Welcome to mathimatikoi.org; a site of university mathematics! Enjoy your stay here!

Hermite - Hadamard's inequality

General Mathematics
Tolaso J Kos Articles: 2
Posts: 853
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

Hermite - Hadamard's inequality

Here is a classic inequality that is not that difficult to prove.

Let $f:[a, b] \rightarrow \mathbb{R}$ be a continuous and convex function. Prove that:

$$f\left ( \frac{a+b}{2} \right )\leq \frac{1}{b-a}\int_{a}^{b}f(x)\, {\rm d}x \leq \frac{f(a)+f(b)}{2}$$
(Hermite - Hadamard's Inequality) Title corrected as informed that the name of the inequality was not quite correct.
Imagination is much more important than knowledge.
Papapetros Vaggelis
Team Member
Articles: 0
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Hermite - Hadamard's inequality

If $\displaystyle{x\in\left[a,b\right]}$, then $\displaystyle{a+b-x\in\left[a,b\right]}$.

Since the function $\displaystyle{f}$ is convex, we get :

$\displaystyle{f(x)+f(a+b-x)\geq 2\,f\,\left(\dfrac{x+a+b-x}{2}\right)=2\,f\,\left(\dfrac{a+b}{2}\right)\,,\forall\,x\in\left[a,b\right]}$ .

The function $\displaystyle{f}$ is continuous, so by integration we have that :

$\displaystyle{\int_{a}^{b}f(x)\,\mathrm{d}x+\int_{a}^{b}f(a+b-x)\,\mathrm{d}x\geq 2\,(b-a)\,f\,\left(\dfrac{a+b}{2}\right)}$

or :

$\displaystyle{\int_{a}^{b}f(x)\,\mathrm{d}x-\int_{b}^{a}f(a+b-x)\,\mathrm{d}(a+b-x)\geq 2\,(b-a)\,f\,\left(\dfrac{a+b}{2}\right)}$

or :

$\displaystyle{2\,\int_{a}^{b}f(x)\,\mathrm{d}x\geq 2\,(b-a)\,f\,\left(\dfrac{a+b}{2}\right)}$

which means that :

$\displaystyle{f\,\left(\dfrac{a+b}{2}\right)\leq \dfrac{1}{b-a}\,\int_{a}^{b}f(x)\,\mathrm{d}x}$ .

On the other hand, from the first mean value theorem,

$\displaystyle{\dfrac{1}{b-a}\,\int_{a}^{b}f(x)\,\mathrm{d}x=f(\xi)}$ for some $\displaystyle{\xi\in\left(a,b\right)}$ .

Please, give a hint for the right-hand inequality.

Geometrical comment

We have that :

$\displaystyle{\int_{a}^{b}f\,\left(\dfrac{a+b}{2}\right)\,\mathrm{d}x\leq \int_{a}^{b}f(x)\,\mathrm{d}x\leq \int_{a}^{b}\dfrac{f(a)+f(b)}{2}\,\mathrm{d}x}$

which means that the area of the parallelogram defined by

$\displaystyle{A(a,0),B(b,0),C\left(a,f\left(\dfrac{a+b}{2}\right)\right),D\left(b,f\left(\dfrac{a+b}{2}\right)\right)}$

is less or equal to the area of the region defined by the graph of the function $\displaystyle{f}$ and

the lines $\displaystyle{x=a\,,x=b}$ and this area is less or equal to the area of the table

defined by $\displaystyle{E(a,0),F(a,f(a)),G(b,0),H(b,f(b))}$ .
Riemann
Articles: 0
Posts: 166
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

Re: Hermite - Hadamard's inequality

To finish off the exercise it remains to prove the first inequality. Well,

\begin{align*}
\frac{1}{b-a} \int_{a}^{b}f(x)\,dx &=\frac{1}{b-a} \left [ \int_{a}^{(a+b)/2} f(x)\,dx + \int_{(a+b)/2}^{b} f(x)\,dx \right ] \\
&=\frac{1}{2}\int_{0}^{1}\left [ f\left ( \frac{a+b- t (b-a)}{2} \right ) + f\left ( \frac{a+b+t(b-a)}{2} \right ) \right ]\,dt \\
&> f \left ( \frac{a+b}{2} \right )
\end{align*}

that is the LHS and the exercise is complete. Equality holds if and only if $f$ is affine.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$