A new version of Jensen’s inequality and related results

Abstract

In this paper we expand Jensen’s inequality to two-variable convex functions and find the lower bound of the Hermite-Hadamard inequality for a convex function on the bounded area from the plane.

1 Introduction

Let μ be a positive measure on X such that $\mu (X)=1$. If f is a real-valued function in $L^1(\mu )$, $a<f(x)<b$ for all $x\in X$ and φ is convex on $(a,b)$, then

$$\phi \left({\int }_{X}fd\mu \right)\le {\int }_X(\phi \circ f)d\mu .$$

(1)

The inequality (1) is known as Jensen’s inequality [1].

In recent years, there have been many extensions, refinements and similar results of the inequality (1). Recall that the function $F:\mathrm{\Delta }=[a,b]\times [c,d]\to \mathbb{R}$ is convex on Δ if

$$F(\lambda x+(1-\lambda )z,\lambda y+(1-\lambda )w)\le \lambda F(x,y)+(1-\lambda )F(z,w)$$

holds for all $(x,y),(z,w)\in \mathrm{\Delta }$ and $\lambda \in [0,1]$. A function $F:\mathrm{\Delta }\to \mathbb{R}$ is called co-ordinated convex on Δ if the partial functions $F_y:[a,b]\to \mathbb{R}$, $F_y(u)=F(u,y)$ and $F_x:[c,d]\to \mathbb{R}$, $F_x(v)=F(x,v)$ are convex for all $x\in [a,b]$ and $y\in [c,d]$. Note that every convex function $F:\mathrm{\Delta }\to \mathbb{R}$ is co-ordinated convex, but the converse is not generally true; see [2]. Also note that if F is a convex function on ${\mathbb{R}}^2$ and g, h are real-valued functions such that $D_g=D_h=\mathbb{R}$, then $f(t)=F(g(t),h(t))$ may be not convex on ℝ.

In this paper under suitable conditions, we expand Jensen’s inequality to two-variable convex functions and deduce some further important inequalities. Finally, we find a lower bound for the integral

$$\frac{1}{{\int }_a^b(g(x)-h(x))dx}{\int }_a^b{\int }_{h(x)}^{g(x)}F(x,y)dydx,$$

where F is convex on the convex bounded area by $y=g(x)$, $y=h(x)$ and $x=a$, $x=b$.

2 Main results

Theorem 1 Let p be a non-negative continuous function on $[a,b]$ such that ${\int }_a^{b}p(x)dx>0$. If g and h are real-valued continuous functions on $[a,b]$ and

$$m_1\le g(x)\le M_1,m_2\le h(x)\le M_2$$

for all $x\in [a,b]$, and F is convex on

$$\mathrm{\Delta }=[m_1,M_1]\times [m_2,M_2],$$

then

$$F(\frac{{\int }_a^{b}g(t)p(t)dt}{{\int }_a^{b}p(t)dt},\frac{{\int }_a^{b}h(t)p(t)dt}{{\int }_a^{b}p(t)dt})\le \frac{{\int }_a^{b}F(g(t),h(t))p(t)dt}{{\int }_a^{b}p(t)dt}$$

(2)

and

$$F(\frac{{\int }_a^{b}g(t)dt}{b-a},\frac{{\int }_a^{b}h(t)dt}{b-a})\le \frac{1}{b-a}{\int }_a^{b}F(g(t),h(t))dt.$$

(3)

The inequalities hold in reversed order if f is concave on Δ.

Proof Denote

$$\alpha (x)=\frac{{\int }_a^{x}g(t)p(t)dt}{{\int }_a^{x}p(t)dt}$$

and

$$\beta (x)=\frac{{\int }_a^{x}h(t)p(t)dt}{{\int }_a^{x}p(t)dt}.$$

Then by L’Hospital’s rule, we have ${lim}_{x\to a}\alpha (x)=g(a)$ and ${lim}_{x\to a}\beta (x)=h(a)$. So, α and β are continuous on $[a,b]$. Denote

$$H(x)=F(\alpha (x),\beta (x))-\frac{{\int }_a^{x}F(g(t),h(t))p(t)dt}{{\int }_a^{x}p(t)dt}.$$

