convexity proof

Question

Suppose $X \subset R^n$ is a convexe set and $f : X \rightarrow R^1$ with $f \in C^1/X$.

How to prove that $f$ is a convexe function on $X$ iff ($\nabla f(x)-\nabla f(y))^T(x-y)\geq0$ for any$x,y\in X$

Answer 1

Let $\phi(t) = f(y+t(x-y))$, then $\phi$ is differentiable.

Suppose $f$ is convex. Then $\phi'$ is non decreasing and so $\phi'(1) -\phi' (0) \ge 0$.

Since $\phi'(t) = \langle \nabla f (y +t (x-y)), x-y \rangle $, we have the desired result.

Now suppose the gradient condition holds. This implies that $\phi'$ is non decreasing, and so $\phi$ is convex and hence, since $x,y$ are arbitrary, $f$ is convex.

Note that only differentiability is required.