Extension of an Inequality on Three Intervals and Applications to Csiszár ϕ-Divergence and Landau–Kolmogorov Inequality

Abstract

In this paper, we generalize an inequality for a convex function in one dimension ${\mathbb{R}}^1$ on three intervals to a function with nondecreasing increments in k dimensions ${\mathbb{R}}^k$ on $(2n+1)$ intervals. We prove all the situations when $n=1,2$ and prove a very special case for a general n as well as the discrete version. The proofs are based on a general conclusion for convex functions, and analogues of this conclusion are established. We apply the discrete case of the inequality to Csiszár $\varphi$-divergence $I_{\varphi }(p,q)$ in information theory, and the continuous case $I_{\varphi }(p_1,q_1)\le I_{\varphi }(p_2,q_2)$ on a measurable set is also established. The same inequality for an $ϵ$-approximately convex function on a discrete set is also established and can be used to prove a similar Landau–Kolmogorov-type inequality.

1. Introduction

Recall some background of Lemma 1. We denote by $A\left(a\right),G\left(a\right)$ and $H\left(a\right)$ the unweighted arithmetic, geometric, and harmonic means of positive real numbers $a_1,\dots ,a_n$ with $a=(a_1,\dots ,a_n)\in {\mathbb{R}}_{+}^n$. The fundamental inequalities

$$H\left(a\right)\le G\left(a\right)\le A\left(a\right)$$

have the following refinements:

$${\displaystyle \frac{H\left(a\right)}{H(1-a)}}\le {\displaystyle \frac{G\left(a\right)}{G(1-a)}}\le {\displaystyle \frac{A\left(a\right)}{A(1-a)}}$$

(1)

for $0<a_i\le {\displaystyle \frac{1}{2}}$ and

$${\displaystyle \frac{H\left(a\right)}{H(1+a)}}\le {\displaystyle \frac{G\left(a\right)}{G(1+a)}}\le {\displaystyle \frac{A\left(a\right)}{A(1+a)}}$$

(2)

for $a_i>0$.

The right side of (1) is due to Ky Fan in [1], which is known as Ky Fan inequality in the literature, and the left side, known as Wang-Wang inequality, is proposed in [2] and followed by [3,4,5,6,7,8,9,10,11]. Part of (2) is proved in [12], and the whole chain (2) is proved in [8].

The Levinson inequality [13] was established in 1964 to generalise the Ky Fan inequality under the framework of three-convex functions. However, for the Levinson inequality and its generalisations [14,15,16,17], it is still unable to unify (1) and (2). Thus, in [18], the following conclusion is established for convex functions on the interval divided into three parts to prove the generalised Levinson inequality so that Ky Fan-type inequalities (1) and (2) can be unified. ln in these Ky Fan-type inequalities can eventually be transformed into Lemma 1 for $\varphi \left(x\right)={\displaystyle \frac{1}{x}}$.

Lemma 1.

Let $\alpha <a<b<\beta$ and $x:[\alpha ,\beta ]\to \mathbb{R}$ be a given function defined on $[\alpha ,\beta ]$. Let $p:[\alpha ,a]\cup [b,\beta ]\to {\mathbb{R}}_{+}$ and $q:[a,b]\to {\mathbb{R}}_{+}$. If $x\left(t\right)$ is increasing in $t\in [\alpha ,\beta ]$ and

$$\begin{array}{cc}\hfill & {\int }_a^{b}q\left(t\right)x\left(t\right)dt={\int }_{\alpha }^{a}p\left(t\right)x\left(t\right)dt+{\int }_b^{\beta }p\left(t\right)x\left(t\right)dt,\hfill \\ \hfill & {\int }_a^{b}q\left(t\right)dt={\int }_{\alpha }^{a}p\left(t\right)dt+{\int }_b^{\beta }p\left(t\right)dt,\hfill \end{array}$$

then

$${\int }_a^{b}q\left(t\right)\varphi \left(x\left(t\right)\right)dt\le {\int }_{\alpha }^{a}p\left(t\right)\varphi \left(x\left(t\right)\right)dt+{\int }_b^{\beta }p\left(t\right)\varphi \left(x\left(t\right)\right)dt$$

holds for every ϕ that is convex such that the integrals exist.

As pointed out in [18], it also has some connection with the Lah–Ribarič inequality, and its generalisation can be used to prove the Hermite–Hadamard inequality. For related research on the Hermite–Hadamard inequality, see [19,20,21,22,23,24,25,26,27].

Due to its simple assumption and potential to prove several inequalities concerning convexity, in this paper, we further extend this inequality to several variables, that is, functions with nondecreasing increments. Theorem 1 below [28,29] is essential in our proof, and we also use definitions and notations in Section 4.1 in [29] (p. 107).

Let ${\mathbb{R}}^k$ denote the k-dimensional vector lattice of points $\mathbf{x}=(x_1,\dots ,x_k)$, $x_i$ real for $i=1,\dots ,k$ with the partial ordering $\mathbf{x}=(x_1,\dots ,x_k)\le (y_1,\dots ,y_k)=\mathbf{y}$ if and only if $x_i\le y_i$ for $i=1,\dots ,k$. For $\mathbf{a},\mathbf{b}\in {\mathbb{R}}^k,\mathbf{a}\le \mathbf{b}$, a set $\{\mathbf{x}\in {\mathbb{R}}^k:\mathbf{a}\le \mathbf{x}\le \mathbf{b}\}$ is called an interval $[\mathbf{a},\mathbf{b}]$. We also use a symbol $\mathbf{I}$ for an interval in ${\mathbb{R}}^k$.

By $\mathbf{x}\left(t\right)=(x_1\left(t\right),\dots ,x_k\left(t\right))$ we denote a mapping of an interval from $\mathbb{R}$ into an interval $\mathbf{I}\in {\mathbb{R}}^k$; property for x means a property for all components $x_j$.

A real-valued function f on an interval $\mathbf{I}\in {\mathbb{R}}^k$ will be said to have nondecreasing increments if

$$f(\mathbf{a}+\mathbf{h})-f\left(\mathbf{a}\right)\le f(\mathbf{b}+\mathbf{h})-f\left(\mathbf{b}\right)$$

whenever $\mathbf{a}\in \mathbf{I},\mathbf{b}+\mathbf{h}\in \mathbf{I}$, $\mathbf{0}\le \mathbf{h}\in {\mathbb{R}}^k$, $\mathbf{a}\le \mathbf{b}$.

Theorem 1.

Let $\mathit{I}$ denote an interval in ${\mathbb{R}}^k$, let $\mathit{x}:[\alpha ,\beta ]\to \mathit{I}$ be a nondecreasing continuous map, and let H be a function of bounded variation and continuous from the left on $[\alpha ,\beta ]$ with $H\left(\alpha \right)=0$. Then,

$${\int }_{\alpha }^{\beta }f\left(\mathbf{x}\left(t\right)\right)dH\left(t\right)\ge 0$$

(3)

for every continuous function $f:\mathit{I}\to \mathbb{R}$ with nondecreasing increments if and only if

$$\begin{array}{c}\hfill H\left(\beta \right)=0,\end{array}$$

(4)

$$\begin{array}{c}\hfill {\int }_{\alpha }^{\beta }H\left(u\right)d\mathbf{x}\left(u\right)=0,\end{array}$$

(5)

and

$$\begin{array}{c}\hfill {\int }_{\alpha }^{t}H\left(u\right)d\mathbf{x}\left(u\right)\ge 0for[\alpha ,t\}\subset [\alpha ,\beta ],\end{array}$$

(6)

where $\int Hd\mathit{x}=(\int Hd{x}_1,\dots ,\int Hd{x}_k)$ and the symbol $[\alpha ,t\}$ refer to either of the intervals $[\alpha ,t]$ or $[\alpha ,t)$.

This is a very general inequality concerning convex functions; for further generalisations of Theorem 1, see [30,31,32], [33] (pp. 351–362). As the majorization theorem also holds for functions with nondecreasing increments [32], it largely extends the utility of majorization on functions with several variables. It is impossible to list all the progress [34,35] and significance of majorization in inequality theory as well as applications in other areas in this paper. For a systematic review, see [36].

Apart form using Theorem 1 to establish extensions for Lemma 1 on ${\mathbb{R}}^k$, we also prove the discrete generalization of Lemma 1, several functions’ version of Lemma 1, and Lemma 1 for a more general $ϵ$-approximately convex function. We also apply Lemma 1 to establish an inequality for Csiszár $\varphi$-divergence $I_{\varphi }(p,q)$ in information theory and Landau–Kolmogorov-type inequality.

2. Main Results

In this section, we use Theorem 1 to prove theorems about inequality on an interval divided into several parts. First we prove the three-part situation and its corollary, and then we consider a slightly more general but more complex five-part situation.

Theorem 2.

Let $\mathit{I}$ denote an interval in ${\mathbb{R}}^k$ and $\alpha <a<b<\beta$; let $\mathit{x}:[\alpha ,\beta ]\to \mathit{I}$ be a nondecreasing continuous map. Let $p:[\alpha ,a]\cup [b,\beta ]\to {\mathbb{R}}_{+}$ and $q:[a,b]\to {\mathbb{R}}_{+}$ be continuous functions. If

$$\begin{array}{cc}\hfill & {\int }_a^{b}q\left(t\right)\mathit{x}\left(t\right)dt={\int }_{\alpha }^{a}p\left(t\right)\mathit{x}\left(t\right)dt+{\int }_b^{\beta }p\left(t\right)\mathit{x}\left(t\right)dt,\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill & {\int }_a^{b}q\left(t\right)dt={\int }_{\alpha }^{a}p\left(t\right)dt+{\int }_b^{\beta }p\left(t\right)dt,\hfill \end{array}$$

(8)

then

$${\int }_a^{b}q\left(t\right)\varphi \left(\mathit{x}\left(t\right)\right)dt\le {\int }_{\alpha }^{a}p\left(t\right)\varphi \left(\mathit{x}\left(t\right)\right)dt+{\int }_b^{\beta }p\left(t\right)\varphi \left(\mathit{x}\left(t\right)\right)dt$$

holds for every continuous function $\varphi :\mathit{I}\to \mathbb{R}$ with nondecreasing increments.

Proof. 

The main idea of the proof is to regard three integrals on $[\alpha ,\beta ]$ in this theorem as one integral on $[\alpha ,\beta ]$ in Theorem 1. Set

$$\begin{array}{c}\hfill H\left(t\right)=\left\{\begin{array}{cc}{\int }_{\alpha }^{t}p\left(u\right)du,\hfill & t\in [\alpha ,a],\hfill \\ {\int }_{\alpha }^{a}p\left(u\right)du-{\int }_a^{t}q\left(u\right)du,\hfill & t\in [a,b],\hfill \\ {\int }_{\alpha }^{a}p\left(u\right)du-{\int }_a^{b}q\left(u\right)du+{\int }_b^{t}p\left(u\right)du,\hfill & t\in [b,\beta ],\hfill \end{array}\right.\end{array}$$

it is easy to see from (8) that $H\left(\alpha \right)=0=H\left(\beta \right)$; thus, (4) is satisfied. H is a $BV$ function, as it is continuous on $[\alpha ,\beta ]$, differentiable $a.e.$ on $[\alpha ,\beta ]$, and $|H^{\prime }| \le max\{\left|p\left(t\right)\right|,\left|q\left(t\right)\right|\}$ (as $p,q$ are continuous on the closed interval).

