Integral Jensen–Mercer and Related Inequalities for Signed Measures with Refinements

Abstract

In this paper, we give necessary and sufficient conditions for the integral Jensen–Mercer inequality and closely related inequalities to be satisfied for finite signed measures. As applications, we obtain new special inequalities that are related to the integral Jensen–Steffensen inequality. We also provide refinements of the majorization-type inequality associated with the Jensen–Mercer inequality for finite signed measures. Using the result obtained, we extend a known refinement. The majorization-type inequalities needed for the proofs are interesting in themselves.

1. Introduction

Let $\left(X,\mathcal{A}\right)$ be a measurable space ($\mathcal{A}$ always means a $\sigma$-algebra of subsets of X). The unit mass at $x\in X$ (the Dirac measure at x) is denoted by ${\epsilon }_x$. Assume $\mu$ is a finite signed measure on $\mathcal{A}$. The real vector space of $\mu$-integrable real functions on X is denoted by $L\left(\mu \right)$. The integrable functions are considered to be measurable.

The integral version of Jensen–Mercer inequality (the original inequality is discrete and comes from [1]) is given in [2], and can be stated as follows.

Theorem 1.

Let $\left(X,\mathcal{A}\right)$ be a measurable space, and let μ be a finite measure on $\mathcal{A}$ such that $\mu \left(X\right)>0$. Let $\phi :X\to \left[a,b\right]$ be a measurable function. Then, for any continuous convex function $f:\left[a,b\right]\to \mathbb{R}$,

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

(1)

In [3], there is a detailed analysis of the fact (as can be seen from the proof of the theorem) that Inequality (1) actually consists of two parts: an integral Jensen inequality,

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

(2)

and a majorization-type inequality,

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

(3)

The generalizations of the different Jensen type inequalities (discrete or integral) that allow the use of negative weights are known as Jensen–Steffensen inequalities, since the first inequality of this type was derived by Steffensen [4] (see also [5]). A well-known and general version of the integral Jensen–Steffensen inequality was proved by Boas [6], where X is a compact interval in $\mathbb{R}$ and the Riemann–Stieltjes integral is used. There exist integral Jensen–Mercer inequalities that use these results (see, e.g., [7,8,9]). Using the Lebesgue–Stieltjes integral, a generalization and extension of Boas’ inequality can be found in paper [10]. In the first part of this paper, we obtain necessary and sufficient conditions for all three inequalities (1)–(3) to be satisfied for finite signed measures, so we obtain a detailed and accurate analysis of the integral Jensen–Mercer inequality under these general conditions. An interesting consequence of our results is that, while all three inequalities are satisfied under the conditions of Theorem 1, they are independent for finite signed measures. To the best of my knowledge, this has not been raised in any previous work, and they have not been examined separately. We also emphasize that our results hold for arbitrary finite signed measures, and the previous results are all in some way related to the conditions used in different versions of the integral Jensen–Steffensen inequality. As applications, we generalize a previous result (Theorem 4 of [7]) coming from the integral Jensen–Steffensen inequality, and obtain new inequalities related to it. We illustrate the usefulness of our results with a concrete inequality comparing continuous means derived from signed measures. This generalizes and extends a similar result for classical discrete measures in [1]. In this paper, we concentrate on the integral forms of the inequalities under consideration, but the results obtained also naturally give conditions for the associated discrete inequalities.

There are many results for the refinement of the integral Jensen–Mercer inequality, some of them refining Inequality (2) (see, e.g., [11,12]) and some of them refining Inequality (3) (see, e.g., [2,3,13]). Since there are many refinements of the integral Jensen inequality, they can be used to obtain refinements of the integral Jensen–Mercer inequality via Inequality (2). The refinement of majorization-type inequalities is not so widely studied, so, in this paper, we focus on the refinement of Inequality (3) for finite signed measures. In the second part of the paper, we give a refinement of Inequality (3) for finite signed measures, which is also derived from a majorization-type inequality. We obtain necessary and sufficient conditions for the refinement under consideration. The result gives a new approach to refining Inequality (3), even for finite measures. As an application, we extend an interesting and known refinement of (3) (for measures) from [2] to finite signed measures.

Our results are mainly based on the results of paper [14]. We will also need to extend some of these, which are interesting statements in themselves.

2. Preliminary Results

In the introduction, we have already seen that Inequality (2) is an integral Jensen inequality, while Inequality (3) is a majorization-type inequality. Accordingly, we briefly review the variations of these inequalities with respect to signed measures. Since, in the present paper, we work with convex functions defined on compact intervals, the following results will be formulated for such intervals. Results for arbitrary intervals are also valid, and can be found in papers [10,14].

We start with a few remarks on integrability, which will be used throughout the paper without further reference.

Let $\left(X,\mathcal{A}\right)$ be a measurable space, let $\mu$ be a finite signed measure on $\mathcal{A}$, and let a, $b\in \mathbb{R}$ with $a<b$. It follows that if $\phi :X\to \left[a,b\right]$ is a measurable function, then $\phi \in L\left(\mu \right)$. If $f:\left[a,b\right]\to \mathbb{R}$ is a convex function, then it is bounded and Borel measurable, and therefore $f\circ \phi \in L\left(\mu \right)$.

The exact conditions under which the integral Jensen inequality is satisfied for signed measures can be found in [14].

Theorem 2

(see Theorem 12 of [14]). 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 a, $b\in \mathbb{R}$ with $a<b$, and let $\phi :X\to \left[a,b\right]$ be a measurable function. Then,

(a) If

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

(4)

and

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

(5)

then

$${\displaystyle \frac{1}{\mu \left(X\right)}}\underset{X}{\int }\phi d\mu \in \left[a,b\right].$$

(6)

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

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

(7)

holds if and only if (4) and (5) are satisfied.

To give a very special case of the extension of Boas’ inequality in [10], we need some preparation.

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

Let $\overline{c}$, $\overline{d}$, c, $d\in \mathbb{R}$ with $\overline{c}<c<d<\overline{d}$, and let $\lambda :\left[c,d\right]\to \mathbb{R}$ be of bounded variation. Extend the function $\lambda$ to the open interval $\left]\overline{c},\overline{d}\right[$ by

$$\overline{\lambda }\left(x\right):=\left\{\begin{array}{c}\lambda \left(c\right),x\in \left]\overline{c},c\right[\\ \lambda \left(x\right),x\in \left[c,d\right]\\ \lambda \left(d\right),x\in \left]d,\overline{d}\right[\end{array}\right..$$

We denote by ${\nu }_{\lambda }$ the Lebesgue–Stieltjes signed measure on ${\mathcal{B}}_{\left]\overline{c},\overline{d}\right[}$ associated with $\overline{\lambda }$. It is obvious that the restriction of ${\nu }_{\lambda }$ to ${\mathcal{B}}_{\left[c,d\right]}$ does not depend on the choice of $\overline{c}$ and $\overline{d}$, and ${\nu }_{\lambda }\left(\left[c,d\right]\right)=\lambda \left(d\right)-\lambda \left(c\right)$.

