Refining the integral Jensen inequality for finite signed measures using majorization

Abstract

Among inequalities that use the concept of convexity, Jensen-type inequalities and majorization-type inequalities are significant and fundamental. An important and widely researched area in the study of Jensen-type inequalities is the refinement of such inequalities. In this paper, we provide a general method for refining the integral Jensen inequality for finite signed measures using integral majorization inequalities. Under the conditions considered in the paper, the results are unique, and even for measures, they give a new approach. We also provide interesting specific refinements, some of which relate to Jensen–Steffensen’s inequality. A new and extended version of this inequality is also obtained in the paper.

1 Introduction

The notion of convexity and the results related to convexity are very useful in mathematics and various applications. A real function f defined on an interval $C\subset {\mathbb {R}}$ is called convex if it satisfies

$$\begin{aligned} f\left( \alpha t_{1}+\left( 1-\alpha \right) t_{2}\right) \le \alpha f\left( t_{1}\right) +\left( 1-\alpha \right) f\left( t_{2}\right) \end{aligned}$$

for all points $t_{1}$ and $t_{2}$ in C and all $\alpha \in \left[ 0,1\right]$.

Among inequalities that use the concept of convexity, Jensen-type inequalities and majorization-type inequalities are significant and fundamental (see [6]). For example, many classical inequalities can be derived from them, and they are applicable to many areas outside mathematics.

To start with, we introduce some special function sets: Let $\left( X,{\mathcal {A}}\right)$ be a measurable space (${\mathcal {A}}$ always means a $\sigma$-algebra of subsets of X). If $\mu$ is a finite 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 integrable functions are considered to be measurable. Let $C\subset {\mathbb {R}}$ be an interval with nonempty interior. Let denote $F_{C}$ the set of all convex functions on C. Assume that $\left( X,{\mathcal {A}},\mu \right)$ and $\left( Y,{\mathcal {B}} ,\nu \right)$ are measure spaces, where $\mu$ and $\nu$ are finite signed measures. If $\varphi :X\rightarrow C$, $\psi :Y\rightarrow C$ are functions such that $\varphi \in L\left( \mu \right)$ and $\psi \in L\left( \nu \right)$, then we define $F_{C}\left( \varphi ,\mu ;\psi ,\nu \right)$ as the set of all functions $f\in F_{C}$ for which $f\circ \varphi \in L\left( \mu \right)$ and $f\circ \psi \in L\left( \nu \right)$. For $\left( X,{\mathcal {A}} ,\mu \right) =\left( Y,{\mathcal {B}},\nu \right)$, the shorter notations $F_{C}\left( \varphi ,\psi ,\mu \right)$ and $F_{C}\left( \varphi ,\mu \right)$ (if $\varphi =\psi$) are used.

The usual version of the integral Jensen inequality for measures can be stated in the following way:

Theorem 1

(see [14]) Let $\left( X,{\mathcal {A}}, \mu \right)$ be a probability space, and let $\varphi \in L\left( \mu \right)$ be a function taking values in an interval $C\subset {\mathbb {R}}$. Then ${\displaystyle \int \limits _{X}} \varphi d\mu$ lies in C, and for every $f\in F_{C}\left( \varphi ,\mu \right)$, the inequality

$$\begin{aligned} f\left( {\displaystyle \int \limits _{X}} \varphi d\mu \right) \le {\displaystyle \int \limits _{X}} f\circ \varphi d\mu \end{aligned}$$

(1)

holds.

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

A recent version of the integral majorization inequality for measures is the following (similar to the standard version, see [2]):

Theorem 2

(see Theorem 6 of [9]) Let $C\subset {\mathbb {R}}$ be an interval with nonempty interior. Let $\left[ a,b\right] \subset {\mathbb {R}}$ with $a<b$, and let $\mu$ be a finite measure on ${\mathcal {B}}_{\left[ a,b\right] }$. Assume further that $\varphi :\left[ a,b\right] \rightarrow C$ is a decreasing function, and $\psi \in L\left( \mu \right) \$is a function taking values in C. If

$$\begin{aligned} \int \limits _{\left[ a,x\right] }\varphi d\mu \le \int \limits _{\left[ a,x\right] }\psi d\mu ,\quad x\in \left[ a,b\right] , \end{aligned}$$

and

$$\begin{aligned} \int \limits _{\left[ a,b\right] }\varphi d\mu =\int \limits _{\left[ a,b\right] }\psi d\mu \end{aligned}$$

are satisfied, then for every $f\in F_{C}\left( \varphi ,\psi ,\mu \right)$, the inequality

$$\begin{aligned} \int \limits _{\left[ a,b\right] }f\circ \varphi d\mu \le \int \limits _{\left[ a,b\right] }f\circ \psi d\mu \end{aligned}$$

(2)

holds.

A natural question is how the previous two results can be extended if $\mu$ is a signed measure. In both cases there is a substantial literature on the problem, we will highlight only a few papers: for the integral Jensen inequality, see [1, 3, 8, 9, 11, 15, 18], while for the integral majorization inequality, see [2, 5, 8, 9, 12, 16].

In this paper we rely on recent generalizations of Theorem 1 and Theorem 2 that give necessary and sufficient conditions for inequalities (1) and (2) using finite signed measures. It should be stressed that the two assertions have the same structure and are based on the same foundations.

Let $C\subset {\mathbb {R}}$ be an interval with nonempty interior. The following notations are introduced for some special functions defined on C:

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

where $a^{+}$ and $a^{-}$ mean the positive and negative parts of $a\in {\mathbb {R}}$, respectively.

The following result is a precise formulation of the integral Jensen inequality for finite signed measures. It is proved in Theorem 12 of [8].

Theorem 3

(integral Jensen inequality) Let $\left( X,{\mathcal {A}} \right)$ be a measurable space, and let $\mu$ be a finite signed measure on ${\mathcal {A}}$ such that $\mu \left( X\right) >0$. Let $C\subset {\mathbb {R}}$ be an interval with nonempty interior, and let $\varphi \in L\left( \mu \right) \$be a function taking values in C. Then

• (a) If

  $$\begin{aligned} \int \limits _{X}p_{C,w}\circ \varphi d\mu =\int \limits _{\left\{ \varphi \ge w\right\} }\left( \varphi -w\right) d\mu \ge 0,\quad w\in C^{\circ } \end{aligned}$$

  (3)

  and

  $$\begin{aligned} \int \limits _{X}n_{C,w}\circ \varphi d\mu =\int \limits _{\left\{ \varphi <w\right\} }\left( w-\varphi \right) d\mu \ge 0,\quad w\in C^{\circ } \end{aligned}$$

  (4)

  are satisfied, then

  $$\begin{aligned} t_{\varphi ,\mu }:=\frac{1}{\mu \left( X\right) }\int \limits _{X}\varphi d\mu \in C. \end{aligned}$$

  (5)

• (b) For every function $f\in F_{C}\left( \varphi ,\mu \right)$, the inequality

  $$\begin{aligned} f\left( \frac{1}{\mu \left( X\right) }\int \limits _{X}\varphi d\mu \right) \le \frac{1}{\mu \left( X\right) }\int \limits _{X}f\circ \varphi d\mu \end{aligned}$$

  holds if and only if (3) and (4) are satisfied.

Remark 1

Assume that $t_{\inf }:=\inf \varphi >-\infty$ i.e. $\varphi$ is bounded from below. The previous result implies that inclusion (5) can be made more precise (which is a natural expectation): $t_{\varphi ,\mu }\ge t_{\inf }$, and if $\varphi \left( x\right) >t_{\inf }$ for every $x\in X$, then $t_{\varphi ,\mu }>t_{\inf }$. Similarly, if $\varphi$ is bounded from above, i.e. $t_{\sup }:=\sup \varphi <\infty$, then $t_{\varphi ,\mu }\le t_{\sup }$. Moreover $t_{\varphi ,\mu }<t_{\sup }$ holds if $\varphi \left( x\right) <t_{\sup }$ for every $x\in X$.

The following integral majorization inequality for finite signed measures is contained in Theorem 6 of [8].

Theorem 4

(integral majorization inequality) Let $\left( X,{\mathcal {A}} ,\mu \right)$ and $\left( Y,{\mathcal {B}},\nu \right)$ be measure spaces, where $\mu$ and $\nu$ are finite signed measures. Let $C\subset {\mathbb {R}}$ be an interval with nonempty interior, and let $\varphi :X\rightarrow C$, $\psi :Y\rightarrow C$ be functions such that $\varphi \in L\left( \mu \right)$ and $\psi \in L\left( \nu \right)$.

Then for every $f\in F_{C}\left( \varphi ,\mu ;\psi ,\nu \right)$, the inequality

$$\begin{aligned} \int \limits _{X}f\circ \varphi d\mu \le \int \limits _{Y}f\circ \psi d\nu \end{aligned}$$

holds if and only if

$$\begin{aligned} \mu \left( X\right) =\nu \left( Y\right) ,\quad \int \limits _{X}\varphi d\mu =\int \limits _{Y}\psi d\nu \end{aligned}$$

(6)

and

$$\begin{aligned} \int \limits _{X}p_{C,w}\circ \varphi d\mu \le \int \limits _{Y}p_{C,w}\circ \psi d\nu \quad w\in C^{\circ } \end{aligned}$$

(7)

are satisfied.

