I am given the following problem, but can't figure it out.
Let $f\colon\mathcal{C}\rightarrow\mathbb{R}$ denote a convex function defined on the convex set $\mathcal{C}$. A (global) minimum of $f$ is an $x^*\in\mathcal{C}$ with $f(x^*)\leq f(x)$ for all $x\in\mathcal{C}$. Show that the set $\mathcal{S}$ of all minimum points of $f$ is convex.
I have tried the following but am unable to "finish" the solution (maybe I'm not even close and working in the wrong direction?):
$$\mathcal{S} = \left\{x^*~\middle|~f(x^*)\leq f(x),\, x\in\mathcal{C}\right\}$$
Let $u, v \in\mathcal{C}$, $\lambda\in(0,1)$. Then
$$\lambda u + (1-\lambda)v \overset{!}{\in} \mathcal{S}\\ \Leftrightarrow f(\lambda u + (1-\lambda)v) \overset{!}{\leq} f(x),\quad\forall x\in\mathcal{C}$$
- $f$ is strict convex: the set S has only one element and the set is convex
- $f$ is convex: because $f$ is defined over a convex set the line of points between $u$ and $v$ are in the set. At this point I'm lost.