Theorem 3

(see Theorem 7 of [10]). Let c, $d\in \mathbb{R}$ with $c<d$, and let $\lambda :\left[c,d\right]\to \mathbb{R}$ be of a bounded variation satisfying

$$\lambda \left(c\right)\le \lambda \left(x\right)\le \lambda \left(d\right)$$

and $\lambda \left(c\right)<\lambda \left(d\right)$. Let a, $b\in \mathbb{R}$ with $a<b$, and let $\phi :\left[c,d\right]\to \left[a,b\right]$ be monotonic (in either sense). Suppose further that at any point of $\left[c,d\right]$ at least one of λ and φ is continuous.

Then,

$${\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}{\displaystyle \underset{\left[c,d\right]}{\int }}\phi d{\nu }_{\lambda }\in \left[a,b\right]$$

and

$$f\left({\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}{\displaystyle \underset{\left[c,d\right]}{\int }}\phi d{\nu }_{\lambda }\right)\le {\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}{\displaystyle \underset{\left[c,d\right]}{\int }}f\circ \phi d{\nu }_{\lambda }$$

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

In the next part of this section, we review the required majorization-type inequalities for signed measures.

The following statement can be found in a more general form in [14].

Theorem 4

(see Theorem 6 of [14]). Let $\left(X,\mathcal{A}\right)$ and $\left(Y,\mathcal{B}\right)$ be measurable spaces, let μ be a finite signed measure on a $\mathcal{A}$, and let ν be a finite signed measure on $\mathcal{B}$. Let a, $b\in \mathbb{R}$ with $a<b$, and let $\phi :X\to \left[a,b\right]$, $\psi :Y\to \left[a,b\right]$ be measurable functions. Then, for every convex function $f:\left[a,b\right]\to \mathbb{R}$, the inequality

$$\underset{X}{\int }f\circ \phi d\mu \le \underset{Y}{\int }f\circ \psi d\nu$$

holds if and only if

$$\mu \left(X\right)=\nu \left(Y\right),\underset{X}{\int }\phi d\mu =\underset{Y}{\int }\psi d\nu$$

and

$$\underset{\left\{\phi \ge w\right\}}{\int }\left(\phi -w\right)d\mu \le \underset{\left\{\psi \ge w\right\}}{\int }\left(\psi -w\right)d\nu ,w\in \left]a,b\right[.$$

We need the next result, which is interesting in itself, and is an extension of the previous statement.

Theorem 5.

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 a finite signed measure on ${\mathcal{A}}_i$, and let ${\nu }_j$ be a finite signed measure on ${\mathcal{B}}_j$. Let a, $b\in \mathbb{R}$ with $a<b$, and let ${\phi }_i:X_i\to \left[a,b\right]$$\left(i=1,\dots ,m\right)$, ${\psi }_j:Y_j\to \left[a,b\right]$ $\left(j=1,\dots ,n\right)$ be measurable functions. Then, for every convex function $f:\left[a,b\right]\to \mathbb{R}$, 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

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

and

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

Proof. 

The train of thought for proving Theorem 6 of [14] can be followed. □

We will also use the following generalization of Theorem 12 (c) in [14].

Theorem 6.

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 a, $b\in \mathbb{R}$ with $a<b$, and let $\phi :X\to \left[a,b\right]$ be a measurable function. Introduce the following notation:

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

Then, for any convex function $f:\left[a,b\right]\to \mathbb{R}$, the inequality

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

(8)

holds if and only if

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

(9)

Proof. 

Define the finite signed measure $\nu$ on ${\mathcal{B}}_{\left[a,b\right]}$ by

$$\nu :=\mu \left(X\right){\displaystyle \frac{b-t_{\phi ,\mu }}{b-a}}{\epsilon }_a+\mu \left(X\right){\displaystyle \frac{t_{\phi ,\mu }-a}{b-a}}{\epsilon }_b.$$

Then,

$$\mu \left(X\right)\left({\displaystyle \frac{b-t_{\phi ,\mu }}{b-a}}f\left(a\right)+{\displaystyle \frac{t_{\phi ,\mu }-a}{b-a}}f\left(b\right)\right)={\displaystyle \underset{\left[a,b\right]}{\int }}fd\nu ,$$

and hence Inequality (8) is equivalent to the inequality

$$\underset{X}{\int }f\circ \phi d\mu \le {\displaystyle \underset{\left[a,b\right]}{\int }}fd\nu .$$

By Theorem 4, the previous inequality holds for every convex function $f:\left[a,b\right]\to \mathbb{R}$ if and only if

$$\mu \left(X\right)=\nu \left(\left[a,b\right]\right),\underset{X}{\int }\phi d\mu =\underset{\left[a,b\right]}{\int }sd\nu \left(s\right)$$

and

$$\underset{\left\{\phi \ge w\right\}}{\int }\left(\phi -w\right)d\mu \le \underset{\left[w,b\right]}{\int }\left(s-w\right)d\nu \left(s\right),w\in \left]a,b\right[.$$

(10)

The equality criteria are obvious.

Inequality (10) can be rewritten in the form

$$\underset{\left\{\phi \ge w\right\}}{\int }\left(\phi -w\right)d\mu \le \mu \left(X\right){\displaystyle \frac{t_{\phi ,\mu }-a}{b-a}}\left(b-w\right)={\displaystyle \frac{b-w}{b-a}}\underset{X}{\int }\left(\phi -a\right)d\mu ,$$

and from this, we obtain (9) by elementary calculation. □

Remark 1.

(a) The conditions of Theorem 12 (c) of [14] include the assumption $t_{\phi ,\mu }\in \left[a,b\right]$, and its proof essentially uses this. Condition (9) does not imply $t_{\phi ,\mu }\in \left[a,b\right]$ in general.

(b) The first result in this direction was Theorem 1 in [15], where μ is a finite signed measure on ${\mathcal{B}}_{\left[a,b\right]}$ such that $\mu \left(\left[a,b\right]\right)=1$, and φ is the identity function on $\left[a,b\right]$. Theorem 6 contains this result as a special case.

