convex functions lie above their supporting lines

Let $f:𝐑\to 𝐑$ be a convex, twice differentiablefunction on $[a,b]$. Then $f(x)$ lies above its supporting lines, i.e. it’sgreater 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 $f^{{}^{\prime \prime }}\left(\xi \left(x\right)\right)\ge 0$ byconvexity.∎