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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 $x\in K$?

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  • $\begingroup$ 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? $\endgroup$
    – erz
    Feb 11 '20 at 5:59
  • $\begingroup$ @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 $K$. $\endgroup$ Feb 11 '20 at 7:55
  • $\begingroup$ could you please provide a reference? I've made an answer based on this fact. $\endgroup$
    – erz
    Feb 11 '20 at 14:58
  • $\begingroup$ @erz Yes, of course. This is Corollary I.1.4 in "Compact convex sets and boundary integrals" by E. M. Alfsen. $\endgroup$ Feb 11 '20 at 18:00
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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 $K$.

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