We will show that $H(b)\le 0$. We have

$$\begin{array}{rcl}{H}^{\prime }(x)& =& \frac{\partial F(\alpha (x),\beta (x))}{\partial \alpha }{\alpha }^{\prime }(x)+\frac{\partial F(\alpha (x),\beta (x))}{\partial \beta }{\beta }^{\prime }(x)\\ -\frac{F(g(x),h(x))p(x)}{{\int }_a^{x}p(t)dt}+p(x)\frac{{\int }_a^{x}F(g(t),h(t))p(t)dt}{{({\int }_a^{x}p(t)dt)}^2}.\end{array}$$

By the convexity of F, we obtain

$$\begin{array}{rcl}F(g(x),h(x))-F(\alpha (x),\beta (x))& \ge & \frac{\partial F(\alpha (x),\beta (x))}{\partial \alpha }(g(x)-\alpha (x))\\ +\frac{\partial F(\alpha (x),\beta (x))}{\partial \beta }(h(x)-\beta (x)).\end{array}$$

So, we get

$$\begin{array}{rcl}{H}^{\prime }(x)& \le & \frac{\partial (\alpha (x),\beta (x))}{\partial \alpha }{\alpha }^{\prime }(x)+\frac{\partial (\alpha (x),\beta (x))}{\partial \beta }{\beta }^{\prime }(x)\\ -\frac{p(x)}{{\int }_a^{x}p(t)dt}[F(\alpha (x),\beta (x))+\frac{\partial (\alpha (x),\beta (x))}{\partial \alpha }(g(x)-\alpha (x))\\ +\frac{\partial F(\alpha (x),\beta (x))}{\partial \beta }(h(x)-\beta (x))]+p(x)\frac{{\int }_a^{x}F(g(t),h(t))p(t)dt}{{({\int }_a^{x}p(t)dt)}^2}.\end{array}$$

Hence,

$$\begin{array}{rcl}{H}^{\prime }(x)& \le & \frac{\partial (\alpha (x),\beta (x))}{\partial \alpha }[{\alpha }^{\prime }(x)-\frac{p(x)}{{\int }_a^{x}p(t)dt}(g(x)-\alpha (x))]\\ +\frac{\partial (\alpha (x),\beta (x))}{\partial \beta }[{\beta }^{\prime }(x)-\frac{p(x)}{{\int }_a^{x}p(t)dt}(h(x)-\beta (x))]\\ -\frac{p(x)F(\alpha (x),\beta (x))}{{\int }_a^{x}p(t)dt}+p(x)\frac{{\int }_a^{x}F(g(t),h(t))g(t)dt}{{({\int }_a^{x}p(t)dt)}^2}.\end{array}$$

By easy calculation, we see that

$${\alpha }^{\prime }(x)-\frac{p(x)}{{\int }_a^{x}p(t)dt}(g(x)-\alpha (x))={\beta }^{\prime }(x)-\frac{p(x)}{{\int }_a^{x}p(t)dt}(h(x)-\beta (x))=0.$$

Therefore,

$$H^{\prime }(x)\le -\frac{p(x)}{{\int }_a^{x}p(t)dt}[F(\alpha (x),\beta (x))-\frac{{\int }_a^{x}F(g(t),h(t))p(t)dt}{{\int }_a^{x}p(t)dt}]=-\frac{p(x)}{{\int }_a^{x}P(t)dt}H(x).$$

Thus,

$$({\int }_a^{x}p(t)dt)H^{\prime }(x)+p(x)H(x)\le 0\Rightarrow {[({\int }_a^{x}p(t)dt)H(x)]}^{\prime }\le 0.$$

So,

$$({\int }_a^{b}p(t)dt)H(b)\le 0\Rightarrow H(b)\le 0.$$

The proof is complete. For the proof of (3), set $p(x)=1$.

Note the inequalities (2) and (3) are sharp because $F(x,y)=1$. □

Corollary 1 Let g and h be real-valued continuous functions. Then we have

• (i) for $\frac{1}{p}+\frac{1}{q}=1$, $p,q>1$,

  $${\int }_a^b|g(t)||h(t)|dt\le {\left({\int }_a^b{|g(t)|}^{p}dt\right)}^{\frac{1}{p}}{\left({\int }_a^b{|h(t)|}^{q}dt\right)}^{\frac{1}{q}}\mathit{\text{Holder’s inequality}},$$

• (ii) for $p\ge 1$,

  

