ·) is (n−1)-convex. It follows that for pairwise distinct y 0 , . . . , y n ∈ J the divided difference [y 0 , . . . , y n ; ϕ(x, ·)] = x α [y 0 , . . . , y n ; f (t, ·)]dt is non-negative for α ≤ x and non-positive for α ≥ x. In either case, ϕ(x, ·) is (n − 1)-convex or (n − 1)-concave. Therefore, ϕ(x, ·) is a linear combination of (n − 1)-convex functions. This shows, that ϕ is (m + 1, n)-regular.
In the following theorems, we give the integral representations of box-(m, n)-convex functions.  (14) is an immediate consequence of Lemma 9 (we skip a trivial induction on m + n).
If f is not (m, n)-regular, then we first use Lemma 7 (obtaining an appropriate pseudo-polynomial V of order (m − 1, n − 1)), and then we use the above argument for the (m, n)-regular function f − V .
As an immediate consequence of Theorem 16, we obtain the following theorem, which is a generalization of Theorem 2 on integral representation of box-monotone functions. Proof. (⇒) Let f be box-(m, n)-convex. Then by Theorem 17, f is of the form (16). Then, for the function P (x, y) as below, we have where k = 1, . . . , m and l = 1, . . . , n. Then, by Theorem 17, the function The proof is obvious. The theorem is proved.

The Box-(m, n)-Convex Orders
By the analogy to the n-convex orders, we define the box-(m, n)-convex orders as follows.
Taking into account that  Theorems 19, 20 give a characterization of the box-(m, n)-convex ordering. They are a generalization of the Levin-Stečkin theorems as well as the Denuit-Lefèvre-Shaked theorems concerning n-convex orderings (see [18] and the references therein).

The Raşa Type Inequality for Box-(m, n)-Convex Functions
Let μ and ν be two signed Borel measures on R such that R ϕ(x)μ(dx) ≤ R ϕ(x)ν(dx) for all n-convex functions ϕ : R → R, provided the integrals exist. Then μ is said to be smaller then ν in the nconvex order (denoted as μ ≤ n−cx ν ). Then the Raşa type inequality (5) can be written equivalently where μ = B(n, x) and ν = B(n, y).
In [9,10], we gave some useful sufficient condition and necessary and sufficient conditions for Borel measures μ and ν to satisfy the following generalized Raşa inequality: Then Theorems 4 and 5 on the necessary and sufficient condition for probability distributions that satisfy the Raşa type inequality for box-(1, 1)convex functions, can be rewritten in terms of box-(1, 1)-convex order.
Note that inequality (22) is a probabilistic version of the Raşa inequality. On the other hand, in [5,6], an analytical version of inequality (22) for binomial distributions has been recently proved. We will prove some necessary and sufficient condition for a more general version of the Raşa inequality than the version (22). We will need two lemmas. Then for all A ∈ R ∞ −∞ x − A q−1 + (q − 1)! τ 1 * · · · * τ q (dx) = F τ1 * F τ2 * · · · * F τq (A).