= Solution
Use zero-interest cash as the one-period <numéraire>, consistent with the stated expectation-price formula. For $p>1$, differentiate the proposed call-price curve twice. The first derivative and the candidate density are
$$
\begin{aligned}
C'(u)&=u^{p-1}(1+u^p)^{1/p-1}-1,\\
\boxed{f_p(u)=C''(u)}&=\boxed{(p-1)u^{p-2}(1+u^p)^{1/p-2},\qquad u>0.}
\end{aligned}
$$
This is the <power call-curve pricing density>. It is strictly positive. Its integral is $C'(\infty)-C'(0+)=1$, and its survival function is $-C'(u)$. Since $C(0+)=1$ and $C(\infty)=0$,
$$
\int_0^\infty u f_p(u)\,du
=\int_0^\infty[-C'(u)]\,du=1.
$$
Likewise, integrating the survival function from $K$ onwards gives
$$
\int_0^\infty(u-K)^+f_p(u)\,du=C(K).
$$
Thus a market whose terminal <stock> has this law under an <equivalent martingale measure> prices the <stock> at one, every proposed call at $C(K)$, and any <integrable> claim $g(S_1)$ at $\int g(u)f_p(u)du$. The finite-market <fundamental theorem of asset pricing> says that an equivalent measure pricing every traded discounted payoff by <expectation> excludes <arbitrage>. This proves the intended conclusion when such an equivalent pricing law is part of the model. For example, take the canonical terminal state space $(0,\infty)$ with <stock> equal to its coordinate and physical law equivalent to the positive density $f_p$.
There is, however, a genuine insufficiency in the literal finite-strike formulation: a finite list of call prices and no-arbitrage alone do not force this pricing law, nor even a continuous terminal distribution. Here is an explicit counterexample. Take $p=2$, one strike $K=1$, and two terminal <stock> values
$$
a=\frac12,\qquad b=2+\sqrt2,
\qquad\mathbb Q(S_1=b)=3-2\sqrt2.
$$
Give the lower <stock> value the remaining strictly positive probability and take this as the physical measure too. Direct calculation gives $\mathbb E S_1=1$ and
$$
\mathbb E(S_1-1)^+=\sqrt2-1=C(1),
$$
so the <stock>/cash/call market is arbitrage-free. Now let
$$
g(u)=(u-a)^2(u-b)^2e^{-u}.
$$
This bounded nonnegative function is zero at both actual <stock> values, so $g(S_1)=0$ almost surely. Yet $\int g(u)f_2(u)du>0$. Charging that positive amount for the identically zero payoff creates an <arbitrage> by selling it. In fact no Lebesgue probability density can price every claim correctly on this two-state market.
Therefore \b[the displayed $f_p$ is the intended continuous pricing density, but the promised no-arbitrage extension requires an equivalent pricing measure with this terminal law; a full call curve identifies that law if such a measure exists, but it does not follow from the printed finite-strike hypotheses alone]. This is the <finite-strike nonidentification of a pricing density>. The counterexample and the corrected sufficient hypothesis account for the literal and intended readings separately.
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