 Solution

Use positive strikes, the natural real-power domain of this price curve. For any such $K$, the stock-minus-call payoff isalmost surely, since $S_1>0$. Therefore the call price must be strictly below the stock price $S_0=1$: if $C(K)\ge 1$, buying stock and selling the call has nonpositive initial cost and strictly positive terminal payoff. Also a negative call price is an immediate arbitrage by buying the call.

For $p<0$, $1+K^p>K^p$ and raising to the negative power $1/p$ reverses the inequality, giving $(1+K^p{)}^{1/p}<K$ and $C(K)<0$. The expression is undefined at $p=0$. For $0<p<1$, strict concavity of the power implies $(1+K{)}^p<1+K^p$, whenceAt $p=1$, $C(K)=1$. Every defined case with $p\le 1$ therefore violates the necessary no-arbitrage bounds. ConsequentlyThis argument does not require a dense family of strikes; even one positive-strike call gives the contradiction. The strict stock-minus-call payoff explains why the borderline $p=1$ is also excluded.