Second, from (7), we have

$${\int }_{\alpha }^{\beta }H\left(u\right)d\mathbf{x}\left(u\right)={\int }_{\alpha }^{\beta }\mathbf{x}\left(u\right)dH\left(u\right)=0,$$

Thus, (5) is satisfied.

Third,

$$\begin{array}{c}\hfill {\int }_{\alpha }^{t}H\left(u\right)d\mathbf{x}\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{\alpha }^t\mathbf{x}\left(u\right)dH\left(u\right),\end{array}$$

Then, we need to prove (6); this should be divided into several situations to discuss, as in each situation, it is easier to identify (6). Specifically in this theorem,

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{\alpha }^t\mathbf{x}\left(u\right)dH\left(u\right)\ge 0,\end{array}$$

holds.

If $\alpha <t\le a$, then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{\alpha }^t\mathbf{x}\left(u\right)dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right){\int }_{\alpha }^{t}dH\left(u\right)=0.\end{array}$$

If $a<t<b$ and $H\left(t\right)\ge 0$, then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{\alpha }^t\mathbf{x}\left(u\right)dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{\alpha }^t\mathbf{x}\left(a\right)dH\left(u\right)\\ \hfill =H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a\right){\int }_{\alpha }^{t}dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right){\int }_{\alpha }^{t}dH\left(u\right)=0.\end{array}$$

If $a<t\le b$ and $H\left(t\right)\le 0$, according to (7) and the definition of H, we have

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{\alpha }^t\mathbf{x}\left(u\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{\beta }\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{\beta }\mathbf{x}\left(b\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+\mathbf{x}\left(b\right){\int }_t^{\beta }dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+\mathbf{x}\left(t\right){\int }_t^{\beta }dH\left(u\right)=0.\end{array}$$

If $b<t<\beta$, according to (7), then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{\alpha }^t\mathbf{x}\left(u\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{\beta }\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{\beta }\mathbf{x}\left(t\right)dH\left(u\right)=0.\end{array}$$

From the above discussion, we conclude that (6) is also satisfied; thus, from Theorem 1, we prove Theorem 2. □

Set $k=2$ and $\varphi (x_1,x_2)=x_1{x}_2$ in Theorem 2; it is easy to see that $\varphi$ is a function with nondecreasing increments. Thus, we have a Chebyshev-type inequality below.

Corollary 1.

Let I denote an interval in $\mathbb{R}$ and $\alpha <a<b<\beta$; let $x_j:[\alpha ,\beta ]\to I,(j=1,2)$ be nondecreasing continuous functions. Let $p:[\alpha ,a]\cup [b,\beta ]\to {\mathbb{R}}_{+}$ and $q:[a,b]\to {\mathbb{R}}_{+}$ be continuous functions. If

$$\begin{array}{cc}\hfill & {\int }_a^{b}q\left(t\right)x_j\left(t\right)dt={\int }_{\alpha }^{a}p\left(t\right)x_j\left(t\right)dt+{\int }_b^{\beta }p\left(t\right)x_j\left(t\right)dt(j=1,2),\hfill \\ \hfill & {\int }_a^{b}q\left(t\right)dt={\int }_{\alpha }^{a}p\left(t\right)dt+{\int }_b^{\beta }p\left(t\right)dt,\hfill \end{array}$$

then

$${\int }_a^{b}q\left(t\right)x_1\left(t\right)x_2\left(t\right)dt\le {\int }_{\alpha }^{a}p\left(t\right)x_1\left(t\right)x_2\left(t\right)dt+{\int }_b^{\beta }p\left(t\right)x_1\left(t\right)x_2\left(t\right)dt.$$

(9)

Example 1.

Take $\alpha =a-\Delta ,\beta =b+\Delta ,(\Delta >0)$ in Corollary 1, and let $x_j\left(t\right)=t,p\left(t\right)=1,q\left(t\right)={\displaystyle \frac{2\Delta }{b-a}}$; it is easy to see that all the conditions are satisfied. Thus, from (9) we have

$${\displaystyle \frac{2\Delta (b^3-a^3)}{b-a}}\le a^3-{(a-\Delta )}^3+{(b+\Delta )}^3-b^3.$$

This inequality holds for all $\Delta >0$ and $b>a$.

For example, take $\Delta =1,a={\displaystyle \frac{\sqrt{1349}-1}{2}},b={\displaystyle \frac{\sqrt{1349}+1}{2}}$; we get

$$2024\le 2030.$$

Set $\varphi (x_1,\dots ,x_k)={\prod }_{j=1}^k{x}_j$ and further suppose $x_j\ge 0$ in Theorem 2; it is easy to see that $\varphi$ is a function with nondecreasing increments. Thus, we have another Chebyshev-type inequality below.

Corollary 2.

Let I denote an interval in ${\mathbb{R}}_{+}$ and $\alpha <a<b<\beta$; let $x_j:[\alpha ,\beta ]\to I,(j=1,\dots ,k)$ be nondecreasing continuous functions. Let $p:[\alpha ,a]\cup [b,\beta ]\to {\mathbb{R}}_{+}$ and $q:[a,b]\to {\mathbb{R}}_{+}$ be continuous functions. If

$$\begin{array}{cc}\hfill & {\int }_a^{b}q\left(t\right)x_j\left(t\right)dt={\int }_{\alpha }^{a}p\left(t\right)x_j\left(t\right)dt+{\int }_b^{\beta }p\left(t\right)x_j\left(t\right)dt(j=1,\dots ,k),\hfill \\ \hfill & {\int }_a^{b}q\left(t\right)dt={\int }_{\alpha }^{a}p\left(t\right)dt+{\int }_b^{\beta }p\left(t\right)dt,\hfill \end{array}$$

then

$${\int }_a^{b}q\left(t\right)\prod _{j=1}^k{x}_j\left(t\right)dt\le {\int }_{\alpha }^{a}p\left(t\right)\prod _{j=1}^k{x}_j\left(t\right)dt+{\int }_b^{\beta }p\left(t\right)\prod _{j=1}^k{x}_j\left(t\right)dt.$$

Theorem 2 divides the interval into three parts; then, we consider an interval $[a_0,a_5]$ divided into five parts $a_0<a_1<a_2<a_3<a_4<a_5$.

Theorem 3.

Let $\mathit{I}$ denote an interval in ${\mathbb{R}}^k$ and $\mathit{x}:[a_0,a_5]\to \mathit{I}$ be a nondecreasing continuous map. Let $p:[a_0,a_1]\cup [a_2,a_3]\cup [a_4,a_5]\to {\mathbb{R}}_{+}$ and $q:[a_1,a_2]\cup [a_3,a_4]\to {\mathbb{R}}_{+}$ be continuous functions. If

$$\begin{array}{c}\hfill \sum _{i=1}^2{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)\mathit{x}\left(t\right)dt=\sum _{i=0}^2{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)\mathit{x}\left(t\right)dt,\end{array}$$

(10)

$$\begin{array}{c}\hfill \sum _{i=1}^2{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)dt=\sum _{i=0}^2{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)dt,\end{array}$$

(11)

$$\begin{array}{c}\hfill {\int }_{a_0}^{a_1}p\left(t\right)dt\le {\int }_{a_1}^{a_2}q\left(t\right)dt\le {\int }_{a_0}^{a_1}p\left(t\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)dt,\end{array}$$

(12)

and

$$\begin{array}{c}\hfill {\int }_{a_1}^{a_2}q\left(t\right)\mathit{x}\left(t\right)dt\ge {\int }_{a_0}^{a_1}p\left(t\right)\mathit{x}\left(t\right)dt+{\int }_{a_2}^{c}p\left(t\right)\mathit{x}\left(t\right)dt,\end{array}$$

(13)

where c is defined in equation ${\int }_{a_1}^{a_2}q\left(t\right)dt={\int }_{a_0}^{a_1}p\left(t\right)dt+{\int }_{a_2}^{c}p\left(t\right)dt$, then

$$\sum _{i=1}^2{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)\varphi \left(\mathit{x}\left(t\right)\right)dt\le \sum _{i=0}^2{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)\varphi \left(\mathit{x}\left(t\right)\right)dt$$

holds for every continuous function $\varphi :\mathit{I}\to \mathbb{R}$ with nondecreasing increments.

Proof. 

The main idea of the proof is to regard five integrals on $[a_0,a_5]$ in this theorem as one integral on $[a,b]$ in Theorem 1. Set

$$\begin{array}{cc}\hfill & H\left(t\right)=\hfill \\ \hfill & \left\{\begin{array}{cc}{\int }_{a_0}^{t}p\left(u\right)du,\hfill & t\in [a_0,a_1],\hfill \\ {\int }_{a_0}^{a_1}p\left(u\right)du-{\int }_{a_1}^{t}q\left(u\right)du,\hfill & t\in [a_1,a_2],\hfill \\ {\int }_{a_0}^{a_1}p\left(u\right)du-{\int }_{a_1}^{a_2}q\left(u\right)du+{\int }_{a_2}^{t}p\left(u\right)du,\hfill & t\in [a_2,a_3],\hfill \\ {\int }_{a_0}^{a_1}p\left(u\right)du-{\int }_{a_1}^{a_2}q\left(u\right)du+{\int }_{a_2}^{a_3}p\left(u\right)du-{\int }_{a_3}^{t}q\left(u\right)du,\hfill & t\in [a_3,a_4],\hfill \\ {\int }_{a_0}^{a_1}p\left(u\right)du-{\int }_{a_1}^{a_2}q\left(u\right)du+{\int }_{a_2}^{a_3}p\left(u\right)du-{\int }_{a_3}^{a_4}q\left(u\right)du+{\int }_{a_4}^{t}p\left(u\right)du,\hfill & t\in [a_4,a_5],\hfill \end{array}\right.\hfill \end{array}$$

it is easy to see from (11) that $H\left(a_0\right)=0=H\left(a_5\right)$; thus, (4) is satisfied.

Second, from (10), we have

$${\int }_{a_0}^{a_5}H\left(u\right)d\mathbf{x}\left(u\right)={\int }_{a_0}^{a_5}\mathbf{x}\left(u\right)dH\left(u\right)=0,$$

Thus, (5) is satisfied.

Third,

$$\begin{array}{c}\hfill {\int }_{a_0}^{t}H\left(u\right)d\mathbf{x}\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right),\end{array}$$

Then, we need to prove (6); this should be divided into several situations to discuss, as in each situation, it is easier to identify (6); specifically in this theorem,

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\ge 0,\end{array}$$

holds.

