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A strictly increasing differentiable concave utility function on all of $\mathbb{R}$ has $U^{\prime }(x)>0$ everywhere. Otherwise monotonicity and the decreasing derivative would make it constant on a right half-line. Its tangent at zero also shows $U(x)\to -\infty$ as $x\to -\infty$.

If $X$ has only one sign, the stated second alternative already holds. Otherwise both $\mathbb{P}(X<0)$ and $\mathbb{P}(X>0)$ are positive. As $\theta \to \infty$, $-U(\theta X)$ tends to infinity on $\{X<0\}$; as $\theta \to -\infty$, it does so on $\{X>0\}$. Since $-U\ge 0$, Fatou lemma implies $F(\theta )\to -\infty$ at both ends. Continuity gives a finite maximizer ${\theta }_{\ast }$, and differentiability gives $F^{\prime }({\theta }_{\ast })=0$.

For optimal marginal utility as a one-period pricing density, put $Z=U^{\prime }({\theta }_{\ast }X)$. It is strictly positive. On $\{|X|\ge 1\}$ it is bounded by $|X|U^{\prime }({\theta }_{\ast }X)$, whose expectation is finite by part ii. On $\{|X|<1\}$ continuity bounds $U^{\prime }$ on the compact interval $[-|{\theta }_{\ast }|,|{\theta }_{\ast }|]$. Thus $Z$ is integrable and $XZ$ is absolutely integrable. Consequently $\boxed{\mathbb{E}[XZ]=0}$. Normalizing $Z$ by its positive expectation, if desired, makes it a probability density with the same zero pricing expectation.

For marginal utility pricing with proportional transaction costs, if $X-\epsilon \ge 0$ almost surely or $-X-\epsilon \ge 0$ almost surely, the source's second alternative holds. Otherwise $\mathbb{P}(X<\epsilon )>0$ and $\mathbb{P}(X>-\epsilon )>0$. The Fatou lemma argument now gives $G(\theta )\to -\infty$ at both ends, since a large positive holding loses on the first event and a large negative holding loses on the second. Let ${\theta }_{\ast }$ be a finite maximizer.

If ${\theta }_{\ast }>0$, set $Z_0=U^{\prime }({\theta }_{\ast }(X-\epsilon ))$. The derivative condition gives $\mathbb{E}[X{Z}_0]=\epsilon \mathbb{E}{Z}_0$. As before, split according to $|X-\epsilon |\ge 1$ to obtain integrability of $Z_0$ from the absolutely integrable derivative, with a compact-set bound on the complement. If ${\theta }_{\ast }<0$, use $Z_0=U^{\prime }({\theta }_{\ast }(X+\epsilon ))$ and obtain $\mathbb{E}[X{Z}_0]=-\epsilon \mathbb{E}{Z}_0$. In either case normalize $Z=Z_0/\mathbb{E}{Z}_0$; then $Z>0$, $\mathbb{E}Z=1$, and $XZ$ is absolutely integrable.

If ${\theta }_{\ast }=0$, a concave maximum has $G_{-}^{\prime }(0)\ge 0\ge G_{+}^{\prime }(0)$, so $\mathbb{E}X\in [-\epsilon ,\epsilon ]$. Choose $Z=1$. In every case of this first alternative, $\boxed{\mathbb{E}[XZ]\in [-\epsilon ,\epsilon ]}$. The normalization establishes a stronger, economically meaningful statement than the literal unnormalized condition: the expected gain under the resulting density lies inside the transaction-cost spread.

The stopped processes are nonnegative martingales. Their expectations equal the finite deterministic value $X_0$. The Fatou lemma gives, for each fixed integer $t$,Thus every $X_t$ is integrable. Apply the preceding discrete-time criterion to conclude $X$ is a martingale, not merely a supermartingale. This is the nonnegative discrete-time local martingale is a martingale result; its conclusion does not extend to arbitrary continuous-time nonnegative local martingales.

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.

Write $V_t=⟨X{⟩}_t$ for the quadratic variation and setThe Hölder factorization of stochastic exponentials follows by adding exponents:ThusThe subtraction inside the numerator is $\sqrt{pq}-1$, outside the square root.

The stochastic exponential $\mathcal{E}(\sqrt{pq}X)$ starts at one and is a nonnegative local martingale, hence a supermartingale. The optional sampling theorem for a supermartingale gives expectation at most one at bounded stopping times. For a finite, possibly unbounded, stopping time $T$, apply this to $T\wedge n$, then use continuity and the Fatou lemma:The Holder inequality with exponents $q$ and $q/(q-1)$ now givesThe inequality also holds when the right side is infinite.

