proof of Hermite-Hadamard integral inequality

First of all, let’s recall that a convex function on a open interval (a,b) is continuousMathworldPlanetmath on (a,b) and admits left and right derivative f+(x) and f-(x) for any x(a,b). For this reason, it’s always possible to construct at least one supporting line ( for f(x) at any x0(a,b) : if f(x0) is differentiableMathworldPlanetmathPlanetmath in x0, one has r(x)=f(x0)+f(x0)(x-x0); if not, it’s obvious that all r(x)=f(x0)+c(x-x0) are supporting lines for any c[f-(x0),f+(x0)].
Let now r(x)=f(a+b2)+c(x-a+b2) be a supporting line of f(x) in x=a+b2(a,b). Then, r(x)f(x). On the other side, by convexity definition, having defined s(x)=f(a)+f(b)-f(a)b-a(x-a) the line connecting the points (a,f(a)) and (b,f(b)) , one has f(x)s(x). Shortly,


Integrating both inequalitiesMathworldPlanetmath between a and b

= ab[f(a+b2)+c(x-a+b2)]𝑑x
= f(a+b2)(b-a)+cab(x-a+b2)𝑑x
= f(a+b2)(b-a)
= ab[f(a)+f(b)-f(a)b-a(x-a)]𝑑x
= f(a)(b-a)+f(b)-f(a)b-aab(x-a)𝑑x
= f(a)+f(b)2(b-a)

and so


which is the thesis.

Title proof of Hermite-Hadamard integral inequality
Canonical name ProofOfHermiteHadamardIntegralInequality
Date of creation 2013-03-22 16:59:22
Last modified on 2013-03-22 16:59:22
Owner Andrea Ambrosio (7332)
Last modified by Andrea Ambrosio (7332)
Numerical id 7
Author Andrea Ambrosio (7332)
Entry type Proof
Classification msc 26D10
Classification msc 26D15