Suppose we have a bounded set $C \subset \mathbb{R}^n$ that is convex and non-empty.
And suppose the family of linear functions $(f_{x})_{x \in \mathbb{R}^n}$ given by $f_{x}: C \rightarrow \mathbb{R}$, $\ f_{x}(c) = x.c$ for $c \in C$ attain their maximum and minimum in the set $C$.
Does this mean $C$ is closed (and hence compact) in $\mathbb{R}^n$?
My idea: I think this does imply $C$ is closed but I am not sure how to write my argument "properly". For every vector $x \in \mathbb{R}^n$, there is a point in $C$ that is "furthest" in the direction of $x$. Then because we are in a convex set we can just "join up all these points" and our set is closed. But how do I write this formally.
Remark: Also is it true that if a linear function on a convex set attains its maximum/minimum it does so on the boundary?
Best Answer
I found a counterexample as I was trying to construct a proof. Let me explain my thought process.
Let $x'$ be on the boundary and assume $x'\not\in C$. Since $x'$ is on the boundary, it has a separating hyperplane. Let $c$ be perpendicular to that plane and consider the function $f(x)=c^Tx$. Since $x'$ is on the boundary, function values can get arbitrarily close to $c^Tx'$, but by construction, cannot exceed $c^Tx'$ (the separating hyperplane is an isocurve). Since $f$ attains its maximum on $C$, there is a point $x^* \in C$ for which $c^Tx' = c^Tx^*$. So, $x^*$ and $x'$ share the same separating hyperplane.
The next step in my proof would be to construct a point on the other side of $x'$ that is in $C$ and reach a contradiction with convexity. This is where I had the aha moment.
Consider the following set, which is a closed rectangle and the first quadrant of a closed circle, minus the point where the rectangle meets the circle.