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Hermite - Hadamard's inequality

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Tolaso J Kos
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Hermite - Hadamard's inequality


Post by Tolaso J Kos » Sun Dec 13, 2015 9:25 pm

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.
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Papapetros Vaggelis
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Re: Hermite - Hadamard's inequality


Post by Papapetros Vaggelis » Sat Dec 26, 2015 12:26 pm

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


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 \(\displaystyle{E(a,0),F(a,f(a)),G(b,0),H(b,f(b))}\) .
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Re: Hermite - Hadamard's inequality


Post by Riemann » Thu Apr 07, 2016 7:46 am

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

\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 )

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