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This smooth utility has $U^{\prime }(x)=x^{-1}-x^{-2}+x^{-3}/3>0$ and $U^{\prime \prime }(x)=-(x-1{)}^2/x^4$. Its derivative is strictly decreasing, diverges at zero and tends to zero at infinity, so the utility is strictly concave and satisfies the Inada conditions. Yet its curvature vanishes at one. The inverse marginal utility cannot have a finite derivative at $y=1/3$, since differentiating $U^{\prime }(I(y))=y$ there would give zero equal to one. Its dual is differentiable but not twice differentiable at that price.

The Inada conditions give $U^{\prime }(0+)=\infty$ and $U^{\prime }(\infty )=0$. Differentiability and strict concavity make $U^{\prime }$ continuous and strictly decreasing, with range $(0,\infty )$. Hence the inverse marginal utility $I=(U^{\prime }{)}^{-1}$ exists. The unique maximizer in the utility conjugate is $x=I(y)$, andThe derivative formula does not require differentiability of $I$: compare the optimizing values at $y$ and $y+h$ to squeeze the difference quotient between $-I(y)$ and $-I(y+h)$, and use continuity of $I$. Thus the dual is continuously differentiable, strictly decreasing, and strictly convex, since $-I$ is strictly increasing.

For the requested second-derivative assertion, a curvature hypothesis is missing. Under the intended nondegeneracy $U^{\prime \prime }(x)<0$ for every $x>0$, inverse differentiation givesThis proves the intended dual differentiability with nonvanishing utility curvature. If $U$ is only twice differentiable, it gives pointwise twice differentiability of the dual; continuous second derivatives additionally follow when $U\in C^2$.

Literal strict concavity does not imply nonvanishing curvature. An explicit Inada utility with vanishing curvature isIts derivative is positive, strictly decreasing, tends to infinity at zero and to zero at infinity. It is smooth and strictly concave, but $U^{\prime \prime }(1)=0$. At $y_0=U^{\prime }(1)=1/3$, a finite derivative of $I$ would contradict differentiation of $U^{\prime }(I(y))=y$, giving $0\cdot I^{\prime }(y_0)=1$. Hence ${\hat{U}}^{\prime }$ is not differentiable there. As printed, the twice-differentiable-dual claim is false; it is valid with $U^{\prime \prime }<0$. The remaining differentiability, monotonicity and strict-convexity conclusions above hold under the printed hypotheses.

Available portfolio wealth and the ruin boundary. Put $h=\bar{r}D$, the constant interest payment on the fixed loan. The loan principal is already included in available portfolio wealth; it is not a growing portfolio holding. ThereforeIn particular the interest outflow is $\bar{r}D$, not $(\bar{r}-r)D$. Writing $w=x+D$ would instead give net portfolio wealth drift $rx+(\mu -r)\theta -(\bar{r}-r)D-c$, which explains the distinction.

Let ${\tau }_0$ denote the ruin time, to avoid confusing it with a fixed terminal horizon. The objective stops at ${\tau }_0$; consequently the absorbing boundary is $V(0)=0$, without an obligation to keep financing the loan after ruin. Dynamic programming gives, for $w>0$,For increasing strictly concave value, put $z=V^{\prime }(w)>0$ and use inverse marginal utility $I$. The optimal controls and the optimized HJB equation arewhere $\tilde{U}(z)={\sup }_{c\ge 0}[U(c)-zc]$. For CRRA utility with $0<R<1$, write $p=1-R$ and $q=1-1/R<0$. Then

Dualization and the printed constant. Use the convex wealth-variable Legendre dualAt an interior maximizing portfolio wealth, $J^{\prime }=-w$, $J^{\prime \prime }=-1/V^{\prime \prime }>0$ and $V=J-z{J}^{\prime }$. The dual HJB equation is the linear Euler differential equationA trial power $z^t$ givesDirect substitution givesThe PDF prints an additional factor $1/2$ before ${\kappa }^2/(2R)$ in its definition of ${\gamma }_M$. That printed definition is inconsistent with its own identity for $Q(q)$. The expression above is the one used here; assume this corrected ${\gamma }_M>0$.

Solution when $\kappa \ne 0$. Assume positive discount $\rho$, nonzero $\sigma$, and $h>0$. Let $m>0$ and $n<0$ be the two roots of $Q$:Since $Q(q)<0$, $n<q<0<m$. Put $C=R/(p{\gamma }_M)$. For $r\ne 0$, the general interior solution isThe appropriate large-wealth condition is the Merton consumption-investment problem boundIndeed any original control consumes $c+h$ in the debt-free comparison model until ruin, and $U(c+h)\ge U(c)$. Dualizing this bound gives $0\le J(z)\le C{z}^q$. Because $n<q$, convexity and this upper bound force $B=0$ as $z\downarrow 0$.