If $a_0\le t\le a_1$, then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right){\int }_{a_0}^{t}dH\left(u\right)=0.\end{array}$$

If $a_1<t<a_2$ and $H\left(t\right)\ge 0$, then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(a_1\right)dH\left(u\right)\\ \hfill =H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_1\right){\int }_{a_0}^{t}dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right){\int }_{a_0}^{t}dH\left(u\right)=0.\end{array}$$

If $a_1<t\le a_2$ and $H\left(t\right)<0$, according to (13) and the definition of c, then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)-{\int }_t^{a_2}q\left(u\right)\mathbf{x}\left(u\right)du+{\int }_{a_2}^{c}p\left(u\right)\mathbf{x}\left(u\right)du\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_2\right){\int }_t^{a_2}q\left(u\right)du+\mathbf{x}\left(a_2\right){\int }_{a_2}^{c}p\left(u\right)du\\ \hfill =H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_2\right)H\left(t\right)\ge 0.\end{array}$$

If $a_2<t\le c$, according to (13) and the definition of c, then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{c}p\left(u\right)\mathbf{x}\left(u\right)du\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+\mathbf{x}\left(t\right){\int }_t^{c}p\left(u\right)du=H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right)H\left(t\right)=0.\end{array}$$

If $c<t\le a_3$, according to (13) and the definition of c, then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)-{\int }_c^{t}p\left(u\right)\mathbf{x}\left(u\right)du\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right){\int }_c^{t}p\left(u\right)du=H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right)H\left(t\right)=0.\end{array}$$

If $a_3<t<a_4$ and $H\left(t\right)\ge 0$, according to (13) and the definition of c, then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)-{\int }_c^{a_3}p\left(u\right)\mathbf{x}\left(u\right)du+{\int }_{a_3}^{t}q\left(u\right)\mathbf{x}\left(u\right)du\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_3\right){\int }_c^{a_3}p\left(u\right)du+\mathbf{x}\left(a_3\right){\int }_{a_3}^{t}q\left(u\right)du\\ \hfill =H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_3\right)H\left(t\right)\ge 0.\end{array}$$

If $a_3<t\le a_4$ and $H\left(t\right)<0$, according to (10), then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_5}\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_5}\mathbf{x}\left(a_4\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+\mathbf{x}\left(a_4\right){\int }_t^{a_5}dH\left(u\right)\\ \hfill =H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_4\right)H\left(t\right)\ge 0.\end{array}$$

If $a_4<t\le a_5$, according to (10), then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_5}\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+\mathbf{x}\left(t\right){\int }_t^{a_5}dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right)H\left(t\right)=0.\end{array}$$

From the above discussion, we conclude that (6) is also satisfied; thus, from Theorem 1, we prove Theorem 3. □

Remark 1.

In Theorem 3, the additional condition (12) is to fulfil the expression in (13); actually, there are some other situations than (12): see Theorem 4.

Remark 2.

Let $a_2$ tend to $a_3$ in Theorem 3. We actually get Theorem 2 as a special case because (12) and (13) are naturally satisfied, and an integral on interval $[a_2,a_3]$ disappears, where $[a_1,a_2],[a_3,a_4]$ merge.

Then, we consider a supplementary theorem for Theorem 3, that is, other situations than (12). The two lemmas below that point out two intrinsic inequalities are needed first.

Lemma 2.

Let $\mathit{I}$ denote an interval in ${\mathbb{R}}^k$ and $\mathit{x}:[a_0,a_5]\to \mathit{I}$ be a nondecreasing continuous map. Let $p:[a_0,a_1]\cup [a_2,a_3]\cup [a_4,a_5]\to {\mathbb{R}}_{+}$ and $q:[a_1,a_2]\cup [a_3,a_4]\to {\mathbb{R}}_{+}$ be continuous functions. If

$$\begin{array}{c}\hfill \sum _{i=1}^2{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)\mathit{x}\left(t\right)dt=\sum _{i=0}^2{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)\mathit{x}\left(t\right)dt,\end{array}$$

(14)

$$\begin{array}{c}\hfill \sum _{i=1}^2{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)dt=\sum _{i=0}^2{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)dt,\end{array}$$

(15)

and

$$\begin{array}{c}\hfill {\int }_{a_1}^{a_2}q\left(t\right)dt>{\int }_{a_0}^{a_1}p\left(t\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)dt,\end{array}$$

then

$$\begin{array}{c}\hfill {\int }_{a_1}^{a_2}q\left(t\right)\mathit{x}\left(t\right)dt\ge {\int }_{a_0}^{a_1}p\left(t\right)\mathit{x}\left(t\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)\mathit{x}\left(t\right)dt+{\int }_{a_4}^{c}p\left(t\right)\mathit{x}\left(t\right)dt,\end{array}$$

where c is defined in equation ${\int }_{a_1}^{a_2}q\left(t\right)dt={\int }_{a_0}^{a_1}p\left(t\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)dt+{\int }_{a_4}^{c}p\left(t\right)dt$.

Proof. 

From the definition of c and (15), we have

$${\int }_{a_3}^{a_4}q\left(t\right)dt={\int }_c^{a_5}p\left(t\right)dt,$$

Thus,

$${\int }_{a_3}^{a_4}q\left(t\right)\mathbf{x}\left(t\right)dt\le {\int }_c^{a_5}p\left(t\right)\mathbf{x}\left(t\right)dt,$$

Combine (14) and we get the desired result. □

Lemma 3.

Let $\mathit{I}$ denote an interval in ${\mathbb{R}}^k$ and $\mathit{x}:[a_0,a_5]\to \mathit{I}$ be a nondecreasing continuous map. Let $p:[a_0,a_1]\cup [a_2,a_3]\cup [a_4,a_5]\to {\mathbb{R}}_{+}$ and $q:[a_1,a_2]\cup [a_3,a_4]\to {\mathbb{R}}_{+}$ be continuous functions. If

$$\begin{array}{c}\hfill \sum _{i=1}^2{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)\mathit{x}\left(t\right)dt=\sum _{i=0}^2{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)\mathit{x}\left(t\right)dt,\end{array}$$

(16)

$$\begin{array}{c}\hfill \sum _{i=1}^2{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)dt=\sum _{i=0}^2{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)dt,\end{array}$$

(17)

and

$$\begin{array}{c}\hfill {\int }_{a_0}^{a_1}p\left(t\right)dt>{\int }_{a_1}^{a_2}q\left(t\right)dt,\end{array}$$

then

$$\begin{array}{c}\hfill {\int }_{a_3}^{a_4}q\left(t\right)\mathit{x}\left(t\right)dt\le {\int }_c^{a_1}p\left(t\right)\mathit{x}\left(t\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)\mathit{x}\left(t\right)dt+{\int }_{a_4}^{a_5}p\left(t\right)\mathit{x}\left(t\right)dt,\end{array}$$

where c is defined in equation ${\int }_{a_3}^{a_4}q\left(t\right)dt={\int }_c^{a_1}p\left(t\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)dt+{\int }_{a_4}^{a_5}p\left(t\right)dt$.

Proof. 

From the definition of c and (17), we have

$${\int }_{a_1}^{a_2}q\left(t\right)dt={\int }_{a_0}^{c}p\left(t\right)dt,$$

Thus,

$${\int }_{a_1}^{a_2}q\left(t\right)\mathbf{x}\left(t\right)dt\ge {\int }_{a_0}^{c}p\left(t\right)\mathbf{x}\left(t\right)dt,$$

Combine (16) and we get the desired result. □

Theorem 4.

Let $\mathit{I}$ denote an interval in ${\mathbb{R}}^k$ and let $\mathit{x}:[a_0,a_5]\to \mathit{I}$ be a nondecreasing continuous map. Let $p:[a_0,a_1]\cup [a_2,a_3]\cup [a_4,a_5]\to {\mathbb{R}}_{+}$ and $q:[a_1,a_2]\cup [a_3,a_4]\to {\mathbb{R}}_{+}$ be continuous functions. If

$$\begin{array}{c}\hfill \sum _{i=1}^2{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)\mathit{x}\left(t\right)dt=\sum _{i=0}^2{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)\mathit{x}\left(t\right)dt,\end{array}$$

(18)

$$\begin{array}{c}\hfill \sum _{i=1}^2{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)dt=\sum _{i=0}^2{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)dt,\end{array}$$

(19)

and one of the following conditions holds:

$$\begin{array}{c}\hfill {\int }_{a_0}^{a_1}p\left(t\right)dt>{\int }_{a_1}^{a_2}q\left(t\right)dt,\end{array}$$

(20)

or

$$\begin{array}{c}\hfill {\int }_{a_1}^{a_2}q\left(t\right)dt>{\int }_{a_0}^{a_1}p\left(t\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)dt,\end{array}$$

(21)

then

$$\sum _{i=1}^2{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)\varphi \left(\mathit{x}\left(t\right)\right)dt\le \sum _{i=0}^2{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)\varphi \left(\mathit{x}\left(t\right)\right)dt$$

holds for every continuous function $\varphi :\mathit{I}\to \mathbb{R}$ with nondecreasing increments.

Proof. 

The main idea of the proof is to regard five integrals on $[a_0,a_5]$ in this theorem as one integral on $[a,b]$ in Theorem 1. Set

$$\begin{array}{cc}\hfill & H\left(t\right)=\hfill \\ \hfill & \left\{\begin{array}{cc}{\int }_{a_0}^{t}p\left(u\right)du,\hfill & t\in [a_0,a_1],\hfill \\ {\int }_{a_0}^{a_1}p\left(u\right)du-{\int }_{a_1}^{t}q\left(u\right)du,\hfill & t\in [a_1,a_2],\hfill \\ {\int }_{a_0}^{a_1}p\left(u\right)du-{\int }_{a_1}^{a_2}q\left(u\right)du+{\int }_{a_2}^{t}p\left(u\right)du,\hfill & t\in [a_2,a_3],\hfill \\ {\int }_{a_0}^{a_1}p\left(u\right)du-{\int }_{a_1}^{a_2}q\left(u\right)du+{\int }_{a_2}^{a_3}p\left(u\right)du-{\int }_{a_3}^{t}q\left(u\right)du,\hfill & t\in [a_3,a_4],\hfill \\ {\int }_{a_0}^{a_1}p\left(u\right)du-{\int }_{a_1}^{a_2}q\left(u\right)du+{\int }_{a_2}^{a_3}p\left(u\right)du-{\int }_{a_3}^{a_4}q\left(u\right)du+{\int }_{a_4}^{t}p\left(u\right)du,\hfill & t\in [a_4,a_5],\hfill \end{array}\right.\hfill \end{array}$$

it is easy to see from (19) that $H\left(a_0\right)=0=H\left(a_5\right)$; thus, (4) is satisfied.

Second, from (18), we have

$${\int }_{a_0}^{a_5}H\left(u\right)d\mathbf{x}\left(u\right)={\int }_{a_0}^{a_5}\mathbf{x}\left(u\right)dH\left(u\right)=0,$$

Thus, (5) is satisfied.

Third,

$$\begin{array}{c}\hfill {\int }_{a_0}^{t}H\left(u\right)d\mathbf{x}\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right),\end{array}$$

Then, we need to prove (6); this should be divided into several situations to discuss, as in each situation, it is easier to identify (6); specifically in this theorem,

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\ge 0,\end{array}$$

holds.

