= Solution
Let $A_\delta=\sup_{K\ge0}K^\delta C(K)<\infty$. The <call-price decay and moment threshold> follows by splitting the preceding <integral> at one. Since $(S-K)_+\le S$, for $0<\varepsilon<\delta$,
$$
\int_0^1 K^{\varepsilon-1}C(K)\,dK\le\frac{\mathbb ES}{\varepsilon}.
$$
For $K\ge1$, the decay bound gives
$$
\int_1^\infty K^{\varepsilon-1}C(K)\,dK
\le A_\delta\int_1^\infty K^{\varepsilon-1-\delta}\,dK
=\frac{A_\delta}{\delta-\varepsilon}.
$$
The <power payoff static call representation> therefore yields
$$
\boxed{M(1+\varepsilon)\le(1+\varepsilon)\mathbb ES
+\frac{\varepsilon(1+\varepsilon)A_\delta}{\delta-\varepsilon}<\infty.}
$$
The $\varepsilon=0$ case is the given finite first moment. The strict endpoint matters: a <Pareto distribution> with $\mathbb P(S>x)=x^{-(1+\delta)}$ for $x\ge1$ has $C(K)=K^{-\delta}/\delta$ for $K\ge1$, but its moment of order $1+\delta$ is infinite. Thus the stated decay condition does not generally imply the endpoint moment.
Back to article page