An important and widely researched area in the study of Jensen-type inequalities is the refinement of such inequalities. Among the numerous papers on this subject, we mention only the book [7] and references therein. A significant part of the refinements to the integral Jensen inequality has the form

$$\begin{aligned} f\left( \frac{1}{\nu \left( Y\right) }\int \limits _{Y}\psi d\nu \right) \le \frac{1}{\mu \left( X\right) }\int \limits _{X}f\circ \varphi d\mu \le \frac{1}{\nu \left( Y\right) }\int \limits _{Y}f\circ \psi d\nu , \end{aligned}$$

(8)

although this is often not apparent at first glance. In the second inequality, however, we can identify an integral majorization inequality. In general, this is also not easy to recognize not only from the shape of the refinement, but also from its proof, since if $\nu$ and $\mu$ are measures (this is the case in most refinements), the proofs can usually be done using only Jensen-type inequalities, majorization does not come up. We illustrate our previous remarks with a simple and known refinement of the integral Jensen inequality and two different proofs of it.

Let $\left( X,{\mathcal {A}}\right)$ be a measurable space. The unit mass at $x\in X$ (the Dirac measure at x) is denoted by $\varepsilon _{x}$. The set of all subsets of X is denoted by $P\left( X\right)$.

Example 1

Let $\left( Y,{\mathcal {B}},\nu \right)$ be a probability space, and let $A_{1}$, $A_{2}\in {\mathcal {B}}$ such that $A_{1}\cup A_{2}=Y$, $A_{1}\cap A_{2}=\emptyset$ and $\nu \left( A_{i}\right) >0$ $\left( i=1,2\right)$. If $\psi \in L\left( \nu \right)$, then for every $f\in F_{{\mathbb {R}}}\left( \psi ,\nu \right)$ inequalities

$$\begin{aligned} f\left( \int \limits _{Y}\psi d\nu \right) \le \frac{1}{\nu \left( A_{1}\right) }\int \limits _{A_{1}}\psi d\nu +\frac{1}{\nu \left( A_{2}\right) }\int \limits _{A_{2}}\psi d\nu \le \int \limits _{Y}f\circ \psi d\nu \end{aligned}$$

(9)

hold.

Obviously, (9) is a refinement of the integral Jensen inequality, and the "usual" proof (relying only on Jensen-type inequalities) is very simple:

$$\begin{aligned} f\left( \int \limits _{Y}\psi d\nu \right)= & {} f\left( \nu \left( A_{1}\right) \left( \frac{1}{\nu \left( A_{1}\right) }\int \limits _{A_{1}}\psi d\nu \right) +\nu \left( A_{2}\right) \left( \frac{1}{\nu \left( A_{2}\right) }\int \limits _{A_{2}}\psi d\nu \right) \right) \nonumber \\\le & {} \nu \left( A_{1}\right) f\left( \frac{1}{\nu \left( A_{1}\right) } \int \limits _{A_{1}}\psi d\nu \right) +\nu \left( A_{2}\right) f\left( \frac{1}{\nu \left( A_{2}\right) }\int \limits _{A_{2}}\psi d\nu \right) \nonumber \\\le & {} \int \limits _{A_{1}}f\circ \psi d\nu +\int \limits _{A_{2}}f\circ \psi d\nu =\int \limits _{Y}f\circ \psi d\nu . \end{aligned}$$

(10)

We now show another approach that uses majorization.

Consider the measure space $\left( X,P\left( X\right) ,\mu \right)$, where $X:=\left\{ 1,2\right\}$, $\mu :=\nu \left( A_{1}\right) \varepsilon _{1}+\nu \left( A_{2}\right) \varepsilon _{2}$ (it is also a probability measure), and let the function $\varphi$ be defined on X by

$$\begin{aligned} \varphi \left( 1\right) :=\frac{1}{\nu \left( A_{1}\right) }\int \limits _{A_{1}}\psi d\nu \quad \text {and}\quad \varphi \left( 2\right) :=\frac{1}{\nu \left( A_{2}\right) }\int \limits _{A_{2}}\psi d\nu . \end{aligned}$$

Using the equality $\int \limits _{Y}\psi d\nu = {\int \limits _{X}} \varphi d\mu$ and the integral Jensen inequality we obtain that

$$\begin{aligned} f\left( \int \limits _{Y}\psi d\nu \right) =f\left( {\displaystyle \int \limits _{X}} \varphi d\mu \right) \le {\displaystyle \int \limits _{X}} f\circ \varphi d\mu , \end{aligned}$$

(11)

and the integral on the right is identical to (10).

We have seen that conditions (6) are satisfied. If the condition (7) is also satisfied, then we can apply Theorem 4 which implies

$$\begin{aligned} {\displaystyle \int \limits _{X}} f\circ \varphi d\mu \le \int \limits _{Y}f\circ \psi d\nu , \end{aligned}$$

(12)

so (9) will be proved.

To verify that condition (7) is met, we calculate the left hand integral. It can be supposed that $\varphi \left( 1\right) \le \varphi \left( 2\right)$. Then

$$\begin{aligned} {\displaystyle \int \limits _{X}} pr_{{\mathbb {R}},w}\circ \varphi d\mu =\left\{ \begin{array}{ll} \int \limits _{Y}\left( \psi -w\right) d\nu , &{} w<\varphi \left( 1\right) \\ \int \limits _{A_{2}}\left( \psi -w\right) d\nu , &{} \varphi \left( 1\right) \le w<\varphi \left( 2\right) \\ 0, &{} \varphi \left( 2\right) \le w \end{array} \right. . \end{aligned}$$

From this (7) easily follows.

The first proof is simpler than the second and easy to understand, it is a natural approach when using measures. However, it does not work for signed measures. The second proof is more complicated, but it shows a way of proving refinement (8) which, as we shall see, will also work for signed measures: by (11), the integral Jensen inequality is applied for $\varphi$ and by (12), an integral majorization inequality is applied between $\varphi$ and $\psi$. For measures (in specific environments), this idea can also be found in some previous papers, see e.g. [4, 17]. Its significance will be shown by the fact that we can give necessary and sufficient conditions for the fulfilment of refinement (8) for signed measures, which is one of the main results of this paper. Even for measures, it provides a new approach. To illustrate the applicability of this result, we use the Boas version of the integral Jensen–Steffensen inequality, for which we also provide an essential generalization. Finally, some concrete refinements are given using signed measures. On the one hand, we obtain significant generalizations of previous results, and on the other hand we will see how the lack of monotonicity of the integral complicates the problem. Theorems 3 and 4 have only recently been published, so in such a general framework the refinements in this paper are unique. Even for the integral Jensen–Steffensen inequality, relatively few refinements are known (see [10, 13]).

2 Preliminary results

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 [18]. The following version of the integral Jensen–Steffensen inequality was proved by Boas [3].

Theorem 5

Let a, $b\in {\mathbb {R}}$ with $a<b$, and let $\lambda :\left[ a,b\right] \rightarrow {\mathbb {R}}$ be either continuous or of bounded variation satisfying

$$\begin{aligned} \lambda \left( a\right) \le \lambda \left( s_{1}\right) \le \lambda \left( t_{1}\right) \le \cdots \le \lambda \left( t_{n-1}\right) \le \lambda \left( s_{n}\right) \le \lambda \left( b\right) \end{aligned}$$

and $\lambda \left( a\right) <\lambda \left( b\right)$, where $a=t_{0}<t_{1}<\cdots<t_{n-1}<t_{n}=b$, $n\in {\mathbb {N}}_{+}$ and $s_{i} \in \left[ t_{i-1},t_{i}\right]$ $\left( i=1,\ldots ,n\right)$. Let $C\subset {\mathbb {R}}$ be an interval with nonempty interior, and let $\psi :\left[ a,b\right] \rightarrow C$ be a continuous function such that it is monotonic (in either sense) in each of the intervals $\left[ t_{i-1} ,t_{i}\right]$ $\left( i=1,\ldots ,n\right)$.

Then

$$\begin{aligned} {\displaystyle \int \limits _{a}^{b}} \psi d\lambda \diagup {\displaystyle \int \limits _{a}^{b}} 1d\lambda \in C \end{aligned}$$

(13)

and for every continuous function $f\in F_{C}$, the inequality

$$\begin{aligned} f\left( {\displaystyle \int \limits _{a}^{b}} \psi d\lambda \diagup {\displaystyle \int \limits _{a}^{b}} 1d\lambda \right) \le {\displaystyle \int \limits _{a}^{b}} f\circ \psi d\lambda \diagup {\displaystyle \int \limits _{a}^{b}} 1d\lambda \end{aligned}$$

(14)

holds.

Remark 2

In paper [3] Riemann–Stieltjes integral is used, so the integrals in (14) should be considered in the Riemann–Stieltjes sense.

