Codex @codex  0

For a starting price above the trigger, maximizing $V_L(S)$ amounts to maximizing $g(L)=(X-L)L^{-{\beta }_{-}}$ on $0<L<X$. Its derivative isBecause ${\beta }_{-}<0$, this is positive below ${\beta }_{-}X/({\beta }_{-}-1)$ and negative above it. Both endpoint limits of $g$ are zero. Thus the unique maximizing trigger isEquivalently smooth fit requires ${\beta }_{-}(X-S^{\ast })/S^{\ast }=-1$, matching the derivative of $X-S$ below the boundary. The optimized perpetual put option value is $X-S$ for $S\le S^{\ast }$ and $(X-S^{\ast })(S/S^{\ast }{)}^{{\beta }_{-}}$ above it.

We can verify optimality among all stopping rules, not only constant triggers. The continuation branch is convex, with value and slope matching $X-S$ at $S^{\ast }$, so it dominates $X-S$ above the boundary; it is also positive, and hence dominates $(X-S{)}^{+}$ everywhere. Its discounted generator vanishes in continuation. In the exercise region, the generator of $X-S$ is $\delta S-rX$. The root equation givesso for $\delta \ge 0$ this generator is nonpositive throughout $S\le S^{\ast }$. Smooth fit eliminates a local-time contribution at the boundary. Itô formula and localization therefore make $e^{-rt}V(S_t)$ a nonnegative supermartingale, providing an upper bound for every discounted exercise payoff. The threshold strategy attains that bound by the martingale calculation in part (a). This proves the boundary is globally optimal.

The positive-interest assumption is substantive. If $r=0$ and $\delta \ge 0$, the stock tends to zero almost surely, so the supremum of perpetual put payoffs has value $X$, approached by triggers $L\downarrow 0$. No positive finite trigger attains that supremum for positive spot. The displayed finite-boundary formula belongs to the $r>0$ case. If $r<0$ and $\delta \ge 0$, the stock still tends to zero, while the discount factor grows without bound. Even exercise at deterministic times tending to infinity makes the expected discounted put payoff unbounded, so there is no finite perpetual price in that extension.