If $a_0\le t\le a_1$, then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right){\int }_{a_0}^{t}dH\left(u\right)=0.\end{array}$$

If $a_1<t\le a_2$, and (20) holds, then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(a_1\right)dH\left(u\right)\\ \hfill =H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_1\right){\int }_{a_0}^{t}dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right){\int }_{a_0}^{t}dH\left(u\right)=0.\end{array}$$

If $a_1<t<a_2$, (21) holds, and $H\left(t\right)\ge 0$, then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(a_1\right)dH\left(u\right)\\ \hfill =H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_1\right){\int }_{a_0}^{t}dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right){\int }_{a_0}^{t}dH\left(u\right)=0.\end{array}$$

If $a_1<t\le a_2$, (21) holds, and $H\left(t\right)<0$, according to Lemma 2, then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)-{\int }_t^{a_2}q\left(u\right)\mathbf{x}\left(u\right)du+{\int }_{a_2}^{a_3}p\left(u\right)\mathbf{x}\left(u\right)du+{\int }_{a_4}^{c}p\left(u\right)\mathbf{x}\left(u\right)du\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_2\right){\int }_t^{a_2}q\left(u\right)du+\mathbf{x}\left(a_2\right){\int }_{a_2}^{a_3}p\left(u\right)du+\mathbf{x}\left(a_2\right){\int }_{a_4}^{c}p\left(u\right)du\\ \hfill =H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_2\right)H\left(t\right)\ge 0.\end{array}$$

If $a_2<t\le a_3$ and (21) holds, it is clear that $H\left(t\right)<0$, according to Lemma 2; then,

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_3}p\left(u\right)\mathbf{x}\left(u\right)du+{\int }_{a_4}^{c}p\left(u\right)\mathbf{x}\left(u\right)du\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+\mathbf{x}\left(t\right){\int }_t^{a_3}p\left(u\right)du+\mathbf{x}\left(t\right){\int }_{a_4}^{c}p\left(u\right)du\\ \hfill =H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right)H\left(t\right)=0.\end{array}$$

If $a_2<t\le a_3$ and (20) holds, according to (18) and Lemma 3, we have

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_5}\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_2}^{t}p\left(u\right)\mathbf{x}\left(u\right)du-{\int }_c^{a_1}p\left(u\right)\mathbf{x}\left(u\right)du\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right){\int }_{a_2}^{t}p\left(u\right)du-\mathbf{x}\left(t\right){\int }_c^{a_1}p\left(u\right)du\\ \hfill =H\left(t\right)\mathbf{x}\left(t\right)+\mathbf{x}\left(t\right){\int }_t^{a_5}dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right)H\left(t\right)=0.\end{array}$$

If $a_3<t\le a_4$ and (21) holds, it is clear that $H\left(t\right)<0$; thus,

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_5}\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_5}\mathbf{x}\left(a_4\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_4\right)H\left(t\right)\ge 0.\end{array}$$

If $a_3<t\le a_4$, (20) holds, and $H\left(t\right)<0$; then,

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_5}\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_5}\mathbf{x}\left(a_4\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_4\right)H\left(t\right)\ge 0.\end{array}$$

If $a_3<t<a_4$, (20) holds, and $H\left(t\right)\ge 0$, according to (18) and Lemma 3, we have

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_5}\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+{\int }_{a_3}^{t}q\left(u\right)\mathbf{x}\left(u\right)du-{\int }_{a_2}^{a_3}p\left(u\right)\mathbf{x}\left(u\right)du-{\int }_c^{a_1}p\left(u\right)\mathbf{x}\left(u\right)du\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+\mathbf{x}\left(a_3\right){\int }_{a_3}^{t}q\left(u\right)du-\mathbf{x}\left(a_3\right){\int }_{a_2}^{a_3}p\left(u\right)du-\mathbf{x}\left(a_3\right){\int }_c^{a_1}p\left(u\right)du\\ \hfill =H\left(t\right)\mathbf{x}\left(t\right)+\mathbf{x}\left(a_3\right){\int }_t^{a_5}dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_3\right)H\left(t\right)\ge 0.\end{array}$$

If $a_4\le t\le a_5$, according to (18), we have

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_5}\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+\mathbf{x}\left(t\right){\int }_t^{a_5}dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right)H\left(t\right)=0.\end{array}$$

From the above discussion, we conclude that (6) is also satisfied; thus, from Theorem 1 we prove Theorem 4. □

Remark 3.

Following Theorems 2–4, is there any further conclusion if we divide an interval $[a,b]$ into more parts like seven, nine, …, $(2n+1)$?

In general, theorems in this section are all special cases of Theorem 1. However, there are no apparent special cases, and the proofs in this section are to point out that these theorems naturally satisfy conditions (4)–(6). The advantage is that for general convex functions (or even some specific convex functions), we cannot examine each $t\in [a,b]$ to satisfy condition (6) in Theorem 1, but in theorems in this section, we only need to examine several points for some equality or inequality to make the conclusion hold.

3. Further Extension

In this section, as a part answer to Remark 3, we prove a very special case when an interval $[a_0,a_{2n+1}]$ is divided into $(2n+1)$ parts $a_0<a_1<\dots <a_{2n}<a_{2n+1}$ for $n\ge 2$.

The lemma below that points out intrinsic inequality is needed first; this can be seen as an extension of Lemma 3 and explain why we need (27) in Theorem 5.

Lemma 4.

Let $\mathit{I}$ denote an interval in ${\mathbb{R}}^k$ and $\mathit{x}:[a_0,a_{2n+1}]\to \mathit{I}$ be a nondecreasing continuous map. Let $p:{\bigcup }_{i=0}^n[a_{2i},a_{2i+1}]\to {\mathbb{R}}_{+}$ and $q:{\bigcup }_{i=1}^n[a_{2i-1},a_{2i}]\to {\mathbb{R}}_{+}$ be continuous functions. If

$$\begin{array}{c}\hfill \sum _{i=1}^n{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)\mathit{x}\left(t\right)dt=\sum _{i=0}^n{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)\mathit{x}\left(t\right)dt,\end{array}$$

(22)

$$\begin{array}{c}\hfill \sum _{i=1}^n{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)dt=\sum _{i=0}^n{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)dt,\end{array}$$

(23)

and

$$\begin{array}{c}\hfill {\int }_{a_0}^{a_1}p\left(t\right)dt\ge \sum _{i=1}^{n-1}{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)dt,\end{array}$$

(24)

then for every $\theta \in {\bigcup }_{i=1}^{n-1}[a_{2i-1},a_{2i}]$, we have (suppose $\theta \in [a_{2m-1},a_{2m}]$)

$$\begin{array}{c}\hfill {\int }_{\theta }^{a_{2m}}q\left(t\right)\mathit{x}\left(t\right)dt+\sum _{i=m+1}^n{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)\mathit{x}\left(t\right)dt\\ \hfill \le {\int }_c^{a_1}p\left(t\right)\mathit{x}\left(t\right)dt+\sum _{i=1}^n{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)\mathit{x}\left(t\right)dt,\end{array}$$

where c is defined in equation

$$\begin{array}{c}\hfill {\int }_{\theta }^{a_{2m}}q\left(t\right)dt+\sum _{i=m+1}^n{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)dt={\int }_c^{a_1}p\left(t\right)dt+\sum _{i=1}^n{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)dt.\end{array}$$

Proof. 

From (23) and (24), we affirm that $c\in [a_0,a_1]$, as shown in the lemma. From the definition of c and (23), we have (note that if $\theta \in [a_1,a_2]$, then the following ${\sum }_{i=1}^{m-1}$ term disappears)

$$\begin{array}{c}\hfill {\int }_{a_{2m-1}}^{\theta }q\left(t\right)dt+\sum _{i=1}^{m-1}{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)dt={\int }_{a_0}^{c}p\left(t\right)dt,\end{array}$$

Thus,

$$\begin{array}{c}\hfill {\int }_{a_{2m-1}}^{\theta }q\left(t\right)\mathbf{x}\left(t\right)dt+\sum _{i=1}^{m-1}{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)\mathbf{x}\left(t\right)dt\ge {\int }_{a_0}^{c}p\left(t\right)\mathbf{x}\left(t\right)dt,\end{array}$$

Combine (22) and we get the desired result. □

Theorem 5.

Let $\mathit{I}$ denote an interval in ${\mathbb{R}}^k$ and $\mathit{x}:[a_0,a_{2n+1}]\to \mathit{I}$ be a nondecreasing continuous map. Let $p:{\bigcup }_{i=0}^n[a_{2i},a_{2i+1}]\to {\mathbb{R}}_{+}$ and $q:{\bigcup }_{i=1}^n[a_{2i-1},a_{2i}]\to {\mathbb{R}}_{+}$ be continuous functions. If

$$\begin{array}{c}\hfill \sum _{i=1}^n{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)\mathit{x}\left(t\right)dt=\sum _{i=0}^n{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)\mathit{x}\left(t\right)dt,\end{array}$$

(25)

$$\begin{array}{c}\hfill \sum _{i=1}^n{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)dt=\sum _{i=0}^n{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)dt,\end{array}$$

(26)

and

$$\begin{array}{c}\hfill {\int }_{a_0}^{a_1}p\left(t\right)dt\ge \sum _{i=1}^{n-1}{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)dt,\end{array}$$

(27)

then

$$\sum _{i=1}^n{\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)\varphi \left(\mathit{x}\left(t\right)\right)dt\le \sum _{i=0}^n{\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)\varphi \left(\mathit{x}\left(t\right)\right)dt$$

holds for every continuous function $\varphi :\mathit{I}\to \mathbb{R}$ with nondecreasing increments.

Proof. 

The main idea of the proof is to regard $(2n+1)$ integrals on $[a_0,a_{2n+1}]$ in this theorem as one integral on $[a,b]$ in Theorem 1. Set

$$\begin{array}{cc}\hfill & H\left(t\right)=\hfill \\ \hfill & \left\{\begin{array}{cc}{\int }_{a_0}^{t}p\left(u\right)du,\hfill & t\in [a_0,a_1],\hfill \\ {\int }_{a_0}^{a_1}p\left(u\right)du-{\int }_{a_1}^{t}q\left(u\right)du,\hfill & t\in [a_1,a_2],\hfill \\ \sum _{i=0}^{m-1}{\int }_{a_{2i}}^{a_{2i+1}}p\left(u\right)du-\sum _{i=1}^m{\int }_{a_{2i-1}}^{a_{2i}}q\left(u\right)du+{\int }_{a_{2m}}^{t}p\left(u\right)du,\hfill & t\in [a_{2m},a_{2m+1}],\hfill \\ \sum _{i=0}^{m-1}{\int }_{a_{2i}}^{a_{2i+1}}p\left(u\right)du-\sum _{i=1}^{m-1}{\int }_{a_{2i-1}}^{a_{2i}}q\left(u\right)du-{\int }_{a_{2m-1}}^{t}q\left(u\right)du,\hfill & t\in [a_{2m-1},a_{2m}],\hfill \end{array}\right.\hfill \end{array}$$

it is easy to see from (26) that $H\left(a_0\right)=0=H\left(a_{2n+1}\right)$; thus, (4) is satisfied.

Second, from (25) we have

$${\int }_{a_0}^{a_{2n+1}}H\left(u\right)d\mathbf{x}\left(u\right)={\int }_{a_0}^{a_{2n+1}}\mathbf{x}\left(u\right)dH\left(u\right)=0,$$

Thus, (29) is satisfied.

Third,

$$\begin{array}{c}\hfill {\int }_{a_0}^{t}H\left(u\right)d\mathbf{x}\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right),\end{array}$$

Then, we need to prove (6); this should be divided into several situations to discuss, as in each situation, it is easier to identify (6); specifically in this theorem,

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\ge 0,\end{array}$$

holds.

Here we suppose $n\ge 3$ because the case $n=2$ is proven by Theorem 4.

