Identity Extension for Function on Three Intervals and Application to Csiszar Divergence, Levinson and Ky Fan Inequalities

Abstract

Using Taylor-type expansions, we obtain identity expressions for functions on three intervals and differences for two pairs of Csiszár $\varphi$-divergence. With some more assumptions in these identities, inequalities for functions on three intervals and Csiszár $\varphi$-divergence can be obtained as special cases. They can also deduce the known generalized trapezoid type inequality. Furthermore, we use the identity to obtain a new extension for Levinson inequality; thus, new refinements and reverses for Ky Fan-type inequalities are established, which can be used to compare or estimate the yields in investments. Special cases of Csiszár $\varphi$-divergence are given, and we obtain new inequalities concerning different pairs of Kullback–Leibler distance, Hellinger distance, $\alpha$-order entropy and ${\chi }^2$-distance.

1. Introduction

How can some classic inequalities and results can be extended, refined or estimated, if we extend more intrinsic theorems under the perspective of more general functions? This paper shows an approach. Recall some classic inequalities and existing theorems first.

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 $a=(a_1,\dots ,a_n)\in {\mathbb{R}}_{+}^n$. Arithmetic mean is the most commonly used average mean in daily life. Geometric mean is applied to data with exponential growth property, like investment growth [1], production growth, and population growth. The compound annual growth rate (CAGR) $r·100\%$ in n years is calculated by the following:

$$\begin{array}{c}\hfill r={\left(\prod _{i=1}^n(1+r_i)\right)}^{{\displaystyle \frac{1}{n}}}-1\end{array}$$

regarding the growth rate $r_i·100\%(r_i\ge -1)$ in each i-th year.

The equivalent resistance value of two parallel resistance, is two times the harmonic means of a resitance value of two. The overall velocity v for a distance $2l$, is the harmonic mean of velocity $v_1$ in the first half distance l and velocity $v_2$ in the second half distance l, calculated as follows:

$$\begin{array}{c}\hfill v={\displaystyle \frac{2l}{t_1+t_2}}={\displaystyle \frac{2}{\frac{t_1}{l}+\frac{t_2}{l}}}={\displaystyle \frac{2}{\frac{1}{v_1}+\frac{1}{v_2}}}.\end{array}$$

The fundamental inequalities are written as follows:

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

These formulas have Ky Fan-type refinements. The second inequality on the right side of (1) is due to Ky Fan in [2] (p. 5), which is known as Ky Fan inequality in the literature, and the first inequality on the left side, known as Wang–Wang inequality, is proposed in [3]. Inequality is written as follows:

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

(1)

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

Part of (2) is proved in [4], and the whole chain (2) is proved in [5]. Inequality is written as follows:

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

(2)

holds for $a_i>0$.

They are followed by [5,6,7,8,9,10,11,12,13,14,15].

In [16], the following Conclusion 1 is established for convex functions on three intervals, to generalize Levinson inequality and unify the right side of Ky Fan-type inequalities (1) and (2). As pointed out in the same paper, it also has some connection with the Lah–Ribarič inequality [17], and its extension can be used to prove the Hermite–Hadamard inequality [18].

Conclusion 1.

Let $a_0<a_1<a_2<a_3$ and $x:[a_0,a_3]\to \mathbb{R}$ be a given function. Let $p:[a_0,a_1]\cup [a_2,a_3]\to {\mathbb{R}}_{+}$ and $q:[a_1,a_2]\to {\mathbb{R}}_{+}$ be two nonnegative functions. If $x\left(t\right)$ is increasing in $t\in [a_0,a_3]$ and

$$\begin{array}{cc}\hfill the following:& {\int }_{a_1}^{a_2}q\left(t\right)x\left(t\right)dt={\int }_{a_0}^{a_1}p\left(t\right)x\left(t\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)x\left(t\right)dt,\hfill \end{array}$$

(3)

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

(4)

then the following:

$${\int }_{a_1}^{a_2}q\left(t\right)\varphi \left(x\left(t\right)\right)dt\le {\int }_{a_0}^{a_1}p\left(t\right)\varphi \left(x\left(t\right)\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)\varphi \left(x\left(t\right)\right)dt$$

(5)

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

Then in [19], this Conclusion 1 is further extended and is also used to establish inequality for Csiszár $\varphi$-divergence. Note that this divergence can quantify the difference between two probability distributions, making it a useful tool for hypothesis testing and statistical inference; quantify the difference between the true distribution and the estimated distribution, making it a useful tool for data compression; and quantify the difference between the true distribution and the predicted distribution, making it a useful tool for classification problems and machine learning.

Recall the notion of Csiszár $\varphi$-divergence [20,21,22]. Given a convex function $\varphi :{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$, the $\varphi$-divergence functional, written as follows:

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

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 [22] and Chapter 9.2 in [20] as well as related references. In [19], the following comparison between two different $I_{\varphi }(p_1,q_1)$ and $I_{\varphi }(p_2,q_2)$ is established.

Theorem 1.

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 the following:

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

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

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

(6)

This theorem also holds for the probability distribution function (see [19]). This inequality looks similar to data processing inequality in a discrete case, but we do not have an assumption for some $P_{XY}=P_X{P}_{Y|X}$, further supposing the ratio of $dP$.

In this paper, we will give identity expressions for these two inequalities (5) and (6), and by letting the function be convex, we obtain inequalities in previous papers as special cases. The identity can also establish the known trapezoid-type inequality (7) (which can be seen as an estimation for the Hermite–Hadamard-type inequality). As the first application, we use these identities to extend Levinson inequality and give new refinements and reverses of Ky Fan-type inequalities (1) and (2). As the second application, we use the identity to obtain new inequalities for special examples of distance functions. Throughout this paper, we use the following notation:

$$\begin{array}{c}\hfill {(x-s)}_{+}=\left\{\begin{array}{cc}0,\hfill & x<s,\hfill \\ x-s,\hfill & x\ge s.\hfill \end{array}\right.\end{array}$$

2. Identity Extension

In this section, we will give identity expressions for Conclusion 1 and Theorem 1. Some special cases and corollary are also mentioned.

Theorem 2.

Let $a_0<a_1<a_2<a_3$ and $x:[a_0,a_3]\to \mathbb{R}$ be a given function. Let $p:[a_0,a_1]\cup [a_2,a_3]\to {\mathbb{R}}_{+}$ and $q:[a_1,a_2]\to {\mathbb{R}}_{+}$ be two nonnegative functions. If $x\left(t\right)$ is increasing in $t\in [a_0,a_3]$ and (3), (4) are satisfied, then for $\varphi :I\to \mathbb{R}$ such that ${\varphi }^{\prime }$ is absolutely continuous, we have the following:

$$\begin{array}{c}\hfill {\int }_{a_0}^{a_1}p\left(t\right)\varphi \left(x\left(t\right)\right)dt-{\int }_{a_1}^{a_2}q\left(t\right)\varphi \left(x\left(t\right)\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)\varphi \left(x\left(t\right)\right)dt={\int }_{x\left(a_0\right)}^{x\left(a_3\right)}{\varphi }^{″}\left(s\right)k\left(s\right)ds,\end{array}$$

where the following is obtained:

$$\begin{array}{c}\hfill k\left(s\right)={\int }_{a_0}^{a_1}p\left(t\right){(x\left(t\right)-s)}_{+}dt-{\int }_{a_1}^{a_2}q\left(t\right){(x\left(t\right)-s)}_{+}dt+{\int }_{a_2}^{a_3}p\left(t\right){(x\left(t\right)-s)}_{+}dt.\end{array}$$

Furthermore, we have $k\left(s\right)\ge 0$.

Proof. 

In (14) of Theorem 14, set $f=\varphi ,n=2,\alpha =a_0,\beta =a_3,a=x\left(a_0\right),b=x\left(a_3\right),g\left(x\right)=x\left(t\right)$ and the following:

$$\begin{array}{c}\hfill p\left(x\right)=\left\{\begin{array}{cc}p\left(t\right),\hfill & t\in [a_0,a_1],\hfill \\ -q\left(t\right),\hfill & t\in (a_1,a_2),\hfill \\ p\left(t\right),\hfill & t\in [a_2,a_3],\hfill \end{array}\right.\end{array}$$

in Theorem 2, taking (3) and (4) into consideration, we obtain the identity.

