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:

${\displaystyle f\left({\frac {a+b}{2}}\right)\leq {\frac {1}{b-a}}\int _{a}^{b}f(x)\,dx\leq {\frac {f(a)+f(b)}{2}}.}$

Generalisations - The concept of a 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.:

{\displaystyle {\begin{aligned}F^{(0)}(s)&:=f(s),\\F^{(1)}(s)&:=\int _{a}^{s}F^{(0)}(u)du=\int _{a}^{s}f(u)du,\\F^{(2)}(s)&:=\int _{a}^{s}F^{(1)}(u)du=\int _{a}^{s}\left(\int _{a}^{t}f(u)du\right)\,dt,\\&\ \ \vdots \\F^{(n)}(s)&:=\int _{a}^{s}F^{(n-1)}(u)\,du,\\&{}\ \ \vdots \end{aligned}}}

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

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

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

{\displaystyle {\begin{aligned}F^{(0)}(s)&=1,\\F^{(1)}(s)&=\int _{-1}^{s}F^{(0)}(u)\,du=\int _{-1}^{s}1du=s+1,\\F^{(2)}(s)&=\int _{-1}^{s}F^{(1)}(u)du=\int _{-1}^{s}(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{aligned}}}

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

{\displaystyle {\begin{aligned}F^{(0)}(s)&=e^{s},\\F^{(1)}(s)&=\int _{0}^{s}F^{(0)}(u)du=\int _{0}^{s}e^{u}du=e^{s}-1,\\F^{(2)}(s)&=\int _{0}^{s}F^{(1)}(u)du=\int _{0}^{s}(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{aligned}}}

Theorem

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:

${\displaystyle \sum _{i=1}^{n}{\frac {F^{(n-1)}(x_{i})}{\Pi _{i}(x_{1},\dots ,x_{n})}}\leq {\frac {1}{n!}}\sum _{i=1}^{n}f(x_{i})}$

where

${\displaystyle \Pi _{i}(x_{1},\dots ,x_{n}):=(x_{i}-x_{1})(x_{i}-x_{2})\cdots (x_{i}-x_{i-1})(x_{i}-x_{i+1})\cdots (x_{i}-x_{n}),\ \ i=1,\dots ,n.}$

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 ${\displaystyle {\underline {x}}=(x_{1},\ldots ,x_{n}),\ {\underline {\alpha }}=(\alpha ,\ldots ,\alpha )}$ and ${\displaystyle \ a<\alpha
Then the limit of the left side exists and

${\displaystyle \lim _{{\underline {x}}\to {\underline {\alpha }}}\sum _{i=1}^{n}{\frac {F^{(n-1)}(x_{i})}{\Pi _{i}(x_{1},\ldots ,x_{n})}}=\lim _{{\underline {x}}\to {\underline {\alpha }}}{\frac {1}{n!}}\sum _{i=1}^{n}f(x_{i})={\frac {f(\alpha )}{(n-1)!}}}$