 Solution

Let $A_{\delta }={\sup }_{K\ge 0}{K}^{\delta }C(K)<\infty$. The call-price decay and moment threshold follows by splitting the preceding integral at one. Since $(S-K{)}_{+}\le S$, for $0<\epsilon <\delta$,For $K\ge 1$, the decay bound givesThe power payoff static call representation therefore yieldsThe $\epsilon =0$ case is the given finite first moment. The strict endpoint matters: a Pareto distribution with $\mathbb{P}(S>x)=x^{-(1+\delta )}$ for $x\ge 1$ has $C(K)=K^{-\delta }/\delta$ for $K\ge 1$, but its moment of order $1+\delta$ is infinite. Thus the stated decay condition does not generally imply the endpoint moment.