= Solution
First we prove the <Krein-Milman theorem> in the required setting. If $K$ is empty the conclusion is immediate, so assume it is nonempty. A <face of a convex set> is a convex subset $F$ with the property that an interior point of a segment in $K$ belongs to $F$ only if both endpoints do. Consider all nonempty weakly compact faces of $K$, ordered by reverse inclusion. A chain has a nonempty intersection by <compactness> and the finite intersection property; that intersection is again a compact face. The <Zorn lemma> gives a minimal compact face $F$.
If $F$ contained distinct points, a bounded <linear functional> separating them would be nonconstant on $F$. Its maximizer set is a nonempty proper compact face of $F$, hence a face of $K$, contradicting minimality. Therefore $F$ is a singleton and its member is an <extreme point> of $K$. The same reasoning applies to each nonempty compact face of $K$, so each such face contains an <extreme point> of $K$.
Let $H$ be the norm-<closed convex hull> of the <extreme points> of $K$. A weakly compact subset of a Banach space is weakly closed, hence norm closed, so $H\subseteq K$. If $x\in K\setminus H$, the <Hahn-Banach separation theorem> gives $\ell\in E'$ with $\ell(x)>\sup_H\ell$. The maximizer face of $\ell$ on $K$ contains an <extreme point> $e$ of $K$. Then $\ell(e)\geq\ell(x)>\sup_H\ell$, contradicting $e\in H$. We conclude
$$
\boxed{K=\overline{\operatorname{conv}}\operatorname{Ex}(K).}
$$
Norm and weak closed convex hulls coincide by the same separation theorem. The printed phrase “closed convex cover” is understood in this standard closed-convex-hull sense.
Now work in the real <space of continuous functions on a compact space> $C(C)$ on the <Cantor set>. The <extreme points of a real continuous-function unit ball> are exactly the continuous sign functions:
$$
\boxed{\operatorname{Ex}(B)=\{f\in C(C):f(x)\in\{-1,1\}\text{ for every }x\in C\}.}
$$
If $|f(x_0)|<1$, continuity supplies a neighbourhood where $|f|\leq1-\varepsilon$. The <Cantor cylinder> sets form a <clopen> base, so choose a nonempty cylinder $D$ inside that neighbourhood. Its <indicator function> is continuous, and $f\pm\varepsilon1_D$ are distinct members of $B$ with midpoint $f$. Hence $f$ is not extreme. Conversely, if $f$ is pointwise sign-valued and $f=(g+h)/2$ with $g,h\in B$, equality at the endpoint of the scalar interval $[-1,1]$ forces $g(x)=h(x)=f(x)$ at every point.
To prove the closed-hull assertion without assuming weak <compactness>, take $f\in B$ and $\varepsilon>0$. A sufficiently fine finite partition of $C$ into <Cantor cylinders> has oscillation of $f$ less than $\varepsilon$ on each cell, by <uniform continuity>. Choose a value $a_j\in[-1,1]$ on each cell and let $h$ be the corresponding continuous step function. Then $\|f-h\|_\infty<\varepsilon$. For each sign vector $s\in\{-1,1\}^m$, let $e_s$ take value $s_j$ on cell $j$, and assign the weight
$$
w_s=\prod_{j=1}^m\frac{1+s_ja_j}{2}.
$$
These weights are nonnegative, sum to one, and satisfy $\sum_s w_s s_j=a_j$. Thus $h=\sum_s w_se_s$ is a <convex combination> of <extreme points>. This is <clopen sign approximation in the real Cantor unit ball>, proving
$$
\boxed{B=\overline{\operatorname{conv}}\operatorname{Ex}(B).}
$$
Nevertheless $B$ is not weakly compact. The suggested functions $f_n(x)=\min(3^nx,1)$ lie in $B$ and converge pointwise to the function equal to zero at $0$ and one at every other point of $C$. It is discontinuous at zero, since $2\cdot3^{-j}\in C$ tends to zero. If $B$ were weakly compact, the sequence, viewed as a net, would have a weakly convergent <subnet> with limit in $C(C)$. Point evaluations are <bounded linear functionals>, so that <subnet> would have the same pointwise limit; its indices are cofinal in the original sequence. The limit would therefore be the discontinuous function just described, a contradiction. This <discontinuous pointwise limit obstruction to weak compactness> gives \b[the required failure of weak <compactness>], despite the closed-hull equality.
Back to article page
