 Solution

We prove the contrapositive. LetThe function $F$ is constant along $N$, so minimize it on $N^{\perp }$. Suppose there is no $H$ with $H\cdot X\ge 0$ almost surely and strict inequality with positive probability. If a sequence $h_k\in N^{\perp }$ satisfies $\Vert h_k\Vert \to \infty$, pass to a subsequence withBecause $H\notin N$ and there is no arbitrage direction, $\mathbb{P}(H\cdot X<0)>0$. On that event, $e^{-h_k\cdot X}\zeta \to \infty$, and Fatou lemma gives ${\mathrm{liminf}}_{k}F(h_k)=\infty$. Hence every finite sublevel set of $F$ in $N^{\perp }$ is bounded. It is also closed, so $F$ attains its infimum there and has a bounded minimizing sequence. This contradicts the assumption. Therefore there is a unit vector $H$ satisfying