Majorization-Type Integral Inequalities Related to a Result of Bennett with Applications

Abstract

In this paper, starting from abstract versions of a result of Bennett given by Niculescu, we derive new majorization-type integral inequalities for convex functions using finite signed measures. The proof of the main result is based on a generalization of a recently discovered majorization-type integral inequality. As applications of the results, we give simple proofs of the integral Jensen and Lah–Ribarič inequalities for finite signed measures, generalize and extend known results, and obtain an interesting new refinement of the Hermite–Hadamard–Fejér inequality.

1. Introduction

In Bennett’s paper [1], a number of interesting discrete inequalities and their consequences are studied. These results are the starting point for several papers (see e.g., [2,3,4,5,6,7,8,9]).

The origin of the following inequality also goes back to Bennett’s paper [1]. It is an abstract version of an inequality obtained by Bennett.

We say that the real intervals $I_1$, $I_2$, and $I_3$ are nonoverlapping if any two have at most one point in common.

Theorem 1 

(Theorem of 1 [8]). Suppose that C is an interval carrying a positive Borel measure μ and $I_1$, $I_2$, and $I_3$ are three nonoverlapping compact subintervals of C of a positive measure. Then

$$\mu \left(I_2\right)=\mu \left(I_1\right)+\mu \left(I_3\right)$$

and

$${\displaystyle \underset{I_2}{\int }}td\mu \left(t\right)={\displaystyle \underset{I_1}{\int }}td\mu \left(t\right)+{\displaystyle \underset{I_3}{\int }}td\mu \left(t\right)$$

provide a set of necessary and sufficient conditions under which every convex function f defined on C verifies the inequality

$${\displaystyle \underset{I_2}{\int }}fd\mu \le {\displaystyle \underset{I_1}{\int }}fd\mu +{\displaystyle \underset{I_3}{\int }}fd\mu .$$

(1)

A similar result is found in [2] for six disjointed compact subintervals of C and for three-convex functions at a point.

Theorem 1 is generalized in [8] by using the concepts of Steffensen–Popoviciu and dual Steffensen–Popoviciu measures, which play an important role in the theory of convex functions.

The $\sigma$-algebra of Borel sets on an interval $C\subset \mathbb{R}$ is denoted by ${\mathcal{B}}_C$.

Definition 1. 

Let $C\subset \mathbb{R}$ be an interval with a nonempty interior, and let μ be a finite signed measure on ${\mathcal{B}}_C$.

(a) μ is called a Steffensen–Popoviciu measure on C if $\mu \left(C\right)>0$ and

$${\displaystyle \underset{C}{\int }}fd\mu \ge 0$$

(2)

for every nonnegative, convex, and μ-integrable function $f:C\to \mathbb{R}$.

(b) μ is called a dual Steffensen–Popoviciu measure on C if $\mu \left(C\right)>0$ and inequality (2) holds for every nonnegative, concave, and μ-integrable function $f:C\to \mathbb{R}$.

The generalization of Theorem 1 is as follows.

Theorem 2 

(Theorem 4 of [8]). The statement of Theorem 1 remains valid if μ is a finite signed Borel measure on C and $I_1$, $I_2$, and $I_3$ are three nonoverlapping subintervals of C such that the restriction of μ to each of the intervals $I_1$ and $I_3$ is a Steffensen–Popoviciu measure and the restriction of μ to $I_2$ is a dual Steffensen–Popoviciu measure.

In this paper, we first extend Theorem 1 in a more general framework. Instead of inequality (1), we consider the new inequality

$${\displaystyle \underset{X_2}{\int }}f\circ {\phi }_{2}d{\mu }_2\le {\displaystyle \underset{X_1}{\int }}f\circ {\phi }_{1}d{\mu }_1+{\displaystyle \underset{X_3}{\int }}f\circ {\phi }_{3}d{\mu }_3,$$

(3)

where $\left(X_i,{\mathcal{A}}_i\right)$ is a measurable space with a finite signed measure ${\mu }_i$ on ${\mathcal{A}}_i$, ${\phi }_i:X_i\to I_i$ is a ${\mu }_i$-integrable function, $I_1$, $I_2$, and $I_3$ are three nonoverlapping and nonempty subintervals of the real interval C, and $f:C\to \mathbb{R}$ is a convex function for which $f\circ {\phi }_i$ is ${\mu }_i$-integrable $\left(i=1,2,3\right)$. It is worth stressing that the intervals can be of any type, not necessarily compact. Starting from a generalization of a majorization-type integral inequality from the recent paper [10], we give necessary and sufficient conditions for inequality (3) to be satisfied. The result includes Theorems 1 and 2 as special cases.

As an application, we first give a simple proof of the integral Jensen and Lah–Ribarič inequalities for finite signed measures which results come from [10]. The formulation of the statements is also sharper than in [10]. As a second application, a common generalization of some of the main results of [5] is obtained. Our result uses finite signed measures instead of the classical Lebesgue measure and gives a joint proof of the theorems considered from [5], which are proven by specific methods there. Finally, we give interesting refinements of the Hermite–Hadamard–Fejér inequality. It is worth noting that the result includes an inequality that can also be used to estimate the weighted arithmetic mean over a bounded open interval. This is a novel extension of the Hermite–Hadamard–Fejér inequality.

2. Preliminary Results

Let $\left(X,\mathcal{A}\right)$ be a measurable space ($\mathcal{A}$ always means a $\sigma$-algebra of subsets of X). If $\mu$ is either a measure or a signed measure on $\mathcal{A}$, then the real vector space of $\mu$-integrable real functions on X is denoted by $L\left(\mu \right)$.

The interior of an interval $C\subset \mathbb{R}$ is denoted by $C^{\circ }$.

The following generalization of Theorem 6 of [10] plays a crucial role in proving the main result of this paper. It is interesting in itself and well applicable.

Theorem 3. 

Let $\left(X_i,{\mathcal{A}}_i\right)$ $\left(i=1,\dots ,m\right)$ and $\left(Y_j,{\mathcal{B}}_j\right)$ $\left(j=1,\dots ,n\right)$ be measurable spaces, let ${\mu }_i$ be finite signed measures on ${\mathcal{A}}_i$, and let ${\nu }_j$ be finite signed measures on ${\mathcal{B}}_j$. Let $C\subset \mathbb{R}$ be an interval with a nonempty interior, and let ${\phi }_i:X_i\to C$, ${\psi }_j:Y_j\to C$ be functions such that ${\phi }_i\in L\left({\mu }_i\right)$ and ${\psi }_j\in L\left({\nu }_j\right)$ $\left(i=1,\dots ,m,j=1,\dots ,n\right)$.

Then, for every convex function $f:C\to \mathbb{R}$ for which $f\circ {\phi }_i\in L\left({\mu }_i\right)$ and $f\circ {\psi }_j\in L\left({\nu }_j\right)$ $\left(i=1,\dots ,m,j=1,\dots ,n\right)$, the inequality

$${\displaystyle \sum _{i=1}^m}\underset{X_i}{\int }f\circ {\phi }_{i}d{\mu }_i\le {\displaystyle \sum _{j=1}^n}\underset{Y_j}{\int }f\circ {\psi }_{j}d{\nu }_j$$

holds if and only if

$${\displaystyle \sum _{i=1}^m}{\mu }_i\left(X_i\right)={\displaystyle \sum _{j=1}^n}{\nu }_j\left(Y_j\right),{\displaystyle \sum _{i=1}^m}\underset{X_i}{\int }{\phi }_{i}d{\mu }_i={\displaystyle \sum _{j=1}^n}\underset{Y_j}{\int }{\psi }_{j}d{\nu }_j$$

(4)

and

$${\displaystyle \sum _{i=1}^m}\underset{\left\{{\phi }_i\ge w\right\}}{\int }\left({\phi }_i-w\right)d{\mu }_i\le {\displaystyle \sum _{j=1}^n}\underset{\left\{{\psi }_j\ge w\right\}}{\int }\left({\psi }_j-w\right)d{\nu }_j,w\in C^{\circ }.$$

(5)

Proof. 

