= 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 $H_{\gamma_j}\to H^*\in\mathbb R^n$. Each $H_{\gamma_j}$ is deterministic, hence so is $H^*$. The first exponential term gives
$$
H_{\gamma_j}\cdot p-x\leq\frac{\log M}{\gamma_j},
$$
and passing to the limit gives $H^*\cdot p\leq x$.
To obtain the <superhedging> inequality, suppose instead that $\mathbb P(X>H^*\cdot P)>0$. There is then an $\varepsilon>0$ such that the event $E=\{X-H^*\cdot P\geq\varepsilon\}$ has positive <probability>. Let $\|P\|\leq B$ <almost surely>. For all sufficiently large $j$,
$$
|(H_{\gamma_j}-H^*)\cdot P|\leq B\|H_{\gamma_j}-H^*\|\leq\varepsilon/2.
$$
Consequently $X-H_{\gamma_j}\cdot P\geq\varepsilon/2$ on $E$, giving
$$
M\geq F_{\gamma_j}(H_{\gamma_j})
\geq\mathbb P(E)e^{\gamma_j\varepsilon/2}\longrightarrow\infty,
$$
a contradiction. \b[The limiting portfolio meets both constraints]:
$$
\boxed{H^*\cdot p\leq x,\qquad H^*\cdot P\geq X\quad\text{almost surely}.}
$$
Bounded $P$ is used precisely to turn convergence of deterministic holdings into a uniform bound on the payoff error.
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