We now introduce a notation used in this paper. Let ${\overline{a}}$, ${\overline{b}}$, a, $b\in {\mathbb {R}}$ with ${\overline{a}}<a<b<{\overline{b}}$, and let $\lambda :\left[ a,b\right] \rightarrow {\mathbb {R}}$ be of bounded variation. Extend the function $\lambda$ to the open interval $\big ] {\overline{a}},{\overline{b}}\big [$ by

$$\begin{aligned} {\overline{\lambda }}\left( s\right) :=\left\{ \begin{array}{ll} \lambda \left( a\right) , &{} s\in \big ] {\overline{a}},a\big [ \\ \lambda \left( s\right) , &{} s\in \left[ a,b\right] \\ \lambda \left( b\right) , &{} s\in \big ] b,{\overline{b}}\big [ \end{array} \right. . \end{aligned}$$

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

Remark 3

Let $\lambda :\left[ a,b\right] \rightarrow {\mathbb {R}}$ be of bounded variation in Theorem 5. Under the conditions of the assertion

$$\begin{aligned} {\displaystyle \int \limits _{a}^{b}} \psi d\lambda = {\displaystyle \int \limits _{\left[ a,b\right] }} \psi d\nu _{\lambda },\quad {\displaystyle \int \limits _{a}^{b}} f\circ \psi d\lambda = {\displaystyle \int \limits _{\left[ a,b\right] }} f\circ \psi d\nu _{\lambda }\quad \text {and\quad } {\displaystyle \int \limits _{a}^{b}} 1d\lambda = {\displaystyle \int \limits _{\left[ a,b\right] }} 1d\nu _{\lambda }, \end{aligned}$$

and

$$\begin{aligned} {\displaystyle \int \limits _{a}^{b}} 1d\lambda = {\displaystyle \int \limits _{\left[ a,b\right] }} 1d\nu _{\lambda }=\nu _{\lambda }\left( \left[ a,b\right] \right) =\lambda \left( b\right) -\lambda \left( a\right) , \end{aligned}$$

where on the left-hand sides are Riemann–Stieltjes integrals and on the right-hand sides are Lebesgue–Stieltjes integrals.

Remark 4

(a) It follows that condition (13) and inequality (14) can be written as

$$\begin{aligned} \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} \psi d\nu _{\lambda }\in C \end{aligned}$$

and

$$\begin{aligned} f\left( \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} \psi d\nu _{\lambda }\right) \le \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} f\circ \psi d\nu _{\lambda }. \end{aligned}$$

(b) We use Lebesgue integral with respect to a finite signed measure in this paper, and therefore we consider Theorem 5 and its subsequent generalization only if $\lambda$ is of bounded variation (we do not deal with the case where $\lambda$ is continuous but it is not of bounded variation).

We need the following two lemmas.

The first result is contained in Lemma 3.1 of [1]. Because only one specific case of this lemma is important for us, and Lebesgue-Stieltjes integral is used (Riemann–Stieltjes integral is considered in Lemma 3.1), we give the proof for completeness.

Lemma 1

Let ${\overline{c}}$, ${\overline{d}}$, c, $d\in {\mathbb {R}}$ with ${\overline{c}}<c<d<{\overline{d}}$, let $\lambda :\left[ c,d\right] \rightarrow {\mathbb {R}}$ be of bounded variation satisfying $\lambda \left( c\right) \le \lambda \left( d\right)$, let $\psi :\left[ c,d\right] \rightarrow {\mathbb {R}}$ be nonnegative, and suppose that at any point of $\left[ c,d\right]$ at least one of $\lambda$ and $\psi$ is continuous.

If either $\psi$ is increasing and

$$\begin{aligned} \lambda \left( s\right) \le \lambda \left( d\right) ,\quad s\in \left[ c,d\right] , \end{aligned}$$

(15)

or $\psi$ is decreasing and

$$\begin{aligned} \lambda \left( c\right) \le \lambda \left( s\right) ,\quad s\in \left[ c,d\right] , \end{aligned}$$

(16)

then

$$\begin{aligned} {\displaystyle \int \limits _{\left[ c,d\right] }} \psi d\nu _{\lambda }\ge 0. \end{aligned}$$

(17)

Proof

Assume that $\psi$ is increasing and (15) holds.

By applying the formula for integration by parts of Lebesgue-Stieltjes integrals, we obtain that

$$\begin{aligned} {\displaystyle \int \limits _{\left[ c,d\right] }} \psi d\nu _{\lambda }=\psi \left( d\right) \lambda \left( d\right) -\psi \left( c\right) \lambda \left( c\right) - {\displaystyle \int \limits _{\left[ c,d\right] }} \lambda d\nu _{\psi }. \end{aligned}$$

(18)

By (15),

$$\begin{aligned} {\displaystyle \int \limits _{\left[ c,d\right] }} \lambda d\nu _{\psi }\le \lambda \left( d\right) \left( \psi \left( d\right) -\psi \left( c\right) \right) , \end{aligned}$$

and therefore it follows from (18), $\psi \left( t\right) \ge 0$ and $\lambda \left( c\right) \le \lambda \left( d\right)$ that

$$\begin{aligned} {\displaystyle \int \limits _{\left[ c,d\right] }} \psi d\nu _{\lambda }\ge & {} \psi \left( d\right) \lambda \left( d\right) -\psi \left( c\right) \lambda \left( c\right) -\lambda \left( d\right) \left( \psi \left( d\right) -\psi \left( c\right) \right) \\= & {} \psi \left( c\right) \left( \lambda \left( d\right) -\lambda \left( c\right) \right) \ge 0. \end{aligned}$$

Assume that $\psi$ is decreasing and (16) holds.

As in the increasing case we can obtain that

$$\begin{aligned} {\displaystyle \int \limits _{\left[ c,d\right] }} \psi d\nu _{\lambda }= & {} \psi \left( d\right) \lambda \left( d\right) -\psi \left( c\right) \lambda \left( c\right) - {\displaystyle \int \limits _{\left[ c,d\right] }} \lambda d\nu _{\psi }\\\ge & {} \psi \left( d\right) \lambda \left( d\right) -\psi \left( c\right) \lambda \left( c\right) -\lambda \left( c\right) \left( \psi \left( d\right) -\psi \left( c\right) \right) =\psi \left( d\right) \left( \lambda \left( d\right) -\lambda \left( c\right) \right) \ge 0. \end{aligned}$$

The proof is complete. $\square$

Remark 5

Obviously, if

$$\begin{aligned} \lambda \left( c\right) \le \lambda \left( s\right) \le \lambda \left( d\right) ,\quad s\in \left[ c,d\right] , \end{aligned}$$

then inequality (17) is satisfied for both increasing and decreasing $\psi$.

Next, we generalize the previous result.

Lemma 2

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

$$\begin{aligned} \lambda \left( a\right) \le \lambda \left( t_{1}\right) \le \cdots \le \lambda \left( t_{n-1}\right) \le \lambda \left( b\right) , \end{aligned}$$

where $a=t_{0}<t_{1}<\cdots<t_{n-1}<t_{n}=b$ $\left( n\in {\mathbb {N}} _{+}\right)$. Assume further that in any interval $\left[ t_{i-1} ,t_{i}\right]$ $\left( i=1,\ldots ,n\right)$ either

$$\begin{aligned} \lambda \left( s\right) \le \lambda \left( t_{i}\right) ,\quad s\in \left[ t_{i-1},t_{i}\right] \end{aligned}$$

(19)

or

$$\begin{aligned} \lambda \left( t_{i-1}\right) \le \lambda \left( s\right) ,\quad s\in \left[ t_{i-1},t_{i}\right] \end{aligned}$$

(20)

is satisfied.

Let $\psi :\left[ a,b\right] \rightarrow {\mathbb {R}}$ be nonnegative, and suppose that at any point of $\left[ a,b\right]$ at least one of $\lambda$ and $\psi$ is continuous. If $\psi$ is increasing on intervals $\left[ t_{i-1},t_{i}\right]$ where (19) is satisfied and decreasing on intervals $\left[ t_{i-1},t_{i}\right]$ where (20) is satisfied, then

$$\begin{aligned} {\displaystyle \int \limits _{\left[ a,b\right] }} \psi d\nu _{\lambda }\ge 0. \end{aligned}$$

Proof

Define the function $\overline{\lambda _{i}}:\big ] {\overline{a}},{\overline{b}}\big [ \rightarrow {\mathbb {R}}$ by

$$\begin{aligned} \overline{\lambda _{i}}\left( s\right) :=\left\{ \begin{array}{ll} \lambda \left( t_{i-1}\right) , &{} s\in \big ] {\overline{a}},t_{i-1}\big [\\ \lambda \left( s\right) , &{} s\in \left[ t_{i-1},t_{i}\right] \\ \lambda \left( t_{i}\right) , &{} s\in \big ] t_{i},{\overline{b}}\big [ \end{array} \right. ,\quad i=1,\ldots ,n, \end{aligned}$$

and consider the Lebesgue–Stieltjes measure $\nu _{\lambda _{i}}$ on ${\mathcal {B}}_{\big ] {\overline{a}},{\overline{b}}\big [ }$ associated with $\overline{\lambda _{i}}$.

