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 Author: Tolaso J Kos [ Sun Dec 13, 2015 9:25 pm ] Post subject: 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.

 Author: Papapetros Vaggelis [ Sat Dec 26, 2015 12:26 pm ] Post subject: 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 commentWe 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 $$\displaystyle{f}$$ andthe lines $$\displaystyle{x=a\,,x=b}$$ and this area is less or equal to the area of the tabledefined by $$\displaystyle{E(a,0),F(a,f(a)),G(b,0),H(b,f(b))}$$ .

 Author: Riemann [ Thu Apr 07, 2016 7:46 am ] Post subject: 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.

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