As for the nonnegativity of $k\left(s\right)$, this can be divided into three situations.

1. If $x\left(a_2\right)\le s\le x\left(a_3\right)$, then we obtain the following:

$$\begin{array}{c}\hfill k\left(s\right)={\int }_{a_2}^{a_3}p\left(t\right){(x\left(t\right)-s)}_{+}dt\ge 0.\end{array}$$

2. If $x\left(a_1\right)<s<x\left(a_2\right)$, then we obtain the following:

$$\begin{array}{c}\hfill k\left(s\right)=-{\int }_{a_1}^{a_2}q\left(t\right){(x\left(t\right)-s)}_{+}dt+{\int }_{a_2}^{a_3}p\left(t\right){(x\left(t\right)-s)}_{+}dt\\ \hfill =-{\int }_c^{a_2}q\left(t\right)(x\left(t\right)-s)dt+{\int }_{a_2}^{a_3}p\left(t\right)(x\left(t\right)-s)dt\end{array}$$

for a certain $c\in [a_1,a_2]$.

2.1. If ${\int }_c^{a_2}q\left(t\right)dt\le {\int }_{a_2}^{a_3}p\left(t\right)dt$, then we obtain the following:

$$\begin{array}{c}\hfill -{\int }_c^{a_2}q\left(t\right)(x\left(t\right)-s)dt+{\int }_{a_2}^{a_3}p\left(t\right)(x\left(t\right)-s)dt\\ \hfill \ge -{\int }_c^{a_2}q\left(t\right)(x\left(a_2\right)-s)dt+{\int }_{a_2}^{a_3}p\left(t\right)(x\left(a_2\right)-s)dt\ge 0.\end{array}$$

2.2. If ${\int }_c^{a_2}q\left(t\right)dt>{\int }_{a_2}^{a_3}p\left(t\right)dt$, then according to (3) and (4) we have the following:

$$\begin{array}{c}\hfill -{\int }_c^{a_2}q\left(t\right)(x\left(t\right)-s)dt+{\int }_{a_2}^{a_3}p\left(t\right)(x\left(t\right)-s)dt\\ \hfill ={\int }_{a_1}^{c}q\left(t\right)(x\left(t\right)-s)dt-{\int }_{a_0}^{a_1}p\left(t\right)(x\left(t\right)-s)dt\\ \hfill \ge {\int }_{a_1}^{c}q\left(t\right)(x\left(a_1\right)-s)dt-{\int }_{a_0}^{a_1}p\left(t\right)(x\left(a_1\right)-s)dt\ge 0.\end{array}$$

3. If $x\left(a_0\right)\le s\le x\left(a_1\right)$, then we obtain the following:

$$\begin{array}{c}\hfill k\left(s\right)={\int }_{a_0}^{a_1}p\left(t\right){(x\left(t\right)-s)}_{+}dt-{\int }_{a_1}^{a_2}q\left(t\right)(x\left(t\right)-s)dt+{\int }_{a_2}^{a_3}p\left(t\right)(x\left(t\right)-s)dt\\ \hfill ={\int }_c^{a_1}p\left(t\right)(x\left(t\right)-s)dt-{\int }_{a_1}^{a_2}q\left(t\right)(x\left(t\right)-s)dt+{\int }_{a_2}^{a_3}p\left(t\right)(x\left(t\right)-s)dt\end{array}$$

for a certain $c\in [a_0,a_1]$, according to (3) and (4) we have the following:

$$\begin{array}{c}\hfill {\int }_c^{a_1}p\left(t\right)(x\left(t\right)-s)dt-{\int }_{a_1}^{a_2}q\left(t\right)(x\left(t\right)-s)dt+{\int }_{a_2}^{a_3}p\left(t\right)(x\left(t\right)-s)dt\\ \hfill =-{\int }_{a_0}^{c}p\left(t\right)(x\left(t\right)-s)dt\ge 0.\end{array}$$

Hence, in all situations we affirm $k\left(s\right)\ge 0$. □

From Theorem 2 we can prove the following theorem, which gives the upper and lower bounds.

Theorem 3.

Let $a_0<a_1<a_2<a_3$ and $x:[a_0,a_3]\to \mathbb{R}$ be a given function. Let $p:[a_0,a_1]\cup [a_2,a_3]\to {\mathbb{R}}_{+}$ and $q:[a_1,a_2]\to {\mathbb{R}}_{+}$ be two nonnegative functions. If $x\left(t\right)$ is increasing in $t\in [a_0,a_3]$ and (3), (4) are satisfied, then for $\varphi :I\to \mathbb{R}$ such that $m\le {\varphi }^{″}\le M$, we have the following:

$$\begin{array}{c}\hfill {\displaystyle \frac{m}{2}}\left({\int }_{a_0}^{a_1}p\left(t\right)x^2\left(t\right)dt-{\int }_{a_1}^{a_2}q\left(t\right)x^2\left(t\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)x^2\left(t\right)dt\right)\\ \hfill \le {\int }_{a_0}^{a_1}p\left(t\right)\varphi \left(x\left(t\right)\right)dt-{\int }_{a_1}^{a_2}q\left(t\right)\varphi \left(x\left(t\right)\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)\varphi \left(x\left(t\right)\right)dt\\ \hfill \le {\displaystyle \frac{M}{2}}\left({\int }_{a_0}^{a_1}p\left(t\right)x^2\left(t\right)dt-{\int }_{a_1}^{a_2}q\left(t\right)x^2\left(t\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)x^2\left(t\right)dt\right).\end{array}$$

Proof. 

Utilize Theorem 2 to obtain the following expression:

$$\begin{array}{c}\hfill {\int }_{a_0}^{a_1}p\left(t\right)\varphi \left(x\left(t\right)\right)dt-{\int }_{a_1}^{a_2}q\left(t\right)\varphi \left(x\left(t\right)\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)\varphi \left(x\left(t\right)\right)dt={\int }_{x\left(a_0\right)}^{x\left(a_3\right)}{\varphi }^{″}\left(s\right)k\left(s\right)ds,\end{array}$$

then from $k\left(s\right)\ge 0$ in Theorem 2 and $m\le {\varphi }^{″}\le M$ we have

$$\begin{array}{c}\hfill m{\int }_{x\left(a_0\right)}^{x\left(a_3\right)}k\left(s\right)ds\le {\int }_{x\left(a_0\right)}^{x\left(a_3\right)}{\varphi }^{″}\left(s\right)k\left(s\right)ds\le M{\int }_{x\left(a_0\right)}^{x\left(a_3\right)}k\left(s\right)ds.\end{array}$$

Use Fubini’s theorem to obtain the following:

$$\begin{array}{c}\hfill {\int }_{x\left(a_0\right)}^{x\left(a_3\right)}k\left(s\right)ds={\int }_{a_0}^{a_1}p\left(t\right){\displaystyle \frac{{(x\left(t\right)-x\left(a_0\right))}^2}{2}}dt\\ \hfill -{\int }_{a_1}^{a_2}q\left(t\right){\displaystyle \frac{{(x\left(t\right)-x\left(a_0\right))}^2}{2}}dt+{\int }_{a_2}^{a_3}p\left(t\right){\displaystyle \frac{{(x\left(t\right)-x\left(a_0\right))}^2}{2}}dt\\ \hfill ={\displaystyle \frac{1}{2}}\left({\int }_{a_0}^{a_1}p\left(t\right)x^2\left(t\right)dt-{\int }_{a_1}^{a_2}q\left(t\right)x^2\left(t\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)x^2\left(t\right)dt\right),\end{array}$$

the last step is due to (3) and (4). □

Remark 1.

It is easy to observe the following:

$$\begin{array}{c}\hfill {\int }_{a_0}^{a_1}p\left(t\right)x^2\left(t\right)dt-{\int }_{a_1}^{a_2}q\left(t\right)x^2\left(t\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)x^2\left(t\right)dt\ge 0\end{array}$$

as it transforms from $k\left(s\right)\ge 0$.

By putting $m=0$ in Theorem 3, we obtain Conclusion 1 for twice differentiable convex functions ϕ.

From Theorem 2 another upper bound can also be given as the following.

Theorem 4.

