= Solution
{c}
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> is
$$
\boxed{\widetilde C(S)=(\bar S-K)(S/\bar S)^p,\qquad 0\leq S\leq\bar S.}
$$
It 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 gives
$$
\frac{\partial\widetilde C}{\partial\bar S}
=S^p\bar S^{-p-1}\bigl[pK-(p-1)\bar S\bigr].
$$
This 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 $\bar S>\hat S$, then
$$
\widetilde C'(\bar S-)=p\left(1-\frac K{\bar S}\right)>1.
$$
Since $\widetilde C(\bar S)=\bar S-K$, the <derivative> comparison gives $\widetilde C(S)<S-K$ for $S$ just below $\bar S$. Therefore \b[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.
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