Then

$$\begin{aligned} \nu _{\lambda }= {\displaystyle \sum \limits _{i=1}^{n}} \nu _{\lambda _{i}}, \end{aligned}$$

and hence

$$\begin{aligned} {\displaystyle \int \limits _{\left[ a,b\right] }} \psi d\nu _{\lambda }= {\displaystyle \sum \limits _{i=1}^{n}} {\displaystyle \int \limits _{\left[ a,b\right] }} \psi d\nu _{\lambda _{i}}= {\displaystyle \sum \limits _{i=1}^{n}} {\displaystyle \int \limits _{\left[ t_{i-1},t_{i}\right] }} \psi d\nu _{\lambda _{i}}. \end{aligned}$$

We can now apply Lemma 1 to each integral

$$\begin{aligned} {\displaystyle \int \limits _{\left[ t_{i-1},t_{i}\right] }} \psi d\nu _{\lambda _{i}}, \end{aligned}$$

so the result follows.

The proof is complete. $\square$

Remark 6

If both conditions (19) and (20) are satisfied on an interval $\left[ t_{i-1},t_{i}\right]$, then $\psi$ can be either increasing or decreasing on $\left[ t_{i-1},t_{i}\right]$.

We will need the following modification of Theorem 4.

Proposition 6

Suppose that the conditions of Theorem 4 are satisfied, with the additional conditions that $\mu \left( X\right)$, $\nu \left( Y\right) >0$.

Then for every $f\in F_{C}\left( \varphi ,\mu ;\psi ,\nu \right)$, the inequality

$$\begin{aligned} \frac{1}{\mu \left( X\right) }\int \limits _{X}f\circ \varphi d\mu \le \frac{1}{\nu \left( Y\right) }\int \limits _{Y}f\circ \psi d\nu \end{aligned}$$

holds if and only if

$$\begin{aligned} \frac{1}{\mu \left( X\right) }\int \limits _{X}\varphi d\mu =\frac{1}{\nu \left( Y\right) }\int \limits _{Y}\psi d\nu \end{aligned}$$

(21)

and

$$\begin{aligned} \frac{1}{\mu \left( X\right) }\int \limits _{X}p_{C,w}\circ \varphi d\mu \le \frac{1}{\nu \left( Y\right) }\int \limits _{Y}p_{C,w}\circ \psi d\nu \quad w\in C^{\circ } \end{aligned}$$

(22)

are satisfied.

Proof

Introduce the set functions ${\widehat{\mu }}:{\mathcal {A}}\rightarrow {\mathbb {R}}$ and ${\widehat{\nu }}:{\mathcal {B}}\rightarrow {\mathbb {R}}$ defined by

$$\begin{aligned} {\widehat{\mu }}\left( A\right) :=\frac{\mu \left( A\right) }{\mu \left( X\right) }\quad \text {and\quad }{\widehat{\nu }}\left( B\right) :=\frac{\nu \left( B\right) }{\nu \left( Y\right) }. \end{aligned}$$

Then ${\widehat{\mu }}$ and ${\widehat{\nu }}$ are finite signed measures with ${\widehat{\mu }}\left( X\right) ={\widehat{\nu }}\left( Y\right) =1$, $L\left( {\widehat{\mu }}\right) =L\left( \mu \right)$, $L\left( {\widehat{\nu }}\right) =L\left( \nu \right)$, for every $\chi \in L\left( \mu \right)$ and $\omega \in L\left( \nu \right)$ we have that

$$\begin{aligned} \int \limits _{X}\chi d{\widehat{\mu }}=\frac{1}{\mu \left( X\right) } \int \limits _{X}\chi d\mu \quad \text {and\quad }\int \limits _{Y}\omega d{\widehat{\nu }}=\frac{1}{\nu \left( Y\right) }\int \limits _{Y}\omega d\nu , \end{aligned}$$

and $F_{C}\left( \varphi ,{\widehat{\mu }};\psi ,{\widehat{\nu }}\right) =F_{C}\left( \varphi ,\mu ;\psi ,\nu \right)$.

Based on the above, the result follows immediately from Theorem 4.

The proof is complete. $\square$

3 Generalization of Jensen–Steffensen inequality

As an application of Theorem 3, we generalize the Boas’ version of Jensen–Steffensen inequality.

Theorem 7

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

$$\begin{aligned} \lambda \left( a\right) \le \lambda \left( t_{1}\right) \le \cdots \le \lambda \left( t_{n-1}\right) \le \lambda \left( b\right) \end{aligned}$$

and $\lambda \left( a\right) <\lambda \left( b\right)$, where $a=t_{0}<t_{1}<\cdots<t_{n-1}<t_{n}=b$ $\left( n\in {\mathbb {N}}_{+}\right)$. Assume further that in any interval $\left[ t_{i-1},t_{i}\right]$ $\left( i=1,\ldots ,n\right)$ either (19) or (20) holds. Let $C\subset {\mathbb {R}}$ be an interval with nonempty interior, and let $\psi :\left[ a,b\right] \rightarrow C$ be increasing on intervals $\left[ t_{i-1},t_{i}\right]$ where (19) is satisfied and decreasing on intervals $\left[ t_{i-1},t_{i}\right]$ where (20) is satisfied. Suppose further that at any point of $\left[ a,b\right]$ at least one of $\lambda$ and $\psi$ is continuous.

Then

$$\begin{aligned} \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} \psi d\nu _{\lambda }\in C \end{aligned}$$

and

$$\begin{aligned} f\left( \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} \psi d\nu _{\lambda }\right) \le \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} f\circ \psi d\nu _{\lambda } \end{aligned}$$

for every $f\in F_{C}$.

Proof

Since $\psi$ is monotonic in each of the intervals $\left[ t_{i-1} ,t_{i}\right]$ $\left( i=1,\ldots ,n\right)$, there is a compact subinterval ${\overline{C}}$ of C such that the range of $\psi$ is contained in ${\overline{C}}$, and therefore $\psi \in L\left( \nu _{\lambda }\right)$ and $F_{C}\left( \psi ,\nu _{\lambda }\right) =F_{C}$.

By Theorem 3, it is enough to show that

$$\begin{aligned} \int \limits _{\left[ a,b\right] }p_{C,w}\circ \psi d\nu _{\lambda }\ge 0,\quad \text {and\quad }\int \limits _{\left[ a,b\right] }n_{C,w}\circ \psi d\nu _{\lambda }\ge 0,\quad w\in C^{\circ } \end{aligned}$$

(23)

are satisfied.

Fix $w\in C^{\circ }$. It is obvious that $p_{C,w}\circ \psi$ and $n_{C,w} \circ \psi$ are nonnegative functions. If $\psi$ is increasing on $\left[ t_{i-1},t_{i}\right]$, then $p_{C,w}\circ \psi$ is also increasing and $n_{C,w}\circ \psi$ is decreasing, while if $\psi$ is decreasing on $\left[ t_{i-1},t_{i}\right]$, then $p_{C,w}\circ \psi$ is also decreasing and $n_{C,w}\circ \psi$ is increasing $\left( i=1,\ldots ,n\right)$.

Inequalities in (23) now follows from Lemma 2.

The proof is complete. $\square$

Remark 7

• (a) The preceding theorem is a substantial generalization of Theorem 5 (see Remark 2), since neither the continuity of $\psi$ nor that of f is required, and the condition on $\lambda$ is weaker. We also have a new proof of Theorem 5 (completely different from Boas’ proof), and this proof clearly shows the origin of the conditions on $\lambda$ and $\psi$.

• (b) We can see that Theorem 7 is a special case of Theorem 3, the significance being that its conditions are usually easy to check.

4 Refining the integral Jensen inequality using majorization

The next result is a refinement of the integral Jensen inequality using finite signed measures.

Theorem 8

Let $\left( X,{\mathcal {A}},\mu \right)$ and $\left( Y,{\mathcal {B}},\nu \right)$ be measure spaces, where $\mu$ and $\nu$ are finite signed measures and $\mu \left( X\right)$, $\nu \left( Y\right) >0$. Let $C\subset {\mathbb {R}}$ be an interval with nonempty interior, and let $\varphi :X\rightarrow C$, $\psi :Y\rightarrow C$ be functions such that $\varphi \in L\left( \mu \right)$ and $\psi \in L\left( \nu \right)$. Then inequalities

$$\begin{aligned} f\left( \frac{1}{\nu \left( Y\right) }\int \limits _{Y}\psi d\nu \right) \le \frac{1}{\mu \left( X\right) }\int \limits _{X}f\circ \varphi d\mu \le \frac{1}{\nu \left( Y\right) }\int \limits _{Y}f\circ \psi d\nu . \end{aligned}$$

(24)

hold for every $f\in F_{C}\left( \varphi ,\mu ;\psi ,\nu \right)$ if and only if (21), (22), (3) and (4) are satisfied.

Proof

We first show the sufficiency of the conditions.