Let $a_0<a_1<a_2<a_3$ and $x:[a_0,a_3]\to \mathbb{R}$ be a given function. Let $p\left(t\right):[a_0,a_1]\cup [a_2,a_3]\to {\mathbb{R}}_{+}$ and $q\left(t\right):[a_1,a_2]\to {\mathbb{R}}_{+}$ be two nonnegative functions. If $x\left(t\right)$ is increasing in $t\in [a_0,a_3]$ and the following:

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

then for $\varphi :I\to \mathbb{R}$ such that ${\varphi }^{″}\in L^p(x\left(a_0\right),x\left(a_3\right))$, we have the following:

$$\begin{array}{c}\hfill \left|{\int }_{a_0}^{a_1}p\left(t\right)\varphi \left(x\left(t\right)\right)dt-{\int }_{a_1}^{a_2}q\left(t\right)\varphi \left(x\left(t\right)\right)dt+{\int }_{a_2}^{a_3}p\left(t\right)\varphi \left(x\left(t\right)\right)dt\right|\\ \hfill \le \parallel {\varphi }^{″}{\parallel }_p{\left({\int }_{x\left(a_0\right)}^{x\left(a_3\right)}{k}^q\left(s\right)ds\right)}^{1/q},\end{array}$$

for $p>1,1/p+1/q=1$, where we obtain the following:

$$\begin{array}{c}\hfill k\left(s\right)={\int }_{a_0}^{a_1}p\left(t\right){(x\left(t\right)-s)}_{+}dt-{\int }_{a_1}^{a_2}q\left(t\right){(x\left(t\right)-s)}_{+}dt+{\int }_{a_2}^{a_3}p\left(t\right){(x\left(t\right)-s)}_{+}dt.\end{array}$$

Proof. 

Take absolute value and use Hölder inequality for Theorem 2. □

From Theorem 4 we can deduce the following extension for trapezoid inequality, see chapter 2 in [18].

Corollary 1.

Let twice differentiable $\varphi :[a,b]\to \mathbb{R}$ such that ${\varphi }^{″}\in L^p(a,b)$, we have the following:

$$\begin{array}{c}\hfill \left|{\displaystyle \frac{\varphi \left(a\right)+\varphi \left(b\right)}{2}}-{\displaystyle \frac{1}{b-a}}{\int }_a^b\varphi \left(t\right)dt\right|\\ \hfill \le {\displaystyle \frac{1}{2}}{(b-a)}^{{\displaystyle \frac{q+1}{q}}}{\left[B(q+1,q+1)\right]}^{{\displaystyle \frac{1}{q}}}{\parallel {\varphi }^{″}\parallel }_p\end{array}$$

(7)

for $p>1,1/p+1/q=1$, where B is Euler’s Beta-function.

Proof. 

In Theorem 4, set $a_0=a-\epsilon ,a_1=a,a_2=b,a_3=b+\epsilon$, $x\left(t\right)=t,p\left(t\right)\equiv {\displaystyle \frac{1}{2\epsilon }},q\left(t\right)\equiv {\displaystyle \frac{1}{b-a}}$, it is clear all the conditions are satisfied.

Then, taking $\epsilon \to 0$, we have the following:

$$\begin{array}{c}\hfill k\left(s\right)={\displaystyle \frac{(b-s)(s-a)}{2(b-a)}}.\end{array}$$

With some calculation, we obtain the following:

$$\begin{array}{c}\hfill {\left({\int }_a^b{\left({\displaystyle \frac{(b-s)(s-a)}{2(b-a)}}\right)}^{q}ds\right)}^{1/q}={\displaystyle \frac{1}{2}}{(b-a)}^{{\displaystyle \frac{q+1}{q}}}{\left[B(q+1,q+1)\right]}^{{\displaystyle \frac{1}{q}}}\end{array}$$

from which we obtain the desired result. □

Taking $p=\infty$ as the original trapezoid inequality, written as follows:

$$\begin{array}{c}\hfill \left|{\int }_a^b\varphi \left(t\right)dt-{\displaystyle \frac{b-a}{2}}[\varphi \left(a\right)+\varphi \left(b\right)]\right|\le {\displaystyle \frac{{(b-a)}^3}{12}}{\parallel {\varphi }^{″}\parallel }_{\infty }.\end{array}$$

It is very useful in numerical integration, as we cut the interval into more smaller intervals (suppose evenly n intervals), the remainder terms (error) will be much smaller, because of the following:

$$n{\left({\displaystyle \frac{b-a}{n}}\right)}^3\le {(b-a)}^3.$$

Remark 2.

Another corollary is $a_0=a-{\displaystyle \frac{\epsilon }{2}},a_1={\displaystyle \frac{a+b-\epsilon }{2}},a_2={\displaystyle \frac{a+b+\epsilon }{2}},a_3=b+{\displaystyle \frac{\epsilon }{2}}$, $x\left(t\right)=t,p\left(t\right)\equiv {\displaystyle \frac{1}{b-a}},q\left(t\right)\equiv {\displaystyle \frac{1}{\epsilon }}$, and take $\epsilon \to 0$, we can obtain estimation for

$$\begin{array}{c}\hfill \left|\varphi \left({\displaystyle \frac{a+b}{2}}\right)-{\displaystyle \frac{1}{b-a}}{\int }_a^b\varphi \left(t\right)dt\right|,\end{array}$$

which is more difficult to calculate.

Then we give the identity expression for Theorem 1, without assuming $\varphi$ to be convex first.

Theorem 5.

If $p_1,q_1,p_2,q_2\in {\mathbb{P}}_n$, then for $\varphi :[0,\infty )\to \mathbb{R}$ such that ${\varphi }^{\prime }$ is absolutely continuous, we have the following:

$$\begin{array}{c}\hfill I_{\varphi }(p_2,q_2)-I_{\varphi }(p_1,q_1)=\\ \hfill {\int }_0^{\infty }{\varphi }^{″}\left(s\right)\left(\sum _{i=1}^n{(p_{2,i}-s{q}_{2,i})}_{+}-\sum _{i=1}^n{(p_{1,i}-s{q}_{1,i})}_{+}\right)ds.\end{array}$$

Proof. 

In (13) of Theorem 14, set $f=\varphi ,n=2,N=n,a=0,b=\infty ,p_i=q_{1,i},x_i={\displaystyle \frac{p_{1,i}}{q_{1,i}}}$ in Theorem 5, we obtain the expression for $I_{\varphi }(p_1,q_1)$.

Similarly, in (13) of Theorem 14, set $f=\varphi ,n=2,N=n,a=0,b=\infty ,p_i=q_{2,i},x_i={\displaystyle \frac{p_{2,i}}{q_{2,i}}}$ in Theorem 5, we obtain the expression for $I_{\varphi }(p_2,q_2)$.

Take into consideration $p_1,q_1,p_2,q_2\in {\mathbb{P}}_n$, we complete the proof. □

Then we state two lemmas below, in which, if $p_1,q_1,p_2,q_2$ satisfy some further conditions, we have the following:

$$\begin{array}{c}\hfill \sum _{i=1}^n{(p_{2,i}-s{q}_{2,i})}_{+}-\sum _{i=1}^n{(p_{1,i}-s{q}_{1,i})}_{+}\ge 0.\end{array}$$

The proofs are in the Some Lemmas section.

Lemma 1.

In Theorem 5, if we further suppose the following:

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

for some $m\le 1\le M$, then we obtain the following:

$$\begin{array}{c}\hfill \sum _{i=1}^n{(p_{2,i}-s{q}_{2,i})}_{+}-\sum _{i=1}^n{(p_{1,i}-s{q}_{1,i})}_{+}\ge 0\end{array}$$

for all $s\in [0,\infty )$.

Lemma 2.

In Theorem 5, if we further suppose $q_1=q_2$, which means $q_{1,i}=q_{2,i},i=1,\dots ,n$, and (here we use q to refer either $q_1$ or $q_2$), then we obtain the following:

$${\displaystyle \frac{p_2}{q}}{\succ }_q{\displaystyle \frac{p_1}{q}},$$

(the sequence ${\displaystyle \frac{p_2}{q}}$ majorizes the sequence ${\displaystyle \frac{p_1}{q}}$ with the weight q), then we obtain the following:

$$\begin{array}{c}\hfill \sum _{i=1}^n{(p_{2,i}-s{q}_{2,i})}_{+}-\sum _{i=1}^n{(p_{1,i}-s{q}_{1,i})}_{+}\ge 0\end{array}$$

for all $s\in [0,\infty )$.