If $a_0\le t\le a_1$, then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right){\int }_{a_0}^{t}dH\left(u\right)=0.\end{array}$$

If $a_1<t\le a_2$, then

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(a_1\right)dH\left(u\right)\\ \hfill =H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_1\right){\int }_{a_0}^{t}dH\left(u\right)\ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right){\int }_{a_0}^{t}dH\left(u\right)=0.\end{array}$$

If $a_{2m}<t\le a_{2m+1},1\le m\le n-1$ (if $m=1$, then the following ${\sum }_{i=1}^{m-1}$ term disappears), according to (25) and Lemma 4 (a special case when $\theta$ is the end point of an interval $[a_{2i-1},a_{2i}]$), we have

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_{2n+1}}\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_{2m}}^{t}p\left(u\right)\mathbf{x}\left(u\right)du-\sum _{i=1}^{m-1}{\int }_{a_{2i}}^{a_{2i+1}}p\left(u\right)\mathbf{x}\left(u\right)du-{\int }_c^{a_1}p\left(u\right)\mathbf{x}\left(u\right)du\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right){\int }_{a_{2m}}^{t}p\left(u\right)du-\mathbf{x}\left(t\right)\sum _{i=1}^{m-1}{\int }_{a_{2i}}^{a_{2i+1}}p\left(u\right)du-\mathbf{x}\left(t\right){\int }_c^{a_1}p\left(u\right)du\\ \hfill =H\left(t\right)\mathbf{x}\left(t\right)+\mathbf{x}\left(t\right){\int }_t^{a_{2n+1}}dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right)H\left(t\right)=0,\end{array}$$

where c is defined in equation

$$\begin{array}{c}\hfill \sum _{i=m+1}^n{\int }_{a_{2i-1}}^{a_{2i}}q\left(u\right)du={\int }_c^{a_1}p\left(u\right)du+\sum _{i=1}^n{\int }_{a_{2i}}^{a_{2i+1}}p\left(u\right)du.\end{array}$$

If $a_{2m-1}<t\le a_{2m},2\le m\le n-1$, according to (25) and let $\theta =t$ in Lemma 4, we have

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_{2n+1}}\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)-\sum _{i=1}^{m-1}{\int }_{a_{2i}}^{a_{2i+1}}p\left(u\right)\mathbf{x}\left(u\right)du-{\int }_c^{a_1}p\left(u\right)\mathbf{x}\left(u\right)du\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right)\sum _{i=1}^{m-1}{\int }_{a_{2i}}^{a_{2i+1}}p\left(u\right)du-\mathbf{x}\left(t\right){\int }_c^{a_1}p\left(u\right)du\\ \hfill =H\left(t\right)\mathbf{x}\left(t\right)+\mathbf{x}\left(t\right){\int }_t^{a_{2n+1}}dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right)H\left(t\right)=0,\end{array}$$

where c is defined in equation

$$\begin{array}{c}\hfill {\int }_t^{a_{2m}}q\left(u\right)du+\sum _{i=m+1}^n{\int }_{a_{2i-1}}^{a_{2i}}q\left(u\right)du={\int }_c^{a_1}p\left(u\right)du+\sum _{i=1}^n{\int }_{a_{2i}}^{a_{2i+1}}p\left(u\right)du.\end{array}$$

If $a_{2n-1}<t\le a_{2n}$ and $H\left(t\right)\ge 0$, according to (25) and let $\theta =a_{2n-2}$ in Lemma 4, we have

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_{2n+1}}\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+{\int }_{a_{2n-1}}^{t}q\left(u\right)\mathbf{x}\left(u\right)du-\sum _{i=1}^{n-1}{\int }_{a_{2i}}^{a_{2i+1}}p\left(u\right)\mathbf{x}\left(u\right)du-{\int }_c^{a_1}p\left(u\right)\mathbf{x}\left(u\right)du\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+\mathbf{x}\left(a_{2n-1}\right){\int }_{a_{2n-1}}^{t}q\left(u\right)du-\mathbf{x}\left(a_{2n-1}\right)\sum _{i=1}^{n-1}{\int }_{a_{2i}}^{a_{2i+1}}p\left(u\right)du\\ \hfill -\mathbf{x}\left(a_{2n-1}\right){\int }_c^{a_1}p\left(u\right)du\\ \hfill =H\left(t\right)\mathbf{x}\left(t\right)+\mathbf{x}\left(a_{2n-1}\right){\int }_t^{a_{2n+1}}dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_{2n-1}\right)H\left(t\right)\ge 0,\end{array}$$

where c is defined in equation

$$\begin{array}{c}\hfill {\int }_{a_{2n-1}}^{a_{2n}}q\left(u\right)du={\int }_c^{a_1}p\left(u\right)du+\sum _{i=1}^n{\int }_{a_{2i}}^{a_{2i+1}}p\left(u\right)du.\end{array}$$

If $a_{2n-1}<t\le a_{2n}$ and $H\left(t\right)<0$, according to (25), we have

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_{2n+1}}\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_{2n+1}}\mathbf{x}\left(a_{2n}\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(a_{2n}\right)H\left(t\right)\ge 0.\end{array}$$

If $a_{2n}\le t\le a_{2n+1}$, according to (25), we have

$$\begin{array}{c}\hfill H\left(t\right)\mathbf{x}\left(t\right)-{\int }_{a_0}^t\mathbf{x}\left(u\right)dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)+{\int }_t^{a_{2n+1}}\mathbf{x}\left(u\right)dH\left(u\right)\\ \hfill \ge H\left(t\right)\mathbf{x}\left(t\right)+\mathbf{x}\left(t\right){\int }_t^{a_{2n+1}}dH\left(u\right)=H\left(t\right)\mathbf{x}\left(t\right)-\mathbf{x}\left(t\right)H\left(t\right)=0.\end{array}$$

From the above discussion, we conclude that (6) is also satisfied; thus, from Theorem 1, we prove Theorem 5. □

Remark 4.

For an interval $[a_0,a_{2n+1}]$ with $a_0<a_1<\dots <a_{2n}<a_{2n+1}$ and $p:{\bigcup }_{i=0}^n[a_{2i},a_{2i+1}]\to {\mathbb{R}}_{+}$ and $q:{\bigcup }_{i=1}^n[a_{2i-1},a_{2i}]\to {\mathbb{R}}_{+}$, the situation (27) in Theorem 5 is only a special case. It is natural to ask: what is the general case (or other cases than (27)) of Theorem 5 under the assumptions (25) and (26)?

Then, we consider the discrete case of Theorem 5. Though $\mathbf{x}$ in Theorem 5 is supposed to be continuous, we can still let $\mathbf{x}\left(t\right)=(x_1\left(t\right),\dots ,x_k\left(t\right))$, being a simple function ${\mathbf{x}}_i=(x_{1,i},\dots ,x_{k,i})$ on $t\in (a_{i-1},a_i]$, for $i=1,\dots ,2n+1$. This is because step functions ${\mathbf{x}}_i$ can be approximated by continuous functions $\mathbf{x}{\left(t\right)}_{\delta }$ (the only error is in each $(a_i-\delta ,a_i+\delta )$) when $\delta \to 0$, so the conclusion of Theorem 5 can also be approximated.

Further, in Theorem 6, let

$$\begin{array}{c}\hfill q_{2i}={\int }_{a_{2i-1}}^{a_{2i}}q\left(t\right)dt,p_{2i+1}={\int }_{a_{2i}}^{a_{2i+1}}p\left(t\right)dt\end{array}$$

in Theorem 5.

Theorem 6.

Let $\mathit{I}$ denote an interval in ${\mathbb{R}}^k$ and let ${\mathit{x}}_i\in \mathit{I},i=1,\dots ,2n+1$ be a nondecreasing sequence with i. Let $p_{2i+1}\in {\mathbb{R}}_{+},i=0,\dots ,n$ and $q_{2i}\in {\mathbb{R}}_{+},i=1,\dots ,n$. If

$$\begin{array}{c}\hfill \sum _{i=1}^n{q}_{2i}{\mathit{x}}_{2i}=\sum _{i=0}^n{p}_{2i+1}{\mathit{x}}_{2i+1},\\ \hfill \sum _{i=1}^n{q}_{2i}=\sum _{i=0}^n{p}_{2i+1},\end{array}$$

and

$$\begin{array}{c}\hfill p_1\ge \sum _{i=1}^{n-1}{q}_{2i},\end{array}$$

then

$$\sum _{i=1}^n{q}_{2i}\varphi \left({\mathit{x}}_{2i}\right)\le \sum _{i=0}^n{p}_{2i+1}\varphi \left({\mathit{x}}_{2i+1}\right)$$

(28)

holds for every continuous function $\varphi :\mathit{I}\to \mathbb{R}$ with nondecreasing increments.

Remark 5.

Inequality (28) in ${\mathbb{R}}^1$ for a convex function is a slightly similar to another type in [37] (p. 824) but actually different. That type [28,37,38,39,40,41,42,43,44,45] is an analogue of the Jensen–Steffensen inequality [46,47,48].

Another way to prove Theorem 6 is to use the discrete case of Theorem 1; here we only prove the sufficiency part of the discrete case.

Theorem 7.

Let $\mathit{I}$ denote an interval in ${\mathbb{R}}^k$, let ${\mathit{X}}_i\in \mathit{I},i=0,\dots ,n$ be a nondecreasing map with i, and let $h_i\in \mathbb{R},i=0,\dots ,n$ be a sequence. Then,

$$\sum _{i=0}^{n}f\left({\mathit{X}}_i\right)h_i\ge 0$$

(29)

for every continuous function $f:\mathit{I}\to \mathbb{R}$ with nondecreasing increments if

$$\begin{array}{c}\hfill \sum _{i=0}^n{h}_i=0,\end{array}$$

(30)

$$\begin{array}{c}\hfill \sum _{i=1}^n({\mathit{X}}_i-{\mathit{X}}_{i-1})\sum _{r=0}^{i-1}{h}_r=0,\end{array}$$

(31)

and

$$\begin{array}{c}\hfill \sum _{i=1}^m({\mathit{X}}_i-{\mathit{X}}_{i-1})\sum _{r=0}^{i-1}{h}_r\ge 0form\in \bigcup _{i=1}^{n-1}i.\end{array}$$

(32)

Proof. 

Since f may be approximated uniformly on $\mathbf{I}$ by functions with continuous nondecreasing first partial derivatives, we may assume that the first partials $f_j,(j=1,\dots ,k)$ exist and are continuous and nondecreasing.

