proof of Hermite-Hadamard integral inequality
First of all, let’s recall that a convex function on a open
interval is continuous on and admits left and right
derivative and for any . For this reason,
it’s always possible to construct at least one supporting line (http://planetmath.org/ConvexFunctionsLieAboveTheirSupportingLines) for at any : if is
differentiable in , one has ; if not, it’s obvious that all are supporting lines for
any .
Let now be a supporting line of in .
Then, . On the other side, by
convexity definition, having defined the line connecting the points and , one has . Shortly,
Integrating both inequalities between and
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 |