I'm really confused on this question. The question states that given the problem $\min_{x \in S} f(x)$, where S is a closed convex set and f a smooth convex function. Let $\bar{S} \subset S$ be the set of points that achieve the global minimum. That is for all $\bar{x} \in \bar{S}$ we have that $f(x) \geq f(\bar{x})$ $\forall x \in S$. Show that $\bar{S}$ is the convex set.
I have no idea how to approach solving this problem. I know from a theorem that:
Let S be a convex set in $\mathbb{R^n}$, f be a continuously differentiable function. Then:
(a) If $\bar{x}$ is a local solution to $\min f(x) \text{ s. t }$ to $x \in S$, then $\bar{x}$ satisfies $\bar{x} \in S$ and $\nabla f(\bar{x})' (x-\bar{x}) \geq 0 \forall x \in S$.
(b) If $\bar{x}$ satisfies (a) and f is convex, then $\bar{x}$ is a global solution of $\min f(x) \text {s. t. } x \in S$.
Here's what I know:
- A set S is convex if for all $x,y \in S$, and $\forall \lambda \in [0,1]$: $\lambda x + (1-\lambda)y \in S$
And I'm guessing I have to prove this for $\bar{S}$, but I'm stuck at how to do this. Help would be much appreciated!
Let $x,y\in\bar S$ and $0\le\lambda\le 1$. As $f$ is convex, we have $$f(\lambda x+(1-\lambda)y)\le \lambda f(x)+(1-\lambda)f(y) = \lambda \min f+(1-\lambda)\min f=\min f, $$ but trivially also $$f(\lambda x+(1-\lambda)y)\ge\min f.$$
(Note that smoothness is not used in this)