At the other endpoint portfolio wealth reaches zero. If $z_{\ast }=V^{\prime }(0+)$, the dual ruin boundary with debt service requiresSolving these two equations givesFor $r>0$, $m>1$, and for $r<0$, $0<m<1$, so the quantity defining $z_{\ast }$ is positive in either case. These formulas determine the entire value. For each $w>0$, choose the unique $z\in (0,z_{\ast })$ satisfyingThenBoth terms in the bracket are positive, even when $r<0$ and $A<0$. Thus $J^{\prime \prime }>0$, $-J^{\prime }$ decreases from infinity to zero as $z$ increases, and the portfolio wealth inversion really is unique. The extended dual is continuously differentiable and convex.

There is no additional condition $J^{\prime \prime }(z_{\ast }^{-})=0$. In factAvailable portfolio wealth is killed at zero; the portfolio can have a nonzero limiting volatility immediately before ruin. Imposing a reflecting-boundary or zero-curvature condition would solve a different problem.

For $r=0$, the linear forcing resonates with the root $m=1$. Put $b=\rho +{\kappa }^2/2=Q^{\prime }(1)$. The dual and boundary constants instead areExtend by zero for $z\ge z_{\ast }$. Hereand the controls remain $c=z^{-1/R}$ and $\theta =\kappa z{J}^{\prime \prime }/\sigma$, with $J^{\prime \prime }=h/(bz)+Cq(q-1)z^{q-2}>0$.

Verification and transversality. The candidate is nonnegative, increasing, strictly concave and zero at ruin. Its HJB equation makes the discounted value plus accrued utility a local supermartingale for every admissible control, and a local martingale for the stated feedback. Localization at positive lower and finite upper portfolio wealth levels gives the finite-horizon comparison. The investment value transversality condition follows from the same debt-free bound: applying the Itô formula to $w^p$ and maximizing its risky term givesThe nonnegative consumption and debt-service drifts only decrease this bound. Thus the expected terminal candidate tends to zero. The feedback has at most linear growth, including a finite limit as portfolio wealth decreases to zero; stopping it at ruin gives an admissible policy. Letting localization levels and then the horizon tend to their limits proves that the candidate is the value, rather than just a formal dual solution. If $D=0$, the absorbing-debt boundary disappears and the ordinary Merton consumption-investment problem formula is recovered.

Zero market price of risk. With $\sigma \ne 0$ and $\kappa =0$, the dual equation is first order. If $\rho >r$, the preceding formulas remain valid with $m=\rho /(\rho -r)$ for $r\ne 0$, and the logarithmic formula with $b=\rho$ for $r=0$; there is no $B$ term and ${\theta }^{\ast }=0$. The same boundary and transversality argument verifies this deterministic consumption policy.

The remaining finite-value case has $r>0$ and $pr<\rho \le r$. Put ${\gamma }_0=(\rho -pr)/R$, $w_c=h/(rR)$ andThe correct convex dual and its corresponding value areThe two value branches have the same value and derivative at $w_c$. Above $w_c$, hold no stock and consume ${\gamma }_0(w-h/r)$; the surplus over $h/r$ grows at rate $r-{\gamma }_0\ge 0$. If $\rho =r$, the lower branch is attained by zero stock holding and constant consumption $ph/R$: portfolio wealth solves $\dot{w}=r(w-w_c)$ until ruin, and direct integration gives $z_{c}w$.

If $\rho <r$ and $0<w<w_c$, the lower branch is a supremum attained in a limit of increasingly rapid fair stock lotteries between zero and $w_c$, followed by the upper-branch policy on success. The success probability tends to $w/w_c$ and the fixed service cost during the lottery tends to zero. This is possible because unrestricted dollar holdings in the nonzero-volatility stock produce a fair Brownian motion exposure even when its excess drift is zero. The supporting linear branch has optimized waiting residual $(r-\rho )z_c(w-w_c)<0$; the fast lotteries, rather than a finite feedback optimizer, supply the missing control limit. The piecewise candidate is concave, has nonpositive waiting residual everywhere, and the preceding moment bound still supplies an upper-bound verification. This degenerate case can have a supremum without an ordinary maximizing strategy.

Pricing kernel and replication. In the nondegenerate Black-Scholes model, put $\kappa =(\mu -r)/\sigma$ and normalize the state-price density by ${\zeta }_0=1$. The process isThe density $e^{rt}{\zeta }_t$ changes probability to the risk-neutral measure. Under that measure $W_t+\kappa t$ is a Brownian motion and the stock drift is $r$. Thus an integrable contingent claim $H$ has time-$t$ priceIn the usual augmented natural Brownian filtration, the Brownian martingale representation theorem supplies a replicating strategy; this is the complete market assumption. For a nonnegative admissible trading strategy without intermediate consumption, the state-price budget constraint is $\mathbb{E}[{\zeta }_T{w}_T]\le w_0$, with equality for a fully invested replicated claim. In particular

Feasibility and the largest slope. First take the intended regime $r>0$, $T>0$, $w_0,S_0>0$, and the usual nonnegative portfolio wealth constraint. The terminal wealth floor isFor any feasible claim the state-price budget constraint impliesTherefore $\alpha \le w_0/S_0$. Conversely, when $0\le \alpha \le w_0/S_0$, hold $\alpha$ shares and put the remaining $w_0-\alpha S_0$ in the continuous-time bank account. Its terminal portfolio wealth isHenceAt equality, $\xi =\bar{\alpha }{S}_T$ has cost exactly $w_0$. Positivity of the state-price density forces $w_T=\bar{\alpha }{S}_T$ almost surely: any strict improvement would cost more. Invest all initial portfolio wealth in $w_0/S_0$ shares and hold them until $T$.

