convex functions lie above their supporting lines

Let $f:𝐑\to 𝐑$ 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 $r(x)=f\left(x_0\right)+f^{\prime }\left(x_0\right)\left(x-x_0\right)$ be the tangent of $f(x)$ in $x=x_0\in [a,b].$

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

$$f\left(x\right)=f\left(x_0\right)+f^{\prime }\left(x_0\right)\left(x-x_0\right)+\frac{1}{2}{f}^{{}^{\prime \prime }}\left(\xi \left(x\right)\right){\left(x-x_0\right)}^2$$

with $\xi \left(x\right)\in [a,b]$. Then

$$f\left(x\right)-r(x)=\frac{1}{2}{f}^{{}^{\prime \prime }}\left(\xi \left(x\right)\right){\left(x-x_0\right)}^2\ge 0$$

since $f^{{}^{\prime \prime }}\left(\xi \left(x\right)\right)\ge 0$ 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