 Solution

For a fixed exercise threshold $\bar{S}>K$, the continuation value solves the same stationary equation but is matched to $\bar{S}-K$ at that threshold. Boundedness at zero removes the negative power. Thus the fixed-threshold value of a perpetual call isIt is also $(\bar{S}-K){\mathbb{E}}_S^{\mathbb{Q}}[e^{-r{\tau }_{\bar{S}}};{\tau }_{\bar{S}}<\infty ]$, where ${\tau }_{\bar{S}}$ is the first up-crossing stopping time. Differentiating while keeping the current $S$ fixed givesThis is positive for $K<\bar{S}<\hat{S}=pK/(p-1)$, so increasing such an early threshold increases its continuation value. The comparison assumes $S<\bar{S}$; a policy already exercised has no remaining continuation value to vary.

If $\bar{S}>\hat{S}$, thenSince $\tilde{C}(\bar{S})=\bar{S}-K$, the derivative comparison gives $\tilde{C}(S)<S-K$ for $S$ just below $\bar{S}$. Therefore quoting this policy value as the price of an unrestricted American call permits arbitrage: buy the option below its immediate exercise value and exercise it at once. A contract that actually commits its holder to the threshold is a different claim; its lower value is a suboptimal policy value, not by itself an arbitrage. The perpetual American call option optimizes over all exercise policies and remains above its exercise payoff.