• (iii) for $p\ge 1$,

  $$\begin{array}{rcl}{\left({\int }_a^b{|g(t)+h(t)|}^{p}dt\right)}^{\frac{1}{p}}& \le & {\left({\int }_a^b{|g(t)|}^{p}dt\right)}^{\frac{1}{p}}\\ +{\left({\int }_a^b{|h(t)|}^{p}dt\right)}^{\frac{1}{p}}\mathit{\text{Minkowski’s inequality}},\end{array}$$

  (iv)

  $$ln(e^{\frac{1}{b-a}{\int }_a^{b}g(t)dt}+e^{\frac{1}{b-a}{\int }_a^{b}h(t)dt})\le \frac{1}{b-a}{\int }_a^{b}ln(e^{g(t)}+e^{h(t)})dt.$$

Proof (i) The function

$$F(x,y)={|x|}^{\frac{1}{p}}{|y|}^{\frac{1}{q}}(\frac{1}{p}+\frac{1}{q}=1),$$

is concave, so by the inequality (3), we have

$$\frac{{({\int }_a^b|g(t)|dt)}^{\frac{1}{p}}}{{(b-a)}^{\frac{1}{p}}}\times \frac{{({\int }_a^b|h(t)|dt)}^{\frac{1}{q}}}{{(b-a)}^{\frac{1}{q}}}\ge \frac{{\int }_a^b{|g(t)|}^{\frac{1}{p}}{|h(t)|}^{\frac{1}{q}}dt}{b-a}.$$

Hence,

$${\int }_a^b{|g(t)|}^{\frac{1}{p}}{|h(t)|}^{\frac{1}{q}}dt\le {\left({\int }_a^b|g(t)|dt\right)}^{\frac{1}{p}}{\left({\int }_a^b|h(t)|dt\right)}^{\frac{1}{q}}.$$

Now, set $|g(t)|\to {|g(t)|}^p$ and $|h(t)|\to {|h(t)|}^q$. We obtain

$${\int }_a^b|g(t)||h(t)|dt\le {\left({\int }_a^b{|g(t)|}^{p}dt\right)}^{\frac{1}{p}}+{\left({\int }_a^b{|h(t)|}^{q}dt\right)}^{\frac{1}{q}}.$$

• (ii) The function $F(x,y)={({|x|}^p+{|y|}^p)}^{\frac{1}{p}}$ is convex for $p\ge 1$ and is concave for $p<1$. So, by the inequality (3), we have

  $${[\frac{{({\int }_a^b|g(t)|dt)}^p}{{(b-a)}^p}+\frac{{({\int }_a^b|h(t)|dt)}^p}{{(b-a)}^p}]}^{\frac{1}{p}}\le \frac{{\int }_a^b{({|g(t)|}^p+{|h(t)|}^p)}^{\frac{1}{p}}dt}{b-a}$$

so

$${\int }_a^b{({|g(t)|}^p+{|h(t)|}^p)}^{\frac{1}{p}}dt\ge {[{\left({\int }_a^b|g(t)|dt\right)}^p+{\left({\int }_a^b|h(t)|dt\right)}^p]}^{\frac{1}{p}}.$$

Now, set $|g(t)|\to {|g(t)|}^{\frac{1}{p}}$ and $|h(t)|\to {|h(t)|}^{\frac{1}{p}}$. We get

$${\int }_a^b{(|g(t)|+|h(t)|)}^{\frac{1}{p}}dt\ge {[{\left({\int }_a^b{|g(t)|}^{\frac{1}{p}}dt\right)}^p+{\left({\int }_a^b{|h(t)|}^{\frac{1}{p}}dt\right)}^p]}^{\frac{1}{p}}.$$

So,

$${\left({\int }_a^b{(|g(t)|+|h(t)|)}^{\frac{1}{p}}dt\right)}^p\ge {\left({\int }_a^b{|g(t)|}^{\frac{1}{p}}dt\right)}^p+{\left({\int }_a^b{|h(t)|}^{\frac{1}{p}}dt\right)}^p.$$

The proof of (iii) is similar to that of (ii) and can be omitted. For the proof of (iv), note $f(x,y)=ln(e^x+e^y)$ is convex on ${\mathbb{R}}^2$. Now, apply the inequality (3). □

Remark 1 By similar assumptions, we can prove Theorem 1 for an n-variable convex function F on ${\mathbb{R}}^n$ and obtain the inequality

$$F(\frac{{\int }_a^b{g}_1(t)dt}{b-a},\dots ,\frac{{\int }_a^b{g}_n(t)dt}{b-a})\le \frac{1}{b-a}{\int }_a^{b}F(g_1(t),\dots ,g_n(t))dt.$$