(c) If μ is a measure, (8) is the well-known Lah–Ribarič inequality (see [16]), for which we obtain a new proof.

Finally, we need the following result for the Lebesgue–Stieltjes integrals.

Lemma 1

(see Lemma 1 of [10]). Let c, $d\in \mathbb{R}$ with $c<d$, let $\lambda :\left[c,d\right]\to \mathbb{R}$ be of a bounded variation satisfying $\lambda \left(c\right)\le \lambda \left(d\right)$, let $\phi :\left[c,d\right]\to \mathbb{R}$ be nonnegative, and suppose that at any point of $\left[c,d\right]$, at least one of λ and φ is continuous.

If either φ is increasing and

$$\lambda \left(s\right)\le \lambda \left(d\right),s\in \left[c,d\right],$$

or φ is decreasing and

$$\lambda \left(c\right)\le \lambda \left(s\right),s\in \left[c,d\right],$$

then

$${\displaystyle \underset{\left[c,d\right]}{\int }}\phi d{\nu }_{\lambda }\ge 0.$$

3. Extension of Integral Jensen–Mercer and Related Inequalities to Signed Measures

In our first statement, we give exact conditions for any of the inequalities (1)–(3) to hold for signed measures.

Theorem 7.

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 a, $b\in \mathbb{R}$ with $a<b$, and let $\phi :X\to \left[a,b\right]$ be a measurable function. Then,

(a) Assume $t_{\phi ,\mu }:={\displaystyle \frac{1}{\mu \left(X\right)}}\underset{X}{\int }\phi d\mu \in \left[a,b\right]$.

The integral Jensen–Mercer inequality

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

(11)

holds for any convex function $f:\left[a,b\right]\to \mathbb{R}$ if and only if

$${\displaystyle \frac{1}{\mu \left(X\right)}}\underset{\left\{\phi <w\right\}}{\int }\left(w-\phi \right)d\mu \le w-a,w\in \left]a,t_{\phi ,\mu }\right]$$

(12)

and

$${\displaystyle \frac{1}{\mu \left(X\right)}}\underset{\left\{\phi \ge w\right\}}{\int }\left(\phi -w\right)d\mu \le b-w,w\in \left]t_{\phi ,\mu },b\right[.$$

(13)

(b) The integral Jensen inequality

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

(14)

holds for any convex function $f:\left[a,b\right]\to \mathbb{R}$ if and only if

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

(15)

and

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

(16)

(c) The majorization-type inequality

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

(17)

holds for any convex function $f:\left[a,b\right]\to \mathbb{R}$ if and only if

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

$$\le \underset{X}{\int }\left(w-a\right)d\mu ,w\in \left]a,{\displaystyle \frac{a+b}{2}}\right].$$

(18)

Proof. 

(a) Since $t_{\phi ,\mu }\in \left[a,b\right]$, $a+b-t_{\phi ,\mu }$ also belongs to $\left[a,b\right]$.

Define the finite signed measure $\nu$ on ${\mathcal{B}}_{\left[a,b\right]}$ by

$$\nu :=\mu \left(X\right)\left({\epsilon }_a+{\epsilon }_b\right).$$

Then, Inequality (11) is equivalent to

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

By Theorem 5, this inequality is fulfilled for any convex function $f:\left[a,b\right]\to \mathbb{R}$ exactly, if

$$2\mu \left(X\right)=\nu \left(\left[a,b\right]\right),\underset{X}{\int }\phi d\mu +\underset{X}{\int }\left(a+b-{\displaystyle \frac{1}{\mu \left(X\right)}}\underset{X}{\int }\phi d\mu \right)d\mu =\underset{\left[a,b\right]}{\int }sd\nu \left(s\right)$$

(19)

and

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

(20)

$$\le \underset{\left[w,b\right]}{\int }\left(s-w\right)d\nu \left(s\right)=\mu \left(X\right)\left(b-w\right),w\in \left]a,b\right[.$$

Conditions (19) are clearly satisfied.

If $w\le t_{\phi ,\mu }$, then (20) means

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

which is exactly (12).

If $w>t_{\phi ,\mu }$, then (20) means

$$\underset{\left\{\phi \ge w\right\}}{\int }\left(\phi -w\right)d\mu \le \mu \left(X\right)\left(b-w\right),$$

which is just (13).

(b) Since

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

it follows from Theorem 2 that Inequality (14) holds for any convex function $f:\left[a,b\right]\to \mathbb{R}$ if and only if

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

and

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

By introducing the notation $w:=a+b-u$, Inequalities (15) and (16) are obtained.

(c) By $\mu \left(X\right)>0$, Inequality (17) is equivalent to

$$\underset{X}{\int }f\circ \phi d\mu +\underset{X}{\int }f\circ \left(a+b-\phi \right)d\mu \le \underset{X}{\int }f\left(a\right)d\mu +\underset{X}{\int }f\left(b\right)d\mu .$$

We can apply Theorem 5, which states that the above inequality holds for any convex function $f:\left[a,b\right]\to \mathbb{R}$ exactly if

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

$$\le \underset{X}{\int }\left(b-w\right)d\mu ,w\in \left]a,b\right[,$$

which is the same as

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

$$\le \underset{X}{\int }\left(w-a\right)d\mu ,w\in \left]a,b\right[.$$

(21)

It is sufficient to show that (18) implies (21).

To prove this, let ${\displaystyle \frac{a+b}{2}}<w<b$. Then by using (18), we obtain

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

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

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

$$\le \underset{X}{\int }\left(2w-\left(a+b\right)\right)d\mu +\underset{X}{\int }\left(\left(a+b-w\right)-a\right)d\mu =\underset{X}{\int }\left(w-a\right)d\mu .$$

□

A direct consequence of the previous statement is the following.

Corollary 1.

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 a, $b\in \mathbb{R}$ with $a<b$, and let $\phi :X\to \left[a,b\right]$ be a measurable function. Then inequalities

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

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

hold for any convex function $f:\left[a,b\right]\to \mathbb{R}$ if and only if (15), (16) and (18) are satisfied.

Remark 2.

(a) To the best of the author’s knowledge, the inequalities in Theorem 7 have not been studied under such general conditions, and the result provides a new approach to the subject. Moreover, it gives precise conditions for their fulfillment.

(b) If μ is a finite measure, then Conditions (15), (16) and (18) are obviously satisfied, and therefore Corollary 1 contains Theorem 1 in the stronger form, such that Inequalities (2) and (3) are satisfied.

