# Inequalities for upper semi-continuous affine functions on compact sets by using extreme points

Suppose $$f_1\colon K\to [0,\infty)$$ and $$f_2\colon K\to[0,\infty)$$ are two upper semi-continuous affine functions,
$$f_i(\lambda x+(1-\lambda)y)=\lambda f_i(x)+(1-\lambda)f_i(y)\ \mbox{ for all }\ x,y\in K\ \mbox{and}\ 0<\lambda<1,$$ on a compact convex set $$K$$ in a locally convex Hausdorff space $$E$$. If $$f_1(x)\le f_2(x)$$ for all extreme points $$x\in K$$, is it then true that $$f_1(x)\le f_2(x)$$ for all $$x\in K$$?

• In a Tychonoff space an upper semi-continuous function is a pointwise infinum of a collection of continuous functions. Perhaps an upper semi-continuous affine function is a pointwise infinum of a collection of continuous affine functions?
– erz
Feb 11 '20 at 5:59
• @erz Yes, such an upper semi-continuous affine function on a convex compact Hausdorff space $K$ is the pointwise limit of a decreasing net of continuous affine functions on $K$. Feb 11 '20 at 7:55
First, let us assume that $$f_2$$ is continuous, and let $$h=f_1-f_2$$, which is an upper semi-continuous affine function, negative on the extreme points of $$K$$.
Let $$c=\sup_K h$$. Since an upper semi-continuous function attains its maximal value on the compact set, $$L=h^{-1}(c)$$ is non-empty and convex. Since $$c$$ is the maximum, $$L=h^{-1}[c,+\infty)$$ is closed, and therefore compact.
Let $$x$$ be an extreme point of $$L$$. Assume that there are $$y,z\in K$$ such that $$x=\frac{1}{2}(y+z)$$. Then, $$h(x)\ge h(y), h(z)$$ and $$h(x)=\frac{1}{2}(h(y)+ h(z))$$ lead to $$h(x)=h(y)=h(z)$$ which contradicts the fact that $$x$$ is extreme in $$L$$. Hence, $$x$$ is also extreme in $$K$$, and so $$c=h(x)\le 0$$. Thus, $$h\le c\le 0$$ on $$K$$.
Now, to pass to the general case, use the fact that $$f_2$$ is an infinum of a set of continuous affine functions $$\{g_i\}$$. For each of these $$g_i$$ we have $$f_1\le g_i$$ on the extreme points of $$K$$, and so from the previous part, on the whole $$K$$. But then $$f_1\le\inf_i g_i=f_2$$ on $$K$$.