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 $x\in K$?

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 $K$.

affinefunction is a pointwise infinum of a collection of continuousaffinefunctions? $\endgroup$