According to (30), we have

$$\begin{array}{c}\hfill \sum _{i=0}^{n}f\left({\mathbf{X}}_i\right)h_i\\ \hfill =\sum _{i=1}^n(f\left({\mathbf{X}}_i\right)-f\left({\mathbf{X}}_{i-1}\right))\sum _{r=i}^n{h}_r\\ \hfill =\sum _{i=1}^n{\int }_0^1▽f({\mathbf{X}}_{i-1}+\lambda ({\mathbf{X}}_i-{\mathbf{X}}_{i-1}))d\lambda \sum _{r=i}^n{h}_r\\ \hfill =\sum _{i=1}^n{\int }_0^1\sum _{j=1}^k{f}_j({\mathbf{X}}_{i-1}+\lambda ({\mathbf{X}}_i-{\mathbf{X}}_{i-1}))(x_{i,j}-x_{i-1,j})d\lambda \sum _{r=i}^n{h}_r\\ \hfill ={\int }_0^1\sum _{j=1}^k\sum _{i=1}^n{f}_j({\mathbf{X}}_{i-1}+\lambda ({\mathbf{X}}_i-{\mathbf{X}}_{i-1}))(x_{i,j}-x_{i-1,j})\sum _{r=i}^n{h}_{r}d\lambda ,\end{array}$$

in which, according to (30) and (31), we have

$$\begin{array}{c}\hfill \sum _{i=1}^n{f}_j({\mathbf{X}}_{i-1}+\lambda ({\mathbf{X}}_i-{\mathbf{X}}_{i-1}))(x_{i,j}-x_{i-1,j})\sum _{r=i}^n{h}_r\\ \hfill =-\sum _{i=1}^n{f}_j({\mathbf{X}}_{i-1}+\lambda ({\mathbf{X}}_i-{\mathbf{X}}_{i-1}))(x_{i,j}-x_{i-1,j})\sum _{r=0}^{i-1}{h}_r\\ \hfill =\sum _{i=1}^n[f_j({\mathbf{X}}_{n-1}+\lambda ({\mathbf{X}}_n-{\mathbf{X}}_{n-1}))-f_j({\mathbf{X}}_{i-1}+\lambda ({\mathbf{X}}_i-{\mathbf{X}}_{i-1}))](x_{i,j}-x_{i-1,j})\sum _{r=0}^{i-1}{h}_r\\ \hfill =\sum _{s=2}^n[f_j({\mathbf{X}}_{s-1}+\lambda ({\mathbf{X}}_s-{\mathbf{X}}_{s-1}))-f_j({\mathbf{X}}_{s-2}+\lambda ({\mathbf{X}}_{s-1}-{\mathbf{X}}_{s-2}))]\\ \hfill \times \sum _{i=1}^{s-1}(x_{i,j}-x_{i-1,j})\sum _{r=0}^{i-1}{h}_r.\end{array}$$

As ${\mathbf{X}}_{s-1}+\lambda ({\mathbf{X}}_s-{\mathbf{X}}_{s-1})\ge {\mathbf{X}}_{s-2}+\lambda ({\mathbf{X}}_{s-1}-{\mathbf{X}}_{s-2})$, from the basic property of a function with nondecreasing increments, we have

$$f_j({\mathbf{X}}_{s-1}+\lambda ({\mathbf{X}}_s-{\mathbf{X}}_{s-1}))-f_j({\mathbf{X}}_{s-2}+\lambda ({\mathbf{X}}_{s-1}-{\mathbf{X}}_{s-2}))\ge 0,$$

while from (32), we have

$$\sum _{i=1}^{s-1}(x_{i,j}-x_{i-1,j})\sum _{r=0}^{i-1}{h}_r\ge 0,s=2,\dots ,n-1.$$

Thus, (29) is proven. □

Then, we consider an analogue of Theorem 1, in which the convex function f is only defined on ${\mathbb{R}}^1$ but each $X_j$ has different $H_j$. The advantage of Theorem 8 is that similar conditions like (5) and (6) in Theorem 1 are unnecessary to hold for each $X_j,j=1,\dots ,k$, but only a “sum” of them as (35) and (36), which reduces k conditions to one condition.

Theorem 8.

Let $I_j$ denote an interval in $\mathbb{R}$, let $X_j:[a,b]\to I_j$ be a nondecreasing continuous function, and let $H_j$ be a function of bounded variation and continuous from the left on $[a,b]$ with $H_j\left(a\right)=0$, $j=1,\dots ,k$. Then,

$$\sum _{j=1}^k{\int }_a^{b}f\left(X_j\left(u\right)\right)d{H}_j\left(u\right)\ge 0$$

(33)

for every continuous convex function $f:[A,B]\to \mathbb{R}$ if

$$\begin{array}{c}\hfill H_j\left(b\right)=0,j=1,\dots ,k,\end{array}$$

(34)

$$\begin{array}{c}\hfill \sum _{j=1}^k{\int }_a^b{H}_j\left(u\right)d{X}_j\left(u\right)=0,\end{array}$$

(35)

and

$$\begin{array}{c}\hfill \sum _{j=1}^k{\int }_{u:X_j\left(u\right)<t}{H}_j\left(u\right)d{X}_j\left(u\right)\ge 0fort\subset [A,B],\end{array}$$

(36)

where $[A,B]=[min{X}_j\left(a\right),max{X}_j\left(b\right)]$.

Proof. 

Since f may be approximated uniformly on $[A,B]$ by functions with a continuous nonnegative second derivative, we may assume that the second derivative exists and is continuous and nonnegative.

Using (34) and (35), we have

$$\begin{array}{c}\hfill \sum _{j=1}^k{\int }_a^{b}f\left(X_j\left(u\right)\right)d{H}_j\left(u\right)\\ \hfill =-\sum _{j=1}^k{\int }_a^b{H}_j\left(u\right)f^{\prime }\left(X_j\left(u\right)\right)d{X}_j\left(u\right)\\ \hfill =-\sum _{j=1}^k{\int }_a^b{H}_j\left(u\right){\int }_A^{X_j\left(u\right)}{f}^{″}\left(t\right)dtd{X}_j\left(u\right)\\ \hfill =-\sum _{j=1}^k{\int }_A^{X_j\left(b\right)}{f}^{″}\left(t\right){\int }_{u:X_j\left(u\right)\ge t}{H}_j\left(u\right)d{X}_j\left(u\right)dt\\ \hfill =-\sum _{j=1}^k{\int }_A^B{f}^{″}\left(t\right){\int }_{u:X_j\left(u\right)\ge t}{H}_j\left(u\right)d{X}_j\left(u\right)dt\end{array}$$

$$\begin{array}{c}\hfill ={\int }_A^B{f}^{″}\left(t\right)\left(-\sum _{j=1}^k{\int }_{u:X_j\left(u\right)\ge t}{H}_j\left(u\right)d{X}_j\left(u\right)\right)dt\\ \hfill ={\int }_A^B{f}^{″}\left(t\right)\sum _{j=1}^k{\int }_{u:X_j\left(u\right)<t}{H}_j\left(u\right)d{X}_j\left(u\right)dt.\end{array}$$

Since (36) holds, each term in the last integral is nonnegative, so (33) is verified. □

The special case $k=1$ of Theorem 8 is equivalent to the special case $k=1$ of Theorem 1.

4. Application

Recall the notion of Csiszár $\varphi$-divergence [49,50,51]. Given a convex function $\varphi :{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$, the $\varphi$-divergence functional

$$I_{\varphi }(p,q):=\sum _{i=1}^n{q}_i\varphi \left({\displaystyle \frac{p_i}{q_i}}\right)$$

is a generalized measure of information, a “distance function” on the set of probability distributions ${\mathbb{P}}_n$. By appropriately defining this convex function $\varphi$, various divergences are derived; see Chapter 1 in [50] and Chapter 9.2 in [51] as well as related references.

In this section, we further compare two different $I_{\varphi }(p_1,q_1)$ and $I_{\varphi }(p_2,q_2)$. The following discrete case of Lemma 1 in [18] is needed.

Conclusion 1.

Let $1\le r<s<m$ and the m-tuple x such that $x_1\le \dots \le x_m$. Let $p_i,(i=1,\dots ,r,s+1,\dots ,m)$ and $q_i,(i=r+1,\dots ,s)$ be two nonnegative sequences. If

$$\begin{array}{cc}\hfill & \sum _{i=r+1}^s{q}_i{x}_i=\sum _{i=1}^r{p}_i{x}_i+\sum _{i=s+1}^m{p}_i{x}_i,\hfill \end{array}$$

(37)

$$\begin{array}{cc}\hfill & \sum _{i=r+1}^s{q}_i=\sum _{i=1}^r{p}_i+\sum _{i=s+1}^m{p}_i,\hfill \end{array}$$

(38)

then

$$\sum _{i=r+1}^s{q}_i\varphi \left(x_i\right)\le \sum _{i=1}^r{p}_i\varphi \left(x_i\right)+\sum _{i=s+1}^m{p}_i\varphi \left(x_i\right)$$

(39)

holds for every $\varphi$ that is convex.

Theorem 9.

Let $\varphi :[0,\infty )\to \mathbb{R}$ be a convex function. If $p_1,q_1,p_2,q_2\in {\mathbb{P}}_n$ and

$$\begin{array}{c}\hfill {\displaystyle \frac{p_{1,i}}{q_{1,i}}}\in [m,M],i=1,\dots ,n;\\ \hfill {\displaystyle \frac{p_{2,i}}{q_{2,i}}}\in [0,m]\cup [M,\infty ),i=1,\dots ,n\end{array}$$

(40)

for some $m\le 1\le M$, then we have

$$I_{\varphi }(p_1,q_1)\le I_{\varphi }(p_2,q_2).$$

(41)

Proof. 

According to (40), we can set $x_1\le \dots \le x_m$ in Conclusion 1 as ${\displaystyle \frac{p_{2,\left[1\right]}}{q_{2,\left[1\right]}}}\le \dots \le {\displaystyle \frac{p_{1,\left[1\right]}}{q_{1,\left[1\right]}}}\le \dots \le {\displaystyle \frac{p_{1,\left[n\right]}}{q_{1,\left[n\right]}}}\le \dots \le {\displaystyle \frac{p_{2,\left[n\right]}}{q_{2,\left[n\right]}}}$, where ${\displaystyle \frac{p_{2,\left[i\right]}}{q_{2,\left[i\right]}}},{\displaystyle \frac{p_{1,\left[i\right]}}{q_{1,\left[i\right]}}}$ are the increasing rearrangement of the ${\displaystyle \frac{p_{2,i}}{q_{2,i}}},{\displaystyle \frac{p_{1,i}}{q_{1,i}}}$. Further, set $q_i$ in Conclusion 1 as $q_{1,i}$; set $p_i$ in Conclusion 1 as $q_{2,i}$, $s-r=n,m=2n$. Since $p_1,q_1,p_2,q_2\in {\mathbb{P}}_n$, conditions (37) and (38) are also satisfied; according to (39) we get (41). □

Remark 6.

Let ${\displaystyle \frac{p_{1,i}}{q_{1,i}}}\equiv 1$; thus, $m=M=1$; we get one special case, which is also the special case of Csiszár–Körner inequality [51] (p. 261). Let ${\displaystyle \frac{p_{2,i}}{q_{2,i}}}=m$ or M, and selecting a proper $q_{2,i}$ to satisfy the condition (sometimes we do not have such selection), we get another special case about upper bounds for $I_{\varphi }(p_1,q_1)$. These kinds of inequalities are considered in Chapter 1 in [50].

Assume that a set $\chi$ and the $\sigma$-finite measure $\mu$ are given. Consider the set of all probability densities on $\sigma$ to be $\Omega :=\left\{p\right|p:\chi \to \mathbb{R},p\left(x\right)\ge 0,{\int }_{\chi }p\left(x\right)d\mu \left(x\right)=1\}$. Define continuous

$$I_{\varphi }(p,q):={\int }_{\chi }q\left(x\right)\varphi \left[{\displaystyle \frac{p\left(x\right)}{q\left(x\right)}}\right]d\mu \left(x\right),p,q\in \Omega .$$

In order to prove the continuous case of Theorem 9, the conclusion below in Chapter 1.4 in [50] is needed.

Theorem 10.

