A Note on the Jensen-Hadamard Inequality

Revue d’analyse numerique et de theorie de l’approximation

Tome 19, No.1, 1990, pp.29-34

1. J.L.W.V.Jensen [4], [5] was the first mathematician who discovered the great importance and perspective of convex functions. among many important general inequalities (See also [9], [10]) his results contain as a special case the following relations:

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

for a continuous and convex function $f:[a,b]\to ℝ$.

Inequality (1) was proves independently by J. Hadamard [2], under a slightly stronger condition: by supposing that $f$ has an increasing derivative on $[a,b]$. We note that the left side of (1) is called sometimes as the ”Hadamard inequality”, while the right side as the ”Jensen inequality” (Or vice-versa). There are also some papers which attribute inequality (1) completely to J. Hadamard. In our opinion, a more penetrating study on the history and priority related to convex functions justifies to call (1) as the ”Jensen-Hadamard inequality”.

In [3] I.B. Lacković has obtained the following generalization of (1):

$f\left({\displaystyle \frac{1}{2n}}{\displaystyle \sum _{k=1}^{2n}}{a}_k\right)$ $\le {\displaystyle \frac{1}{n}}({\displaystyle \frac{1}{a_2-a_1}}{\displaystyle {\int }_{a_1}^{a_1}}f\left(x\right)dx+\mathrm{\dots }+$ (2)

$+{\displaystyle \frac{1}{a_{2^n}-a_{2_{n-1}}}}{\displaystyle {\int }_{2^{n-1}}^{a_{2^n}}})$ $\le {\displaystyle \frac{1}{2n}}{\displaystyle \sum _{k=1}^{2n}}f\left(a_k\right),$

where $f:[a,b]\to ℝ$ has an increasing derivative on $[a,b]$ and $a_i\in [a,b]\left(i=1,2,\mathrm{\dots },2n\right)$ with .

We notice that, in fact (2) is valid for $f$ continuous and convex, since the proof is based essentially on (1).

A very general extension for the right-habd side of (1) was proved by A. Lupaş [8]. Let $C[a,b]$ denote the normed linear space of all funcitons $f:[a,b]\to ℝ$ which are continuous on $[a,b]$ and let $F:C[a,b]\to ℝ$ be a positive linear functional with $F\left(e_0\right)=1$ (which $e_0\left(t\right)=1,t\to [a,b]$). If $f\in C[a,b]$ is a convex funciton on $[a,b]$ then $x_0\in [a,b]$, where $x_0=F\left(e_1\right)$ (with $e_1\left(t\right)=t\in [a,b]$) and

$$f_0\le F\left(f\right)$$ (3)

When we consider the functional of the form $F_1\left(f\right)={\sum }_{j=1}^n{w}_{j}f\left(x_j\right)$ with $w_j\ge 0,F_1\left(e_0\right)=1$ and one finds:

$$f\left(\sum _{j=1}^n{w}_j{x}_j\right)\le \sum _{j=1}^n{w}_{j}f\left(x_j\right)$$ (4)

for $x_j\in [a,b],w_j\ge 0,(j=1,2,\mathrm{\dots },n),{\sum }_{j=1}^n{w}_j=1$. This is the classical Jensen inequality.

By considering $F_2\left(f\right)=\frac{1}{y-x}{\int }_x^{y}f\left(t\right)𝑑t,x,y$ being arbitrary distinct points from $[a,b]$, one of us [11] has proved the following generalization, important in applications. Let $f:[a,b]\to R$ be a $2k$-times differentiable funciton having continuous $2kth$ derivative on $[a,b]$ and satisfying $f^{\left(2^k\right)}\left(t\right){\ge }_{\left(>\mathrm{?}\mathrm{?}\mathrm{?}\mathrm{?}\mathrm{?}\right)}0$ for $t\in (a,b)$. Then one has the inequality:

$${\int }_a^{b}f\left(x\right)𝑑s{\ge }_{\left(>\mathrm{?}\mathrm{?}\right)}\sum _{p=1}^k\frac{{\left(b-a\right)}^{2p-1}}{2^{2p-2}\cdot \left(2p-1\right)!}{f}^{\left(2p-1\right)}\left(\frac{a+b}{2}\right)$$ (5)

For other generalizations and new proofs, see e.g. [7], [14], [15]. For applications in analysis and number theory, see e.g. [2], [11], [12], [13].

