Proving that a function has a maximizer

Question

I am trying to prove this:

Let $X$ be a nonempty, convex, and compact subset of $\mathbb{R}^n$ and $f:X \rightarrow \mathbb{R}$ a concave function. Then $f$ has a maximizer.

Now this is trivial (somewhat) if f is continuous because of Weistrass' theorem. But f isn't necessarily continuous here.

I have been stuck on this for a couple of days now. I can sort of imagine why this would be true if $f:\mathbb{R} \rightarrow \mathbb{R}$. But I haven't gotten any idea further than that yet.

Any help would be great!

Answer 1

This is not true. For (my) convenience, I answer the equivalent question of the existence of a minimizer of a convex function.

Let $X = [0,1]$ and $$ f(x) = \begin{cases} x & \text{for } x \in (0,1], \\ 1 &\text{for } x = 0. \end{cases}$$ Then, $f(x) > 0$ for all $x \in X$ and $f(x) \to 0$ as $x \searrow 0$. Hence, $f$ does not possess a minimizer on $X$.

Answer 2

As @gerw already answered, the claim is not true. I'd like to add that there is a little additional assumption which will lead to a true statement:

A concave and upper semi-continuos function on a convex and compact domain has a maximizer.