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The terminal density is strictly positive and has expectation one under the usual deterministic initial bond-price convention. Its density process isFor $0\le s\le t\le T$, the Bayes formula for conditional expectation under a change of measure givesThe same calculation at $s=0$ gives the finite expectation $1/P(0,T)$, so this is a true martingale, not just a formal conditional identity. Thus the continuous-time bank account measured in units of the maturity-$T$ bond is a ${\mathbb{Q}}_T$-martingale. This is the forward measure change of numéraire. If the initial bond price were random rather than given, integrability of its reciprocal would need to be included for this true-martingale assertion.

The integral against the local-martingale vector $YP$ is a local martingale, provided the predictable holdings are stochastically integrable. Integrating part (b) givesBoth $Y_t{X}_t$ and the cumulative deflated consumption are nonnegative. Hence $M_t\ge -Y_0{X}_0$. With the usual finite deterministic initial capital, the shifted processis a nonnegative local martingale, and is therefore a supermartingale. For completeness, a localizing sequence turns it into true martingales; conditional Fatou lemma for their nonnegative stopped values gives the supermartingale inequality and ordinary Fatou gives integrability at each time. Subtracting the initial constant provesThis is supermartingale control of deflated consumption gains. It uses $c\ge 0$ as well as $X\ge 0$; an unrestricted stochastic integral is not necessarily a true supermartingale.

Concavity gives the supporting-tangent inequalityMultiply by the discount factor $e^{-bt}$ and use the first-order condition $e^{-bt}{U}^{\prime }(c_t)=Y_t$. This gives the pointwise marginal-utility verification of optimal consumption inequalityApply the budget inequality from part (d) to the competing admissible strategy, and use equality for the proposed one:The two weighted consumption integrals are finite, so their difference is integrable. Under the usual positive discount-rate assumption $b>0$, the utility integrals are also integrable: $U(0)\le U(x)\le L$ for a finite upper bound $L$, and ${\int }_0^{\infty }{e}^{-bt}dt=1/b$. Integrating the tangent inequality and taking expectations therefore givesEconomically, both consumers face the same state-price budget, and the candidate spends it exactly where its discounted marginal utility equals the state price. This is utility duality with martingale deflators in its consumption form.

A positive $b$, or another hypothesis making the infinite-horizon objectives well defined and permitting this integration, is needed. The printed question does not specify the sign of $b$. Bounded utility alone does not ensure that an undiscounted infinite time integral exists: a bounded integrand can have both infinite positive and negative parts. Thus the conclusion is established under the standard discount convention $b>0$, and also whenever the displayed objectives satisfy the stated integrability conditions; without either convention the literal infinite-horizon comparison need not be a defined mathematical expression.

This can occur within an admissible financial model, not just for an abstract bounded integrand. Take $U(x)=1-2e^{-x}$ and $b=0$. Choose a deterministic smooth nonnegative $c$ with successive unit-length plateaus alternating between zero and the integer $n$, and connect them over intervals whose lengths have finite sum. Then ${\int }_0^{\infty }2c_t{e}^{-c_t}dt<\infty$: the high-plateau contributions sum to ${\sum }_{n\ge 1}2n{e}^{-n}$ and the transition integrand is bounded by $2/e$. SetThe bank account is a positive deterministic Itô process and $Y_t{B}_t=Y_0$, so $Y$ is a state-price density. Holding $X_t/B_t$ bank units finances consumption with nonnegative wealth and exact budget equality. Also $U^{\prime }(c_t)=Y_t$. Nevertheless every zero plateau contributes $-1$ to the utility integral, while each sufficiently high plateau contributes a fixed positive amount. Both its negative and positive parts are infinite, so the undiscounted objective is undefined. This establishes why the missing discount or objective-integrability hypothesis is substantive.

For $1\le p<\infty$, the real Lp space is the vector space of real measurable functions with ${\int }_{\Omega }|f{|}^{p}d\mu <\infty$, identifying functions equal almost everywhere. Its Lp norm is $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. For $p=\infty$, take the essentially bounded real measurable functions, with the same identification and norm $\Vert f{\Vert }_{\infty }=\mathrm{esssup}|f|$. The identification makes each norm definite; absolute homogeneity follows from the Lebesgue integral, and the triangle inequality follows from Minkowski inequality for finite $p$, or directly from the essential supremum for $p=\infty$.