Let $\varphi :[0,\infty )\to \mathbb{R}$ be a convex mapping on the interval $[r,R]\subset [0,\infty )$ with $r\le 1\le R$. If $p,q\in \Omega$ and $r\le {\displaystyle \frac{p\left(y\right)}{q\left(y\right)}}\le R$ for all $y\in \chi$, then we have the inequality

$$I_{\varphi }(p,q)\le {\displaystyle \frac{R-1}{R-r}}·\varphi \left(r\right)+{\displaystyle \frac{1-r}{R-r}}·\varphi \left(R\right).$$

(42)

Then, we state our main theorem.

Theorem 11.

Let $\varphi :[0,\infty )\to \mathbb{R}$ be a convex function. If $p_1,q_1,p_2,q_2\in \Omega$ and

$$\begin{array}{c}\hfill {\displaystyle \frac{p_1\left(x\right)}{q_1\left(x\right)}}\in [m,M],\\ \hfill {\displaystyle \frac{p_2\left(x\right)}{q_2\left(x\right)}}\in [0,m]\cup [M,\infty )\end{array}$$

(43)

for all $x\in \chi$ for some $m\le 1\le M$, then we have

$$I_{\varphi }(p_1,q_1)\le I_{\varphi }(p_2,q_2).$$

(44)

Proof. 

The case $m=M=1$ is easy to see from the Jensen inequality; thus, we suppose $M>m$.

$$\begin{array}{c}\hfill I_{\varphi }(p_2,q_2)={\int }_{\chi }{q}_2\left(x\right)\varphi \left[{\displaystyle \frac{p_2\left(x\right)}{q_2\left(x\right)}}\right]d\mu \left(x\right)\\ \hfill ={\int }_{{\chi }_1}{q}_2\left(x\right)\varphi \left[{\displaystyle \frac{p_2\left(x\right)}{q_2\left(x\right)}}\right]d\mu \left(x\right)+{\int }_{{\chi }_2}{q}_2\left(x\right)\varphi \left[{\displaystyle \frac{p_2\left(x\right)}{q_2\left(x\right)}}\right]d\mu \left(x\right),\end{array}$$

where ${\chi }_1=\{x:{\displaystyle \frac{p_2\left(x\right)}{q_2\left(x\right)}}\le m\}$, ${\chi }_2=\{x:{\displaystyle \frac{p_2\left(x\right)}{q_2\left(x\right)}}\ge M\}$. □

Due to the Jensen inequality, we have

$$\begin{array}{c}\hfill {\int }_{{\chi }_1}{q}_2\left(x\right)\varphi \left[{\displaystyle \frac{p_2\left(x\right)}{q_2\left(x\right)}}\right]d\mu \left(x\right)+{\int }_{{\chi }_2}{q}_2\left(x\right)\varphi \left[{\displaystyle \frac{p_2\left(x\right)}{q_2\left(x\right)}}\right]d\mu \left(x\right)\\ \hfill \ge \varphi \left({\displaystyle \frac{{\int }_{{\chi }_1}{p}_2\left(x\right)d\mu \left(x\right)}{{\int }_{{\chi }_1}{q}_2\left(x\right)d\mu \left(x\right)}}\right){\int }_{{\chi }_1}{q}_2\left(x\right)d\mu \left(x\right)+\varphi \left({\displaystyle \frac{{\int }_{{\chi }_2}{p}_2\left(x\right)d\mu \left(x\right)}{{\int }_{{\chi }_2}{q}_2\left(x\right)d\mu \left(x\right)}}\right){\int }_{{\chi }_2}{q}_2\left(x\right)d\mu \left(x\right),\end{array}$$

And as

$$\begin{array}{c}\hfill {\displaystyle \frac{{\int }_{{\chi }_1}{p}_2\left(x\right)d\mu \left(x\right)}{{\int }_{{\chi }_1}{q}_2\left(x\right)d\mu \left(x\right)}}\le m,{\displaystyle \frac{{\int }_{{\chi }_2}{p}_2\left(x\right)d\mu \left(x\right)}{{\int }_{{\chi }_2}{q}_2\left(x\right)d\mu \left(x\right)}}\ge M,\end{array}$$

we can let

$$\begin{array}{c}\hfill {\displaystyle \frac{{\int }_{{\chi }_1}{p}_2\left(x\right)d\mu \left(x\right)}{{\int }_{{\chi }_1}{q}_2\left(x\right)d\mu \left(x\right)}}=r,{\displaystyle \frac{{\int }_{{\chi }_2}{p}_2\left(x\right)d\mu \left(x\right)}{{\int }_{{\chi }_2}{q}_2\left(x\right)d\mu \left(x\right)}}=R\end{array}$$

like Theorem 10 for $I_{\varphi }(p_1,q_1)$.

Further, from $p_1,q_1,p_2,q_2\in \Omega$ we have

$$\begin{array}{c}\hfill {\int }_{{\chi }_1}{q}_2\left(x\right)d\mu \left(x\right)={\displaystyle \frac{R-1}{R-r}},{\int }_{{\chi }_2}{q}_2\left(x\right)d\mu \left(x\right)={\displaystyle \frac{1-r}{R-r}},\end{array}$$

Thus,

$$I_{\varphi }(p_2,q_2)\ge {\displaystyle \frac{R-1}{R-r}}·\varphi \left(r\right)+{\displaystyle \frac{1-r}{R-r}}·\varphi \left(R\right),$$

while Theorem 10 indicates that (because ${\displaystyle \frac{p_1\left(x\right)}{q_1\left(x\right)}}\in [m,M]\subseteq [r,R]$)

$$I_{\varphi }(p_1,q_1)\le {\displaystyle \frac{R-1}{R-r}}·\varphi \left(r\right)+{\displaystyle \frac{1-r}{R-r}}·\varphi \left(R\right),$$

Combining the two inequalities above, we get (44).

5. Further Application

In this section, we first extend Lemma 1 to an $ϵ$-approximately convex function [52] on a discrete set [29] (p. 48), and then we use this theorem to establish a similar Landau–Kolmogorov-type inequality.

Definition 1.

For a discrete set $E_n:=\{x_1,\dots ,x_n\},x_1<\dots <x_n$ and a fixed nonnegative number ϵ, a function $\varphi :E_n\to \mathbb{R}$ is called ϵ-approximately convex on a discrete set $E_n$ if

$$\varphi \left(x_{i_2}\right)\le {\displaystyle \frac{x_{i_2}-x_{i_1}}{x_{i_3}-x_{i_1}}}\varphi \left(x_{i_1}\right)+{\displaystyle \frac{x_{i_3}-x_{i_2}}{x_{i_3}-x_{i_1}}}\varphi \left(x_{i_3}\right)+ϵ$$

for all $x_{i_3}>x_{i_2}>x_{i_1}$ and $x_{i_j}\in E_n$.

If $ϵ=0$, then it reduces to a convex function defined on a discrete set.

In order to prove Theorem 12, the two lemmas below are needed.

Lemma 5.

Let $1<k<n$ and $E_n:=\{x_1,\dots ,x_n\},x_1<\dots <x_n$. Let $p_i,(i=1,\dots ,k-1,k+1,\dots ,n)$ and $q_k$ be nonnegative. If

$$\begin{array}{cc}\hfill & q_k{x}_k=\sum _{i=1}^{k-1}{p}_i{x}_i+\sum _{i=k+1}^n{p}_i{x}_i,\hfill \end{array}$$

(45)

$$\begin{array}{cc}\hfill & q_k=\sum _{i=1}^{k-1}{p}_i+\sum _{i=k+1}^n{p}_i,\hfill \end{array}$$

(46)

then

$$q_k\varphi \left(x_k\right)\le \sum _{i=1}^{k-1}{p}_i\varphi \left(x_i\right)+\sum _{i=k+1}^n{p}_i\varphi \left(x_i\right)+q_kϵ$$

(47)

holds for every ϕ that is ϵ-approximately convex on a discrete set $E_n$.

Proof. 

Use definition and induction for n. □

Lemma 6.

Let $1\le r<s<m$ and $x_1<\dots <x_m$. Let $p_i,(i=1,\dots ,r,s+1,\dots ,m)$ and $q_i,(i=r+1,\dots ,s)$ be two nonnegative sequences. If

$$\begin{array}{c}\hfill \sum _{i=r+1}^s{q}_i{x}_i=\sum _{i=1}^r{p}_i{x}_i+\sum _{i=s+1}^m{p}_i{x}_i,\end{array}$$

(48)

$$\begin{array}{c}\hfill \sum _{i=r+1}^s{q}_i=\sum _{i=1}^r{p}_i+\sum _{i=s+1}^m{p}_i,\end{array}$$

(49)

then $\exists p_i^{\prime }$ such that $0\le p_i^{\prime }\le p_i$ for $i=1,\dots ,r,s+1,\dots ,m$, and

$$\begin{array}{c}\hfill q_s{x}_s=\sum _{i=1}^r{p}_i^{\prime }{x}_i+\sum _{i=s+1}^m{p}_i^{\prime }{x}_i,\\ \hfill q_s=\sum _{i=1}^r{p}_i^{\prime }+\sum _{i=s+1}^m{p}_i^{\prime }.\end{array}$$

Proof. 

It is obvious for $r+1=s$; then, we only prove general situations $s>r+1$. Consider the continuous function

$$\begin{array}{c}\hfill f(p_1^{\prime },\dots ,p_r^{\prime },p_{s+1}^{\prime },\dots ,p_m^{\prime }):=\sum _{i=1}^r{p}_i^{\prime }{x}_i+\sum _{i=s+1}^m{p}_i^{\prime }{x}_i\end{array}$$

defined on $\prod [0,p_i]$ under the constraint

$$\begin{array}{c}\hfill \sum _{i=1}^r{p}_i^{\prime }+\sum _{i=s+1}^m{p}_i^{\prime }=q_s.\end{array}$$

It is sufficient to prove Lemma 6 by proving

$$\begin{array}{c}\hfill f_{min}\le q_s{x}_s\le f_{max}.\end{array}$$

This can be divided into four situations.

1. If

$$q_s\le \sum _{i=s+1}^m{p}_i,q_s\le \sum _{i=1}^r{p}_i,$$

then

$$\begin{array}{c}\hfill f(p_1^{\prime },\dots ,p_r^{\prime },0,\dots ,0)\le q_s{x}_s\le f(0,\dots ,0,p_{s+1}^{\prime },\dots ,p_m^{\prime })\end{array}$$

for a sequence of $p_i^{\prime }$ such that $p_1^{\prime }+\dots +p_r^{\prime }+0+\dots +0=q_s$, and for another sequence of $p_i^{\prime }$ such that $0+\dots +0+p_{s+1}^{\prime }+\dots +p_m^{\prime }=q_s$.

2. If

$$q_s\le \sum _{i=s+1}^m{p}_i,q_s>\sum _{i=1}^r{p}_i,$$

then

$$\begin{array}{c}\hfill q_s{x}_s\le f(0,\dots ,0,p_{s+1}^{\prime },\dots ,p_m^{\prime }).\end{array}$$

And according to (49),

$$\sum _{i=r+1}^{s-1}{q}_i<\sum _{i=s+1}^m{p}_i,$$

Thus, $\exists (p_i-p_i^{\prime })$ such that $0\le (p_i-p_i^{\prime })\le p_i$ for $i=s+1,\dots ,m$, and

$$\sum _{i=r+1}^{s-1}{q}_i=\sum _{i=s+1}^m(p_i-p_i^{\prime }),$$

so

$$\sum _{i=r+1}^{s-1}{q}_i{x}_i<\sum _{i=s+1}^m(p_i-p_i^{\prime })x_i,$$

Combining (48), we have

$$\begin{array}{c}\hfill q_s{x}_s>\sum _{i=1}^r{p}_i^{\prime }{x}_i+\sum _{i=s+1}^m{p}_i^{\prime }{x}_i=f(p_1^{\prime },\dots ,p_r^{\prime },p_{s+1}^{\prime },\dots ,p_m^{\prime }).\end{array}$$

The proofs for the other two situations are similar.