(c) It can be illustrated by simple examples that Conditions (12), (13), (15), (16) and (18) are independent, and even

$${\displaystyle \frac{1}{\mu \left(X\right)}}\underset{X}{\int }\phi d\mu \in \left[a,b\right]$$

does usually not follow from any of Conditions (12), (13) and (18).

(d) We show only one interesting example, which will be used in the next application. Consider the measure space $\left(\left[0,2\right],{\mathcal{B}}_{\left[0,2\right]}\right)$, and define the finite signed measure μ on ${\mathcal{B}}_{\left[0,2\right]}$ as follows:

$$\mu :=2{\epsilon }_0-{\epsilon }_1+{\epsilon }_2.$$

Let $\phi :\left[0,2\right]\to \left[0,1\right]$, $\phi \left(x\right):={\displaystyle \frac{1}{2}}x$.

Then,

$$f\left(0+1-{\displaystyle \frac{1}{\mu \left(\left[0,2\right]\right)}}\underset{\left[0,2\right]}{\int }\phi d\mu \right)=f\left({\displaystyle \frac{3}{4}}\right),$$

$${\displaystyle \frac{1}{\mu \left(\left[0,2\right]\right)}}\underset{\left[0,2\right]}{\int }f\circ \left(0+1-\phi \right)d\mu =f\left(0\right)-{\displaystyle \frac{1}{2}}f\left({\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{1}{2}}f\left(1\right)$$

and

$$f\left(0\right)+f\left(1\right)-{\displaystyle \frac{1}{\mu \left(\left[0,2\right]\right)}}\underset{\left[0,2\right]}{\int }f\circ \phi d\mu ={\displaystyle \frac{1}{2}}f\left({\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{1}{2}}f\left(1\right).$$

It can be easily checked that the integral Jensen–Mercer inequality

$$f\left(0+1-{\displaystyle \frac{1}{\mu \left(\left[0,2\right]\right)}}\underset{\left[0,2\right]}{\int }\phi d\mu \right)\le f\left(0\right)+f\left(1\right)-{\displaystyle \frac{1}{\mu \left(\left[0,2\right]\right)}}\underset{\left[0,2\right]}{\int }f\circ \phi d\mu$$

and the integral Jensen inequality

$$f\left(0+1-{\displaystyle \frac{1}{\mu \left(\left[0,2\right]\right)}}\underset{\left[0,2\right]}{\int }\phi d\mu \right)\le {\displaystyle \frac{1}{\mu \left(\left[0,2\right]\right)}}\underset{\left[0,2\right]}{\int }f\circ \left(0+1-\phi \right)d\mu$$

hold for any convex function $f:\left[0,1\right]\to \mathbb{R}$, but the majorization-type inequality is not satisfied for all convex functions $f:\left[0,1\right]\to \mathbb{R}$, since for $f\left(0\right)<f\left({\displaystyle \frac{1}{2}}\right)$,

$${\displaystyle \frac{1}{\mu \left(\left[0,2\right]\right)}}\underset{\left[0,2\right]}{\int }f\circ \left(0+1-\phi \right)d\mu <f\left(0\right)+f\left(1\right)-{\displaystyle \frac{1}{\mu \left(\left[0,2\right]\right)}}\underset{\left[0,2\right]}{\int }f\circ \phi d\mu ,$$

while for $f\left(0\right)>f\left({\displaystyle \frac{1}{2}}\right)$, the opposite inequality is true.

As an application, using the previous theorem and the integral Jensen–Steffensen inequality in Theorem 3, we give more easily verifiable sufficient conditions for the integral Jensen–Mercer inequality (11).

Theorem 8.

Let c, $d\in \mathbb{R}$ with $c<d$, and let $\lambda :\left[c,d\right]\to \mathbb{R}$ be of a bounded variation satisfying

$$\lambda \left(c\right)\le \lambda \left(x\right)\le \lambda \left(d\right)$$

and $\lambda \left(c\right)<\lambda \left(d\right)$. Let a, $b\in \mathbb{R}$ with $a<b$, and let $\phi :\left[c,d\right]\to \left[a,b\right]$ be monotonic (in either sense). Suppose further that at any point of $\left[c,d\right]$, at least one of λ and φ is continuous.

Then, inequality

$$f\left(a+b-{\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}{\displaystyle \underset{\left[c,d\right]}{\int }}\phi d{\nu }_{\lambda }\right)\le f\left(a\right)+f\left(b\right)-{\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}\underset{\left[c,d\right]}{\int }f\circ \phi d{\nu }_{\lambda }$$

(22)

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

Proof. 

By Theorem 3,

$$t_{\phi ,\mu }:={\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}{\displaystyle \underset{\left[c,d\right]}{\int }}\phi d{\nu }_{\lambda }\in \left[a,b\right].$$

It follows from Theorem 7(a) that (22) holds for any convex function $f:\left[a,b\right]\to \mathbb{R}$ if

$${\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}\underset{\left\{\phi <w\right\}}{\int }\left(w-\phi \right)d{\nu }_{\lambda }\le w-a,w\in \left]a,t_{\phi ,\mu }\right]$$

(23)

and

$${\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}\underset{\left\{\phi \ge w\right\}}{\int }\left(\phi -w\right)d{\nu }_{\lambda }\le b-w,w\in \left]t_{\phi ,\mu },b\right[.$$

(24)

Inequality (23) is equivalent to

$$0\le \underset{\left\{\phi <w\right\}}{\int }\left(\phi -a\right)d{\nu }_{\lambda }+\underset{\left\{\phi \ge w\right\}}{\int }\left(w-a\right)d{\nu }_{\lambda }=\underset{\left[c,d\right]}{\int }{\chi }_{1}d{\nu }_{\lambda },w\in \left]a,t_{\phi ,\mu }\right],$$

where

$${\chi }_1\left(x\right):=\left\{\begin{array}{c}\phi \left(x\right)-a\mathrm{if}x\in \left\{\phi <w\right\}\\ w-a\mathrm{if}x\in \left\{\phi \ge w\right\}\end{array}\right.,$$

while Inequality (24) is equivalent to

$$0\le \underset{\left\{\phi <w\right\}}{\int }\left(b-w\right)d{\nu }_{\lambda }+\underset{\left\{\phi \ge w\right\}}{\int }\left(b-\phi \right)d{\nu }_{\lambda }=\underset{\left[c,d\right]}{\int }{\chi }_{2}d{\nu }_{\lambda },w\in \left]t_{\phi ,\mu },b\right[,$$

where

$${\chi }_2\left(x\right):=\left\{\begin{array}{c}b-w\mathrm{if}x\in \left\{\phi <w\right\}\\ b-\phi \left(x\right)\mathrm{if}x\in \left\{\phi \ge w\right\}\end{array}\right..$$

Let $w\in \left]a,b\right[$ be fixed, and assume $\phi$ is increasing. Then, the sets $\left\{\phi <w\right\}$ and $\left\{\phi \ge w\right\}$ are disjoint intervals, whose union is $\left[c,d\right]$, the function ${\chi }_1$ is nonnegative and increasing, while the function ${\chi }_2$ is nonnegative and decreasing, and therefore Lemma 1 can be applied.

