It is well known that the set of minimizers of a convex function over a convex set is convex.
It is also true that it is closed. But I have not been able to show this result.
Let $f$ be a convex function on a convex set $C$, I wish to show that the set of minimizers of $f$, denoted as $S$, is closed.
To show $S$ is closed, I need to show $\mathbb{R}^n \backslash S$ is open.
To show $\mathbb{R}^n \backslash S$ is open, I need to show that there exists some $r > 0$ such that $\{y \in \mathbb{R}^n | \|x - y\|_2 < r, x \in \mathbb{R}^n \backslash S\} \subseteq\mathbb{R}^n \backslash S$.
In other words, any $y$ from this ball needs to satisfy $f(y) > f_\min$, where $f_\min$ is the minimum value of $f(x)$.
I am not sure how to proceed from this point. I've tried this: let $y = x + (r/2)u$, $u \in \mathbb{R}^n \backslash S$, then $f(y) = f(x+(r/2)u) = f(r/2(x + u) + r/2x) \leq (r/2) f(x+u) + r/2f(x)$, but this is not getting me anywhere.
Let $a_n$ be a sequence of minimizers that converges to $a$. To show the set is closed we need to show $a$ is also a minimizer.
For all $n\in\mathbb{N}$ you have $a_n\in argmin_{x\in S} f(x)$ where both $f$ and $S$ are convex. This means we have, $f(a_n)\leq f(x)$ for all $x\in S$.
Since all convex functions $f:\mathbb{R}^n\rightarrow\mathbb{R}$ are continuous, we have:
$f(a) = f(\lim_{n\rightarrow\infty} a_n) = \lim_{n\rightarrow\infty} f(a_n)$.
Since $f(a_n)\leq f(x)$ for all $x\in S$ and $n\in\mathbb{N}$, it follows that $\lim_{n\rightarrow\infty}f(a_n)\leq \lim_{n\rightarrow\infty}f(x)= f(x)$ for all $x\in S$. Therefore, by the preceding chain of equalities, it follows that $f(a)\leq f(x)$ for all $x\in S$, so that $a\in argmin_{x\in S} f(x)$.