Here is a completeness proof valid on any measure space. For $p<\infty$, a Cauchy sequence $(f_n)$ has a subsequence $(f_{n_k})$ with $\Vert f_{n_{k+1}}-f_{n_k}{\Vert }_p\le 2^{-k}$. Choose measurable function representatives and put $h_k=f_{n_{k+1}}-f_{n_k}$. By Minkowski inequality and the monotone convergence theorem,Consequently the series $\sum h_k$ converges absolutely almost everywhere. Define $f=f_{n_1}+\sum h_k$ there, and define it to be zero on the measurable exceptional null set. Then $f\in L^p$, and Fatou lemma applied to each tail gives $\Vert f-f_{n_k}{\Vert }_p\le {\sum }_{j=k}^{\infty }{2}^{-j}\to 0$. The original Cauchy sequence also converges in Lp norm, by the triangle inequality. For $p=\infty$, choose the same subsequence using the essential supremum norm. Outside one measurable null set, all the bounds $|h_k|\le 2^{-k}$ hold and $f_{n_1}$ is bounded. The series then converges uniformly there, with an essentially bounded measurable function limit and the same tail estimate in essential supremum norm. Thus all these spaces are Banach spaces.

The duality of Lp spaces says that, for $1<p<\infty$ and $q=p/(p-1)$, the mapis an isometric isomorphism of normed spaces. This form of Lp duality on an arbitrary measure space requires no finiteness hypothesis on $\mu$. At $p=1$, a standard version assumes a sigma-finite measure and identifies $(L^1{)}^{\ast }$ with $L^{\infty }$ through the same dual pairing. That endpoint assertion must not be made without a suitable measure-space hypothesis.

We first prove the required duality of Lp spaces for a finite measure. Let $F\in (L^p{)}^{\ast }$ and set $\nu (E)=F(1_E)$. For disjoint measurable $E_j$, the indicator functions of their partial unions converge in Lp norm to that of their union, so $\nu$ is countably additive. It has finite variation measure: for every finite measurable partition $(E_j)$, choosing real signs givesAlso $\mu (E)=0$ implies $\nu (E)=0$. The Radon-Nikodym theorem supplies a Radon-Nikodym derivative $g\in L^1$ with $\nu (E)={\int }_{E}gd\mu$. Linearity gives $F(f)=\int fgd\mu$ for simple functions. Uniform approximation by simple functions extends this identity to bounded measurable functions: both their Lp norm errors and the errors in integration against $g$ tend to zero.

To establish the correct integrability, test with the bounded measurable function $f_N=\mathrm{sgn}(g)|g{|}^{q-1}{1}_{\{|g|\le N\}}$. Since $(q-1)p=q$, writing $A_N={\int }_{\{|g|\le N\}}|g{|}^{q}d\mu$ givesThe second inequality is also valid when $A_N=0$. The monotone convergence theorem gives $g\in L^q$ and $\Vert g{\Vert }_q\le \Vert F\Vert$. Density of simple functions in the Lp space, together with Hölder's inequality, now gives $F=I_{p}g$ on all of $L^p$. Conversely Hölder's inequality gives $\Vert I_{p}g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, testing againstgives $\Vert f{\Vert }_p=1$ and $(I_{p}g)(f)=\Vert g{\Vert }_q$, so $\boxed{\Vert I_{p}g\Vert =\Vert g{\Vert }_q}$. This also proves uniqueness of the representing Radon-Nikodym derivative.

For completeness, the passage to an arbitrary measure space can be made without losing a hypothesis in the reflexive Banach space argument below. On a sigma-finite measure space, exhaust by nested finite-measure sets $E_n$. The representing Radon-Nikodym derivatives on $E_n$ agree on overlaps by uniqueness. Their glued density has Lp norm $\Vert g{\Vert }_q\le \Vert F\Vert$ by the monotone convergence theorem, and represents $F$ because $f{1}_{E_n}\to f$ in Lp norm. Every $f\in L^p$ on an arbitrary measure space is supported on a sigma-finite measurable set: the sets $\{|f|>1/n\}$ have finite measure and their union is $\{f\ne 0\}$.