Combine Theorem 5, Lemma 1 and Lemma 2, we have the following estimation about $I_{\varphi }(p_2,q_2)-I_{\varphi }(p_1,q_1)$.

Theorem 6.

Under the assumptions of Theorem 5, if condition in Lemma 1 or Lemma 2 is further satisfied, then we obtain the following:

$$\begin{array}{c}\hfill r{\int }_0^{\infty }\left(\sum _{i=1}^n{(p_{2,i}-s{q}_{2,i})}_{+}-\sum _{i=1}^n{(p_{1,i}-s{q}_{1,i})}_{+}\right)ds\\ \hfill \le I_{\varphi }(p_2,q_2)-I_{\varphi }(p_1,q_1)\\ \hfill \le R{\int }_0^{\infty }\left(\sum _{i=1}^n{(p_{2,i}-s{q}_{2,i})}_{+}-\sum _{i=1}^n{(p_{1,i}-s{q}_{1,i})}_{+}\right)ds\end{array}$$

for $\varphi :[0,\infty )\to \mathbb{R}$ such that $r\le {\varphi }^{″}\le R$.

Proof. 

Under the assumptions of Lemma 1 or Lemma 2, we have the following:

$$\begin{array}{c}\hfill \sum _{i=1}^n{(p_{2,i}-s{q}_{2,i})}_{+}-\sum _{i=1}^n{(p_{1,i}-s{q}_{1,i})}_{+}\ge 0,\end{array}$$

from Theorem 5 we prove the conclusion. □

Remark 3.

Set $r=0$, we obtain Theorem 1 for twice differentiable convex functions ϕ.

Remark 4.

If $p_1,q_1,p_2,q_2>0$, then we have a more direct expression, written as follows:

$$\begin{array}{c}\hfill {\int }_0^{\infty }\left(\sum _{i=1}^n{(p_{2,i}-s{q}_{2,i})}_{+}-\sum _{i=1}^n{(p_{1,i}-s{q}_{1,i})}_{+}\right)ds=\sum _{i=1}^n\left({\displaystyle \frac{p_{2,i}^2}{2q_{2,i}}}-{\displaystyle \frac{p_{1,i}^2}{2q_{1,i}}}\right)\end{array}$$

for $\varphi :[min{\displaystyle \frac{p_{2,i}}{q_{2,i}}},max{\displaystyle \frac{p_{2,i}}{q_{2,i}}}]\to \mathbb{R}$ such that $r\le {\varphi }^{″}\le R$.

3. Application

In this section, we first use Theorem 3 to establish Levinson and Ky Fan-type inequalities, and then use Theorem 6 and Remark 4 to give several special examples of distance functions in mathematical statistics and information theory.

In [16], the following general inequality for 3-convex function is established, which can be used to prove Jensen inequality, Levinson inequality, refine power means inequality and unify two Ky Fan-type inequalities.

Conclusion 2.

Suppose that $f:[\alpha ,\beta ]\to \mathbb{R}$ is a continuous 3-convex function, for $i=1,\dots ,n:a_i,b_i\in [\alpha ,\beta ];b_i>a_i$, the inequality is written as follows:

$$\begin{array}{cc}\hfill & f\left({\displaystyle \frac{{\sum }_{i=1}^n(b_i^2-a_i^2)}{2{\sum }_{i=1}^n(b_i-a_i)}}+{\displaystyle \frac{{\sum }_{i=1}^n(b_i-a_i)}{2n}}\right)\hfill \\ \hfill -& f\left({\displaystyle \frac{{\sum }_{i=1}^n(b_i^2-a_i^2)}{2{\sum }_{i=1}^n(b_i-a_i)}}-{\displaystyle \frac{{\sum }_{i=1}^n(b_i-a_i)}{2n}}\right)\hfill \\ \hfill \le & {\displaystyle \frac{1}{n}}\sum _{i=1}^n(f\left(b_i\right)-f\left(a_i\right))\hfill \end{array}$$

hold.

In this section, based on Theorem 3, we will further extend this conclusion and use the extension to refine Ky Fan-type inequalities. These results are generalizations of Levinson, Jensen, and Ky Fan inequalities. The condition $m\le f^{\prime ″}\le M$ is more general than $f^{\prime ″}\ge 0$ as $m,M$ may be $-\infty ,\infty$.

Theorem 7.

Let $f:[\alpha ,\beta ]\to \mathbb{R}$ be a continuous function such that $m\le f^{\prime ″}\le M$. For $a_i,b_i\in [\alpha ,\beta ];b_i>a_i,i=1,\dots ,n$, we have the following:

$$\begin{array}{c}\hfill {\displaystyle \frac{m}{6n}}\left(\sum _{i=1}^n(b_i^3-a_i^3)-{\displaystyle \frac{3{\left[{\sum }_{i=1}^n(b_i^2-a_i^2)\right]}^2}{4{\sum }_{i=1}^n(b_i-a_i)}}-{\displaystyle \frac{{\left[{\sum }_{i=1}^n(b_i-a_i)\right]}^3}{4n^2}}\right)\\ \hfill \le {\displaystyle \frac{1}{n}}\sum _{i=1}^n(f\left(b_i\right)-f\left(a_i\right))-f\left({\displaystyle \frac{{\sum }_{i=1}^n(b_i^2-a_i^2)}{2{\sum }_{i=1}^n(b_i-a_i)}}+{\displaystyle \frac{{\sum }_{i=1}^n(b_i-a_i)}{2n}}\right)\end{array}$$

(8)

$$\begin{array}{c}\hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em} +f\left({\displaystyle \frac{{\sum }_{i=1}^n(b_i^2-a_i^2)}{2{\sum }_{i=1}^n(b_i-a_i)}}-{\displaystyle \frac{{\sum }_{i=1}^n(b_i-a_i)}{2n}}\right)\end{array}$$

(9)

$$\begin{array}{c}\hfill \le {\displaystyle \frac{M}{6n}}\left(\sum _{i=1}^n(b_i^3-a_i^3)-{\displaystyle \frac{3{\left[{\sum }_{i=1}^n(b_i^2-a_i^2)\right]}^2}{4{\sum }_{i=1}^n(b_i-a_i)}}-{\displaystyle \frac{{\left[{\sum }_{i=1}^n(b_i-a_i)\right]}^3}{4n^2}}\right).\end{array}$$

(10)

And we also have the follwoing:

$$\begin{array}{c}\hfill \sum _{i=1}^n(b_i^3-a_i^3)\ge {\displaystyle \frac{3{\left[{\sum }_{i=1}^n(b_i^2-a_i^2)\right]}^2}{4{\sum }_{i=1}^n(b_i-a_i)}}+{\displaystyle \frac{{\left[{\sum }_{i=1}^n(b_i-a_i)\right]}^3}{4n^2}}.\end{array}$$

Proof. 

The proof is divided into two parts.

1. Term (9) satisfies all the corresponding condition in Theorem 3.

It is easy to observe the following:

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{n}}\sum _{i=1}^n(f\left(b_i\right)-f\left(a_i\right))-f\left({\displaystyle \frac{{\sum }_{i=1}^n(b_i^2-a_i^2)}{2{\sum }_{i=1}^n(b_i-a_i)}}+{\displaystyle \frac{{\sum }_{i=1}^n(b_i-a_i)}{2n}}\right)\\ \hfill +f\left({\displaystyle \frac{{\sum }_{i=1}^n(b_i^2-a_i^2)}{2{\sum }_{i=1}^n(b_i-a_i)}}-{\displaystyle \frac{{\sum }_{i=1}^n(b_i-a_i)}{2n}}\right)\\ \hfill ={\int }_{\alpha }^{\beta }{C}_1\left(t\right)f^{\prime }\left(t\right)dt-{\int }_{\overline{a}}^{\overline{b}}{C}_2\left(t\right)f^{\prime }\left(t\right)dt\\ \hfill ={\int }_{\alpha }^{\overline{a}}{C}_1\left(t\right)f^{\prime }\left(t\right)dt-{\int }_{\overline{a}}^{\overline{b}}(C_2\left(t\right)-C_1\left(t\right))f^{\prime }\left(t\right)dt+{\int }_{\overline{b}}^{\beta }{C}_1\left(t\right)f^{\prime }\left(t\right)dt,\end{array}$$

where

$$\overline{a}={\displaystyle \frac{{\sum }_{i=1}^n(b_i^2-a_i^2)}{2{\sum }_{i=1}^n(b_i-a_i)}}-{\displaystyle \frac{{\sum }_{i=1}^n(b_i-a_i)}{2n}},\overline{b}={\displaystyle \frac{{\sum }_{i=1}^n(b_i^2-a_i^2)}{2{\sum }_{i=1}^n(b_i-a_i)}}+{\displaystyle \frac{{\sum }_{i=1}^n(b_i-a_i)}{2n}},$$

and $C_1\left(t\right)={\displaystyle \frac{k}{n}}$, if t belongs to and only belongs to k interval(s) of all $[a_i,b_i],(i=1,\dots ,n)$, written as follows:

$$\begin{array}{c}\hfill C_2\left(t\right)=\left\{\begin{array}{cc}1,\hfill & t\in [\overline{a},\overline{b}],\hfill \\ 0,\hfill & t\notin [\overline{a},\overline{b}].\hfill \end{array}\right.\end{array}$$