We can follow the proof of Theorem 6 of [10], with appropriate modifications.

The proof is complete. □

Remark 1. 

Analogously to Theorem 6 in [10], the previous statement could be formulated for (nonnegative) increasing, (nonnegative) decreasing, or nonnegative convex functions only.

We now give the characterization of the Steffensen–Popoviciu and the dual Steffensen–Popoviciu measures.

A characterization of Steffensen–Popoviciu measures is presented in [10]. Characterization of Steffensen–Popoviciu measures for compact intervals occurred much earlier, independently due to Popoviciu [11] and Fink [12].

The identity function on an interval $C\subset \mathbb{R}$ is denoted by $i{d}_C$.

Theorem 4 

(see Theorem 11 of [10]). Let $C\subset \mathbb{R}$ be an interval with a nonempty interior, and let μ be a finite signed measure on ${\mathcal{B}}_C$ such that $i{d}_C\in L\left(\mu \right)$. Then

(a) For every nonnegative, convex and μ-integrable function $f:C\to \mathbb{R}$ the inequality

$$\underset{C}{\int }fd\mu \ge 0$$

holds if and only if

$$\mu \left(C\right)\ge 0$$

and

$$\underset{C\cap \left]-\infty ,w\right]}{\int }\left(w-t\right)d\mu \left(t\right)\ge 0,\underset{C\cap \left[w,\infty \right[}{\int }\left(t-w\right)d\mu \left(t\right)\ge 0,w\in C^{\circ }.$$

(6)

(b) μ is a Steffensen–Popoviciu measure if and only if $\mu \left(C\right)>0$ and (6) is satisfied.

Characterization of dual Steffensen–Popoviciu measures for compact intervals and continuous concave functions can be found in a recent paper [13]. We generalize the result slightly by dropping the continuity condition.

Theorem 5. 

Let $\left[a,b\right]\subset \mathbb{R}$ be a compact interval with a nonempty interior, and let μ be a finite signed measure defined on ${\mathcal{B}}_{\left[a,b\right]}$. Then,

(a) For every nonnegative concave function $f:\left[a,b\right]\to \mathbb{R}$, the inequality

$${\displaystyle \underset{\left[a,b\right]}{\int }}fd\mu \ge 0$$

holds if and only if

$${\displaystyle \frac{1}{w-a}}{\displaystyle \underset{\left[a,w\right[}{\int }}\left(t-a\right)d\mu \left(t\right)+\mu \left(\left\{w\right\}\right)+{\displaystyle \frac{1}{b-w}}{\displaystyle \underset{\left]w,b\right]}{\int }}\left(b-t\right)d\mu \left(t\right)\ge 0,w\in \left[a,b\right],$$

(7)

where the first term is taken to be 0 for $w=a$, while the third term is taken to be 0 for $w=b$.

(b) μ is a dual Steffensen–Popoviciu measure if and only if $\mu \left(\left[a,b\right]\right)>0$ and (7) is satisfied.

Proof. 

(a)If f is continuous, we can apply Theorem 3.4 of [13].

If f is not continuous, then there exists an increasing sequence ${\left(f_n\right)}_{n=1}^{\infty }$ of nonnegative concave functions such that $f_n$ is continuous $\left(n\in {\mathbb{N}}_{+}\right)$ and $\left(f_n\right)$ converges pointwise to f on C, and therefore the result follows from the first part of the proof and B. Levi’s theorem.

(b) It is proven in [13] that (7) implies $\mu \left(\left[a,b\right]\right)\ge 0$.

The proof is complete. □

We often use the following simple statement in this paper.

Lemma 1. 

Let $\left(X,\mathcal{A}\right)$ be a measurable space, and let μ be a finite signed measure on $\mathcal{A}$. Let $C\subset \mathbb{R}$ be an interval with a nonempty interior, and let $\phi :X\to C$ be a function such that $\phi \in L\left(\mu \right)$.

(a) Suppose that $t\in C$ and t is not the left endpoint of C. If

$${\displaystyle \underset{\left\{\phi <w\right\}}{\int }}\left(w-\phi \right)d\mu \ge 0,w\in \left]-\infty ,t\right[{\displaystyle \bigcap }C,$$

then

$${\displaystyle \underset{\left\{\phi <t\right\}}{\int }}\left(t-\phi \right)d\mu \ge 0.$$

(b) Suppose that $t\in C$ and t is not the right endpoint of C. If

$${\displaystyle \underset{\left\{\phi \ge w\right\}}{\int }}\left(\phi -w\right)d\mu \ge 0,w\in \left]t,\infty \right[{\displaystyle \bigcap }C,$$

then

$${\displaystyle \underset{\left\{\phi \ge t\right\}}{\int }}\left(\phi -t\right)d\mu \ge 0.$$

Proof. 

Only (a) is proved, and (b) can be treated similarly.

Let ${\left(w_n\right)}_{n=1}^{\infty }$ be an increasing sequence in $\left]-\infty ,t\right[\cap C$ converging to t. Then, the sequence of functions

$${\left(\left(w_n-\phi \right)·{\chi }_{\left\{\phi <w_n\right\}}\right)}_{n=1}^{\infty }$$

is increasing and converges pointwise to the function

$$\left(t-\phi \right)·{\chi }_{\left\{\phi <t\right\}}.$$

Since $\mu$ is finite, the monotone convergence theorem implies that

$${\displaystyle \underset{\left\{\phi <t\right\}}{\int }}\left(t-\phi \right)d\mu =\underset{n\to \infty }{lim}{\displaystyle \underset{\left\{\phi <w_n\right\}}{\int }}\left(w_n-\phi \right)d\mu \ge 0.$$

The proof is complete. □

We need the following result. This is a simplified version of Lemma 4 of [14].

Lemma 2 

(see Lemma 4 of [14]). Let $\left[a,b\right]\subset \mathbb{R}$ with $a<b$, and let μ be a finite signed measure on ${\mathcal{B}}_{\left[a,b\right]}$ such that

$$\mu \left(A\right)=\mu \left(a+b-A\right),A\in {\mathcal{B}}_{\left[a,b\right]}.$$

Then,

$$\underset{\left[a,b\right]}{\int }td\mu \left(t\right)={\displaystyle \frac{a+b}{2}}\mu \left(\left[a,b\right]\right).$$

Finally, for completeness, we give the Hermite–Hadamard inequality for measures and its special case, the Hermite–Hadamard–Fejér inequality.

Theorem 6 

(see [15]). Let a, $b\in \mathbb{R}$, $a<b$.

(a) Let μ be a finite measure on ${\mathcal{B}}_{\left[a,b\right]}$ with $\mu \left(\left[a,b\right]\right)>0$. Then,

$$f\left(t_{\mu }\right)\le {\displaystyle \frac{1}{\mu \left(\left[a,b\right]\right)}}\underset{\left[a,b\right]}{\int }f\left(t\right)d\mu \left(t\right)\le {\displaystyle \frac{b-t_{\mu }}{b-a}}f\left(a\right)+{\displaystyle \frac{t_{\mu }-a}{b-a}}f\left(b\right)$$

for every convex function $f:\left[a,b\right]\to \mathbb{R}$, where

$$t_{\mu }:={\displaystyle \frac{1}{\mu \left(\left[a,b\right]\right)}}\underset{\left[a,b\right]}{\int }td\mu \left(t\right)$$

represents the barycenter of μ.

