Not to be confused with Hadamard's inequality.

In mathematics, the Hermite–Hadamard inequality, named after Charles Hermite and Jacques Hadamard and sometimes also called Hadamard's inequality, states that if a function ƒ : [ab] → R is convex, then the following chain of inequalities hold:

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

One of the most natural extensions of the right side of this classical inequality is due to Zoltán Retkes. For to formulate this result one has to introduce the notion of iterated integrals.

## The concept of the sequence of iterated integrals

Suppose that −∞ < a < b < ∞, and let f:[a, b] → be an integrable real function. Under the above conditions the following sequence of functions is called the sequence of iterated integrals of f,where asb.:

\begin{align} F^{(0)}(s) & := f(s), \\ F^{(1)}(s) & := \int^s_a F^{(0)}(u)du=\int^s_a f(u)du, \\ F^{(2)}(s) & := \int^s_a F^{(1)}(u)du=\int^s_a \left( \int^t_a f(u)du \right ) \, dt, \\ & \ \ \vdots \\ F^{(n)}(s) & := \int^s_a F^{(n-1)}(u) \, du, \\ & {}\ \ \vdots \end{align}

### Example 1

Let [a, b] = [0, 1] and f(s) ≡ 1. Then the sequence of iterated integrals of 1 is defined on [0, 1], and

\begin{align} F^{(0)}(s) & = 1, \\ F^{(1)}(s) & = \int^s_0 F^{(0)}(u) \, du=\int^s_0 1 \, du=s, \\ F^{(2)}(s) & = \int^s_0 F^{(1)}(u)du=\int^s_0 u \, du={s^2 \over 2}, \\ & {} \ \ \vdots \\ F^{(n)}(s) & := \int^s_0 {u^{n-1}\over (n-1)!}du={s^n \over n!}, \\ & {} \ \ \vdots \end{align}

### Example 2

Let [a,b] = [−1,1] and f(s) ≡ 1. Then the sequence of iterated integrals of 1 is defined on [−1, 1], and

\begin{align} F^{(0)}(s) & = 1, \\ F^{(1)}(s) & = \int^s_{-1} F^{(0)}(u) \, du=\int^s_{-1} 1 du=s+1, \\ F^{(2)}(s) & = \int^s_{-1} F^{(1)}(u)du=\int^s_{-1} (u+1) \, du={s^2 \over 2!}+{s \over 1!}+{1 \over 2!}={(s+1)^2 \over 2!}, \\ & {} \ \vdots \\ F^{(n)}(s) & = {s^n \over n!}+{s^{n-1}\over {(n-1)!1!}}+{s^{n-2} \over (n-2)!2!}+ \dots +{1 \over n!} ={(s+1)^n \over n!}, \\ & {} \ \vdots \end{align}

### Example 3

Let [a, b] = [0, 1] and f(s) = es. Then the sequence of iterated integrals of f is defined on [0, 1], and

\begin{align} F^{(0)}(s) & = e^s, \\ F^{(1)}(s) & = \int^s_0 F^{(0)}(u)du=\int^s_0 e^u du=e^s-1, \\ F^{(2)}(s) & = \int^s_0 F^{(1)}(u)du=\int^s_0 (e^u-1) du=e^s-s-1, \\ & {} \ \vdots \\ F^{(n)}(s) & = e^s-\sum_{i=0}^{n-1}\frac {s^i}{i!} \\ & {}\ \vdots \end{align}

### Theorem (Retkes inequality)

Suppose that −∞ < a < b < ∞, and let f:[a,b]→R be a convex function, a < xi < b, i = 1, ..., n, such that xixj, if ij. Then the following holds:

$\sum_{i=1}^n \frac {F^{(n-1)}(x_i)}{\Pi_i(x_1,\dots,x_n)}\leq \frac {1}{n!} \sum_{i=1}^n f(x_i)$

where

$\Pi_i(x_1,\dots,x_n):=(x_i-x_1)(x_i-x_2)\cdots(x_i-x_{i-1})(x_i-x_{i+1})\cdots(x_i-x_n),\ \ i=1,\dots,n.$

In the concave case ≤ is changed to ≥.

Remark 1. If f is convex in the strict sense then ≤ is changed to < and equality holds iff f is linear function.

Remark 2. The inequality is sharp in the following limit sense: let $\underline x =(x_1,\ldots,x_n),\ \underline \alpha = (\alpha, \ldots ,\alpha)$ and $\ a<\alpha
Then the limit of the left side exists and

$\lim_{\underline x \to \underline \alpha} \sum_{i=1}^n \frac{F^{(n-1)}(x_i)}{\Pi_i(x_1,\ldots,x_n)}=\lim_{\underline x \to \underline \alpha}\frac{1}{n!}\sum_{i=1}^n f(x_i)= \frac{f(\alpha)}{(n-1)!}$