Optimal payoff below the feasibility limit. For $\alpha <\bar{\alpha }$ the floor is strictly positive, and its price is strictly less than $w_0$. Assume the utility function is increasing, differentiable and strictly concave, satisfies the Inada conditions, and has the integrability needed for the finite-budget optimization. These are the usual hypotheses implicit in using inverse marginal utility. For each positive multiplier $\lambda$, maximizeseparately in every state. Its derivative decreases through zero at $I(\lambda {\zeta }_T)$, so the floored marginal utility optimizer isThe multiplier is characterized byUnder the stated integrability hypotheses the left side is continuous and decreasing, tends to the floor cost as $\lambda \uparrow \infty$, and tends to infinity as $\lambda \downarrow 0$. It is strictly decreasing wherever it exceeds the floor cost: on the event where the inverse-marginal-utility payoff exceeds the floor, a larger multiplier strictly reduces that payoff. Therefore the budget determines a unique finite multiplier. For CRRA utility the inverse marginal utility is $I(y)=y^{-1/R}$; lognormal moments provide the needed integrability.

For completeness, pointwise maximality gives, for any feasible competing terminal portfolio wealth $Y$,Taking expectations and using $\mathbb{E}[{\zeta }_{T}Y]\le w_0=\mathbb{E}[{\zeta }_T{w}_T^{\ast }]$ proves optimality. Strict concavity gives uniqueness of the terminal claim. Its price processis nonnegative and starts from $w_0$; claim replication therefore turns the payoff optimizer into an admissible portfolio.

What the missing interest-rate hypothesis changes. The PDF does not explicitly assume $r>0$ or give the utility and admissibility hypotheses above. These omissions matter. At $r=0$, under nonnegative admissibility, the largest feasible slope remains $w_0/S_0$: for larger slopes the positive-part floor costs strictly more than $w_0$, since $S_T$ has support $(0,\infty )$. But every $\alpha \le w_0/S_0$ already gives floor cost exactly $w_0$, so the only feasible terminal claim is $\xi$. There is no spare budget for an inverse-marginal-utility improvement. For example $\alpha =0$, $\kappa \ne 0$ and CRRA utility make $\mathbb{E}[{\zeta }_T\max \{w_0,I(\lambda {\zeta }_T)\}]>w_0$ for every finite $\lambda$. The prescribed positive finite multiplier then does not exist.

For negative $r$, nonnegative admissibility requires the effective floor ${\xi }_{+}=\max \{\xi ,0\}$. Define its costThis is continuous and convex. For $\alpha \le w_0/S_0$ it equals $w_0{e}^{-rT}+\alpha S_0(1-e^{-rT})$, exceeding $w_0$ below the endpoint. At ${\alpha }_0=w_0/S_0$, $G({\alpha }_0)=w_0$ and $G^{\prime }({\alpha }_0)=S_0(1-e^{-rT})<0$. Above that endpoint the lognormal stock gives strict convexity, and $G(\alpha )\to \infty$. Thus there is a unique second root ${\alpha }_1>{\alpha }_0$ of $G({\alpha }_1)=w_0$, the feasible slopes are $[{\alpha }_0,{\alpha }_1]$, and the largest is ${\alpha }_1$. Equivalently, for $\alpha >{\alpha }_0$ the floor price is $\alpha$ times the European call option price with strike $S_0-w_0/\alpha$. At the upper endpoint replicate ${\xi }_{+}$; in the interval with strict budget slack the same floored marginal utility optimizer applies with ${\xi }_{+}$.

If instead portfolio wealth may be negative and utility is defined on all real portfolio wealth, the floor itself has its affine replication cost: at $r=0$ every slope is feasible, and at negative $r$ every $\alpha \ge w_0/S_0$ is feasible. There is then no largest finite slope. The intended stock-only endpoint and strict-slack optimizer use positive interest and standard nonnegative admissibility.

For an increasing concave utility on positive wealth, its utility conjugate uses the indicated supremum and is a convex function of the positive price variable. If $F(x)=-U(x)$ on positive $x$, extended by infinity elsewhere, then $\hat{U}(y)=F^{\ast }(-y)$. Under the Inada conditions its optimizer is the inverse marginal utility, and its derivative is the negative of that optimizer. The sign convention is part of the definition.