(b) Let λ be the Lebesgue measure on ${\mathcal{B}}_{\left[a,b\right]}$. Let the weight function $w:\left[a,b\right]\to \mathbb{R}$ be nonnegative, Lebesgue-integrable, and symmetric about the vertical line $t={\displaystyle \frac{a+b}{2}}$, that is,

$$w\left(t\right)=w\left(a+b-t\right),t\in \left[a,b\right].$$

Then

$$f\left({\displaystyle \frac{a+b}{2}}\right){\displaystyle \underset{a}{\overset{b}{\int }}}wd\lambda \le {\displaystyle \underset{a}{\overset{b}{\int }}}fwd\lambda \le {\displaystyle \frac{f\left(a\right)+f\left(b\right)}{2}}{\displaystyle \underset{a}{\overset{b}{\int }}}wd\lambda$$

for every convex function $f:\left[a,b\right]\to \mathbb{R}$.

3. Main Results

The following result is a common generalization of Theorems 1 and 2.

Theorem 7. 

Let $C\subset \mathbb{R}$ be an interval with a nonempty interior. Let $I_1$, $I_2$, and $I_3$ in this order be nonoverlapping and nonempty subintervals of C. Let $a_i\le b_i$ be the endpoints of $I_i$ $\left(i=1,2,3\right)$. Let $\left(X_i,{\mathcal{A}}_i\right)$ be a measurable space, and let ${\mu }_i$ be a finite signed measure on ${\mathcal{A}}_i$ $\left(i=1,2,3\right)$. Let ${\phi }_i:X_i\to I_i$ be a function such that ${\phi }_i\in L\left({\mu }_i\right)$ $\left(i=1,2,3\right)$.

Then, for every convex function $f:C\to \mathbb{R}$ for which $f\circ {\phi }_i\in L\left({\mu }_i\right)$ $\left(i=1,2,3\right)$, the inequality

$${\displaystyle \underset{X_2}{\int }}f\circ {\phi }_{2}d{\mu }_2\le {\displaystyle \underset{X_1}{\int }}f\circ {\phi }_{1}d{\mu }_1+{\displaystyle \underset{X_3}{\int }}f\circ {\phi }_{3}d{\mu }_3$$

holds if and only if

$${\mu }_2\left(X_2\right)={\mu }_1\left(X_1\right)+{\mu }_3\left(X_3\right),$$

(8)

$${\displaystyle \underset{X_2}{\int }}{\phi }_{2}d{\mu }_2={\displaystyle \underset{X_1}{\int }}{\phi }_{1}d{\mu }_1+{\displaystyle \underset{X_3}{\int }}{\phi }_{3}d{\mu }_3$$

(9)

and

$${\displaystyle \underset{\left\{{\phi }_1<w\right\}}{\int }}\left(w-{\phi }_1\right)d{\mu }_1\ge 0,a_1<w\le a_2,$$

(10)

$${\displaystyle \underset{\left\{{\phi }_2\ge w\right\}}{\int }}\left({\phi }_2-w\right)d{\mu }_2\le {\displaystyle \underset{X_3}{\int }}\left({\phi }_3-w\right)d{\mu }_3,a_2<w<b_2$$

(11)

and

$${\displaystyle \underset{\left\{{\phi }_3\ge w\right\}}{\int }}\left({\phi }_3-w\right)d{\mu }_3\ge 0,b_2\le w<b_3.$$

(12)

Proof. 

Theorem 3 is obviously applicable, and only the actual form of conditions (4) and (5) needs to be given.

It is easy to see that the two conditions in (4) are exactly (8) and (9) in the current situation.

In the present case, inequality (5) is equivalent to

$${\displaystyle \underset{\left\{{\phi }_2\ge w\right\}}{\int }}\left({\phi }_2-w\right)d{\mu }_2$$

$$\le {\displaystyle \underset{\left\{{\phi }_1\ge w\right\}}{\int }}\left({\phi }_1-w\right)d{\mu }_1+{\displaystyle \underset{\left\{{\phi }_3\ge w\right\}}{\int }}\left({\phi }_3-w\right)d{\mu }_3,w\in C^{\circ }.$$

(13)

It is sufficient to show that the previous inequality is satisfied exactly if inequalities (10)–(12) hold. We will justify this by taking into account the position of w. The cases considered may include empty cases, depending on the intervals.

In the following, let $w\in C^{\circ }$ be fixed.

(i) If $w\le a_1$, then (13) takes the following form:

$${\displaystyle \underset{X_2}{\int }}\left({\phi }_2-w\right)d{\mu }_2\le {\displaystyle \underset{X_1}{\int }}\left({\phi }_1-w\right)d{\mu }_1+{\displaystyle \underset{X_3}{\int }}\left({\phi }_3-w\right)d{\mu }_3,$$

which is obviously satisfied by (8) and (9).

(ii) Now suppose that $a_1<w<b_1$. Then, (13) is

$${\displaystyle \underset{X_2}{\int }}\left({\phi }_2-w\right)d{\mu }_2\le {\displaystyle \underset{\left\{{\phi }_1\ge w\right\}}{\int }}\left({\phi }_1-w\right)d{\mu }_1+{\displaystyle \underset{X_3}{\int }}\left({\phi }_3-w\right)d{\mu }_3,$$

and this is the same, because of (8) and (9), as

$${\displaystyle \underset{\left\{{\phi }_1<w\right\}}{\int }}\left(w-{\phi }_1\right)d{\mu }_1\ge 0.$$

For $b_1\le w\le a_2$, (13) means that

$${\displaystyle \underset{X_2}{\int }}\left({\phi }_2-w\right)d{\mu }_2\le {\displaystyle \underset{X_3}{\int }}\left({\phi }_3-w\right)d{\mu }_3,$$

and according to (8) and (9), it is the same as

$${\displaystyle \underset{\left\{{\phi }_1<w\right\}}{\int }}\left(w-{\phi }_1\right)d{\mu }_1={\displaystyle \underset{X_1}{\int }}\left(w-{\phi }_1\right)d{\mu }_1\ge 0.$$

These two cases give (10).

(iii) Assume $a_2<w<b_2$. Then, (13) has the form

$${\displaystyle \underset{\left\{{\phi }_2\ge w\right\}}{\int }}\left({\phi }_2-w\right)d{\mu }_2\le {\displaystyle \underset{X_3}{\int }}\left({\phi }_3-w\right)d{\mu }_3,$$

which is identical to (11).

(iv) Now, let $b_2\le w\le a_3$. Then, (13) is equivalent to

$$0\le {\displaystyle \underset{X_3}{\int }}\left({\phi }_3-w\right)d{\mu }_3={\displaystyle \underset{\left\{{\phi }_3\ge w\right\}}{\int }}\left({\phi }_3-w\right)d{\mu }_3.$$

If $a_3<w<b_3$, then (13) can be rewritten as follows:

$$0\le {\displaystyle \underset{\left\{{\phi }_3\ge w\right\}}{\int }}\left({\phi }_3-w\right)d{\mu }_3.$$

We have obtained (12).

(v) Finally, if $b_3\le w$, then (13) is obviously satisfied.

The proof is complete. □

Remark 2. 

If ${\mu }_i$ is a measure in the previous theorem $\left(i=1,2,3\right)$, then conditions (10)–(12) are automatically met. Conditions (10) and (12) are obviously satisfied, and only (11) needs to be justified. Let $a_2<w<b_2$ be fixed. Condition (11) is proved if the following inequalities are true:

$${\displaystyle \underset{\left\{{\phi }_2\ge w\right\}}{\int }}\left({\phi }_2-w\right)d{\mu }_2\le {\displaystyle \underset{X_2}{\int }}{\displaystyle \frac{b_2-w}{b_2-a_2}}\left({\phi }_2-a_2\right)d{\mu }_2$$

$$\le {\displaystyle \underset{X_3}{\int }}{\displaystyle \frac{b_2-w}{b_2-a_2}}\left({\phi }_3-a_2\right)d{\mu }_3\le {\displaystyle \underset{X_3}{\int }}\left({\phi }_3-w\right)d{\mu }_3.$$

The first inequality comes from the easily verifiable facts that the function in the second integral is nonnegative and

$${\phi }_2\left(x\right)-w\le {\displaystyle \frac{b_2-w}{b_2-a_2}}\left({\phi }_2\left(x\right)-a_2\right)\mathrm{if}w\le {\phi }_2\left(x\right).$$

conditions (8) and (9) imply the second inequality.