By using (21), (3) and (4), Theorem 3 (a) shows that

$$\begin{aligned} \frac{1}{\nu \left( Y\right) }\int \limits _{Y}\psi d\nu =\frac{1}{\mu \left( X\right) }\int \limits _{X}\varphi d\mu \in C. \end{aligned}$$

By applying first Theorem 3(b) and then Proposition 6, we obtain that

$$\begin{aligned}{} & {} f\left( \frac{1}{\nu \left( Y\right) }\int \limits _{Y}\psi d\nu \right) =f\left( \frac{1}{\mu \left( X\right) }\int \limits _{X}\varphi d\mu \right) \\{} & {} \quad \le \frac{1}{\mu \left( X\right) }\int \limits _{X}f\circ \varphi d\mu \le \frac{1}{\nu \left( Y\right) }\int \limits _{Y}f\circ \psi d\nu . \end{aligned}$$

Now we come to the proof that the conditions are also necessary.

Since the second inequality in (24) holds for every $f\in F_{C}\left( \varphi ,\mu ;\psi ,\nu \right)$, it follows from Proposition 6 that (21) and (22) must be satisfied. By using the first inequality in (24) and (21), we obtain that

$$\begin{aligned} f\left( \frac{1}{\mu \left( X\right) }\int \limits _{X}\varphi d\mu \right) =f\left( \frac{1}{\nu \left( Y\right) }\int \limits _{Y}\psi d\nu \right) \le \frac{1}{\mu \left( X\right) }\int \limits _{X}f\circ \varphi d\mu \end{aligned}$$

holds for every $f\in F_{C}\left( \varphi ,\mu ;\psi ,\nu \right)$. Since $p_{C,w}$ and $n_{C,w}$ belong to $F_{C}\left( \varphi ,\mu ;\psi ,\nu \right)$ for every $w\in C^{\circ }$, and $\mu \left( X\right) >0$, (3) and (4) follow from the previous inequality.

The proof is complete. $\square$

It is worth giving the version of the previous statement for measures.

Corollary 9

Let $\left( X,{\mathcal {A}},\mu \right)$ and $\left( Y,{\mathcal {B}},\nu \right)$ be measure spaces, where $\mu$ and $\nu$ are finite and not identically zero measures. Let $C\subset {\mathbb {R}}$ be an interval with nonempty interior, and let $\varphi :X\rightarrow C$, $\psi :Y\rightarrow C$ be functions such that $\varphi \in L\left( \mu \right)$ and $\psi \in L\left( \nu \right)$. Then inequality (24) holds for every $f\in F_{C}\left( \varphi ,\mu ;\psi ,\nu \right)$ if and only if (21) and (22) are satisfied.

Remark 8

Assume that $\mu$ and $\nu$ are finite and not identically zero measures. An important and applicable observation can be drawn from the previous corollary: if we have an integral majorization inequality of the form

$$\begin{aligned} \frac{1}{\mu \left( X\right) }\int \limits _{X}f\circ \varphi d\mu \le \frac{1}{\nu \left( Y\right) }\int \limits _{Y}f\circ \psi d\nu , \end{aligned}$$

then we automatically obtain a refinement (described in (24)) of the integral Jensen inequality

$$\begin{aligned} f\left( \frac{1}{\nu \left( Y\right) }\int \limits _{Y}\psi d\nu \right) \le \frac{1}{\nu \left( Y\right) }\int \limits _{Y}f\circ \psi d\nu . \end{aligned}$$

The following result is a simple but illustrative example of the previous observation.

Proposition 10

Under the conditions of Theorem 2, inequalities

$$\begin{aligned} f\left( \int \limits _{\left[ a,b\right] }\psi d\mu \right) \le \int \limits _{\left[ a,b\right] }f\circ \varphi d\mu \le \int \limits _{\left[ a,b\right] }f\circ \psi d\mu \end{aligned}$$

holds.

5 Specific refinements

As a first application, we give a concrete refinement of the general version of integral Jensen inequality for signed measures.

Theorem 11

Let $\left( Y,{\mathcal {B}},\nu \right)$ be a measure space, where $\nu$ is a finite signed measure and $\nu \left( Y\right) >0$. Let $C\subset {\mathbb {R}}$ be an interval with nonempty interior, and let $\psi :Y\rightarrow C$ be a function such that $\psi \in L\left( \nu \right)$. Let the index set I denote either $\left\{ 1,\ldots ,n\right\}$ for some $n\ge 1$ or ${\mathbb {N}}_{+}$. Suppose we are given a sequence $\left( Y_{i}\right) _{i\in I}$ of pairwise disjoint sets $Y_{i}\in {\mathcal {B}}$ such that $\nu \left( Y_{i}\right) >0$ for every $i\in I$, $\bigcup \limits _{i\in I}Y_{i}=Y$ and

$$\begin{aligned} \int \limits _{Y_{i}}p_{C,w}\circ \psi d\nu= & {} \int \limits _{\left\{ \psi \ge w\right\} \cap Y_{i}}\left( \psi -w\right) d\nu \nonumber \\\ge & {} \max \left( 0,\int \limits _{Y_{i}}\left( \psi -w\right) d\nu \right) ,\quad w\in C^{\circ },\quad i\in I. \end{aligned}$$

(25)

Then inequalities

$$\begin{aligned} f\left( \frac{1}{\nu \left( Y\right) }\int \limits _{Y}\psi d\nu \right) \le \frac{1}{\nu \left( Y\right) }\int \limits _{Y}f\circ \varphi d\nu \le \frac{1}{\nu \left( Y\right) }\int \limits _{Y}f\circ \psi d\nu . \end{aligned}$$

hold for every $f\in F_{C}\left( \psi ,\nu \right)$, where the function $\varphi :Y\rightarrow C$ is defined by

$$\begin{aligned} \varphi \left( y\right) :=\frac{1}{\nu \left( Y_{i}\right) }\int \limits _{Y_{i}}\psi d\nu ,\quad y\in Y_{i},\quad i\in I. \end{aligned}$$

Proof

Step by step, we check that the conditions in Theorem 8 are satisfied.

• (i) It is obvious that $F_{C}\left( \varphi ,\nu \right) =F_{C}$ and

  $$\begin{aligned} \int \limits _{Y}\varphi d\nu =\int \limits _{Y}\psi d\nu . \end{aligned}$$

• (ii) Next we show that

  $$\begin{aligned} \int \limits _{Y}p_{C,w}\circ \varphi d\nu =\int \limits _{\left\{ \varphi \ge w\right\} }\left( \varphi -w\right) d\nu \ge 0,\quad w\in C^{\circ }. \end{aligned}$$

  (26)

  Let $w\in C^{\circ }$ be fixed. It follows from the definition of $\varphi$ that either $\left\{ \varphi \ge w\right\} =\emptyset$ or there exists a subsequence $\left( Y_{i_{j}}\right) _{j\in J}$ of $\left( Y_{i}\right) _{i\in I}$ such that

  $$\begin{aligned} \left\{ \varphi \ge w\right\} = {\displaystyle \bigcup \limits _{j\in J}} Y_{i_{j}}. \end{aligned}$$

  (27)

  If $\left\{ \varphi \ge w\right\} =\emptyset$, then (26) is obvious. In the other case

  $$\begin{aligned} \int \limits _{Y}p_{C,w}\circ \varphi d\nu = {\displaystyle \sum \limits _{j\in J}} \int \limits _{Y_{i_{j}}}\left( \varphi -w\right) d\nu = {\displaystyle \sum \limits _{j\in J}} \int \limits _{Y_{i_{j}}}\left( \psi -w\right) d\nu , \end{aligned}$$

  (28)

  and therefore inequality (26) follows if

  $$\begin{aligned} \int \limits _{Y_{i_{j}}}\left( \psi -w\right) d\nu \ge 0,\quad j\in J. \end{aligned}$$

  Since $\varphi \ge w$ on $Y_{i_{j}}$ and $\nu \left( Y_{i_{j}}\right) >0$,

  $$\begin{aligned} \int \limits _{Y_{i_{j}}}\left( \psi -w\right) d\nu =\nu \left( Y_{i_{j} }\right) \left( \varphi \left( y\right) -w\right) \ge 0,\quad y\in Y_{i_{j}}. \end{aligned}$$

• (iii) It can be proved similarly that

  $$\begin{aligned} \int \limits _{Y}n_{C,w}\circ \varphi d\nu =\int \limits _{\left\{ \varphi <w\right\} }\left( w-\varphi \right) d\nu \ge 0,\quad w\in C^{\circ }. \end{aligned}$$

• (iv) Finally we show that

  $$\begin{aligned} \int \limits _{Y}p_{C,w}\circ \varphi d\nu \le \int \limits _{Y}p_{C,w}\circ \psi d\nu \quad w\in C^{\circ }. \end{aligned}$$

  Let $w\in C^{\circ }$ be fixed. We have seen in part (ii) that either $\left\{ \varphi \ge w\right\} =\emptyset$ or there exists a subsequence $\left( Y_{i_{j}}\right) _{j\in J}$ of $\left( Y_{i}\right) _{i\in I}$ such that (27) holds. If $\left\{ \varphi \ge w\right\} =\emptyset$, then by (25),

  $$\begin{aligned} \int \limits _{Y}p_{C,w}\circ \varphi d\nu =0\le {\displaystyle \sum \limits _{i\in I}} \int \limits _{Y_{i}}p_{C,w}\circ \psi d\nu =\int \limits _{Y}p_{C,w}\circ \psi d\nu . \end{aligned}$$

  If $\left\{ \varphi \ge w\right\} \ne \emptyset$, then again using (25) we have that

  $$\begin{aligned}{} & {} \int \limits _{Y}p_{C,w}\circ \varphi d\nu = {\displaystyle \sum \limits _{j\in J}} \int \limits _{Y_{i_{j}}}\left( \varphi -w\right) d\nu = {\displaystyle \sum \limits _{j\in J}} \int \limits _{Y_{i_{j}}}\left( \psi -w\right) d\nu \\{} & {} \quad \le {\displaystyle \sum \limits _{j\in J}} \int \limits _{Y_{i_{j}}}p_{C,w}\circ \psi d\nu \le {\displaystyle \sum \limits _{i\in I}} \int \limits _{Y_{i}}p_{C,w}\circ \psi d\nu =\int \limits _{Y}p_{C,w}\circ \psi d\nu . \end{aligned}$$

  It follows from the above that the result can be obtained by applying Theorem 8.

