differentiable at $x\in U$.
We want to prove that if $x\in U$ is a global minimum for $f$, then for all $y\in U$ : $\mathrm{d}_x f.(y-x)=0$.
Using just the fact that $U$ is a convex set and $x\in U$ is in particular a local minimum, I proved that for all $y\in U$ : $\mathrm{d}_x f.(y-x)\ge 0$.
Maybe to get the equality, I have to use the fact that $U$ is an open set of $\mathbb{R}^d$ ?
$\underline{PS}$ : I know that it is a general statement for differentiable functions on an open set which have a local extremum. But here I want a proof which uses convexity.
Thanks in advance !