# Is this bounded convex set in $\mathbb{R}^n$ closed?

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?

• Note that the argument used in this very interesting question math.stackexchange.com/questions/1434741/… may be of some help..... Nov 14 '18 at 16:25
• Is $x.c$ the inner product between $x$ and $c$? The answer to the last question is yes due to the maximum principle. Nov 14 '18 at 18:38
• Yes, it is the inner product. Nov 14 '18 at 23:49

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.