Then, in Theorem 3, we can set $p\left(t\right)=C_1\left(t\right),q\left(t\right)=(C_2\left(t\right)-C_1\left(t\right)),x\left(t\right)=t,\varphi =f^{\prime },a_0=\alpha ,a_1=\overline{a},a_2=\overline{b},a_3=\beta$ in Theorem 7, this is feasible, as we only need to check condition (3) and (4) naturally hold.

First, we obtain the following:

$${\int }_{\alpha }^{\beta }{C}_1\left(t\right)dt={\displaystyle \frac{1}{n}}\sum _{i=1}^n(b_i-a_i)=\overline{b}-\overline{a}={\int }_{\overline{a}}^{\overline{b}}{C}_2\left(t\right)dt,$$

then the following:

$${\int }_{\alpha }^{\overline{a}}{C}_1\left(t\right)dt+{\int }_{\overline{b}}^{\beta }{C}_1\left(t\right)dt={\int }_{\overline{a}}^{\overline{b}}(C_2\left(t\right)-C_1\left(t\right))dt,$$

which satisfies condition (4).

Second, we obtain the following:

$${\int }_{\alpha }^{\beta }{C}_1\left(t\right)tdt={\displaystyle \frac{1}{n}}\sum _{i=1}^n{\int }_{a_i}^{b_i}tdt={\displaystyle \frac{1}{n}}\sum _{i=1}^n{\displaystyle \frac{b_i^2-a_i^2}{2}}={\displaystyle \frac{{\overline{b}}^2-{\overline{a}}^2}{2}}={\int }_{\overline{a}}^{\overline{b}}{C}_2\left(t\right)tdt,$$

then the following:

$${\int }_{\alpha }^{\overline{a}}{C}_1\left(t\right)tdt+{\int }_{\overline{b}}^{\beta }{C}_1\left(t\right)tdt={\int }_{\overline{a}}^{\overline{b}}(C_2\left(t\right)-C_1\left(t\right))tdt,$$

which satisfies condition (3).

Thus, we can use Theorem 3 to obtain the following:

$$\begin{array}{c}\hfill {\displaystyle \frac{m}{2}}\left({\int }_{\alpha }^{\overline{a}}{C}_1\left(t\right)t^{2}dt-{\int }_{\overline{a}}^{\overline{b}}(C_2\left(t\right)-C_1\left(t\right))t^{2}dt+{\int }_{\overline{b}}^{\beta }{C}_1\left(t\right)t^{2}dt\right)\\ \hfill \le {\int }_{\alpha }^{\overline{a}}{C}_1\left(t\right)f^{\prime }\left(t\right)dt-{\int }_{\overline{a}}^{\overline{b}}(C_2\left(t\right)-C_1\left(t\right))f^{\prime }\left(t\right)dt+{\int }_{\overline{b}}^{\beta }{C}_1\left(t\right)f^{\prime }\left(t\right)dt\\ \hfill \le {\displaystyle \frac{M}{2}}\left({\int }_{\alpha }^{\overline{a}}{C}_1\left(t\right)t^{2}dt-{\int }_{\overline{a}}^{\overline{b}}(C_2\left(t\right)-C_1\left(t\right))t^{2}dt+{\int }_{\overline{b}}^{\beta }{C}_1\left(t\right)t^{2}dt\right),\end{array}$$

and

$$\begin{array}{c}\hfill {\int }_{\alpha }^{\overline{a}}{C}_1\left(t\right)t^{2}dt-{\int }_{\overline{a}}^{\overline{b}}(C_2\left(t\right)-C_1\left(t\right))t^{2}dt+{\int }_{\overline{b}}^{\beta }{C}_1\left(t\right)t^{2}dt\ge 0.\end{array}$$

2. Calculate terms (8) and (10), and nonnegativity.

$$\begin{array}{c}\hfill 0\le {\int }_{\alpha }^{\overline{a}}{C}_1\left(t\right)t^{2}dt-{\int }_{\overline{a}}^{\overline{b}}(C_2\left(t\right)-C_1\left(t\right))t^{2}dt+{\int }_{\overline{b}}^{\beta }{C}_1\left(t\right)t^{2}dt\\ \hfill ={\int }_{\alpha }^{\beta }{C}_1\left(t\right)t^{2}dt-{\int }_{\overline{a}}^{\overline{b}}{C}_2\left(t\right)t^{2}dt\\ \hfill ={\displaystyle \frac{1}{n}}\sum _{i=1}^n{\int }_{a_i}^{b_i}{t}^{2}dt-{\int }_{\overline{a}}^{\overline{b}}{t}^{2}dt={\displaystyle \frac{1}{3n}}\sum _{i=1}^n(b_i^3-a_i^3)-{\displaystyle \frac{1}{3}}({\overline{b}}^3-{\overline{a}}^3)\\ \hfill ={\displaystyle \frac{1}{3n}}\sum _{i=1}^n(b_i^3-a_i^3)-{\displaystyle \frac{1}{3}}\left({\displaystyle \frac{3{\left[{\sum }_{i=1}^n(b_i^2-a_i^2)\right]}^2}{4n{\sum }_{i=1}^n(b_i-a_i)}}+{\displaystyle \frac{{\left[{\sum }_{i=1}^n(b_i-a_i)\right]}^3}{4n^3}}\right),\end{array}$$

which is equivalent to the desired conclusion. □

Theorem 8 is powerful enough to deduce several old and new inequalities below.

Remark 5.

Take $m=0$ in Theorem 7, we obtain Conclusion 2, if we further set $b_i=c-a_i,0\le a_i\le {\displaystyle \frac{c}{2}}$ we obtain the Levinson inequality.

We can also deduce a known estimation for Jensen inequality below.

Corollary 2.

Let $g:[\alpha ,\beta ]\to \mathbb{R}$ be a continuous function such that $m\le g^{″}\le M$. For $a_i\in [\alpha ,\beta ];i=1,.\dots n$, we have the following:

$$\begin{array}{c}\hfill {\displaystyle \frac{m}{2}}\left({\displaystyle \frac{{\sum }_{i=1}^n{a}_i^2}{n}}-{\left({\displaystyle \frac{{\sum }_{i=1}^n{a}_i}{n}}\right)}^2\right)\\ \hfill \le {\displaystyle \frac{1}{n}}\sum _{i=1}^{n}g\left(a_i\right)-g\left({\displaystyle \frac{{\sum }_{i=1}^n{a}_i}{n}}\right)\\ \hfill \le {\displaystyle \frac{M}{2}}\left({\displaystyle \frac{{\sum }_{i=1}^n{a}_i^2}{n}}-{\left({\displaystyle \frac{{\sum }_{i=1}^n{a}_i}{n}}\right)}^2\right).\end{array}$$

Proof. 

In Theorem 7, set $b_i=a_i+\Delta (\Delta >0)$, and multiple the inequality with ${\displaystyle \frac{1}{\Delta }}$. Let $\Delta \to 0$, we obtain the following:

$$\begin{array}{c}\hfill {\displaystyle \frac{m}{2}}\left({\displaystyle \frac{{\sum }_{i=1}^n{a}_i^2}{n}}-{\left({\displaystyle \frac{{\sum }_{i=1}^n{a}_i}{n}}\right)}^2\right)\\ \hfill \le {\displaystyle \frac{1}{n}}\sum _{i=1}^n{f}^{\prime }\left(a_i\right)-f^{\prime }\left({\displaystyle \frac{{\sum }_{i=1}^n{a}_i}{n}}\right)\\ \hfill \le {\displaystyle \frac{M}{2}}\left({\displaystyle \frac{{\sum }_{i=1}^n{a}_i^2}{n}}-{\left({\displaystyle \frac{{\sum }_{i=1}^n{a}_i}{n}}\right)}^2\right),\end{array}$$

let $g=f^{\prime }$ we obtain the corollary. □

Then we use Theorem 7 to establish refinements and reverses for Ky Fan-type inequalities (1), (2) mentioned in the introduction section, these special cases are new.