Use support localization of an Lp functional as follows. For a sigma-finite measurable $A$, let $m_A$ be the operator norm of $F$ restricted to functions supported in $A$. The support observation gives ${\sup }_A{m}_A=\Vert F\Vert$. Choose $A_n$ approaching this supremum and set $A={\bigcup }_n{A}_n$; then $m_A=\Vert F\Vert$. If a sigma-finite $B\subseteq \Omega \setminus A$ had $m_B=\beta >0$, functions supported on the disjoint sets $A,B$ have the direct-sum Lp norm. Optimizing their two scalar coefficients by Hölder's inequality, and using functions approaching the two restriction operator norms, would givea contradiction. Thus $F$ vanishes on functions supported outside $A$. The density on $A$, extended by zero, represents $F$ globally. The case $F=0$ simply uses $g=0$. This proves the stated Lp duality on an arbitrary measure space.

Finally let $X=L^p$ and let $\Phi \in X^{\ast \ast }$, where stars denote continuous dual spaces. Compose with $I_p$ to obtain the bounded linear functional $g\mapsto \Phi (I_{p}g)$ on $L^q$. Applying duality of Lp spaces with the exponents reversed gives $f\in L^p$ with $\Phi (I_{p}g)=\int fgd\mu$. For the canonical embedding into the bidual $J_{X}f$, its value on $I_{p}g$ is also $\int fgd\mu$. Since $I_p$ is onto, $\Phi =J_{X}f$. Its norm is $\Vert f{\Vert }_p$ by the same dual pairing norm identity. Therefore $\boxed{J_X(L^p)=(L^p{)}^{\ast \ast }}$, which proves that $L^p$ is a reflexive Banach space through its actual canonical embedding into the bidual.

For the simple predictable process from part (a), define its stochastic integral byThe time-zero value contributes nothing. Write $D_i=M_{t\wedge t_{i+1}}-M_{t\wedge t_i}$. If $i<j$ and $t>t_j$, then $h_i{D}_i{h}_j$ is ${\mathcal{F}}_{t_j}$-measurable and $\mathbb{E}[D_j\mid {\mathcal{F}}_{t_j}]=0$. If $t\le t_j$, then $D_j=0$. Boundedness ensures integrability, so all off-diagonal terms vanish by conditional expectation. ConsequentlyThe quadratic variation theorem says $M^2-[M]$ is a local martingale. A bounded continuous martingale is square-integrable, and localization gives the conditional second-moment identityMultiplication by $h_i^2$ and summation therefore give the Itô isometryFor Brownian motion, $[M{]}_s=s$. More generally the Itô isometry holds for every predictable process with finite right-hand side, against a continuous square-integrable martingale. Indeed, on a fixed horizon the quadratic-variation measure ${\mu }_M(E)=\mathbb{E}{\int }_0^t{1}_{E}d[M]$ is finite. Part (c) approximates the integrand by simple predictable processes; the displayed equality makes the elementary integrals Cauchy in $L^2$ and defines their unique limit. The finite sum is exactly the elementary case of the Itô isometry.

A martingale with the specified terminal value must equal its conditional expectation. If ${\hat{S}}_t=0$, default has already occurred and the terminal European call option payoff is zero. If ${\hat{S}}_t>0$, put $h=T-t$. Conditional on survival to $T$,where $Y$ has the standard normal distribution; the conditional probability of this survival is $e^{-\lambda h}$. The conditional expectation of the payoff in the enlarged filtration used in part (c) is consequentlyThis expression is already measurable with respect to the natural filtration of $\hat{S}$, so the tower property of conditional expectation gives the same value there. The survival factor cancels the compensating growth factor:The separate zero branch avoids division by zero. At $t=T$ it gives the required payoff, using $F(0,m)=(1-m{)}^{+}$; integrability follows from $({\hat{S}}_T-K{)}^{+}\le {\hat{S}}_T$.

If wealth and consumption are nonnegative, this local martingale is bounded below by minus the finite initial deflated capital. Adding that capital gives a nonnegative local martingale, hence a supermartingale by the Conditional Fatou lemma. Consequently expected discounted consumption cannot exceed initial deflated wealth. The consumption sign and predictable stochastic integrability are part of the hypotheses.