 Solution

Suppose $A_0$ lies on the relative boundary of $K$ in its span $L$. There is a nonzero $x\in L$ with $x^{\top }y\ge 0$ 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 $x_n=(p_n-z_n)/\Vert p_n-z_n\Vert$. The minimization inequality for a nearest point gives $x_n^{\top }y\ge 0$ on $K$ and $x_n^{\top }{p}_n=0$. A subsequence of the unit vectors converges to a unit vector $x\in L$. Since $0\le x_n^{\top }{A}_0\le \Vert A_0-z_n\Vert$, it has the asserted support properties.

The test-weight characterization of $K$ now gives $x^{\top }{A}_1\ge 0$ 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 $\Vert x\Vert =1$. ThereforeThis supplies the positive-terminal-payoff alternative. If $K$ has zero-dimensional span, its relative boundary is empty, so that case requires no supporting vector.