In particular, we can obtain a similar inequality for Holder and Minkowski inequalities. For example, by the concavity of

$$F(t_1,t_2,\dots ,t_n)=\prod _{i=1}^n{|t_i|}^{\frac{1}{p_i}}(\sum _{i=1}^n\frac{1}{pi}=1),$$

we can get the inequality

$${\int }_a^b(\prod _{i=1}^n|g_i|)dt\le \prod _{i=1}^n{\left({\int }_a^b{|g_i|}^{p_i}\right)}^{\frac{1}{p_i}}.$$

3 Hermite-Hadamard inequality

Let $f:[a,b]\to \mathbb{R}$ be a convex function, then the following inequality is known as the Hermite-Hadamard inequality [3] and [4]:

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f(x)dx\le \frac{f(a)+f(b)}{2}.$$

(4)

In [5], Dragomir established the following similar inequality (4) for convex functions on the co-ordinates on a rectangle from the plane ${\mathbb{R}}^2$.

Theorem 2 Suppose $f:\mathrm{△}=[a,b]\times [c,d]\to \mathbb{R}$ is a convex function on the co-ordinates on △. Then one has the inequalities

$$\begin{array}{rcl}f(\frac{a+b}{2},\frac{c+d}{2})& \le & \frac{1}{(b-a)(d-c)}{\int }_a^b{\int }_c^{d}f(x,y)dydx\\ \le & \frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}.\end{array}$$

Also Dragomir investigated the Hermite-Hadamard inequality on the disk [6] and [7].

In [8], Matejíčka proved the left-hand side of the Hermite-Hadamard inequality of several variables for a convex function on certain convex compact sets. In the following theorem, we prove the left-hand side of the Hermite-Hadamard inequality in another way and as a result of Theorem 2.

Theorem 3 Let △ be a bounded area by a convex function h and a concave function g on $[a,b]$ such that for any $x\in [a,b]$, $g(x)\ge h(x)$. Also, let F be a two-variable convex function on △. Then one has the inequality

$$F(\frac{{\int }_a^{b}x(g(x)-h(x))dx}{{\int }_a^b(g(x)-h(x))dx},\frac{\frac{1}{2}{\int }_a^b(g^2(x)-h^2(x))dx}{{\int }_a^b(g(x)-h(x))dx})\le \frac{{\int }_a^b{\int }_{h(x)}^{g(x)}F(x,y)dydx}{{\int }_a^b(g(x)-h(x))dx}.$$

Proof Since F is convex on △, hence f is co-ordinated convex on △. So, $F_x:[h(x),g(x)]\to \mathbb{R}$, $F_x(y)=F(x,y)$ is convex on $[h(x),g(x)]$ for all $x\in [a,b]$. By the left-hand side of the Hermite-Hadamard inequality (4), we have

$$(g(x)-h(x))F(x,\frac{g(x)+h(x)}{2})\le {\int }_{h(x)}^{g(x)}F(x,y)dy.$$

Integrating this inequality on $[a,b]$, we obtain

$${\int }_a^b(g(x)-h(x))F(x,\frac{g(x)+h(x)}{2})dx\le {\int }_a^b{\int }_{h(x)}^{g(x)}F(x,y)dydx.$$

So,

$$\frac{{\int }_a^b(g(x)-h(x))F(x,\frac{g(x)+h(x)}{2})dx}{{\int }_a^b(g(x)-h(x))dx}\le \frac{1}{{\int }_a^b(g(x)-h(x))dx}{\int }_a^b{\int }_{h(x)}^{g(x)}F(x,y)dydx.$$

Now, let $p(x)=g(x)-h(x)$. By the inequality (2), we get

$$\begin{array}{rcl}F(\frac{{\int }_a^{b}x(g(x)-h(x))dx}{{\int }_a^b(g(x)-h(x))dx},\frac{\frac{1}{2}{\int }_a^b(g^2(x)-h^2(x))dx}{{\int }_a^b(g(x)-h(x))dx})& \le & \frac{{\int }_a^b(g(x)-h(x))F(x,\frac{g(x)+h(x)}{2})dx}{{\int }_a^b(g(x)-h(x))dx}\\ \le & \frac{{\int }_a^b(g(x)-h(x))F(x,\frac{g(x)+h(x)}{2})dx}{{\int }_a^b(g(x)-h(x))dx}.\end{array}$$

The proof is complete. □