Theorem 8.

Denoting by $A\left(a\right),G\left(a\right)$ the arithmetic and geometric means of positive real numbers $a_1,\dots ,a_n$ with $a=(a_1,\dots ,a_n)\in {\mathbb{R}}_{+}^n$, we have the following:

$${\displaystyle \frac{G\left(a\right)}{G(1-a)}}\le {\displaystyle \frac{G\left(a\right)}{G(1-a)}}·e^{K/{(1-a_{min})}^3}\le {\displaystyle \frac{A\left(a\right)}{A(1-a)}}\le {\displaystyle \frac{G\left(a\right)}{G(1-a)}}·e^{K/a_{min}^3}$$

for $0<a_i\le {\displaystyle \frac{1}{2}}$, where $a_{min}=min\{a_1,\dots ,a_n\}$, and the following:

$$\begin{array}{c}\hfill K={\displaystyle \frac{1}{n}}\sum _{i=1}^n{a}_i^2-{\displaystyle \frac{2}{3n}}\sum _{i=1}^n{a}_i^3-{\left({\displaystyle \frac{1}{n}}\sum _{i=1}^n{a}_i\right)}^2+{\displaystyle \frac{2}{3}}{\left({\displaystyle \frac{1}{n}}\sum _{i=1}^n{a}_i\right)}^3\ge 0.\end{array}$$

Proof. 

Set $f\left(x\right)=lnx$ defined on $[a_{min},1-a_{min}]$ in Theorem 7, for $0<a_i\le {\displaystyle \frac{1}{2}}$ and $b_i=1-a_i$. As $f^{\prime ″}={\displaystyle \frac{2}{x^3}}$ we know the following:

$${\displaystyle \frac{2}{{(1-a_{min})}^3}}\le f^{\prime ″}\le {\displaystyle \frac{2}{a_{min}^3}}.$$

Thus we can use Theorem 7. With some calculation and simplification, then taking exp we obtain the desired results. □

Remark 6.

The Ky Fan inequality is equivalent to the following:

$${\displaystyle \frac{G\left(a\right)}{G(2-a)}}\le {\displaystyle \frac{A\left(a\right)}{A(2-a)}}$$

(11)

for $0<a_i\le 1$, the readers may also establish similar refinement and reverse like the proof above, for the following discussion.

Consider two investment products $X,Y$ that have inverse yield $b_i=(1-a_i)·100\%,-b_i=(a_i-1)·100\%$ in each period i, for $i=1,\dots ,n$ (e. g. ETF and its inverse ETF).

If $b_i$ is nonnegative in each period of time, it is clear that the overall investment yield of X is higher than Y after n periods, which is written as follows:

$$(\prod _{i=1}^n(1+b_i)-1)·100\%=y_X\ge y_Y=(\prod _{i=1}^n(1-b_i)-1)·100\%.$$

With (11) we obtain a more accurate estimation, written as follows:

$$y_X\ge \left[(y_Y+1){\left({\displaystyle \frac{1+A\left(b\right)}{1-A\left(b\right)}}\right)}^n-1\right]·100\%\ge y_Y,$$

where $A\left(b\right)$ is the arithmetic average of X’s yields $b_i$ in n periods.

Apply the refinement and reverse of Ky Fan inequality, we can obtain even more detailed estimations, left to readers.

Theorem 9.

Denote by $A\left(a\right),G\left(a\right)$ the arithmetic and geometric means of positive real numbers $a_1,\dots ,a_n$ with $a=(a_1,\dots ,a_n)\in {\mathbb{R}}_{+}^n$, we have the following:

$${\displaystyle \frac{G\left(a\right)}{G(1+a)}}\le {\displaystyle \frac{G\left(a\right)}{G(1+a)}}·e^{K/{(1+a_{max})}^3}\le {\displaystyle \frac{A\left(a\right)}{A(1+a)}}\le {\displaystyle \frac{G\left(a\right)}{G(1+a)}}·e^{K/a_{min}^3}$$

for $a_i>0$, where the following is formulated:

$$\begin{array}{c}\hfill a_{min}=min\{a_1,\dots ,a_n\},a_{max}=max\{a_1,\dots ,a_n\},\\ \hfill K={\displaystyle \frac{1}{n}}\sum _{i=1}^n{a}_i^2-{\left({\displaystyle \frac{1}{n}}\sum _{i=1}^n{a}_i\right)}^2\ge 0.\end{array}$$

Proof. 

Set $f\left(x\right)=lnx$ defined on $[a_{min},1+a_{max}]$ in Theorem 7, for $a_i>0$ and $b_i=1+a_i$. As $f^{\prime ″}={\displaystyle \frac{2}{x^3}}$ we know the following:

$${\displaystyle \frac{2}{{(1+a_{max})}^3}}\le f^{\prime ″}\le {\displaystyle \frac{2}{a_{min}^3}}.$$

Thus we can use Theorem 7. With some calculation and simplification, taking exp we obtain the desired results. □

Remark 7.

If $\lambda >0$, then we obtain the following:

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

(12)

for $a_i>0$. Readers may also establish similar refinement and reverse like the proof above, replacing 1 with λ, for the following discussion.

Funds with better annual yield performance often have higher management fee rates, so it is important to consider the actual investment yields after paying fees. Here we just consider a simplified model.

Suppose the “better” fund X has the annual yield $(a_i+{\lambda }_i-1)·100\%$, ${\lambda }_i\ge 0$ for each year i, with a fixed annual management fee rate $\theta ·100\%$, and the “worse” fund Y has the annual yield $(a_i-1)·100\%$ for each year i, without management fee. Define $\lambda =min\{{\lambda }_i;i=1,\dots ,n\}$.

From (12) we have the following:

$$\begin{array}{c}\hfill {\left({\displaystyle \frac{A\left(a\right)}{A(\lambda +a)}}\right)}^n\prod _{i=1}^n(a_i+{\lambda }_i)\ge {\left({\displaystyle \frac{A\left(a\right)}{A(\lambda +a)}}\right)}^n\prod _{i=1}^n(a_i+\lambda )\ge \prod _{i=1}^n{a}_i,\end{array}$$

thus, we obtain the following:

$$\begin{array}{c}\hfill \prod _{i=1}^n\left[(a_i+{\lambda }_i)\left(1-{\displaystyle \frac{\lambda }{\lambda +A\left(a\right)}}\right)\right]\ge \prod _{i=1}^n{a}_i.\end{array}$$

It means, if the annual fee rate $\theta \le {\displaystyle \frac{\lambda }{\lambda +A\left(a\right)}}$, then the actual yield of the fund X must be higher than Y after n years.

Sometimes we even don’t need to know the actual value of each $a_i$. For example, if we know the worse Y has a negative yield each year, then we affirm $A\left(a\right)\le 1$; if we further know the annual fee rate of X satisfy $\theta \le {\displaystyle \frac{\lambda }{\lambda +1}}$, then we predict the actual yield of the fund X must be higher than Y.

Apply the refinement and reverse of Ky Fan-type inequality, we have more detailed prediction about, when the X has actual higher or lower yield than Y, concerning the annual fee rate θ. Left to readers.

For similar converses of Ky Fan inequality with the following form:

$${\displaystyle \frac{A\left(a\right)}{A(1-a)}}\le {\displaystyle \frac{G\left(a\right)}{G(1-a)}}·e^C$$

for some C concerning $a_i$, see [9,23].

Then we list some examples of distance functions, and use Theorem 6, Remark 4 to give inequalities for these distance functions.

Definition 1.

For the follwoing:

$$\varphi \left(t\right)=tlnt,t>0$$

the ϕ-divergence is calculated as follows:

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

the Kullback–Leibler distance.

We have the following estimation for two different pairs of Kullback–Leibler distances.

Theorem 10.