A simple calculation shows that

$${\displaystyle \frac{b_2-w}{b_2-a_2}}\left({\phi }_3-a_2\right)\le {\phi }_3-w,$$

which yields the third inequality.

If we assume that ${\mu }_i\left(X_i\right)\ge 0$ $\left(i=1,2,3\right)$ in Theorem 7, then conditions (10)–(12) are further simplified. They can be formulated using only the interior of the intervals $I_1$, $I_2$ and $I_3$.

Proposition 1. 

Suppose that conditions of Theorem 7 are satisfied with the additional constraint that ${\mu }_i\left(X_i\right)\ge 0$ $\left(i=1,2,3\right)$. Then, (10)–(12) are equivalent to the following conditions:

$${\displaystyle \underset{\left\{{\phi }_1<w\right\}}{\int }}\left(w-{\phi }_1\right)d{\mu }_1\ge 0,w\in I_1^{\circ },$$

(14)

$${\displaystyle \underset{\left\{{\phi }_2\ge w\right\}}{\int }}\left({\phi }_2-w\right)d{\mu }_2\le {\displaystyle \underset{X_3}{\int }}\left({\phi }_3-w\right)d{\mu }_3,w\in I_2^{\circ }$$

and

$${\displaystyle \underset{\left\{{\phi }_3\ge w\right\}}{\int }}\left({\phi }_3-w\right)d{\mu }_3\ge 0,w\in I_3^{\circ }.$$

(15)

Proof. 

We only need to show two things.

(i) We first verify that (14) implies

$${\displaystyle \underset{X_1}{\int }}\left(w-{\phi }_1\right)d{\mu }_1\ge 0,b_1\le w\le a_2.$$

(16)

Since

$${\displaystyle \underset{X_1}{\int }}\left(w-{\phi }_1\right)d{\mu }_1={\displaystyle \underset{X_1}{\int }}\left(b_1-{\phi }_1\right)d{\mu }_1+{\displaystyle \underset{X_1}{\int }}\left(w-b_1\right)d{\mu }_1$$

and ${\mu }_1\left(X_1\right)\ge 0$, it is enough to show that (16) holds for $w=b_1$.

If $I_1^{\circ }=\varnothing$, then the integral in (16) is zero for $w=b_1$.

If $I_1^{\circ }\ne \varnothing$, then Lemma 1 (a) can be applied by using (14).

(ii) Second, we confirm that (15) yields

$${\displaystyle \underset{X_3}{\int }}\left({\phi }_3-w\right)d{\mu }_3\ge 0,b_2\le w\le a_3.$$

(17)

Since

$${\displaystyle \underset{X_3}{\int }}\left({\phi }_3-w\right)d{\mu }_3={\displaystyle \underset{X_3}{\int }}\left({\phi }_3-a_3\right)d{\mu }_3+{\displaystyle \underset{X_3}{\int }}\left(a_3-w\right)d{\mu }_3$$

and ${\mu }_3\left(X_3\right)\ge 0$, all we need to prove is that (17) holds for $w=a_3$.

If $I_3^{\circ }=\varnothing$, then the integral in (17) is zero for $w=a_3$.

If $I_3^{\circ }\ne \varnothing$, then Lemma 1 (b) can be applied by using (15).

The proof is complete. □

The following statement is a simple but important consequence of Theorem 7. This is the version that can be easily compared with Theorem 1. We consider only the case related to Proposition 1.

Theorem 8. 

Let $C\subset \mathbb{R}$ be an interval with a nonempty interior. Let $I_1$, $I_2$, and $I_3$ in this order be nonoverlapping and nonempty subintervals of C. Let ${\mu }_i$ be a finite signed measure on ${\mathcal{B}}_{I_i}$ such that ${\mu }_i\left(I_i\right)\ge 0$ $\left(i=1,2,3\right)$. Assume $i{d}_{I_1}\in L\left({\mu }_1\right)$ and $i{d}_{I_3}\in L\left({\mu }_3\right)$.

Then, for every convex function $f:C\to \mathbb{R}$ for which $f\in L\left({\mu }_i\right)$ $\left(i=1,3\right)$, the inequality

$${\displaystyle \underset{I_2}{\int }}fd{\mu }_2\le {\displaystyle \underset{I_1}{\int }}fd{\mu }_1+{\displaystyle \underset{I_3}{\int }}fd{\mu }_3$$

(18)

holds if and only if

$${\mu }_2\left(I_2\right)={\mu }_1\left(I_1\right)+{\mu }_3\left(I_3\right),$$

(19)

$${\displaystyle \underset{I_2}{\int }}td{\mu }_2\left(t\right)={\displaystyle \underset{I_1}{\int }}td{\mu }_1\left(t\right)+{\displaystyle \underset{I_3}{\int }}td{\mu }_3\left(t\right)$$

(20)

and

$${\displaystyle \underset{I_1\cap \left]-\infty ,w\right[}{\int }}\left(w-t\right)d{\mu }_1\left(t\right)\ge 0,w\in I_1^{\circ },$$

(21)

$${\displaystyle \underset{I_2\cap \left[w,\infty \right[}{\int }}\left(t-w\right)d{\mu }_2\left(t\right)\le {\displaystyle \underset{I_3}{\int }}\left(t-w\right)d{\mu }_3\left(t\right),w\in I_2^{\circ }$$

(22)

and

$${\displaystyle \underset{I_3\cap \left[w,\infty \right[}{\int }}\left(t-w\right)d{\mu }_3\left(t\right)\ge 0,w\in I_3^{\circ }.$$

(23)

Proof. 

Since $I_2$ is necessarily bounded, the function $i{d}_{I_2}$ and the restriction of f to $I_2$ are also bounded, and therefore the finiteness of ${\mu }_2$ implies that $i{d}_{I_2}$ and $f{\mid }_{I_2}$ are ${\mu }_2$-integrable. The result then follows directly from Proposition 1.

The proof is complete. □

As a special case of the previous theorem, we obtain an extension of Theorem 2.

The following notations are used for some special functions defined on an interval $C\subset \mathbb{R}$:

$$p_{C,w}\left(t\right):={\left(t-w\right)}^{+},n_{C,w}\left(t\right):={\left(t-w\right)}^{-}t,w\in C.$$

Corollary 1. 

Let $C\subset \mathbb{R}$ be an interval with a nonempty interior. Let $I_1$, $I_2$, and $I_3$ in this order be nonoverlapping subintervals of C, whose interior is not empty. Let $a_2$, $b_2$ be the endpoints of $I_2$. Let ${\mu }_i$ be a finite signed measure on ${\mathcal{B}}_{I_i}$ $\left(i=1,2,3\right)$ such that $i{d}_{I_1}\in L\left({\mu }_1\right)$ and $i{d}_{I_3}\in L\left({\mu }_3\right)$.

If (19) and (20) are satisfied, ${\mu }_1$ and ${\mu }_3$ are Steffensen–Popoviciu measures, and ${\mu }_2$ is a dual Steffensen–Popoviciu measure, then the inequality (18) holds for every convex function $f:C\to \mathbb{R}$ for which $f\in L\left({\mu }_i\right)$ $\left(i=1,3\right)$.

