$x_0$ is a global minimum over $X \Leftrightarrow\nabla f(x_0)^T(y-x_0)\geq 0$ for all $y\in X$

Question

Let $\emptyset\neq X\subseteq \mathbb R^n$ convex and $f:\mathbb R^n\rightarrow \mathbb R \in C^1$ and convex. Show that $x_0$ is a global minimum over $X \Leftrightarrow\nabla f(x_0)^T(y-x_0)\geq 0$ for all $y\in X$.
I already solved $"\Leftarrow"$ but I have problems to show the other implication.

Answer 1

Suppose for some $y^* \in X$ that $\langle \nabla f(x_0), y^*-x_0\rangle < 0$. Let $x(\alpha) = (1-\alpha) x_0 + \alpha y^*$. Note that $x(\alpha) \in X$ for all $\alpha \in [0,1]$ by convexity of $X$. Now, $$ \frac{d}{d\alpha}f\big(x(\alpha)\big) = \big\langle\nabla f(x(\alpha)), y^* - x_0\big\rangle, $$ but evaluating at $\alpha = 0$ then yields: $$ \frac{d}{d\alpha}f\big(x(\alpha)\big)\bigg\vert_{\alpha = 0} = \langle f(x_0), y^* - x_0\rangle < 0. $$ Then, starting at $x_0 = x(0)$, and moving along the path $x(\alpha)$ just a little (i.e. for small $\alpha$) lowers the value of $f$, contradicting the fact that $x_0$ is a global minima.