= Solution
Suppose $A_0$ lies on the relative boundary of $K$ in its span $L$. There is a nonzero $x\in L$ with $x^\top y\geq0$ for every $y\in K$ and $x^\top A_0=0$. Here is a direct construction of the supporting vector. Choose points $z_n\in L\setminus K$ converging to $A_0$, let $p_n$ be their nearest points in $K$, and set $x_n=(p_n-z_n)/\|p_n-z_n\|$. The minimization inequality for a nearest point gives $x_n^\top y\geq0$ on $K$ and $x_n^\top p_n=0$. A subsequence of the unit vectors converges to a unit vector $x\in L$. Since $0\leq x_n^\top A_0\leq\|A_0-z_n\|$, it has the asserted support properties.
The test-weight characterization of $K$ now gives $x^\top A_1\geq0$ almost surely. This random payoff cannot vanish almost surely: otherwise $x$ would annihilate every element of $D$, hence all of $L$, contradicting $x\in L$ and $\|x\|=1$. Therefore
$$
\boxed{x^\top A_0=0,\qquad x^\top A_1\geq0\text{ a.s.},\qquad P(x^\top A_1>0)>0.}
$$
This supplies the positive-terminal-payoff alternative. If $K$ has zero-dimensional span, its relative boundary is empty, so that case requires no supporting vector.
Back to article page