Proof. 

According to Theorem 8, we need to show that conditions (21)–(23) are satisfied.

Since the function $n_{I_1,w}$ is nonnegative and convex for all $w\in I_1^{\circ }$ and the measure ${\mu }_1$ is a Steffensen–Popoviciu measure, (21) follows.

Similarly, since the function $p_{I_3,w}$ is nonnegative and convex for all $w\in I_3^{\circ }$ and the measure ${\mu }_3$ is also a Steffensen–Popoviciu measure, (23) follows too.

For every $w\in I_2^{\circ }$, let $l:\mathbb{R}\to \mathbb{R}$ be the affine function joining to the points $\left(a_2,0\right)$ and $\left(b_2,b_2-w\right)$. Then, $l{\mid }_{I_1}$ is nonpositive, $l-p_{I_2,w}$ is concave and nonnegative, and $l{\mid }_{I_3}\le i{d}_{I_3}-w$. Using this, the properties of the measures, and (19) and (20), we obtain

$${\displaystyle \underset{I_2\cap \left[w,\infty \right[}{\int }}\left(t-w\right)d{\mu }_2\left(t\right)\le {\displaystyle \underset{I_2}{\int }}l\left(t\right){\mu }_2\left(t\right)={\displaystyle \underset{I_1}{\int }}l\left(t\right){\mu }_1\left(t\right)+{\displaystyle \underset{I_3}{\int }}l\left(t\right){\mu }_3\left(t\right)$$

$$\le {\displaystyle \underset{I_3}{\int }}l\left(t\right){\mu }_3\left(t\right)\le {\displaystyle \underset{I_3}{\int }}\left(t-w\right)d{\mu }_3\left(t\right),w\in I_2^{\circ },$$

which gives (22).

The proof is complete. □

Remark 3. 

For a finite signed measure, we can use Theorems 4 and 5 to decide whether it is a Steffensen–Popoviciu measure or a dual Steffensen–Popoviciu measure.

4. Applications

Theorem 12 of [10] is an extension of the classical integral Jensen and Lah–Ribarič inequalities to finite signed measures. As a first application, we give a sharper formulation and simpler proof of these results.

Theorem 9. 

Let $\left(X,\mathcal{A}\right)$ be a measurable space, and let μ be a finite signed measure on $\mathcal{A}$ such that $\mu \left(X\right)>0$. Let $\phi :X\to \mathbb{R}$ be a μ-integrable function. Let $C_{\phi }$ be the narrowest interval containing the range of φ, and let $a\le b$ be the endpoints of $C_{\phi }$. Then,

(a) Let

$$t_{\phi ,\mu }:={\displaystyle \frac{1}{\mu \left(X\right)}}\underset{X}{\int }\phi d\mu ,$$

and assume $t_{\phi ,\mu }\in C_{\phi }$. The inequality

$$f\left({\displaystyle \frac{1}{\mu \left(X\right)}}\underset{X}{\int }\phi d\mu \right)\le {\displaystyle \frac{1}{\mu \left(X\right)}}\underset{X}{\int }f\circ \phi d\mu$$

(24)

holds for every convex function $f:C_{\phi }\to \mathbb{R}$ for which $f\circ \phi \in L\left(\mu \right)$ if and only if

$$\underset{\left\{\phi <w\right\}}{\int }\left(w-\phi \right)d\mu \ge 0,w\in \left]-\infty ,t_{\phi ,\mu }\right[{\displaystyle \bigcap }{C}_{\phi }^{\circ }$$

(25)

and

$$\underset{\left\{\phi \ge w\right\}}{\int }\left(\phi -w\right)d\mu \ge 0,w\in \left]t_{\phi ,\mu },\infty \right[{\displaystyle \bigcap }{C}_{\phi }^{\circ }.$$

(26)

(b) If $a\in \mathbb{R}$ and

$$\underset{\left\{\phi \ge w\right\}}{\int }\left(\phi -w\right)d\mu \ge 0,w\in C_{\phi }^{\circ },$$

(27)

then $t_{\phi ,\mu }\in C_{\phi }\cap \left[a,\infty \right[$.

If $b\in \mathbb{R}$ and

$$\underset{\left\{\phi <w\right\}}{\int }\left(w-\phi \right)d\mu \ge 0,w\in C_{\phi }^{\circ }.$$

(28)

then $t_{\phi ,\mu }\in C_{\phi }\cap \left]-\infty ,b\right]$.

(c) The inequality (24) holds for every convex function $f:C_{\phi }\to \mathbb{R}$ for which $f\circ \phi \in L\left(\mu \right)$ if and only if (27) and (28) are satisfied.

(d) Assume a, $b\in \mathbb{R}$ and $t_{\phi ,\mu }\in \left[a,b\right]$. For every convex function $f:\left[a,b\right]\to \mathbb{R}$, the inequality

$${\displaystyle \frac{1}{\mu \left(X\right)}}\underset{X}{\int }f\circ \phi d\mu \le {\displaystyle \frac{b-t_{\phi ,\mu }}{b-a}}f\left(a\right)+{\displaystyle \frac{t_{\phi ,\mu }-a}{b-a}}f\left(b\right)$$

(29)

holds if and only if

$$\left(b-w\right)\underset{\left\{\phi <w\right\}}{\int }\left(\phi -a\right)d\mu +\left(w-a\right)\underset{\left\{\phi \ge w\right\}}{\int }\left(b-\phi \right)d\mu \ge 0,w\in \left]a,b\right[.$$

(30)

Proof. 

If $\phi$ is a constant function, then the statements are obvious.

In the following, we suppose that $\phi$ is not a constant function, that is, the interior of the interval $C_{\phi }$ is nonempty.

(a) Define the interval $I_2$ by $I_2:=\left\{t_{\phi ,\mu }\right\}$ and the intervals $I_1$ and $I_3$ by

$$I_1:=\left]-\infty ,t_{\phi ,\mu }\right[{\displaystyle \bigcap }{C}_{\phi },I_3:=\left[t_{\phi ,\mu },\infty \right[{\displaystyle \bigcap }{C}_{\phi }$$

when $t_{\phi ,\mu }\ne a$, and

$$I_1:=\left\{t_{\phi ,\mu }\right\},I_3:=\left]t_{\phi ,\mu },\infty \right[{\displaystyle \bigcap }{C}_{\phi }$$

when $t_{\phi ,\mu }=a$.

Then, $I_1$, $I_2$, and $I_3$ are nonempty and nonoverlapping intervals.

Let

$$X_1:=\left\{\phi \in I_1\right\},X_2:=\left\{t_{\phi ,\mu }\right\},X_3:=\left\{\phi \in I_3\right\}.$$

Then, $X_1$ and $X_3$ are nonempty and measurable subsets of X. Let ${\phi }_i$ denote the restriction of $\phi$ to $X_i$, and ${\mu }_i$ the restriction of $\mu$ to ${\mathcal{A}}_i:=X_i\cap \mathcal{A}$ $\left(i=1,3\right)$. Let ${\mathcal{A}}_2:=\left\{\varnothing ,X_2\right\}$, consider the measure ${\mu }_2:=\mu \left(X\right){\epsilon }_{t_{\phi ,\mu }}$ on ${\mathcal{A}}_2$, and let ${\phi }_2:X_2\to \mathbb{R}$, ${\phi }_2\left(t_{\phi ,\mu }\right):=t_{\phi ,\mu }$.

With these notations, the conditions of Theorem 7 are satisfied and inequality (24) can be written as

$${\displaystyle \underset{X_2}{\int }}f\circ {\phi }_{2}d{\mu }_2\le {\displaystyle \underset{X_1}{\int }}f\circ {\phi }_{1}d{\mu }_1+{\displaystyle \underset{X_3}{\int }}f\circ {\phi }_{3}d{\mu }_3.$$

According to Theorem 7, it is sufficient to show that (8) and (9) are satisfied and (10)–(12) are equivalent to (25) and (26). All of these are easy to check.