In conclusion, we can always find some $p_i^{\prime }$ that

$$\begin{array}{c}\hfill q_s{x}_s\ge f(p_1^{\prime },\dots ,p_r^{\prime },p_{s+1}^{\prime },\dots ,p_m^{\prime }),\end{array}$$

and another sequence of $p_i^{\prime }$ that

$$\begin{array}{c}\hfill q_s{x}_s\le f(p_1^{\prime },\dots ,p_r^{\prime },p_{s+1}^{\prime },\dots ,p_m^{\prime })\end{array}$$

for fixed $p_i,q_i$, and $x_i$. As f is continuous, we can find $p_i^{\prime }$ such that

$$\begin{array}{c}\hfill q_s{x}_s=f(p_1^{\prime },\dots ,p_r^{\prime },p_{s+1}^{\prime },\dots ,p_m^{\prime })=\sum _{i=1}^r{p}_i^{\prime }{x}_i+\sum _{i=s+1}^m{p}_i^{\prime }{x}_i,\end{array}$$

which proves the conclusion. □

Then, we state an inequality for an $ϵ$-approximately convex function on a discrete set.

Theorem 12.

Let $1\le r<s<m$ and $E_m:=\{x_1,\dots ,x_m\},x_1<\dots <x_m$. Let $p_i,(i=1,\dots ,r,s+1,\dots ,m)$ and $q_i,(i=r+1,\dots ,s)$ be two nonnegative sequences. If

$$\begin{array}{cc}\hfill & \sum _{i=r+1}^s{q}_i{x}_i=\sum _{i=1}^r{p}_i{x}_i+\sum _{i=s+1}^m{p}_i{x}_i,\hfill \end{array}$$

(50)

$$\begin{array}{cc}\hfill & \sum _{i=r+1}^s{q}_i=\sum _{i=1}^r{p}_i+\sum _{i=s+1}^m{p}_i,\hfill \end{array}$$

(51)

then

$$\sum _{i=r+1}^s{q}_i\varphi \left(x_i\right)\le \sum _{i=1}^r{p}_i\varphi \left(x_i\right)+\sum _{i=s+1}^m{p}_i\varphi \left(x_i\right)+ϵ\sum _{i=r+1}^s{q}_i$$

(52)

holds for every ϕ that is ϵ-approximately convex on a discrete set $E_m$.

Proof. 

The intuition of the proof is that under the assumption of (50) and (51) we can decompose (52) to each $q_i\varphi \left(x_i\right)$, which holds as an inequality as in Lemma 5.

In Lemma 5, we prove the situation for $s-r=1$ (for every n); suppose Theorem 12 holds for $s-r=N$ (for every m), it remains to prove $s-r=N+1$ by induction ($r+N+1<m$).

$$\begin{array}{c}\hfill \sum _{i=r+1}^{r+N+1}{q}_i\varphi \left(x_i\right)=q_{r+N+1}\varphi \left(x_{r+N+1}\right)+\sum _{i=r+1}^{r+N}{q}_i\varphi \left(x_i\right).\end{array}$$

According to Lemma 6, $\exists p_i^{\prime }$ such that $0\le p_i^{\prime }\le p_i$ for $i=1,\dots ,r,r+N+2,\dots ,m$, and

$$\begin{array}{cc}\hfill & q_{r+N+1}{x}_{r+N+1}=\sum _{i=1}^r{p}_i^{\prime }{x}_i+\sum _{i=r+N+2}^m{p}_i^{\prime }{x}_i,\hfill \\ \hfill & q_{r+N+1}=\sum _{i=1}^r{p}_i^{\prime }+\sum _{i=r+N+2}^m{p}_i^{\prime }.\hfill \end{array}$$

Thus, from Lemma 5, we have

$$\begin{array}{cc}\hfill & q_{r+N+1}\varphi \left(x_{r+N+1}\right)\le \sum _{i=1}^r{p}_i^{\prime }\varphi \left(x_i\right)+\sum _{i=r+N+2}^m{p}_i^{\prime }\varphi \left(x_i\right)+q_{r+N+1}ϵ.\hfill \end{array}$$

(53)

Similarly, $0\le (p_i-p_i^{\prime })\le p_i$ for $i=1,\dots ,r,r+N+2,\dots ,m$, and according to (50) and (51),

$$\begin{array}{cc}\hfill & \sum _{i=r+1}^{r+N}{q}_i{x}_i=\sum _{i=1}^r(p_i-p_i^{\prime })x_i+\sum _{i=r+N+2}^m(p_i-p_i^{\prime })x_i,\hfill \\ \hfill & \sum _{i=r+1}^{r+N}{q}_i=\sum _{i=1}^r(p_i-p_i^{\prime })+\sum _{i=r+N+2}^m(p_i-p_i^{\prime }).\hfill \end{array}$$

Thus, from the assumption $s-r=N$ of induction, we have

$$\begin{array}{cc}\hfill & \sum _{i=r+1}^{r+N}{q}_i\varphi \left(x_i\right)\le \sum _{i=1}^r(p_i-p_i^{\prime })\varphi \left(x_i\right)+\sum _{i=r+N+2}^m(p_i-p_i^{\prime })\varphi \left(x_i\right)+ϵ\sum _{i=r+1}^{r+N}{q}_i.\hfill \end{array}$$

(54)

Combining (53) and (54), we prove the situation for $s-r=N+1$, so (52) holds. □

Then, we use Theorem 12 to establish a similar Landau–Kolmogorov-type inequality [53]. Recall some background of the L-K inequality.

Let ${\parallel f\parallel }_{\infty }=ess.sup\left|f\left(t\right)\right|,t\in \mathbb{R}$. Denote $f^{\left(0\right)}=f,f^{\left(1\right)}=f^{\prime },f^{\left(2\right)}=f^{\prime \prime },\dots$, and we suppose $\parallel f^{\left(i\right)}{\parallel }_{\infty }$ all exist and are finite. In 1938, Kolmogorov proved that the best constant for

$$\parallel f^{\left(k\right)}{\parallel }_{\infty }\le {C(n,k,\mathbb{R})\parallel f\parallel }_{\infty }^{1-{\displaystyle \frac{k}{n}}}{\parallel f^{\left(n\right)}\parallel }_{\infty }^{{\displaystyle \frac{k}{n}}}$$

(55)

is $C(n,k,\mathbb{R})=K_{n-k}/K_n^{{\displaystyle \frac{n-k}{n}}}$, where

$$K_n={\displaystyle \frac{4}{\pi }}\sum _{v=0}^{+\infty }{\left({\displaystyle \frac{{(-1)}^v}{2v+1}}\right)}^{n+1}$$

are the Favard constants ($K_0=1,K_1={\displaystyle \frac{\pi }{2}},K_2={\displaystyle \frac{{\pi }^2}{8}},K_3={\displaystyle \frac{{\pi }^3}{24}},K_4={\displaystyle \frac{5{\pi }^4}{384}},K_5={\displaystyle \frac{{\pi }^5}{240}},\dots$). He also showed that $1<C(n,k,\mathbb{R})<{\displaystyle \frac{\pi }{2}}$ for all n and k. Inequality (55) is known as the Landau–Kolmogorov inequality.

Conclusion 2.

For a certain f with $\parallel f^{\left(k\right)}{\parallel }_{\infty }>0$, let $\varphi \left(x\right)=ln\parallel f^{\left(x\right)}{\parallel }_{\infty }$, then $\varphi$ is $ϵ$-approximately convex on a discrete set ${\mathbb{N}}_{+}$ for $ϵ=ln{\displaystyle \frac{\pi }{2}}$.

Proof. 

For $0=x_1<x_2<x_3$, it is easy to see from the L-K inequality, considering $C(n,k,\mathbb{R})<{\displaystyle \frac{\pi }{2}}$. If $x_1\ne 0$, then set

$$h=f^{\left(x_1\right)},h^{(x_2-x_1)}=f^{\left(x_2\right)},h^{(x_3-x_1)}=f^{\left(x_3\right)}$$

and use the L-K inequality. □

Applying the conclusion above to Theorem 12 with a slight footnote change, we get the L-K-type inequality below.

Theorem 13.

Let $0\le r<s<m$, $p_i,(i=0,\dots ,r,s+1,\dots ,m)$, and $q_i,(i=r+1,\dots ,s)$ be two nonnegative sequences. If

$$\begin{array}{cc}\hfill & \sum _{i=r+1}^s{q}_{i}i=\sum _{i=0}^r{p}_{i}i+\sum _{i=s+1}^m{p}_{i}i,\hfill \\ \hfill & \sum _{i=r+1}^s{q}_i=\sum _{i=0}^r{p}_i+\sum _{i=s+1}^m{p}_i,\hfill \end{array}$$

then

$$\prod _{i=r+1}^s\parallel f^{\left(i\right)}{\parallel }_{\infty }^{q_i}\le C\prod _{i=0}^r\parallel f^{\left(i\right)}{\parallel }_{\infty }^{p_i}\prod _{i=s+1}^m{\parallel f^{\left(i\right)}\parallel }_{\infty }^{p_i},$$

where $C<{\left({\displaystyle \frac{\pi }{2}}\right)}^{{\sum }_{i=r+1}^s{q}_i}$.

Remark 7.

For specific $p_i,(i=0,\dots ,r,s+1,\dots ,m)$ and $q_i,(i=r+1,\dots ,s)$, to determine the best

$$C=sup{\displaystyle \frac{{\prod }_{i=r+1}^s{\parallel f^{\left(i\right)}\parallel }_{\infty }^{q_i}}{{\prod }_{i=0}^r\parallel f^{\left(i\right)}{\parallel }_{\infty }^{p_i}{\prod }_{i=s+1}^m{\parallel f^{\left(i\right)}\parallel }_{\infty }^{p_i}}},{\parallel f^{\left(i\right)}\parallel }_{\infty }\ne 0$$

is very difficult for most situations. This direction involving many norms of different orders of derivatives of a function is also related to another so-called ”Kolmogorov’s Problem”; see [54,55].

6. Conclusions

In this paper, we concentrate on an inequality that points out some basic properties for convex functions and has connections with other inequalities for convex functions. We generalise it from a convex function on $\mathbb{R}$ to some analogue of convex functions with several variables, which is called functions with nondecreasing increments, and we prove the whole situation on three intervals, five intervals, and a special situation for $2n+1$ intervals. In these theorems, we all get weaker conditions or conditions that are easier to identify than Theorem 1. As applications of Lemma 1, we first get some inequalities in information theory, compare two different Csiszár $\varphi$-divergence, and then extend Lemma 1 to an $ϵ$-approximately convex function, which can slightly extend the Landau–Kolmogorov inequality.

For future directions, the first focus is the unsolved problem for $2n+1$ intervals. The second is to find out the best constants in Theorem 13. The third is trying to find other inequalities like the theorems in this paper, which can be proven by Theorem 1, and the condition is much simpler and easier to identify than Theorem 1.