Hermite–Hadamard inequality

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 : [a, b] → R is convex, then the following chain of inequalities hold:

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

Generalisations - The concept of a sequence of iterated integrals

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

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

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

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

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

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

Example 3

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

${\begin{aligned}F^{(0)}(s)&=e^{s},\\F^{(1)}(s)&=\int _{0}^{s}F^{(0)}(u)du=\int _{0}^{s}e^{u}du=e^{s}-1,\\F^{(2)}(s)&=\int _{0}^{s}F^{(1)}(u)du=\int _{0}^{s}(e^{u}-1)du=e^{s}-s-1,\\&{}\ \vdots \\F^{(n)}(s)&=e^{s}-\sum _{i=0}^{n-1}{\frac {s^{i}}{i!}}\\&{}\ \vdots \end{aligned}}$

Theorem

Suppose that −∞ < a < b < ∞, and let f:[a,b]→R be a convex function, a < x_i < b, i = 1, ..., n, such that x_i ≠ x_j, if i ≠ j. 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 $\ a<\alpha <b.$
Then the limit of the left side exists and

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

References

• Jacques Hadamard, "Étude sur les propriétés des fonctions entières et en particulier d'une fonction considérée par Riemann", Journal de Mathématiques Pures et Appliquées, volume 58, 1893, pages 171–215.
• Zoltán Retkes, "An extension of the Hermite–Hadamard Inequality", Acta Sci. Math. (Szeged), 74 (2008), pages 95–106.
• Mihály Bessenyei, "The Hermite–Hadamard Inequality on Simplices", American Mathematical Monthly, volume 115, April 2008, pages 339–345.
• Flavia-Corina Mitroi, Eleutherius Symeonidis, "The converse of the Hermite-Hadamard inequality on simplices", Expo. Math. 30 (2012), pp. 389–396. DOI:10.1016/j.exmath.2012.08.011; ISSN 0723-0869