(b) The proof of Theorem 12 (a) in [10] can be copied.

(c) If the inequality (24) holds for every convex function $f:C_{\phi }\to \mathbb{R}$ for which $f\circ \phi \in L\left(\mu \right)$, then $t_{\phi ,\mu }\in C_{\phi }$, and by (a), conditions (25) and (26) are satisfied. We have to prove that (27) and (28) are also satisfied.

Let $w\in \left]-\infty ,t_{\phi ,\mu }\right[\cap C_{\phi }^{\circ }$. Then, by (25),

$$\underset{\left\{\phi \ge w\right\}}{\int }\left(\phi -w\right)d\mu =\underset{X}{\int }\left(\phi -w\right)d\mu +\underset{\left\{\phi <w\right\}}{\int }\left(w-\phi \right)d\mu$$

$$=\mu \left(X\right)\left(t_{\phi ,\mu }-w\right)+\underset{\left\{\phi <w\right\}}{\int }\left(w-\phi \right)d\mu \ge 0.$$

If $w=t_{\phi ,\mu }$, (27) follows from (26) and Lemma 1 (b).

We can prove similarly that if $w\in \left[t_{\phi ,\mu },\infty \right[\cap C_{\phi }^{\circ }$, then (28) holds.

Conversely, if (27) and (28) are satisfied, then (25) and (26) are obviously satisfied, and by (b), $t_{\phi ,\mu }\in C_{\phi }$.

(d) Define the nonoverlapping intervals $I_1$, $I_2$, and $I_3$ by

$$I_1:=\left\{a\right\},I_2:=\left[a,b\right],I_3:=\left\{b\right\}.$$

The measurable spaces $\left(X_i,{\mathcal{A}}_i\right)$ $\left(i=1,2,3\right)$ with the finite signed measures ${\mu }_i$ on ${\mathcal{A}}_i$ $\left(i=1,2,3\right)$ are defined by

$$\left(\left\{a\right\},\left\{\varnothing ,\left\{a\right\}\right\},\mu \left(X\right){\displaystyle \frac{b-t_{\phi ,\mu }}{b-a}}{\epsilon }_a\right),\left(X,\mathcal{A},\mu \right)$$

and

$$\left(\left\{b\right\},\left\{\varnothing ,\left\{b\right\}\right\},\mu \left(X\right){\displaystyle \frac{t_{\phi ,\mu }-a}{b-a}}{\epsilon }_b\right).$$

The functions ${\phi }_i:X_i\to \mathbb{R}$ $\left(i=1,2,3\right)$ are defined by

$${\phi }_1:=i{d}_{\left\{a\right\}},{\phi }_2:=\phi ,{\phi }_3:=i{d}_{\left\{b\right\}}.$$

With these notations, the conditions of Theorem 7 are satisfied and inequality (29) can be written as

$${\displaystyle \underset{X_2}{\int }}f\circ {\phi }_{2}d{\mu }_2\le {\displaystyle \underset{X_1}{\int }}f\circ {\phi }_{1}d{\mu }_1+{\displaystyle \underset{X_3}{\int }}f\circ {\phi }_{3}d{\mu }_3.$$

Now, Theorem 7 implies the result if we show that (8) and (9) are satisfied and (10)–(12) are equivalent to (30). All can be obtained by simple calculations.

The proof is complete. □

Remark 4. 

(a) If there exists a point $t\in C_{\phi }$ such that the inequality

$$f\left(t\right)\le {\displaystyle \frac{1}{\mu \left(X\right)}}\underset{X}{\int }f\circ \phi d\mu$$

holds for every convex function $f:C_{\phi }\to \mathbb{R}$ for which $f\circ \phi \in L\left(\mu \right)$, then $t=t_{\phi ,\mu }$ necessarily.

(b) Assume C is an interval such that $C_{\phi }\subset C$.

If (27) and (28) are satisfied, then by $\mu \left(X\right)>0$, these inequalities also hold for every $w\in C^{\circ }$.

If $f:C\to \mathbb{R}$ is a convex function for which $f\circ \phi \in L\left(\mu \right)$, then $f{\mid }_{C_{\phi }}$ is also a convex function such that $f{\mid }_{C_{\phi }}\circ \phi \in L\left(\mu \right)$. However, if $f:C_{\phi }\to \mathbb{R}$ is a convex function for which $f\circ \phi \in L\left(\mu \right)$, then f may not be extended to a convex function defined on C. It can be seen that the statements (a), (b), and (c) are the sharpest for $C=C_{\phi }$.

A common generalization of some of the main results of [5] is obtained in the following statement. It is formulated in such a way that it can be easily comparable with Theorem 2.2 [5].

Proposition 2. 

Let $C\subset \mathbb{R}$ be an interval, and let $a_1$, $b_1$, $a_2$, $b_2$, $a_3$, $b_3\in C$ such that $a_1<b_1\le a_2<b_2\le a_3<b_3$. Let ${\mu }_i$ be a finite signed measure on ${\mathcal{B}}_{\left[a_i,b_i\right]}$ such that ${\mu }_i\left(\left[a_i,b_i\right]\right)>0$ $\left(i=1,2,3\right)$. Let p, q, and r be positive numbers. Then, the inequality

$${\displaystyle \frac{q}{{\mu }_2\left(\left[a_2,b_2\right]\right)}}{\displaystyle \underset{\left[a_2,b_2\right]}{\int }}fd{\mu }_2\le {\displaystyle \frac{p}{{\mu }_1\left(\left[a_1,b_1\right]\right)}}{\displaystyle \underset{\left[a_1,b_1\right]}{\int }}fd{\mu }_1+{\displaystyle \frac{r}{{\mu }_3\left(\left[a_3,b_3\right]\right)}}{\displaystyle \underset{\left[a_3,b_3\right]}{\int }}fd{\mu }_3$$

holds for every convex function $f:C\to \mathbb{R}$ if and only if

$$q=p+r,$$

(31)

$${\displaystyle \frac{q}{{\mu }_2\left(\left[a_2,b_2\right]\right)}}{\displaystyle \underset{\left[a_2,b_2\right]}{\int }}td{\mu }_2\left(t\right)={\displaystyle \frac{p}{{\mu }_1\left(\left[a_1,b_1\right]\right)}}{\displaystyle \underset{\left[a_1,b_1\right]}{\int }}td{\mu }_1\left(t\right)+{\displaystyle \frac{r}{{\mu }_3\left(\left[a_3,b_3\right]\right)}}{\displaystyle \underset{\left[a_3,b_3\right]}{\int }}td{\mu }_3\left(t\right),$$

(32)

and

$${\displaystyle \underset{\left]a_1,w\right[}{\int }}\left(w-t\right)d{\mu }_1\left(t\right)\ge 0,w\in \left]a_1,b_1\right[,$$

(33)

$${\displaystyle \frac{q}{{\mu }_2\left(\left[a_2,b_2\right]\right)}}{\displaystyle \underset{\left[w,b_2\right[}{\int }}\left(t-w\right)d{\mu }_2\left(t\right)\le {\displaystyle \frac{r}{{\mu }_3\left(\left[a_3,b_3\right]\right)}}{\displaystyle \underset{\left[a_3,b_3\right]}{\int }}\left(t-w\right)d{\mu }_3\left(t\right),w\in \left]a_2,b_2\right[$$

(34)

and

$${\displaystyle \underset{\left[w,b_3\right[}{\int }}\left(t-w\right)d{\mu }_3\left(t\right)\ge 0,w\in \left]a_3,b_3\right[.$$

(35)

Proof. 