2. In what follows, our aim is to prove a discrete analogous of (1) as well as to obtain some refinements of the Jensen-Hadamard inequalitiy. At the end we will give an application of one of these refinements in the theory of Euler’s Gamma function.

Theorem 1

Let: $I\to ℝ\left(I\subset R,\text{interval}\right)$ ve a continuous convex function and let $a,b\in I,n\in N^{\ast }$. Then holds true the following inequality

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{n}\sum _{i=1}^{n}f\left(\frac{1}{n+1}\right)a+(1-\frac{i}{n+1})b)\le \frac{f\left(a\right)+f\left(b\right)}{2}$$ (6)

Proof. Using relation (4) with $w_j=\frac{1}{n},j=1,2,\mathrm{\dots },n$, one finds successively:

$f\left({\displaystyle \frac{\left({\sum }_{i=1}^n{m}_i\right)a+{\sum }_{i=1}^n\left(1-m_i\right)b}{n}}\right)$ $\le {\displaystyle \frac{1}{n}}{\displaystyle \sum _{i=1}^n}f\left(m_{i}a+\left(1-m_i\right)b\right)\le$

$\le {\displaystyle \frac{\left({\sum }_{i=1}^{n}m\right)f\left(a\right)+{\sum }_{i=1}^n\left(1-m_i\right)f\left(b\right)}{n}}$

with $m_i>0\left(i=1,2,\mathrm{\dots },m\right)$ arbitrary positive real nubmers. Select $m_i=\frac{1}{n+1}$ and observe that ${\sum }_{i=1}^n{m}_i={\sum }_{i=1}^n\left(1-m_i\right)=\frac{n}{2}$. This gives immediately (6).

For $n=2$ we obtain:  

Corollary 2

If $f:I\to ℝ$ is continuous and covnex, then for all a, $b\in I$ one has

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{2}\left[f\left(\frac{a+2b}{3}\right)+f\left(\frac{2a+b}{3}\right)\right]\le \frac{f\left(a\right)+f\left(b\right)}{2}$$ (7)

For an application, choose $f\left(t\right)=-\log t,t>0$. We get the following simple refinement of the arithmetic-geometirc inequality:

$$\frac{a+b}{2}\ge \frac{1}{3}\sqrt{\left(a+2b\right)\left(2a+b\right)}\ge \sqrt{ab},a,b>0$$ (8)

For $f\left(t\right)=\frac{1}{t},t>0$ we have:

$$\frac{1}{A}\le \frac{9\left(a+b\right)}{2\left(a+2b\right)\left(2a+b\right)}\le \frac{1}{H}$$ (9)

where $A$ and $H$ denote the arithmetic and harmonic means of $a,b$, respectively.

The next result offers a refinement of the Jensen-Hadamard inequality.

Theorem 3

Let $f:I\to ℝ$ be a continuous convex function and $n\in N^{\ast }$. Then one has the following inequalities:

$f\left({\displaystyle \frac{a+b}{2}}\right)$ $\le {\displaystyle \frac{{\int }_a^b\mathrm{\cdots }{\int }_a^{b}f\left({\sum }_{i=1}^{n+1}{x}_i/\left(n+1\right)\right)𝑑x_1,\mathrm{\dots }d{x}_{n+1}}{{\left(b-a\right)}^{n+1}}}\le$ (10)

$\le {\displaystyle \frac{{\int }_a^b\mathrm{\cdots }{\int }_a^b\left({\sum }_{i=1}^n{x}_i/n\right)𝑑x_1,\mathrm{\dots }d{x}_n}{{\left(b-a\right)}^n}}\le \mathrm{\dots }\le {\displaystyle \frac{{\int }_a^b{\int }_a^{b}f\left(\left(x_1+x_2\right)/2\right)𝑑x_1𝑑x_2}{{\left(b-a\right)}^2}}\le$

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

Proof. By Jensen’s inewquality (See (4) we can write):

${\displaystyle \frac{1}{n+1}}[f\left({\displaystyle \frac{x_1+\mathrm{\dots }+x_n}{n}}\right)+f\left({\displaystyle \frac{x_2+\mathrm{\dots }+x_{n+1}}{n}}\right)+\mathrm{\dots }+$

$+f\left({\displaystyle \frac{x_{n+1}+x_1+\mathrm{\dots }+x_{m-1}}{n}}\right)]$ $\ge$