The proof is complete. $\square$

It is worth formulating the previous statement in terms of Theorem 7.

Corollary 12

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

$$\begin{aligned} \lambda \left( a\right) \le \lambda \left( s_{1}\right) \le \lambda \left( t_{1}\right) \le \cdots \le \lambda \left( t_{n-1}\right) \le \lambda \left( s_{n}\right) \le \lambda \left( b\right) \end{aligned}$$

and

$$\begin{aligned} \lambda \left( t_{i-1}\right) <\lambda \left( t_{i}\right) ,\quad i=1,\ldots ,n, \end{aligned}$$

where $a=t_{0}<t_{1}<\cdots<t_{n-1}<t_{n}=b$, $n\in {\mathbb {N}}_{+}$ and $s_{i}\in \left[ t_{i-1},t_{i}\right]$ $\left( i=1,\ldots ,n\right)$. Let $C\subset {\mathbb {R}}$ be an interval with nonempty interior, and let $\psi :\left[ a,b\right] \rightarrow C$ be monotonic (in either sense) in each of the intervals $\left[ t_{i-1},t_{i}\right]$ $\left( i=1,\ldots ,n\right)$.

Then inequalities

$$\begin{aligned} f\left( \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} \psi d\nu _{\lambda }\right) \le \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} f\circ \varphi d\nu _{\lambda }\le \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} f\circ \psi d\nu _{\lambda } \end{aligned}$$

(29)

hold for every $f\in F_{C}$, where the function $\varphi :\left[ a,b\right] \rightarrow C$ is defined by

$$\begin{aligned} \varphi \left( s\right) :=\left\{ \begin{array}{l} \frac{1}{\nu _{\lambda }\left( [ t_{i-1},t_{i}[ \right) } \int \limits _{[ t_{i-1},t_{i}[ }\psi d\nu _{\lambda },\quad s\in [ t_{i-1},t_{i}[,\quad i=1,\ldots ,n-1\\ \frac{1}{\nu _{\lambda }\left( \left[ t_{n-1},t_{n}\right] \right) } \int \limits _{\left[ t_{n-1},t_{n}\right] }\psi d\nu _{\lambda },\quad s\in \left[ t_{n-1},t_{n}\right] \end{array} \right. . \end{aligned}$$

(30)

Proof

We only need to check that inequality (25) is satisfied using the sets $Y_{i}:=[ t_{i-1},t_{i}[$ $\left( i=1,\ldots ,n-1\right)$, $Y_{n}:=\left[ t_{n-1},b\right]$.

Since $\lambda$ is continuous,

$$\begin{aligned} {\displaystyle \int \limits _{[ t_{i-1},t_{i}[ }} \psi d\nu _{\lambda }= {\displaystyle \int \limits _{\left[ t_{i-1},t_{i}\right] }} \psi d\nu _{\lambda },\quad i=1,\ldots ,n. \end{aligned}$$

Since $\psi$ is monotonic in each of the intervals $\left[ t_{i-1} ,t_{i}\right]$ $\left( i=1,\ldots ,n\right)$, $p_{C,w}\circ \psi$ is also monotonic on $\left[ t_{i-1},t_{i}\right]$ $\left( i=1,\ldots ,n\right)$, and hence we can apply Lemma 1 which implies that

$$\begin{aligned} {\displaystyle \int \limits _{\left[ t_{i-1},t_{i}\right] }} p_{C,w}\circ \psi d\nu _{\lambda }\ge 0,\quad w\in C^{\circ },\quad i=1,\ldots ,n. \end{aligned}$$

It remains to prove

$$\begin{aligned} {\displaystyle \int \limits _{\left[ t_{i-1},t_{i}\right] }} p_{C,w}\circ \psi d\nu _{\lambda }\ge \int \limits _{\left[ t_{i-1},t_{i}\right] }\left( \psi -w\right) d\nu _{\lambda },\quad w\in C^{\circ },\quad i=1,\ldots ,n. \end{aligned}$$

(31)

Fix $w\in C^{\circ }$ and $i\in \left\{ 1,\ldots ,n\right\}$. Then

$$\begin{aligned} \int \limits _{\left[ t_{i-1},t_{i}\right] }\left( \psi -w\right) d\nu _{\lambda }= {\displaystyle \int \limits _{\left[ t_{i-1},t_{i}\right] }} p_{C,w}\circ \psi d\nu _{\lambda }- {\displaystyle \int \limits _{\left[ t_{i-1},t_{i}\right] }} n_{C,w}\circ \psi d\nu _{\lambda } \end{aligned}$$

(32)

Another application of Lemma 1 gives that

$$\begin{aligned} {\displaystyle \int \limits _{\left[ t_{i-1},t_{i}\right] }} n_{C,w}\circ \psi d\nu _{\lambda }\ge 0, \end{aligned}$$

and therefore (31) follows from (32).

The proof is complete. $\square$

Remark 9

Under the conditions of the previous result, the integrals in (29) and (30) can be taken in Riemann–Stieltjes sense, so that (29) and (30) can be written in the following form

$$\begin{aligned} f\left( {\displaystyle \int \limits _{a}^{b}} \psi d\lambda \diagup {\displaystyle \int \limits _{a}^{b}} 1d\lambda \right) \le {\displaystyle \int \limits _{a}^{b}} f\circ \varphi d\lambda \diagup {\displaystyle \int \limits _{a}^{b}} 1d\lambda \le {\displaystyle \int \limits _{a}^{b}} f\circ \psi d\lambda \diagup {\displaystyle \int \limits _{a}^{b}} 1d\lambda \end{aligned}$$

and

$$\begin{aligned} \varphi \left( s\right) :=\int \limits _{t_{i-1}}^{t_{i}}\psi d\lambda \diagup \int \limits _{t_{i-1}}^{t_{i}}1d\lambda ,\quad \left\{ \begin{array}{l} s\in \big [ t_{i-1},t_{i}\big [,\quad i=1,\ldots ,n-1\\ s\in \left[ t_{n-1},t_{n}\right] \end{array}\right. . \end{aligned}$$

The next refinement corresponds to Theorem 7.

Theorem 13

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

$$\begin{aligned} \lambda \left( a\right) \le \lambda \left( t\right) \le \lambda \left( b\right) \end{aligned}$$

and $\lambda \left( a\right) <\lambda \left( b\right)$. Let $C\subset {\mathbb {R}}$ be an interval with nonempty interior, and let $\psi :\left[ a,b\right] \rightarrow C$ be monotonic (in either sense). Suppose further that at any point of $\left[ a,b\right]$ at least one of $\lambda$ and $\psi$ is continuous. Let

$$\begin{aligned} t_{\psi ,\nu }:=\frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} \psi d\nu _{\lambda }. \end{aligned}$$

Then inequalities

$$\begin{aligned}{} & {} f\left( t_{\psi ,\lambda }\right) \le \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} f\circ \left( \alpha \psi +\left( 1-\alpha \right) t_{\psi ,\lambda }\right) d\nu _{\lambda } \\{} & {} \quad \le \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} f\circ \psi d\nu _{\lambda },\quad \alpha \in \left[ 0,1\right] \end{aligned}$$

hold for every $f\in F_{C}$.

Proof

We show that the conditions of Theorem 8 are satisfied by choosing

$$\begin{aligned} Y=X:=\left[ a,b\right] ,\quad \nu =\mu :=\nu _{\lambda }, \end{aligned}$$

and

$$\begin{aligned} \psi :\left[ a,b\right] \rightarrow C,\quad \varphi :=\alpha \psi +\left( 1-\alpha \right) t_{\psi ,\lambda },\quad \alpha \in \left[ 0,1\right] . \end{aligned}$$

By Theorem 7, $t_{\psi ,\lambda }\in C$. It now follows from the definition of $\varphi$ that $\varphi \left( \left[ a,b\right] \right) \subset C$. It is obvious that $\varphi$ is monotonic in the same sense as $\psi$.

Similarly to the first idea in the prof of Theorem 7, we can explain that $\psi$, $\varphi \in L\left( \nu _{\lambda }\right)$ and $F_{C}\left( \varphi ,\nu _{\lambda };\psi ,\nu _{\lambda }\right) =F_{C}$.

