convex functions lie above their supporting lines


Let f:𝐑→𝐑 be a convex, twice differentiable function on [a,b]. Then f⁢(x) lies above its supporting lines, i.e. it’s greater than any tangent line in [a,b].

Proof.

:

Let r⁢(x)=f⁢(x0)+f′⁢(x0)⁢(x-x0) be the tangent of f⁢(x) in x=x0∈[a,b].

By Taylor theorem, with remainder in Lagrange form, one has, for any x∈[a,b]:

f⁢(x)=f⁢(x0)+f′⁢(x0)⁢(x-x0)+12⁢f′′⁢(ξ⁢(x))⁢(x-x0)2

with ξ⁢(x)∈[a,b]. Then

f⁢(x)-r⁢(x)=12⁢f′′⁢(ξ⁢(x))⁢(x-x0)2≥0

since f′′⁢(ξ⁢(x))≥0 by convexity. ∎

Title convex functions lie above their supporting lines
Canonical name ConvexFunctionsLieAboveTheirSupportingLines
Date of creation 2013-03-22 16:59:20
Last modified on 2013-03-22 16:59:20
Owner Andrea Ambrosio (7332)
Last modified by Andrea Ambrosio (7332)
Numerical id 5
Author Andrea Ambrosio (7332)
Entry type Result
Classification msc 52A41
Classification msc 26A51
Classification msc 26B25