We can prove in a similar way if $\phi$ is decreasing. □

Remark 3.

(a) Theorem 8, on the one hand, generalizes Theorem 4 of [7], but also takes a different, more natural approach to the proof. In Theorem 4 of [7], the functions φ and f are continuous, and the Riemann–Stieltjes integral is used.

(b) The example in Remark 1 (d) shows that under the conditions of Theorem 8 the majorization-type inequality

$${\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}\underset{\left[c,d\right]}{\int }f\circ \left(a+b-\phi \right)d{\nu }_{\lambda }\le f\left(a\right)+f\left(b\right)-{\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}\underset{\left[c,d\right]}{\int }f\circ \phi d{\nu }_{\lambda }$$

does not hold for any convex function $f:\left[a,b\right]\to \mathbb{R}$ in general.

The applicability of the previous statement is illustrated by the following specific case, which is related to means.

Corollary 2.

Let c, $d\in \mathbb{R}$ with $c<d$, and let $\lambda :\left[c,d\right]\to \mathbb{R}$ be of bounded variation satisfying

$$\lambda \left(c\right)\le \lambda \left(x\right)\le \lambda \left(d\right)$$

and $\lambda \left(c\right)<\lambda \left(d\right)$. Let a, $b\in \mathbb{R}$ with $0<a<b$, and let $\phi :\left[c,d\right]\to \left[a,b\right]$ be monotonic (in either sense). Suppose further that at any point of $\left[c,d\right]$, at least one of λ and φ is continuous. Then,

$$\tilde{A}:=a+b-{\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}{\displaystyle \underset{\left[c,d\right]}{\int }}\phi d{\nu }_{\lambda }\ge {\displaystyle \frac{ab}{exp\left(\frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}{\displaystyle \underset{\left[c,d\right]}{\int }}ln\circ \phi d{\nu }_{\lambda }\right)}}=:\tilde{G}.$$

(25)

Proof. 

Let f be the restriction of $-ln$ to $\left[a,b\right]$ in Theorem 8. We obtain

$$-ln\left(a+b-{\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}{\displaystyle \underset{\left[c,d\right]}{\int }}\phi d{\nu }_{\lambda }\right)$$

$$\le -ln\left(a\right)-ln\left(b\right)-{\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}{\displaystyle \underset{\left[c,d\right]}{\int }}\left(-ln\circ \phi \right)d{\nu }_{\lambda },$$

which is equivalent to (25). □

Remark 4.

(a) Some clarification is needed to show that $\tilde{A}$ and $\tilde{G}$ are indeed means under the conditions of the statement.

Applying Theorem 3 first to φ and then to $-ln\circ \phi$ instead of φ ($-ln\circ \phi$ is also monotonic), we see that

$$A:={\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}{\displaystyle \underset{\left[c,d\right]}{\int }}\phi d{\nu }_{\lambda }\in \left[a,b\right]$$

and

$${\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}{\displaystyle \underset{\left[c,d\right]}{\int }}\left(ln\circ \phi \right)d{\nu }_{\lambda }\in \left[ln\left(a\right),ln\left(b\right)\right],$$

and, therefore,

$$G:=exp\left({\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}{\displaystyle \underset{\left[c,d\right]}{\int }}ln\circ \phi d{\nu }_{\lambda }\right)\in \left[a,b\right].$$

They imply that

$$\tilde{A}\in \left[a,b\right]and\tilde{G}\in \left[a,b\right].$$

If λ is increasing, i.e., ${\nu }_{\lambda }$ is a measure, then A is the arithmetic mean of φ, while G is the geometric mean of φ.

(b) Corollary 2 is a generalization of the similar result for discrete means in [1]. We emphasize that we have obtained a result for continuous means defined by signed measures.

As another application of Theorem 7, we strengthen the conditions of Theorem 8 so that Inequalities (14) and (17) are satisfied together. The advantage of the result is that easily verifiable conditions are obtained within a familiar framework.

Theorem 9.

Let c, $d\in \mathbb{R}$ with $c<d$, and let $\lambda :\left[c,d\right]\to \mathbb{R}$ be of bounded variation satisfying $\lambda \left(c\right)<\lambda \left(d\right)$. Let a, $b\in \mathbb{R}$ with $a<b$, and let $\phi :\left[c,d\right]\to \left[a,b\right]$ be increasing for which $\phi \left(c\right)=a$ and $\phi \left(d\right)=b$. Suppose that at any point of $\left[c,d\right]$, at least one of λ and φ is continuous. Define

$$u_{{\displaystyle \frac{a+b}{2}}}:=sup\left\{x\in \left[c,d\right]\mid \phi \left(x\right)<{\displaystyle \frac{a+b}{2}}\right\}$$

and

$$v_{{\displaystyle \frac{a+b}{2}}}:=inf\left\{x\in \left[c,d\right]\mid \phi \left(x\right)>{\displaystyle \frac{a+b}{2}}\right\}.$$

Suppose further that one of the following two conditions are fulfilled.

(i) Inequalities $\lambda \left(c\right)\le \lambda \left(x\right)\le \lambda \left(v_{{\displaystyle \frac{a+b}{2}}}\right)$$\left(c\le x\le v_{{\displaystyle \frac{a+b}{2}}}\right)$ hold, and λ is increasing and continuous on $\left[v_{{\displaystyle \frac{a+b}{2}}},d\right]$.

(ii) The function λ is increasing and continuous on $\left[c,u_{{\displaystyle \frac{a+b}{2}}}\right]$, and $\lambda \left(u_{{\displaystyle \frac{a+b}{2}}}\right)\le \lambda \left(x\right)\le \lambda \left(d\right)$$\left(u_{{\displaystyle \frac{a+b}{2}}}\le x\le d\right)$.