For $p_1,q_1,p_2,q_2\in {\mathbb{P}}_n$ with $p_1,q_1,p_2,q_2>0$, if condition in Lemma 1 or Lemma 2 is further satisfied, then the following is calculated:

$$\begin{array}{c}\hfill min\left\{{\displaystyle \frac{q_{2,i}}{p_{2,i}}}\right\}·\sum _{i=1}^n\left({\displaystyle \frac{p_{2,i}^2}{2q_{2,i}}}-{\displaystyle \frac{p_{1,i}^2}{2q_{1,i}}}\right)\\ \hfill \le \sum _{i=1}^n{p}_{2,i}ln\left({\displaystyle \frac{p_{2,i}}{q_{2,i}}}\right)-\sum _{i=1}^n{p}_{1,i}ln\left({\displaystyle \frac{p_{1,i}}{q_{1,i}}}\right)\\ \hfill \le max\left\{{\displaystyle \frac{q_{2,i}}{p_{2,i}}}\right\}·\sum _{i=1}^n\left({\displaystyle \frac{p_{2,i}^2}{2q_{2,i}}}-{\displaystyle \frac{p_{1,i}^2}{2q_{1,i}}}\right).\end{array}$$

Proof. 

As ${\varphi }^{″}\left(t\right)={\displaystyle \frac{1}{t}}$, we have the following:

$$\begin{array}{c}\hfill min\left\{{\displaystyle \frac{q_{j,i}}{p_{j,i}}}\right\}={\displaystyle \frac{1}{max\left\{\frac{p_{j,i}}{q_{j,i}}\right\}}}\le {\varphi }^{″}\le {\displaystyle \frac{1}{min\left\{\frac{p_{j,i}}{q_{j,i}}\right\}}}=max\left\{{\displaystyle \frac{q_{j,i}}{p_{j,i}}}\right\}.\end{array}$$

And under the assumption of Lemma 1 or Lemma 2, we have the following:

$$\begin{array}{c}\hfill min\left\{{\displaystyle \frac{q_{j,i}}{p_{j,i}}}\right\}=min\left\{{\displaystyle \frac{q_{2,i}}{p_{2,i}}}\right\},max\left\{{\displaystyle \frac{q_{j,i}}{p_{j,i}}}\right\}=max\left\{{\displaystyle \frac{q_{2,i}}{p_{2,i}}}\right\}.\end{array}$$

Then we can use Theorem 6 and Remark 4 to obtain the result. □

Definition 2.

Let the following be true:

$$\varphi \left(t\right)={\displaystyle \frac{1}{2}}{(1-\sqrt{t})}^2,t>0$$

the ϕ-divergence is as follows:

$$I_{\varphi }(p,q):={\displaystyle \frac{1}{2}}\sum _{i=1}^n{(\sqrt{q_i}-\sqrt{p_i})}^2,$$

the Hellinger distance.

We have the following estimation for two different pairs of Hellinger distance.

Theorem 11.

For $p_1,q_1,p_2,q_2\in {\mathbb{P}}_n$ with $p_1,q_1,p_2,q_2>0$, if condition in Lemma 1 or Lemma 2 is further satisfied, then the following is calculated:

$$\begin{array}{c}\hfill min\left\{{\displaystyle \frac{1}{4}}{\left({\displaystyle \frac{p_{2,i}}{q_{2,i}}}\right)}^{-3/2}\right\}·\sum _{i=1}^n\left({\displaystyle \frac{p_{2,i}^2}{2q_{2,i}}}-{\displaystyle \frac{p_{1,i}^2}{2q_{1,i}}}\right)\\ \hfill \le {\displaystyle \frac{1}{2}}\sum _{i=1}^n{(\sqrt{q_{2,i}}-\sqrt{p_{2,i}})}^2-{\displaystyle \frac{1}{2}}\sum _{i=1}^n{(\sqrt{q_{1,i}}-\sqrt{p_{1,i}})}^2\\ \hfill \le max\left\{{\displaystyle \frac{1}{4}}{\left({\displaystyle \frac{p_{2,i}}{q_{2,i}}}\right)}^{-3/2}\right\}·\sum _{i=1}^n\left({\displaystyle \frac{p_{2,i}^2}{2q_{2,i}}}-{\displaystyle \frac{p_{1,i}^2}{2q_{1,i}}}\right).\end{array}$$

Proof. 

As ${\varphi }^{″}\left(t\right)={\displaystyle \frac{t^{-3/2}}{4}}$, with similar discussion in Theorem above we obtain the result. □

Definition 3.

For $\alpha >1$, let the following be true:

$$\varphi \left(t\right)=t^{\alpha },t>0$$

the ϕ-divergence is written as follows:

$$I_{\varphi }(p,q):=\sum _{i=1}^n{p}_i^{\alpha }{q}_i^{1-\alpha },$$

the  $\alpha$-order entropy. And Rényi divergence of order α is defined by the following:

$$D_{\alpha }(p,q):={\displaystyle \frac{1}{\alpha -1}}ln\left(\sum _{i=1}^n{p}_i^{\alpha }{q}_i^{1-\alpha }\right).$$

We have the following estimation for two different pairs of $\alpha$-order entropy.

Theorem 12.

For $p_1,q_1,p_2,q_2\in {\mathbb{P}}_n$ with $p_1,q_1,p_2,q_2>0$, if condition in Lemma 1 or Lemma 2 is further satisfied, then the following is calculated:

$$\begin{array}{c}\hfill min\left\{\alpha (\alpha -1){\left({\displaystyle \frac{p_{2,i}}{q_{2,i}}}\right)}^{\alpha -2}\right\}·\sum _{i=1}^n\left({\displaystyle \frac{p_{2,i}^2}{2q_{2,i}}}-{\displaystyle \frac{p_{1,i}^2}{2q_{1,i}}}\right)\\ \hfill \le \sum _{i=1}^n{p}_{2,i}^{\alpha }{q}_{2,i}^{1-\alpha }-\sum _{i=1}^n{p}_{1,i}^{\alpha }{q}_{1,i}^{1-\alpha }\\ \hfill \le max\left\{\alpha (\alpha -1){\left({\displaystyle \frac{p_{2,i}}{q_{2,i}}}\right)}^{\alpha -2}\right\}·\sum _{i=1}^n\left({\displaystyle \frac{p_{2,i}^2}{2q_{2,i}}}-{\displaystyle \frac{p_{1,i}^2}{2q_{1,i}}}\right).\end{array}$$

Proof. 

As ${\varphi }^{″}\left(t\right)=\alpha (\alpha -1)t^{\alpha -2}$, with similar discussion in Theorem above we obtain the result. □

Definition 4.

Let the following be true:

$$\varphi \left(t\right)={(t-1)}^2,t>0$$

the ϕ-divergence is calculated as follows:

$$I_{\varphi }(p,q):=\sum _{i=1}^n{\displaystyle \frac{{(p_i-q_i)}^2}{q_i}},$$

the ${\chi }^2$-distance.

We have the following estimation for two different pairs of ${\chi }^2$-distance. This is a special case of identity.

Theorem 13.

For $p_1,q_1,p_2,q_2\in {\mathbb{P}}_n$ with $p_1,q_1,p_2,q_2>0$, if condition in Lemma 1 or Lemma 2 is further satisfied, then the following is calculated:

$$\begin{array}{c}\hfill \sum _{i=1}^n{\displaystyle \frac{{(p_{2,i}-q_{2,i})}^2}{q_{2,i}}}-\sum _{i=1}^n{\displaystyle \frac{{(p_{1,i}-q_{1,i})}^2}{q_{1,i}}}=\sum _{i=1}^n\left({\displaystyle \frac{p_{2,i}^2}{q_{2,i}}}-{\displaystyle \frac{p_{1,i}^2}{q_{1,i}}}\right).\end{array}$$

Proof. 

As ${\varphi }^{″}\left(t\right)=2$, the inequality in Theorem 6 becomes identity. □

Remark 8.

If we let $p_{1,i}=q_{1,i}$ in the four theorems above, we obtain estimation for these four types of divergence $I_{\varphi }(p_2,q_2)$.

4. Conclusions

In this paper, we give identities for convex functions on three intervals and the difference for Csiszár $\varphi$-divergence, which are Theorem 2 and Theorem 5. These two general identities can deduce some inequalities [19] by letting the outer function be convex. Then they are used to establish the extension for Levinson inequality [24,25,26,27], thus Ky Fan-type inequalities [2,5] can be refined and improved. Some special cases of distance function inequalities [20] for Csiszár $\varphi$-divergence can be proven.

