## Minimizers – 01, a variational inequality

To minimize a differentiable function, we usually know it’s gradient. Consider the function $f \in C^1(K)$ with $K$ be a closed convex set. Let $F(x)$ be gradient of $f$, we have:

Prop: Suppose there exists an $x \in K$ such that $f(x) = \min\limits_{y \in K} f(y)$.

Then $x$ is a solution of the variational inequality $x \in K: (F(x), y-x) \ge 0$ for $y \in K$.

Proof.

$\varphi(t) = f(x + t(y-x)), 0 \le t \le 1$. Minimum at $x$ implies $t=0$. Therefore $\varphi'(0) = 0$, so  $\varphi'(0) = (gradf,y-x)$, we have $(gradf,y-x) \ge 0$ since $\varphi'(0) = (gradf,y-x)$.

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