$f\left({\displaystyle \frac{\frac{x_1+\mathrm{\dots }+x_n}{n}+\mathrm{\dots }+\frac{x_{n+1}+\mathrm{\dots }+x_{n-1}}{n}}{n+1}}\right)$ $=f\left({\displaystyle \frac{x+\mathrm{\dots }+x_{n+1}}{n+1}}\right)$

so by integration on ${[a,n]}^{n+1}$, we easily obtain

${\displaystyle {\int }_a^b}dt{\displaystyle {\int }_a^b}\mathrm{\cdots }{\displaystyle {\int }_a^b}f\left({\displaystyle \frac{x_1+\mathrm{\dots }+x_n}{n}}\right)d{x}_1\mathrm{\dots }d{x}_n\cdot$

$\ge {\displaystyle {\int }_a^b}\mathrm{\cdots }{\displaystyle {\int }_a^b}f\left({\displaystyle \frac{x_1+\mathrm{\dots }+x_{n+1}}{n+1}}\right)𝑑x_1\mathrm{\dots }𝑑x_{n+1},$

that ism the middle inequalities in (10).  

On the other hand, it is well-known that for a continuous convex function $f$, one has $f\left(u\right)-f\left(v\right)\ge \left(u-v\right)f_{+}^{\prime }\left(v\right),u,v\in I$, where $f_{+}^{\prime }$ dentoes the right derivative of $f$. Choose $u=\left(x_1+\mathrm{\dots }+x_{n+1}\right)\backslash \left(n+1\right),v=\left(a+b\right)/2$ and integrate the obtained inequality on ${[a,b]}^{n+1}$. Since

$${\int }_a^b{\int }_a^b\mathrm{\cdots }{\int }_a^b\frac{x_1+\mathrm{\dots }+x_{n+1}}{n+1}𝑑x_1\mathrm{\dots }𝑑x_{n+1}={\left(b-a\right)}^{n+1}\left(\frac{a+b}{2}\right),$$

we have obtained the first inequality of (10), which concludes the proof of Theorem 3 For $n=1$ one gets:

Corollary 4

If $f:I\to R$ is continuous and convex, then for all one has

$$f\left(\frac{\alpha +b}{2}\right)\le {\int }_a^b{\int }_a^b\frac{f\left(\frac{x+y}{2}\right)dxdy}{{\left(b-a\right)}^2}\le {\int }_a^b\frac{f\left(x\right)dx}{\left(b-a\right)}\le \frac{f\left(a\right)+f\left(b\right)}{2}$$ (11)

For an application, set $f\left(t\right)=-{\left(\log \mathrm{\Gamma }\left(t\right)\right)}^{\prime }=-\psi \left(t\right)$, where $\mathrm{\Gamma }$ and $\psi$ are the Euler gamma and digamma functions, respectively. It is well known ([1], [12], [16]) that for $t>0,$ thus $f$ is convex. Clearly,

$${\int }_a^b\psi \left(\frac{x+y}{2}\right)𝑑x=2\log \mathrm{\Gamma }\left(\frac{a+b}{2}\right)-2\log \mathrm{\Gamma }\left(\frac{a+y}{2}\right)$$

so (11) may be written also in the form

$\psi \left({\displaystyle \frac{a+b}{2}}\right)$ $>{\displaystyle \frac{4}{{\left(b-a\right)}^2}}\left[{\displaystyle {\int }_{a+b}^a}\log \mathrm{\Gamma }\left(t\right)𝑑t-{\displaystyle {\int }_a^{\frac{a+b}{2}}}\log \mathrm{\Gamma }\left(t\right)𝑑t\right]>$

$>\log {\displaystyle \frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}}>{\displaystyle \frac{\psi \left(a\right)+\psi \left(b\right)}{2}}.$

For $a=x+s,b=x+1$ this contains a generalization and refinement of a result by D. Kershaw [6].

ACKNOWLEDGEMENTS. The authors wist to thank Prof. Dr. B.Crstici and Conf. Dr. A. Lupaş for some reprints and information related to the Jensen-Hadamard inequality.

Received 30.X.1989        Băile Herculae

Jud. Caraş-Severin

România

Faculty of Technology

Zagreb

Yugoslavia

4336 Forţeni Nr. 79

Jud. Harghita

România

Received 10 II 1993        Department of Mathematics

Timişoara University

B-dul V. Pârvan

R-1900 Timişoara, România