Then, the inequalities

$$f\left(a+b-{\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}{\displaystyle \underset{\left[c,d\right]}{\int }}\phi d{\nu }_{\lambda }\right)\le {\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}{\displaystyle \underset{\left[c,d\right]}{\int }}f\circ \left(a+b-\phi \right)d{\nu }_{\lambda }$$

(26)

$$\le f\left(a\right)+f\left(b\right)-{\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}{\displaystyle \underset{\left[c,d\right]}{\int }}f\circ \phi d{\nu }_{\lambda }$$

(27)

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

Proof. 

Since the conditions of Theorem 3 are satisfied,

$${\displaystyle \frac{1}{{\nu }_{\lambda }\left(\left[c,d\right]\right)}}{\displaystyle \underset{\left[c,d\right]}{\int }}\phi d{\nu }_{\lambda }\in \left[a,b\right]$$

and (26) follow.

By Theorem 7(c), we only need to show that

$${\displaystyle \underset{\left\{\phi <w\right\}}{\int }}\left(w-\phi \right)d{\nu }_{\lambda }+{\displaystyle \underset{\left\{\phi \ge a+b-w\right\}}{\int }}\left(\phi -\left(a+b-w\right)\right)d{\nu }_{\lambda }$$

$$\le {\displaystyle \underset{\left[c,d\right]}{\int }}\left(w-a\right)d{\nu }_{\lambda },w\in \left]a,{\displaystyle \frac{a+b}{2}}\right].$$

(28)

Let $w\in \left]a,{\displaystyle \frac{a+b}{2}}\right]$ be fixed.

We start with some notations.

Since $\phi$ is increasing with $\phi \left(c\right)=a$ and $\phi \left(d\right)=b$, the sets

$$I_w:=\left\{\phi <w\right\},J_{a+b-w}:=\left\{\phi >a+b-w\right\}$$

are nonempty and disjoint subintervals of $\left[c,d\right]$, the starting point of $I_w$ is c, and the ending point of $J_{a+b-w}$ is d. Denote $u_w$ as the ending point of $I_w$, and $v_{a+b-w}$ as the starting point of $J_{a+b-w}$. Obviously, $c\le u_w\le v_{a+b-w}\le d$.

Let

$$K_w:=\left[c,d\right]\setminus \left(I_w{\displaystyle \bigcup }{J}_{a+b-w}\right).$$

Then $K_w$ is also an interval, which can be empty, with starting point $u_w$ and ending point $v_{a+b-w}$.

Using these intervals, (28) is equivalent to

$$0\le {\displaystyle \underset{I_w}{\int }}\left(\phi -a\right)d{\nu }_{\lambda }+{\displaystyle \underset{K_w}{\int }}\left(w-a\right)d{\nu }_{\lambda }+{\displaystyle \underset{J_{a+b-w}}{\int }}\left(b-\phi \right)d{\nu }_{\lambda }.$$

(29)

We consider two cases according to Conditions (i) and (ii).

(a) Assume Condition (i) is satisfied.

Then, $\lambda$ is continuous at $v_{a+b-w}$, $\lambda \left(x\right)\le \lambda \left(v_{a+b-w}\right)$$\left(c\le x\le v_{a+b-w}\right)$ and $\lambda \left(x\right)\ge \lambda \left(v_{a+b-w}\right)$ $\left(v_{a+b-w}\le x\le d\right)$.

Let the function $\widehat{\phi }$ be defined on the interval $I_w{\displaystyle \bigcup }{K}_w{\displaystyle \bigcup }\left\{v_{a+b-w}\right\}$ by

$$\widehat{\phi }\left(x\right):=\left\{\begin{array}{c}\phi \left(x\right),\mathrm{if}x\in I_w\hfill \\ w,\mathrm{if}x\in K_w{\displaystyle \bigcup }\left\{v_{a+b-w}\right\}\hfill \end{array}\right..$$

Due to the continuity of $\lambda$ at $v_{a+b-w}$,

$${\displaystyle \underset{I_w}{\int }}\left(\phi -a\right)d{\nu }_{\lambda }+{\displaystyle \underset{K_w}{\int }}\left(w-a\right)d{\nu }_{\lambda }={\displaystyle \underset{\left[c,v_{a+b-w}\right]}{\int }}\left(\widehat{\phi }-a\right)d{\nu }_{\lambda }.$$

(30)

Since $\widehat{\phi }-a$ is nonnegative and increasing on $\left[c,v_{a+b-w}\right]$ and $\lambda \left(x\right)\le \lambda \left(v_{a+b-w}\right)$$\left(c\le x\le v_{a+b-w}\right)$, we can apply Lemma 1, which states by (30) that

$$0\le {\displaystyle \underset{I_w}{\int }}\left(\phi -a\right)d{\nu }_{\lambda }+{\displaystyle \underset{K_w}{\int }}\left(w-a\right)d{\nu }_{\lambda }.$$

According to the continuity of $\lambda$ at $v_{a+b-w}$,

$${\displaystyle \underset{J_{a+b-w}}{\int }}\left(b-\phi \right)d{\nu }_{\lambda }={\displaystyle \underset{\left[v_{a+b-w},d\right]}{\int }}\left(b-\phi \right)d{\nu }_{\lambda }.$$

Since $b-\phi$ is decreasing on $\left[v_{a+b-w},d\right]$ and $\lambda \left(x\right)\ge \lambda \left(v_{a+b-w}\right)$$\left(v_{a+b-w}\le x\le d\right)$, Lemma 1 can be applied again, and we obtain

$$0\le {\displaystyle \underset{J_{a+b-w}}{\int }}\left(b-\phi \right)d{\nu }_{\lambda }.$$

We can see that, under Condition (i), Inequality (29) is satisfied.

(b) Assume Condition (ii) is satisfied.

Then $\lambda$ is continuous at $u_w$, $\lambda \left(x\right)\le \lambda \left(u_w\right)$$\left(c\le x\le u_w\right)$ and $\lambda \left(x\right)\ge \lambda \left(u_w\right)$$\left(u_w\le x\le d\right)$.