The result is an immediate consequence of Theorem 8.

The proof is complete. □

Remark 5. 

(a) As mentioned in Remark 2, for finite measures, conditions (33)–(35) are automatically satisfied. Therefore, in this case, it is sufficient to consider only conditions (31) and (32).

Theorem 2.2 of [5] is the special case of the previous result, where ${\mu }_i$ is the classical Lebesgue measure on ${\mathcal{B}}_{\left[a_i,b_i\right]}\left(i=1,2,3\right)$. Then, of course, the form of (32) is

$$q\left(a_2+b_2\right)=p\left(a_1+b_1\right)+r\left(a_3+b_3\right).$$

(36)

Theorem 2.1 of [5] is the special case of the previous result, where ${\mu }_i$ is the classical Lebesgue measure on ${\mathcal{B}}_{\left[a_i,b_i\right]}\left(i=1,3\right)$ and

$${\mu }_2:={\displaystyle \frac{q}{2}}\left({\epsilon }_{a_2}+{\epsilon }_{b_2}\right).$$

In this case, (32) can also be written as (36).

Proposition 2 is a generalization of the above theorems even for finite measures.

(b) The proofs of Theorem 2.1 and Theorem 2.2 are unique and direct proofs, and Proposition 2 clearly shows the relationship between the two theorems.

Finally, we give interesting refinements of the Hermite–Hadamard–Fejér inequality.

Proposition 3. 

Let a, $b\in \mathbb{R}$, $a<b$. Let λ be the Lebesgue measure on ${\mathcal{B}}_{\left[a,b\right]}$. Let the weight function $w:\left[a,b\right]\to \mathbb{R}$ be positive, Lebesgue-integrable, and symmetric about the vertical line $t={\displaystyle \frac{a+b}{2}}$, that is,

$$w\left(t\right)=w\left(a+b-t\right),t\in \left[a,b\right].$$

Then, there exists a unique positive number $t_0\in \left]0,{\displaystyle \frac{b-a}{2}}\right[$ such that

(a) inequalities

$$f\left({\displaystyle \frac{a+b}{2}}\right){\displaystyle \underset{a}{\overset{b}{\int }}}wd\lambda$$

$$\le 2{\displaystyle \underset{a+t_0}{\overset{b-t_0}{\int }}}fwd\lambda \le {\displaystyle \underset{a}{\overset{b}{\int }}}fwd\lambda \le 2\left({\displaystyle \underset{a}{\overset{a+t_0}{\int }}}fwd\lambda +{\displaystyle \underset{b-t_0}{\overset{b}{\int }}}fwd\lambda \right)$$

(37)

$$\le {\displaystyle \frac{1}{2}}\left({\displaystyle \frac{a+t_0-t_{a,\mu }}{t_0}}\left(f\left(a\right)+f\left(b\right)\right)+{\displaystyle \frac{t_{a,\mu }-a}{t_0}}\left(f\left(a+t_0\right)+f\left(b-t_0\right)\right)\right){\displaystyle \underset{a}{\overset{b}{\int }}}wd\lambda$$

$$\le {\displaystyle \frac{f\left(a\right)+f\left(b\right)}{2}}{\displaystyle \underset{a}{\overset{b}{\int }}}wd\lambda$$

hold for every convex function $f:\left[a,b\right]\to \mathbb{R}$, where

$$t_{a,\mu }:=4{\displaystyle \underset{a}{\overset{a+t_0}{\int }}}tw\left(t\right)d\lambda \left(t\right)/{\displaystyle \underset{a}{\overset{b}{\int }}}wd\lambda$$

represents the barycenter of $w\lambda$ on $\left[a,a+t_0\right]$.

(b) inequalities (37) also hold for every convex function $f:\left]a,b\right[\to \mathbb{R}$ for which $fw$ is Lebesgue-integrable.

Proof. 

(a) Let the measure $\mu :=w\lambda$, that is,

$$\mu \left(A\right):={\displaystyle \underset{A}{\int }}wd\lambda ,A\in {\mathcal{B}}_{\left[a,b\right]}.$$

Then, $\mu$ is absolutely continuous with respect to $\lambda$. Since w is positive, the function

$$p:\left[a,b\right]\to \mathbb{R},t\to \mu \left(\left[a,t\right]\right)$$

is strictly increasing and $p\left(b\right)>0$. Furthermore, p is continuous, $p\left(a\right)=0$ and $p\left({\displaystyle \frac{a+b}{2}}\right)={\displaystyle \frac{1}{2}}\mu \left(\left[a,b\right]\right)$ due to the symmetry property of w. It now follows that there exists a unique positive number $t_0\in \left]0,{\displaystyle \frac{b-a}{2}}\right[$ such that

$$p\left(a+t_0\right)={\displaystyle \frac{1}{4}}\mu \left(\left[a,b\right]\right).$$

Again, using the symmetry of w, we obtain that

$$p\left(b-t_0\right)={\displaystyle \frac{1}{4}}\mu \left(\left[a,b\right]\right).$$

We introduce the following intervals

$$I_1:=\left[a,a+t_0\right],I_2:=\left[a+t_0,b-t_0\right],I_3:=\left[b-t_0,b\right].$$

These intervals are nonoverlapping and

$$\mu \left(I_2\right)=\mu \left(I_1\right)+\mu \left(I_3\right).$$

By applying Lemma 2 first to the interval $\left[a,b\right]$ and then to the interval $I_2$, we obtain

$${\displaystyle \underset{a}{\overset{b}{\int }}}td\mu \left(t\right)={\displaystyle \frac{a+b}{2}}\mu \left(\left[a,b\right]\right),{\displaystyle \underset{I_2}{\int }}td\mu \left(t\right)={\displaystyle \frac{a+b}{2}}\mu \left(I_2\right)={\displaystyle \frac{a+b}{4}}\mu \left(\left[a,b\right]\right).$$

(38)

Since

$${\displaystyle \underset{I_1}{\int }}td\mu \left(t\right)+{\displaystyle \underset{I_2}{\int }}td\mu \left(t\right)+{\displaystyle \underset{I_3}{\int }}td\mu \left(t\right)={\displaystyle \underset{a}{\overset{b}{\int }}}td\mu \left(t\right),$$

it follows from (38) that

$${\displaystyle \underset{I_2}{\int }}td\mu \left(t\right)={\displaystyle \underset{I_1}{\int }}td\mu \left(t\right)+{\displaystyle \underset{I_3}{\int }}td\mu \left(t\right).$$

From the above, we can apply Theorem 8, from which we obtain the second and the third inequalities using

$${\displaystyle \underset{a}{\overset{b}{\int }}}fd\mu ={\displaystyle \underset{I_1}{\int }}fd\mu +{\displaystyle \underset{I_2}{\int }}fd\mu +{\displaystyle \underset{I_3}{\int }}fd\mu .$$

Applying the Hermite–Hadamard–Fejér inequality to f on $I_2$, we obtain the first inequality.

Now, we prove the fourth inequality.

Let

$$t_{b,w}:={\displaystyle \frac{4}{\mu \left(\left[a,b\right]\right)}}{\displaystyle \underset{b-t_0}{\overset{b}{\int }}}tw\left(t\right)d\lambda \left(t\right)$$

be the barycenter of $\mu$ on $\left[b-t_0,b\right]$.

