Show that the set of global minimizers of $f$ is a convex set. If there can be only one global minimizer, how?

Question

I'm studying non linear optimization and there's the following exercise:

Suppose that $f$ is a convex function. Show that the set of global minimizers of $f$ is a convex set.

A point $x^*$ is a global minimizer if $f(x^*)\le f(x)$ for all $x\in \mathbb{R}^n$. So how is it possible for $2$ or more to exist?

Answer 1

Let $f$ be a convex function and assume that $x_1,x_2$ are global minimizers (that is, $f(x_1)=f(x_2)\le f(x),\forall x$. It follows by definition that, for all $t\in [0,1],$

$$f(tx_1+(1-t)x_2)\le tf(x_1)+(1-t)f(x_2) \le tf(x)+(1-t)f(x)=f(x).$$ This shows that

$$f(tx_1+(1-t)x_2)\le f(x)$$ from where we get that $tx_1+(1-t)x_2$ is also a global minimizer and thus we have shown that the set of global minimizers is convex.

If the function is strictly convex then there is only one global minimizer but it is not the case if we consider convex functions.

For example

$$f(x)=\begin{cases}0 &\quad \text{if}\quad x\le 0\\ x^2 &\quad \text{if}\quad x>0 \end{cases}$$ is a convex function and the set of global minimizers is $(-\infty,0].$

Answer 2

If there are 2 global minimizers, the statement in particular implies that there are infinitely many - namely also all those on the line segment between the two minimizers. So, there is either exactly one global minimizer or infinitely many.