It is easy to check that

$$\begin{aligned} {\displaystyle \int \limits _{\left[ a,b\right] }} \psi d\nu _{\lambda }= {\displaystyle \int \limits _{\left[ a,b\right] }} \varphi d\nu _{\lambda }, \end{aligned}$$

and hence condition (21) holds.

From what we have established so far for $\varphi$, it follows that Theorem 7 can be applied to $\varphi$ (and therefore Theorem 3 implies that conditions (3) and (4) are satisfied). Similarly, it can be proved (we will need it in the proof) that

$$\begin{aligned} {\displaystyle \int \limits _{\left\{ \psi \ge w\right\} }} \left( \psi -w\right) d\nu _{\lambda }\ge 0,\quad {\displaystyle \int \limits _{\left\{ \psi <w\right\} }} \left( w-\psi \right) d\nu _{\lambda }\ge 0,\quad w\in C^{\circ }. \end{aligned}$$

(33)

We can see that only the verification of condition (22) remains to be done. This means that we have to show that the inequality

$$\begin{aligned} {\displaystyle \int \limits _{\left\{ \alpha \psi +\left( 1-\alpha \right) t_{\psi ,\lambda }\ge w\right\} }} \left( \alpha \psi +\left( 1-\alpha \right) t_{\psi ,\lambda }-w\right) d\nu _{\lambda }\le {\displaystyle \int \limits _{\left\{ \psi \ge w\right\} }} \left( \psi -w\right) d\nu _{\lambda },\quad w\in C^{\circ } \end{aligned}$$

(34)

holds for all $\alpha \in \left[ 0,1\right]$.

If $\alpha =0$, then (34) is trivial, so it can be assumed that $0<\alpha \le 1$. In this case (34) can be rewritten in the form

$$\begin{aligned} {\displaystyle \int \limits _{\left\{ \psi \ge \frac{1}{\alpha }\left( w-\left( 1-\alpha \right) t_{\psi ,\lambda }\right) \right\} }} \left( \alpha \psi +\left( 1-\alpha \right) t_{\psi ,\lambda }-w\right) d\nu _{\lambda }\le {\displaystyle \int \limits _{\left\{ \psi \ge w\right\} }} \left( \psi -w\right) d\nu _{\lambda },\quad w\in C^{\circ }. \end{aligned}$$

(35)

(a) We first prove (35) when $\alpha \in ] 0,1]$ and $w\in C^{\circ }$ are fixed and $\psi$ is increasing. For simpler notation, let

$$\begin{aligned} z:=\frac{1}{\alpha }\left( w-\left( 1-\alpha \right) t_{\psi ,\lambda }\right) . \end{aligned}$$

If $t_{\psi ,\lambda }\in C^{\circ }$ and $w=t_{\psi ,\lambda }$, then both sides of inequality (35) are ${\displaystyle \int \limits _{\left\{ \psi \ge t_{\psi ,\lambda }\right\} }} \left( \psi -t_{\psi ,\lambda }\right) d\nu _{\lambda }$, so the inequality is true.

If $w\ne t_{\psi ,\lambda }$, then some easy calculation shows that either $t_{\psi ,\lambda }<w<z$ or $z<w<t_{\psi ,\lambda }$.

(i) Assume first that $t_{\psi ,\lambda }<w<z$. Then $\left\{ \psi \ge z\right\} \subset \left\{ \psi \ge w\right\}$.

If either $\left\{ \psi \ge z\right\} =\emptyset$ or $\left\{ \psi \ge z\right\} =\left\{ \psi \ge w\right\}$, then (35) obviously holds (in the first case (33) can be applied, while in the second case both sides of (35) are identical).

Suppose $\left\{ \psi \ge z\right\} \ne \emptyset$ and $\left\{ \psi \ge z\right\} \ne \left\{ \psi \ge w\right\}$. Since $t_{\psi ,\lambda }<w$, Remark 1 implies that $\left\{ \psi <w\right\} \ne \emptyset$ too. Now, if we define the function $\chi$ on $\left[ a,b\right]$ by

$$\begin{aligned} \chi \left( t\right) :=\left\{ \begin{array}{ll} 0, &{} \text {if} \ t\in \left\{ \psi<w\right\} \\ \psi -w, &{} \text {if} \ t\in \left\{ w\le \psi <z\right\} \\ \psi -w-\left( \varphi -w\right) =\left( 1-\alpha \right) \left( \psi -t_{\psi ,\lambda }\right) , &{} \text {if} \ t\in \left\{ \psi \ge z\right\} \end{array}\right. , \end{aligned}$$

then inequality (35) is equivalent to the following:

$$\begin{aligned} {\displaystyle \int \limits _{\left[ a,b\right] }} \chi d\nu _{\lambda }\ge 0. \end{aligned}$$

(36)

Since $\psi$ is increasing, each of the three sets $\left\{ \psi <w\right\}$, $\left\{ w\le \psi <z\right\}$ and $\left\{ \psi \ge z\right\}$ is a nonempty interval, which form a partition of $\left[ a,b\right]$ in this order. Let $t_{z}$ denote the left endpoint of the interval $\left\{ \psi \ge z\right\}$. Then exactly one of $t_{z}\in \left\{ \psi \ge z\right\}$ and $t_{z}\in \left\{ w\le \psi <z\right\}$ is satisfied.

It follows from the definition of $\chi$ that it is nonnegative.

In the next step we show that $\chi$ is increasing.

Since $\psi$ is increasing, $\chi$ is increasing on $\left\{ \psi<w\right\} \cup \left\{ w\le \psi <z\right\}$ and $\chi$ is increasing on $\left\{ \psi \ge z\right\}$. It follows that we only need to prove the following: if $t_{z}\in \left\{ \psi \ge z\right\}$, then $\chi \left( t\right) \le \chi \left( t_{z}\right)$ for all $t\in \left\{ w\le \psi <z\right\}$, and if $t_{z}\in \left\{ w\le \psi <z\right\}$, then $\chi \left( t_{z}\right) \le \chi \left( t\right)$ for all $t\in \left\{ \psi \ge z\right\}$.

Assume that $t_{z}\in \left\{ \psi \ge z\right\}$, and let $t\in \left\{ w\le \psi <z\right\}$. Then $\psi \left( t_{z}\right) \ge z$, and hence

$$\begin{aligned} \chi \left( t_{z}\right)= & {} \left( 1-\alpha \right) \left( \psi \left( t_{z}\right) -t_{\psi ,\lambda }\right) \ge \left( 1-\alpha \right) \left( z-t_{\psi ,\lambda }\right) \\= & {} \left( 1-\alpha \right) \left( \frac{1}{\alpha }\left( w-\left( 1-\alpha \right) t_{\psi ,\lambda }\right) -t_{\psi ,\lambda }\right) =\frac{1}{\alpha }\left( 1-\alpha \right) \left( w-t_{\psi ,\lambda }\right) \\= & {} z-w\ge \psi \left( t\right) -w=\chi \left( t\right) . \end{aligned}$$

Now assume that $t_{z}\in \left\{ w\le \psi <z\right\}$, and let $t\in \left\{ \psi \ge z\right\}$. Then $\psi (tz) < z$, and hence

$$\begin{aligned} \chi \left( t_{z}\right)= & {} \psi \left( t_{z}\right) -w\le z-w=\frac{1}{\alpha }\left( w-\left( 1-\alpha \right) t_{\psi ,\lambda }\right) -w\\= & {} \frac{1}{\alpha }\left( 1-\alpha \right) \left( w-t_{\psi ,\lambda }\right) =\left( 1-\alpha \right) \left( z-t_{\psi ,\lambda }\right) \le \left( 1-\alpha \right) \left( \psi \left( t\right) -t_{\psi ,\lambda }\right) =\chi \left( t\right) . \end{aligned}$$

We have seen that $\chi$ is nonnegative and increasing, so Lemma 1 gives inequality (36).

(ii) Now assume that $z<w<t_{\psi ,\lambda }$. Then $\left\{ \psi \ge w\right\} \subset \left\{ \psi \ge z\right\}$, and hence $\left\{ \psi<z\right\} \subset \left\{ \psi <w\right\}$.

In this case, we use the fact that

$$\begin{aligned} {\displaystyle \int \limits _{\left[ a,b\right] }} \left( \alpha \psi +\left( 1-\alpha \right) t_{\psi ,\lambda }-w\right) d\nu _{\lambda }= {\displaystyle \int \limits _{\left[ a,b\right] }} \left( \psi -w\right) d\nu _{\lambda },\quad w\in C^{\circ }, \end{aligned}$$

so instead of inequality (35) we can prove the equivalent inequality

$$\begin{aligned} {\displaystyle \int \limits _{\left\{ \psi<\frac{1}{\alpha }\left( w-\left( 1-\alpha \right) t_{\psi ,\lambda }\right) \right\} }} \left( w-\alpha \psi -\left( 1-\alpha \right) t_{\psi ,\lambda }\right) d\nu _{\lambda }\le {\displaystyle \int \limits _{\left\{ \psi <w\right\} }} \left( w-\psi \right) d\nu _{\lambda },\quad w\in C^{\circ }.\nonumber \\ \end{aligned}$$

(37)

If either $\left\{ \psi <z\right\} =\emptyset$ or $\left\{ \psi<z\right\} =\left\{ \psi <w\right\}$, then (37) obviously holds (see the similar cases in part (i)).