Due to the symmetry property of w and the choice of $t_0$,

$$t_{b,w}={\displaystyle \frac{4}{\mu \left(\left[a,b\right]\right)}}{\displaystyle \underset{a}{\overset{a+t_0}{\int }}}\left(a+b-t\right)w\left(t\right)d\lambda \left(t\right)=a+b-t_{a,w}.$$

(39)

By the Hermite–Hadamard inequality,

$${\displaystyle \frac{4}{\mu \left(\left[a,b\right]\right)}}{\displaystyle \underset{a}{\overset{a+t_0}{\int }}}fwd\lambda \le {\displaystyle \frac{a+t_0-t_{a,w}}{t_0}}f\left(a\right)+{\displaystyle \frac{t_{a,w}-a}{t_0}}f\left(a+t_0\right).$$

Again using the Hermite–Hadamard inequality and (39), we obtain

$${\displaystyle \frac{4}{\mu \left(\left[a,b\right]\right)}}{\displaystyle \underset{b-t_0}{\overset{b}{\int }}}tw\left(t\right)d\lambda \left(t\right)\le {\displaystyle \frac{b-t_{b,w}}{t_0}}f\left(b-t_0\right)+{\displaystyle \frac{t_{b,w}-\left(b-t_0\right)}{t_0}}f\left(b\right)$$

$$={\displaystyle \frac{t_{a,w}-a}{t_0}}f\left(b-t_0\right)+{\displaystyle \frac{a+t_0-t_{a,w}}{t_0}}f\left(b\right).$$

From the previous two inequalities, the fourth inequality is implied.

The fifth inequality remains to be verified, which, with rearrangement, is equivalent to the following

$${\displaystyle \frac{t_{a,w}-a}{t_0}}\left(f\left(a+t_0\right)+f\left(b-t_0\right)\right)\le \left(1-{\displaystyle \frac{a+t_0-t_{a,w}}{t_0}}\right)\left(f\left(a\right)+f\left(b\right)\right),$$

that is,

$${\displaystyle \frac{f\left(a+t_0\right)-f\left(a\right)}{t_0}}\le {\displaystyle \frac{f\left(b\right)-f\left(b-t_0\right)}{t_0}}.$$

This inequality follows immediately from the convexity of f.

(b) Using the intervals

$$I_1:=\left]a,a+t_0\right],I_2,I_3:=\left[b-t_0,b\right[,$$

we can prove exactly as in (a).

The proof is complete. □

Remark 6. 

(a) The two inequalities in (37) illustrate very well the phenomenon of concentration convexity to endpoints. In addition, they refine the Hermite–Hadamard–Fejér inequality.

(b) From (a) and (b), the inequalities in (37) can be considered as an extension of the Hermite–Hadamard–Fejér inequality for bounded open intervals. This is particularly interesting for convex functions on $\left]a,b\right[$ which are unbounded but μ-integrable on $\left]a,b\right[$. The restriction of every convex function $f:\left[a,b\right]\to \mathbb{R}$ to $\left]a,b\right[$ is a bounded and convex function. Conversely, any bounded and convex function on $\left]a,b\right[$ can be extended to a convex function defined on $\left[a,b\right]$. However, there may be unbounded and convex functions on $\left]a,b\right[$ which are μ-integrable on $\left]a,b\right[$, and inequality (37) can be applied to such functions.

A simple consequence of the previous result is the following refinement of the classical Hermite–Hadamard inequality.

Corollary 2. 

Let a, $b\in \mathbb{R}$, $a<b$. Then,

(a) inequalities

$$f\left({\displaystyle \frac{a+b}{2}}\right)$$

$$\le {\displaystyle \frac{2}{b-a}}{\displaystyle \underset{\frac{3a+b}{4}}{\overset{\frac{a+3b}{4}}{\int }}}fd\lambda \le {\displaystyle \frac{1}{b-a}}{\displaystyle \underset{a}{\overset{b}{\int }}}fd\lambda \le {\displaystyle \frac{2}{b-a}}\left({\displaystyle \underset{a}{\overset{\frac{3a+b}{4}}{\int }}}fd\lambda +{\displaystyle \underset{\frac{a+3b}{4}}{\overset{b}{\int }}}fd\lambda \right)$$

(40)

$$\le {\displaystyle \frac{1}{4}}\left(f\left(a\right)+f\left({\displaystyle \frac{3a+b}{4}}\right)+f\left({\displaystyle \frac{a+3b}{4}}\right)+f\left(b\right)\right)\le {\displaystyle \frac{f\left(a\right)+f\left(b\right)}{2}}$$

hold for every convex function $f:\left[a,b\right]\to \mathbb{R}$.

(b) The inequalities in (40) also hold for every convex function $f:\left]a,b\right[\to \mathbb{R}$ for which f is Lebesgue-integrable.

Proof. 

In this case, the number $t_0$ is ${\displaystyle \frac{b-a}{4}}$, and therefore the result follows from Proposition 3 with some calculation.

The proof is complete. □

Remark 7. 

(a) The inequality on the right hand side of (40) is proved in [6] by means of a different method. In the same paper, the inequality

$${\displaystyle \frac{3}{b-a}}{\displaystyle \underset{\frac{2a+b}{3}}{\overset{\frac{a+2b}{3}}{\int }}}fd\lambda \le {\displaystyle \frac{1}{b-a}}{\displaystyle \underset{a}{\overset{b}{\int }}}fd\lambda$$

is shown as a lower bound. The left-hand inequality in (40) is sharper, since

$${\displaystyle \frac{3}{b-a}}{\displaystyle \underset{\frac{2a+b}{3}}{\overset{\frac{a+2b}{3}}{\int }}}fd\lambda \le {\displaystyle \frac{2}{b-a}}{\displaystyle \underset{\frac{3a+b}{4}}{\overset{\frac{a+3b}{4}}{\int }}}fd\lambda .$$

This follows from the fact that f is a continuous convex function on $\left[{\displaystyle \frac{3a+b}{4}},{\displaystyle \frac{a+3b}{4}}\right]$, $\left[{\displaystyle \frac{3a+b}{4}},{\displaystyle \frac{a+3b}{4}}\right]\subset \left[{\displaystyle \frac{2a+b}{3}},{\displaystyle \frac{a+2b}{3}}\right]$ and these two intervals have the same midpoint.

(b) If we want to choose the numbers $a<t_1<t_2<b$ such that

$${\displaystyle \underset{t_1}{\overset{t_2}{\int }}}fd\lambda \le {\displaystyle \underset{a}{\overset{t_1}{\int }}}fd\lambda +{\displaystyle \underset{t_2}{\overset{b}{\int }}}fd\lambda$$

for every convex function $f:\left[a,b\right]\to \mathbb{R}$, then according to Theorem 8,

$$t_2-t_1=t_1-a+b-t_2$$

and

$${\displaystyle \underset{t_1}{\overset{t_2}{\int }}}td\lambda \left(t\right)\le {\displaystyle \underset{a}{\overset{t_1}{\int }}}fd\lambda \left(t\right)+{\displaystyle \underset{t_2}{\overset{b}{\int }}}fd\lambda \left(t\right)$$

necessarily. From these conditions, a simple calculation gives that

$$t_1={\displaystyle \frac{3a+b}{4}},t_2={\displaystyle \frac{a+3b}{4}}.$$

We can see that the inequalities in (40) are satisfied only for the intervals

$$\left[a,{\displaystyle \frac{3a+b}{4}}\right],\left[{\displaystyle \frac{3a+b}{4}},{\displaystyle \frac{a+3b}{4}}\right],\left[{\displaystyle \frac{a+3b}{4}},b\right].$$

5. Conclusions

The notion of convexity and the inequalities associated with it are important both from a theoretical and an application point of view. Abstract versions of an inequality from Bennett were formulated and studied by Niculescu. In this paper, we extend these results to finite signed measures in a more general form. New majorization-type integral inequalities are obtained. To prove the main result, we give a generalization of a recent majorization-type integral inequality, and this novel result is interesting in itself. As applications of the results, we give simple proofs of the integral Jensen and Lah–Ribarič inequalities for finite signed measures, generalize and extend known results, and obtain an interesting new refinement of the Hermite–Hadamard–Fejér inequality.