Let the function $\widehat{\phi }$ be defined on the interval $\left\{u_w\right\}{\displaystyle \bigcup }{K}_w{\displaystyle \bigcup }{J}_{a+b-w}$ by

$$\widehat{\phi }\left(x\right):=\left\{\begin{array}{c}b+a-w,\mathrm{if}x\in \left\{u_w\right\}{\displaystyle \bigcup }{K}_w\hfill \\ \phi \left(x\right),\mathrm{if}x\in J_{a+b-w}\hfill \end{array}\right..$$

Due to the continuity of $\lambda$ at $u_w$,

$${\displaystyle \underset{I_w}{\int }}\left(\phi -a\right)d{\nu }_{\lambda }={\displaystyle \underset{\left[c,u_w\right]}{\int }}\left(\phi -a\right)d{\nu }_{\lambda }.$$

Since $\phi -a$ is increasing on $\left[c,u_w\right]$ and $\lambda \left(x\right)\le \lambda \left(u_w\right)$$\left(c\le x\le u_w\right)$, Lemma 1 implies that

$$0\le {\displaystyle \underset{I_w}{\int }}\left(\phi -a\right)d{\nu }_{\lambda }.$$

According to the continuity of $\lambda$ at $u_w$,

$${\displaystyle \underset{K_w}{\int }}\left(w-a\right)d{\nu }_{\lambda }+{\displaystyle \underset{J_{a+b-w}}{\int }}\left(b-\phi \right)d{\nu }_{\lambda }={\displaystyle \underset{\left[u_w,d\right]}{\int }}\left(b-\widehat{\phi }\right)d{\nu }_{\lambda }.$$

(31)

Since $b-\widehat{\phi }$ is nonnegative and decreasing on $\left[u_w,d\right]$ and $\lambda \left(x\right)\ge \lambda \left(u_w\right)$$\left(u_w\le x\le d\right)$, another application of Lemma 1 by (31) gives that

$$0\le {\displaystyle \underset{K_w}{\int }}\left(w-a\right)d{\nu }_{\lambda }+{\displaystyle \underset{J_{a+b-w}}{\int }}\left(b-\phi \right)d{\nu }_{\lambda }.$$

It can be seen that, under Condition (ii), Inequality (29) is also satisfied. □

If $\phi$ is decreasing, then the counterpart of the previous result is the following statement.

Theorem 10.

Let c, $d\in \mathbb{R}$ with $c<d$, and let $\lambda :\left[c,d\right]\to \mathbb{R}$ be of bounded variation satisfying $\lambda \left(c\right)<\lambda \left(d\right)$. Let a, $b\in \mathbb{R}$ with $a<b$, and let $\phi :\left[c,d\right]\to \left[a,b\right]$ be decreasing, for which $\phi \left(c\right)=b$ and $\phi \left(d\right)=a$. Suppose that at any point of $\left[c,d\right]$ at least one of λ and φ is continuous. Define

$$u_{{\displaystyle \frac{a+b}{2}}}:=inf\left\{x\in \left[c,d\right]\mid \phi \left(x\right)<{\displaystyle \frac{a+b}{2}}\right\}$$

and

$$v_{{\displaystyle \frac{a+b}{2}}}:=sup\left\{x\in \left[c,d\right]\mid \phi \left(x\right)>{\displaystyle \frac{a+b}{2}}\right\}.$$

Suppose further that one of the following two conditions is fulfilled.

(i) The function λ is increasing and continuous on $\left[c,v_{{\displaystyle \frac{a+b}{2}}}\right]$, and $\lambda \left(v_{{\displaystyle \frac{a+b}{2}}}\right)\le \lambda \left(x\right)\le \lambda \left(d\right)$$\left(v_{{\displaystyle \frac{a+b}{2}}}\le x\le d\right)$

(ii) Inequalities $\lambda \left(c\right)\le \lambda \left(x\right)\le \lambda \left(u_{{\displaystyle \frac{a+b}{2}}}\right)$$\left(c\le x\le u_{{\displaystyle \frac{a+b}{2}}}\right)$ hold, and λ is increasing and continuous on $\left[u_{{\displaystyle \frac{a+b}{2}}},d\right]$.

Then, Inequalities (26) and (27) hold for any convex function $f:\left[a,b\right]\to \mathbb{R}$.

Remark 5.

I could not find any previous results for the joint analysis of Inequalities (26) and (27) for signed measures.

4. Refinements of the Majorization-Type Inequality Associated with the Jensen–Mercer Inequality to Signed Measures

We start with an interesting result in itself, which we will need in the refinements.

Lemma 2.

Let $\left(X,\mathcal{A}\right)$ and $\left(Y,\mathcal{B}\right)$ be measurable spaces, let μ be a finite signed measure on $\mathcal{A}$, and ν be a finite signed measure on $\mathcal{B}$ such that $\mu \left(X\right)>0$ and $\nu \left(Y\right)>0$. Let a, $b\in \mathbb{R}$ with $a<b$. Let $\phi :X\to \left[a,b\right]$ and $\psi :Y\to \left[a,b\right]$ be measurable functions. Then, inequality

$$\underset{X}{\int }f\circ \phi d\mu \le \underset{Y}{\int }f\circ \psi d\nu$$

(32)

holds for any convex function $f:\left[a,b\right]\to \mathbb{R}$ if and only if inequality

$$\underset{X}{\int }f\circ \left(a+b-\phi \right)d\mu \le \underset{Y}{\int }f\circ \left(a+b-\psi \right)d\nu$$

(33)

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

Proof. 

Since the values of $\phi$ and $\psi$ are in $\left[a,b\right]$, the same is true for the functions $a+b-\phi$ and $a+b-\psi$.

It follows from Theorem 4 that Inequality (32) holds for any convex function $f:\left[a,b\right]\to \mathbb{R}$ if and only if

$$\mu \left(X\right)=\nu \left(Y\right),\underset{X}{\int }\phi d\mu =\underset{Y}{\int }\psi d\nu$$

(34)

and

$$\underset{\left\{\phi \ge w\right\}}{\int }\left(\phi -w\right)d\mu \le \underset{\left\{\psi \ge w\right\}}{\int }\left(\psi -w\right)d\nu ,w\in \left]a,b\right[.$$

(35)

By Theorem 4, the Inequality (33) is true exactly if

$$\mu \left(X\right)=\nu \left(Y\right),\underset{X}{\int }\left(a+b-\phi \right)d\mu =\underset{Y}{\int }\left(a+b-\psi \right)d\nu$$

and

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

(36)

The equality conditions are obviously equivalent.

The equivalence of (35) and (36) follows from the equality conditions, and from the fact that

$$w\in \left]a,b\right[\iff a+b-w\in \left]a,b\right[.$$

For completeness, we show how (36) follows from (35).

