# proof of Hermite-Hadamard integral inequality

First of all, let’s recall that a convex function on a open interval $(a,b)$ is continuous  on $(a,b)$ and admits left and right derivative $f^{+}(x)$ and $f^{-}(x)$ for any $x\in(a,b)$. For this reason, it’s always possible to construct at least one supporting line (http://planetmath.org/ConvexFunctionsLieAboveTheirSupportingLines) for $f\left(x\right)$ at any $x_{0}\in(a,b)$ : if $f\left(x_{0}\right)$ is differentiable   in $x_{0}$, one has $r(x)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)$; if not, it’s obvious that all $r(x)=f\left(x_{0}\right)+c\left(x-x_{0}\right)$ are supporting lines for any $c\in[f^{-}(x_{0}),f^{+}(x_{0})]$.
Let now $r(x)=f\left(\frac{a+b}{2}\right)+c\left(x-\frac{a+b}{2}\right)$ be a supporting line of $f(x)$ in $x=\frac{a+b}{2}\in(a,b)$. Then, $r\left(x\right)\leq f\left(x\right)$. On the other side, by convexity definition, having defined $s\left(x\right)=f\left(a\right)+\frac{f(b)-f(a)}{b-a}(x-a)$ the line connecting the points $(a,f(a))$ and $(b,f(b))$ , one has $f(x)\leq s(x)$. Shortly,

 $r\left(x\right)\leq f\left(x\right)\leq s(x)$

Integrating both inequalities  between $a$ and $b$

 $\int_{a}^{b}r\left(x\right)dx\leq\int_{a}^{b}f\left(x\right)dx\leq\int_{a}^{b}% s(x)dx$
 $\displaystyle\int_{a}^{b}r\left(x\right)dx$ $\displaystyle=$ $\displaystyle\int_{a}^{b}\left[f\left(\frac{a+b}{2}\right)+c\left(x-\frac{a+b}% {2}\right)\right]dx$ $\displaystyle=$ $\displaystyle f\left(\frac{a+b}{2}\right)(b-a)+c\int_{a}^{b}\left(x-\frac{a+b}% {2}\right)dx$ $\displaystyle=$ $\displaystyle f\left(\frac{a+b}{2}\right)(b-a)$ $\displaystyle\int_{a}^{b}s(x)dx$ $\displaystyle=$ $\displaystyle\int_{a}^{b}\left[f\left(a\right)+\frac{f(b)-f(a)}{b-a}(x-a)% \right]dx$ $\displaystyle=$ $\displaystyle f(a)(b-a)+\frac{f(b)-f(a)}{b-a}\int_{a}^{b}(x-a)dx$ $\displaystyle=$ $\displaystyle\frac{f(a)+f(b)}{2}(b-a)$

and so

 $f\left(\frac{a+b}{2}\right)(b-a)\leq\int_{a}^{b}f\left(x\right)dx\leq\frac{f(a% )+f(b)}{2}(b-a)$

which is the thesis.

Title proof of Hermite-Hadamard integral inequality ProofOfHermiteHadamardIntegralInequality 2013-03-22 16:59:22 2013-03-22 16:59:22 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 7 Andrea Ambrosio (7332) Proof msc 26D10 msc 26D15