Suppose $\left\{ \psi <z\right\} \ne \emptyset$ and $\left\{ \psi<z\right\} \ne \left\{ \psi <w\right\}$. Since $w<t_{\psi ,\lambda }$, Remark 1 implies that $\left\{ \psi \ge w\right\} \ne \emptyset$ too. Now, if we define the function $\omega$ on $\left[ a,b\right]$ by

$$\begin{aligned} \omega \left( t\right) :=\left\{ \begin{array}{ll} 0, &{} \text {if} \ t\in \left\{ \psi \ge w\right\} \\ w-\psi , &{} \text {if} \ t\in \left\{ z\le \psi<w\right\} \\ w-\psi -\left( w-\varphi \right) =\left( 1-\alpha \right) \left( t_{\psi ,\lambda }-\psi \right) , &{} \text {if} \ t\in \left\{ \psi <z\right\} \end{array}\right. , \end{aligned}$$

then inequality (37) is equivalent to the following:

$$\begin{aligned} {\displaystyle \int \limits _{\left[ a,b\right] }} \omega d\nu _{\lambda }\ge 0. \end{aligned}$$

(38)

As in case (i), the sets $\left\{ \psi <z\right\}$, $\left\{ z\le \psi <w\right\}$ and $\left\{ \psi \ge w\right\}$ are nonempty intervals, which form a partition of $\left[ a,b\right]$ in this order. Let $t_{z}$ denote the right endpoint of the interval $\left\{ \psi <z\right\}$. Then exactly one of $t_{z}\in \left\{ \psi <z\right\}$ and $t_{z}\in \left\{ z\le \psi <w\right\}$ is satisfied.

The definition of $\omega$ implies that it is nonnegative.

By analogously proving that $\chi$ is increasing, we can show that $\omega$ is decreasing, but we omit the details.

Since $\omega$ is nonnegative and decreasing, Lemma 1 gives inequality (38).

(b) We still need to justify the case where $\alpha \in ] 0,1]$ and $w\in C^{\circ }$ are fixed and $\psi$ is decreasing.

This case can be reduced to the previous one by choosing ${\widehat{C}}:=-C$ and ${\widehat{\psi }}:=-\psi$, and by using that

$$\begin{aligned} F_{{\widehat{C}}}=\left\{ t\rightarrow f\left( -t\right) \mid t\in {\widehat{C}},\quad f\in F_{C}\right\} . \end{aligned}$$

The proof is complete. $\square$

The following is an essential generalization of the previous result with a simple proof.

Corollary 14

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

$$\begin{aligned} \lambda \left( a\right) \le \lambda \left( t\right) \le \lambda \left( b\right) \end{aligned}$$

and $\lambda \left( a\right) <\lambda \left( b\right)$. Let $C\subset {\mathbb {R}}$ be an interval with nonempty interior, and let $\zeta :\left[ a,b\right] \rightarrow C$ be monotonic (in either sense) such that at any point of $\left[ a,b\right]$ at least one of $\lambda$ and $\zeta$ is continuous.

Let $\left( Y,{\mathcal {B}},\nu \right)$ be a measure space, where $\nu$ is a finite signed measure and $\nu \left( Y\right) >0$. Let $\psi :Y\rightarrow C$ be a function such that $\psi \in L\left( \nu \right)$ and

$$\begin{aligned} t_{\zeta ,\lambda }:=\frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} \zeta d\nu _{\lambda }=\frac{1}{\nu \left( Y\right) } {\displaystyle \int \limits _{Y}} \psi d\nu =:t_{\psi ,\nu } \end{aligned}$$

(39)

and

$$\begin{aligned} \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) }\int \limits _{\left[ a,b\right] }p_{C,w}\circ \zeta d\nu _{\lambda }\le \frac{1}{\nu \left( Y\right) }\int \limits _{Y}p_{C,w}\circ \psi d\nu _{\lambda }\quad w\in C^{\circ } \end{aligned}$$

(40)

are satisfied.

Then inequalities

$$\begin{aligned}{} & {} f\left( t_{\psi ,\nu }\right) \le {\displaystyle \int \limits _{\left[ a,b\right] }} f\circ \left( \alpha \zeta +\left( 1-\alpha \right) t_{\psi ,\nu }\right) d\nu _{\lambda }\diagup {\displaystyle \int \limits _{\left[ a,b\right] }} 1d\nu _{\lambda } \\{} & {} \quad \le {\displaystyle \int \limits _{\left[ a,b\right] }} f\circ \zeta d\nu _{\lambda }\diagup {\displaystyle \int \limits _{\left[ a,b\right] }} 1d\nu _{\lambda }\le \frac{1}{\nu \left( Y\right) } {\displaystyle \int \limits _{Y}} f\circ \psi d\nu ,\quad \alpha \in \left[ 0,1\right] \end{aligned}$$

hold for every $f\in F_{C}\left( \psi ,\nu \right)$.

Proof

By Theorem 7, $t_{\zeta ,\nu }\in C$. Using (39) and then applying Theorem 13 with $\zeta$ instead of $\psi$, we obtain that

$$\begin{aligned} f\left( t_{\psi ,\nu }\right)= & {} f\left( t_{\zeta ,\lambda }\right) \le \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} f\circ \left( \alpha \zeta +\left( 1-\alpha \right) t_{\zeta ,\lambda }\right) d\nu _{\lambda }\\= & {} \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} f\circ \left( \alpha \zeta +\left( 1-\alpha \right) t_{\psi ,\lambda }\right) d\nu _{\lambda }\le \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} f\circ \zeta d\nu _{\lambda }. \end{aligned}$$

From this we obtain the statement using Theorem 6 (conditions (39) and (40) ensure the applicability of the theorem).

The proof is complete. $\square$

Remark 10

• (a) Although Corollary 14 is more general than Theorem 13, it can be seen that the essential part of it is the statement in Theorem 13.

• (b) Corollary 14 is a substantial generalization of Theorem 2.1 in [17]. In the mentioned result $\psi$ is also defined in $\left[ a,b\right]$, it uses an absolutely continuous measure (not a signed measure) with respect to the Lebesgue measure on $\left[ a,b\right]$, and finally, condition (40) is replaced by a stronger one. The proof of the inequality in Theorem 13 is very simple in Theorem 2.1, since the monotonicity of the integral can be exploited.

• (c) The discrete version of Theorem 2.1 in [17] can be found in [4]. This can be obtained directly from the integral form, but a direct proof can also be given (see Theorem 5 and Theorem 6 in [4]).

It is worth mentioning the special case of the previous corollary where the function $\zeta$ is also defined on $\left[ a,b\right]$. We do not state the statement in full generality, but instead of condition (40) we use a condition that is close to the usual notion of majorization between functions. We can do this based on the results of the recent paper [9].

Corollary 15

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

$$\begin{aligned} \lambda \left( a\right) \le \lambda \left( t\right) \le \lambda \left( b\right) \end{aligned}$$

and $\lambda \left( a\right) <\lambda \left( b\right)$. Let $C\subset {\mathbb {R}}$ be an interval with nonempty interior, and let $\psi$, $\zeta :\left[ a,b\right] \rightarrow C$ be functions such that either both decreasing and

$$\begin{aligned} \int \limits _{\left[ a,x\right] }\zeta d\nu _{\lambda }\le \int \limits _{\left[ a,x\right] }\psi d\nu _{\lambda },\quad x\in \left[ a,b\right] \end{aligned}$$

or both increasing and

$$\begin{aligned} \int \limits _{\left[ a,x\right] }\psi d\nu _{\lambda }\le \int \limits _{\left[ a,x\right] }\zeta d\nu _{\lambda },\quad x\in \left[ a,b\right] . \end{aligned}$$

Suppose further that at any point of $\left[ a,b\right]$ at least one of $\lambda$ and $\zeta$ is continuous and

$$\begin{aligned} {\displaystyle \int \limits _{\left[ a,b\right] }} \psi d\nu _{\lambda }= {\displaystyle \int \limits _{\left[ a,b\right] }} \zeta d\nu _{\lambda }. \end{aligned}$$

Then inequalities

$$\begin{aligned} f\left( t_{\psi ,\lambda }\right)\le & {} \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} f\circ \left( \alpha \zeta +\left( 1-\alpha \right) t_{\psi ,\lambda }\right) d\nu _{\lambda }\\\le & {} \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} f\circ \zeta d\nu _{\lambda }\le \frac{1}{\nu _{\lambda }\left( \left[ a,b\right] \right) } {\displaystyle \int \limits _{\left[ a,b\right] }} f\circ \psi d\nu _{\lambda },\quad \alpha \in \left[ 0,1\right] \end{aligned}$$

hold for every $f\in F_{C}$.

Proof

It follows from Corollary 14 by using Theorem 6 in [9].

The proof is complete. $\square$