Using (34) and then (35), we obtain

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

$$+\underset{\left\{\phi \ge a+b-w\right\}}{\int }\left(\phi -\left(a+b-w\right)\right)d\mu \le \underset{Y}{\int }\left(a+b-\psi -w\right)d\nu$$

$$+\underset{\left\{\psi \ge a+b-w\right\}}{\int }\left(\psi -\left(a+b-w\right)\right)d\nu$$

$$=\underset{\left\{\psi <a+b-w\right\}}{\int }\left(a+b-\psi -w\right)d\nu ,w\in \left]a,b\right[,$$

which gives (36). □

Using our results so far, we can obtain refinements of the majorization part of the integral Jensen–Mercer Inequality (see (17)).

Theorem 11.

Let $\left(X,\mathcal{A}\right)$ and $\left(Y,\mathcal{B}\right)$ be measurable spaces, let μ be a finite signed measure on $\mathcal{A}$, and ν be a finite signed measure on $\mathcal{B}$ such that $\mu \left(X\right)>0$ and $\nu \left(Y\right)>0$. Let a, $b\in \mathbb{R}$ with $a<b$. Let $\phi :X\to \left[a,b\right]$ and $\psi :Y\to \left[a,b\right]$ be measurable functions.

Then, the inequalities

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

(37)

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

(38)

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

(39)

hold for any convex function $f:\left[a,b\right]\to \mathbb{R}$ if and only if

$${\displaystyle \frac{1}{\mu \left(X\right)}}\underset{X}{\int }\phi d\mu ={\displaystyle \frac{1}{\nu \left(Y\right)}}\underset{Y}{\int }\psi d\nu ,$$

(40)

$${\displaystyle \frac{1}{\mu \left(X\right)}}\underset{\left\{\phi \ge w\right\}}{\int }\left(\phi -w\right)d\mu \le {\displaystyle \frac{1}{\nu \left(Y\right)}}\underset{\left\{\psi \ge w\right\}}{\int }\left(\psi -w\right)d\nu ,w\in \left]a,b\right[$$

(41)

and

$$\underset{\left\{\psi <w\right\}}{\int }\left(w-\psi \right)d\nu +\underset{\left\{\psi \ge a+b-w\right\}}{\int }\left(\psi -\left(a+b-w\right)\right)d\nu$$

$$\le \underset{Y}{\int }\left(w-a\right)d\nu ,w\in \left]a,{\displaystyle \frac{a+b}{2}}\right].$$

Proof. 

Introducing the finite signed measure $\widehat{\mu }:={\displaystyle \frac{\mu }{\mu \left(X\right)}}$ on $\mathcal{A}$ and $\widehat{\nu }:={\displaystyle \frac{\nu }{\nu \left(Y\right)}}$ on $\mathcal{B}$, Inequalities (37)–(39) are equivalent to

$$\underset{X}{\int }f\circ \phi d\widehat{\mu }+\underset{X}{\int }f\circ \left(a+b-\phi \right)d\widehat{\mu }\le \underset{X}{\int }f\circ \phi d\widehat{\mu }+\underset{Y}{\int }f\circ \left(a+b-\psi \right)d\widehat{\nu }$$

$$\le \underset{Y}{\int }f\circ \psi d\widehat{\nu }+\underset{Y}{\int }f\circ \left(a+b-\psi \right)d\widehat{\nu }\le f\left(a\right)+f\left(b\right),$$

and, therefore, the result follows in turn from Lemma 2, Theorem 4 and Theorem 7(c). □

As an application of the previous theorem, an important special refinement is given in the following statement.

Theorem 12.

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 a, $b\in \mathbb{R}$ with $a<b$, and let $\phi :X\to \left[a,b\right]$ be a measurable function. Define

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

Then, the inequalities

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

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

hold for any convex function $f:\left[a,b\right]\to \mathbb{R}$ if and only if

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

(42)

Proof. 

Let $\left(Y,\mathcal{B}\right):=\left(\left[a,b\right],{\mathcal{B}}_{\left[a,b\right]}\right)$, let the finite signed measure $\nu$ on ${\mathcal{B}}_{\left[a,b\right]}$ be defined by

$$\nu :=\mu \left(X\right){\displaystyle \frac{b-t_{\phi ,\mu }}{b-a}}{\epsilon }_a+\mu \left(X\right){\displaystyle \frac{t_{\phi ,\mu }-a}{b-a}}{\epsilon }_b,$$

and let $\psi :\left[a,b\right]\to \left[a,b\right]$, $\psi \left(s\right):=s$. With these parameters, in the proof of Theorem 6, we have seen that conditions in (40) are satisfied, and (41) is equivalent to (42). Therefore, it now follows from Theorem 11 that

$$\underset{X}{\int }f\circ \left(a+b-\phi \right)d\mu \le \underset{\left[a,b\right]}{\int }f\circ \left(a+b-s\right)d\nu \left(s\right)$$

$$\le \underset{\left[a,b\right]}{\int }f\left(s\right)d\nu \left(s\right)+\underset{\left[a,b\right]}{\int }f\circ \left(a+b-s\right)d\nu \left(s\right)-\underset{X}{\int }f\circ \phi d\mu$$

is equivalent to (42).

However, simple calculation shows that

$$\underset{\left[a,b\right]}{\int }f\circ \left(a+b-s\right)d\nu \left(s\right)=\mu \left(X\right)\left({\displaystyle \frac{b-t_{\phi ,\mu }}{b-a}}f\left(b\right)+{\displaystyle \frac{t_{\phi ,\mu }-a}{b-a}}f\left(a\right)\right)$$

and

$$\underset{\left[a,b\right]}{\int }f\left(s\right)d\nu \left(s\right)+\underset{\left[a,b\right]}{\int }f\circ \left(a+b-s\right)d\nu \left(s\right)-\underset{X}{\int }f\circ \phi d\mu$$

$$=\mu \left(X\right)\left(f\left(a\right)+f\left(b\right)\right)-\underset{X}{\int }f\circ \phi d\mu .$$

□

Remark 6.

Theorem 12 is an exact generalization of Corollary 2.3 in [2] to finite signed measures. In the mentioned result, the probability measure space is used.

5. Conclusions

The notion of convexity and the inequalities associated with it are important from both a theoretical and an application point of view. For finite measures, the integral Jensen–Mercer inequality decomposes into an integral Jensen and a majorization-type inequality. All three inequalities are still much studied today. In this paper, we investigate these inequalities for finite signed measures, and give necessary and sufficient conditions for their fulfillment. It is an essential improvement on previous similar results to obtain precise conditions with such generality. The novel methods used in the proofs offer the possibility to achieve further results in the field. Finally, we give a refinement of the studied majorization-type inequality for finite signed measures, which is a new approach to the topic of refinements even for finite measures.