# convex functions lie above their supporting lines

Let $f:\mathbf{R}\rightarrow\mathbf{R}$ 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

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

since $f^{{}^{\prime\prime}}\left(\xi\left(x\right)\right)\geq 0$ by convexity. ∎

Title convex functions lie above their supporting lines ConvexFunctionsLieAboveTheirSupportingLines 2013-03-22 16:59:20 2013-03-22 16:59:20 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 5 Andrea Ambrosio (7332) Result msc 52A41 msc 26A51 msc 26B25