5. Some Lemmas

The generalized Taylor-type expansions [28] are needed in this paper.

Theorem 14.

(i) Let $N,n\in \mathbb{N}$ and $f:I\to \mathbb{R}$ be a function such that $f^{(n-1)}$ is absolutely continuous on $I\subset \mathbb{R}$, $a,b\in I$, $a<b$. Furthermore, let $x_i\in [a,b]$ and $p_i\in \mathbb{R}$ for $i\in \{1,2,\dots ,N\}$. Then the following is calculated:

$$\begin{array}{cc}\hfill \sum _{i=1}^N{p}_{i}f\left(x_i\right)& =\sum _{k=0}^{n-1}{\displaystyle \frac{f^{\left(k\right)}\left(a\right)}{k!}}\sum _{i=1}^N{p}_i{(x_i-a)}^k\hfill \\ \hfill & +{\displaystyle \frac{1}{(n-1)!}}{\int }_a^b{f}^{\left(n\right)}\left(s\right)\left(\sum _{i=1}^N{p}_i{(x_i-s)}_{+}^{n-1}\right)ds.\hfill \end{array}$$

(13)

(ii) Let $p:[\alpha ,\beta ]\to \mathbb{R}$ and $g:[\alpha ,\beta ]\to [a,b]$ be integrable functions. Let f satisfy assumptions from part (i). Then the following is calculated:

$$\begin{array}{cc}\hfill {\int }_{\alpha }^{\beta }p\left(x\right)f\left(g\left(x\right)\right)dx& =\sum _{k=0}^{n-1}{\displaystyle \frac{f^{\left(k\right)}\left(a\right)}{k!}}{\int }_{\alpha }^{\beta }p\left(x\right){(g\left(x\right)-a)}^{k}dx\hfill \\ \hfill & +{\displaystyle \frac{1}{(n-1)!}}{\int }_a^b{f}^{\left(n\right)}\left(s\right){\int }_{\alpha }^{\beta }p\left(x\right){(g\left(x\right)-s)}_{+}^{n-1}dxds.\hfill \end{array}$$

(14)

Proof for Lemma 1

Proof. 

This can be divided into three situations. Notice that in the proof we will use the condition $p_1,q_1,p_2,q_2\in {\mathbb{P}}_n$ for several times, to obtain some identities.

1. For $s\in [0,m]$.

$$\begin{array}{c}\hfill \sum _{i=1}^n{(p_{2,i}-s{q}_{2,i})}_{+}-\sum _{i=1}^n{(p_{1,i}-s{q}_{1,i})}_{+}\\ \hfill =\sum _{i=1}^n(p_{2,i}-s{q}_{2,i})-\sum _{i=1}^n(p_{1,i}-s{q}_{1,i})-\sum _{i:{\displaystyle \frac{p_{2,i}}{q_{2,i}}}<s}(p_{2,i}-s{q}_{2,i})\\ \hfill =0-\sum _{i:{\displaystyle \frac{p_{2,i}}{q_{2,i}}}<s}(p_{2,i}-s{q}_{2,i})\ge 0.\end{array}$$

2. For $s\in [m,M]$.

$$\begin{array}{c}\hfill \sum _{i=1}^n{(p_{2,i}-s{q}_{2,i})}_{+}-\sum _{i=1}^n{(p_{1,i}-s{q}_{1,i})}_{+}\\ \hfill =\sum _{i:{\displaystyle \frac{p_{2,i}}{q_{2,i}}}\ge s}{q}_{2,i}\left({\displaystyle \frac{p_{2,i}}{q_{2,i}}}-s\right)-\sum _{i:{\displaystyle \frac{p_{1,i}}{q_{1,i}}}\ge s}{q}_{1,i}\left({\displaystyle \frac{p_{1,i}}{q_{1,i}}}-s\right).\end{array}$$

2.1. If the following is true:

$$\begin{array}{c}\hfill \sum _{i:{\displaystyle \frac{p_{2,i}}{q_{2,i}}}\ge s}{q}_{2,i}\ge \sum _{i:{\displaystyle \frac{p_{1,i}}{q_{1,i}}}\ge s}{q}_{1,i},\end{array}$$

then the following is calculated:

$$\begin{array}{c}\hfill \sum _{i:{\displaystyle \frac{p_{2,i}}{q_{2,i}}}\ge s}{q}_{2,i}\left({\displaystyle \frac{p_{2,i}}{q_{2,i}}}-s\right)-\sum _{i:{\displaystyle \frac{p_{1,i}}{q_{1,i}}}\ge s}{q}_{1,i}\left({\displaystyle \frac{p_{1,i}}{q_{1,i}}}-s\right)\\ \hfill \ge \sum _{i:{\displaystyle \frac{p_{2,i}}{q_{2,i}}}\ge s}{q}_{2,i}\left(M-s\right)-\sum _{i:{\displaystyle \frac{p_{1,i}}{q_{1,i}}}\ge s}{q}_{1,i}\left(M-s\right)\ge 0.\end{array}$$

2.2. If the following is true:

$$\begin{array}{c}\hfill \sum _{i:{\displaystyle \frac{p_{2,i}}{q_{2,i}}}\ge s}{q}_{2,i}<\sum _{i:{\displaystyle \frac{p_{1,i}}{q_{1,i}}}\ge s}{q}_{1,i},\end{array}$$

which indicates the following:

$$\begin{array}{c}\hfill \sum _{i:{\displaystyle \frac{p_{2,i}}{q_{2,i}}}<s}{q}_{2,i}>\sum _{i:{\displaystyle \frac{p_{1,i}}{q_{1,i}}}<s}{q}_{1,i},\end{array}$$

then the following is calculated:

$$\begin{array}{c}\hfill \sum _{i:{\displaystyle \frac{p_{2,i}}{q_{2,i}}}\ge s}{q}_{2,i}\left({\displaystyle \frac{p_{2,i}}{q_{2,i}}}-s\right)-\sum _{i:{\displaystyle \frac{p_{1,i}}{q_{1,i}}}\ge s}{q}_{1,i}\left({\displaystyle \frac{p_{1,i}}{q_{1,i}}}-s\right)\\ \hfill =-\sum _{i:{\displaystyle \frac{p_{2,i}}{q_{2,i}}}<s}{q}_{2,i}\left({\displaystyle \frac{p_{2,i}}{q_{2,i}}}-s\right)+\sum _{i:{\displaystyle \frac{p_{1,i}}{q_{1,i}}}<s}{q}_{1,i}\left({\displaystyle \frac{p_{1,i}}{q_{1,i}}}-s\right)\\ \hfill \ge \sum _{i:{\displaystyle \frac{p_{2,i}}{q_{2,i}}}<s}{q}_{2,i}\left(s-m\right)-\sum _{i:{\displaystyle \frac{p_{1,i}}{q_{1,i}}}<s}{q}_{1,i}\left(s-m\right)\ge 0.\end{array}$$

3. For $s\in [M,\infty )$.

$$\begin{array}{c}\hfill \sum _{i=1}^n{(p_{2,i}-s{q}_{2,i})}_{+}-\sum _{i=1}^n{(p_{1,i}-s{q}_{1,i})}_{+}=\sum _{i=1}^n{(p_{2,i}-s{q}_{2,i})}_{+}\ge 0.\end{array}$$

□

Proof for Lemma 2:

Proof. 

As ${\displaystyle \frac{p_2}{q}}$ weighted majorizes ${\displaystyle \frac{p_1}{q}}$ with the weight q, an equivalent condition is written as follows:

$$\begin{array}{c}\hfill \sum _{i=1}^n{q}_i·{\displaystyle \frac{p_{2,i}}{q_i}}=\sum _{i=1}^n{q}_i·{\displaystyle \frac{p_{1,i}}{q_i}},\\ \hfill \sum _{i=1}^n{q}_i{\left({\displaystyle \frac{p_{2,i}}{q_i}}-s\right)}_{+}\ge \sum _{i=1}^n{q}_i{\left({\displaystyle \frac{p_{1,i}}{q_i}}-s\right)}_{+}\end{array}$$

for all $s\in \mathbb{R}$, see chapter 4 in [29]. □