 Solution

Let $M={\sup }_{\gamma >1}{F}_{\gamma }(H_{\gamma })<\infty$. Choose ${\gamma }_j\to \infty$. Boundedness of the minimizers and the Bolzano-Weierstrass theorem yield a subsequence, still indexed by $j$, with $H_{{\gamma }_j}\to H^{\ast }\in {\mathbb{R}}^n$. Each $H_{{\gamma }_j}$ is deterministic, hence so is $H^{\ast }$. The first exponential term givesand passing to the limit gives $H^{\ast }\cdot p\le x$.

To obtain the superhedging inequality, suppose instead that $\mathbb{P}(X>H^{\ast }\cdot P)>0$. There is then an $\epsilon >0$ such that the event $E=\{X-H^{\ast }\cdot P\ge \epsilon \}$ has positive probability. Let $\Vert P\Vert \le B$ almost surely. For all sufficiently large $j$,Consequently $X-H_{{\gamma }_j}\cdot P\ge \epsilon /2$ on $E$, givinga contradiction. The limiting portfolio meets both constraints:Bounded $P$ is used precisely to turn convergence of deterministic holdings into a uniform bound on the payoff error.