Sup of a linear function

Question

Let $X$ be a banach space or simply a normed space and $C$ a convex (closed) subset of $X$. It is true that if $x \in C$ is such that $f(x)=\sup f(C)$, (in other words $x$ is a supporting point for $C$) where $f$ is a linear continuous function then $x$ must be in the boundary points of C? It seems logical to me cause the growth of $f$ in each direction is linear, but I can't find a proof or a coutrexample. Can anyone help?

Answer 1

If $f$ is constant, then $x$ need not be a boundary point, it can be any point of $C$. So we can presume that $f$ is not constant.

Note that for any set $S$, a point $x \in S$ is either in the interior or the boundary.

Now consider the point $x \in C$ in the question. Suppose $x$ is not a boundary point then it must be an interior point. Since $f$ is not constant, there is some $h$ such that $f(x+h) > f(x)$ (that is, $f(h) >0$), and since $x$ is an interior point, there is some $\lambda >0$ such that $x+ \lambda h \in C$, and we have $f(x+\lambda h) = f(x) + \lambda f(h) > f(x)$, which is a contradiction.