Use deterministic times tending to infinity and the Fatou lemma:Apply the terminal scaling inequality for stochastic exponentials to $X=aM$ and $r=1/a$. Part (c) gives $\mathbb{E}[\mathcal{E}(aM{)}_{\infty }]=1$, soLetting $a\uparrow 1$ shows that the terminal expectation is at least one. The nonnegative local martingale $\mathcal{E}(M)$ starts at one and is a supermartingale, so the Fatou lemma gives the opposite inequality. Consequently

The terminal expectation criterion for a nonnegative local martingale now closes the argument. Conditional Fatou lemma applied to the supermartingale at times tending to infinity givesThe left side has expectation one and the right side at most one, so equality holds almost surely. Thusand uniform integrability of conditional expectations proves that $\mathcal{E}(M)$ is a uniformly integrable martingale. This establishes the required stopped-moment form of the Kazamaki criterion.

For $S_0>0$, the stochastic exponential solution stays strictly positive. Put $U_t=\sqrt{S_t}$. The Itô formula givesThus $U$ is a nonnegative local supermartingale. To justify the true supermartingale property, stop where $U$ or its stochastic integral exceeds successive bounds. The stopped Itô formula gives $\mathbb{E}[U_{T\wedge {\tau }_m}\mid {\mathcal{F}}_t]\le U_{t\wedge {\tau }_m}$ for $T\ge t$. The conditional Fatou lemma and ${\tau }_m\uparrow \infty$ yieldIn particular $\mathbb{E}{U}_T\le U_0<\infty$. The case $S_0=0$ is the identically zero process. This is the square-root stock supermartingale.

Normalize $B_0=1$ and put ${\tilde{S}}_t=e^{-rt}{S}_t$. Its dynamics are $d{\tilde{S}}_t={\tilde{S}}_t\sigma (t,S_t)d{W}_t$. Bounded volatility makes this stochastic exponential a true martingale on each finite horizon, by the Novikov condition. It also gives a finite second moment: stopping the Itô formula for $S^2$ and applying the Gronwall inequality yields $\mathbb{E}{S}_T^2\le S_0^2{e}^{(2r+L^2)T}$ when $\sigma \le L$.

The discounted payoff $\xi =e^{-rT}(S_T-K{)}^{+}$ is thus square-integrable. Define the nonnegative martingaleThe Brownian martingale representation theorem says that every square-integrable martingale in the Brownian filtration has a representation $M_t=M_0+{\int }_0^t{h}_{u}d{W}_u$ with predictable $h$ and $\mathbb{E}{\int }_0^T{h}_u^{2}du<\infty$. Since $\tilde{S}>0$ and $\sigma >0$, chooseThen discounted gains satisfy $d{M}_t={\pi }_{t}d{\tilde{S}}_t$. Consequently $X_t=B_t{M}_t={\varphi }_t{B}_t+{\pi }_t{S}_t$ is self-financing, nonnegative, and hence an admissible trading strategy. At maturity $X_T=(S_T-K{)}^{+}$, proving claim replication at cost $C(T,K)$.

For minimality, discounted wealth of any admissible self-financing portfolio is a local martingale bounded below, hence a supermartingale by localization and the conditional Fatou lemma. Thus any such replication with initial wealth $x$ obeysTogether with the constructed portfolio, this provesThis is Brownian representation replication in a local volatility market. The given drift $r$ means that the original probability measure already serves as the risk-neutral measure.

The original PDF assumes $u\ge 0$ at the beginning of this question; the TeX conversion omits that hypothesis. First replace a smooth nonnegative solution by $v=u+\epsilon >0$, which solves the same homogeneous divergence-form elliptic equation. For $w=v^{q/2}$, the chain rule gives the favorable termin the distributional sense. Thus $w$ is a weak subsolution. Testing its subsolution inequality with ${\zeta }^{2}w$ yields the same Caccioppoli inequality as before. More precisely, testing the equation for $v$ with ${\zeta }^2{v}^{q-1}$ givesIndeed the main term is at least $(q-1)\lambda \int {\zeta }^2{v}^{q-2}|\nabla v{|}^2$, and the cross term is bounded by twice $\Lambda$ times the square roots of this weighted gradient integral and $\int v^q|\nabla \zeta {|}^2$. Let $\epsilon \downarrow 0$ to obtain the stated power energy estimate.

Apply the fixed-domain Sobolev embedding theorem to $\zeta u^{q/2}$. The crucial constant stays uniform before taking the $2/q$ power:where $D_{\ell }\ge 1$ is independent of $q$. This is uniform power gain for elliptic solutions. The exponent is $q\alpha$, not the TeX conversion's $q^{\alpha }$. The PDF's weaker prefactor $C_{\ell }$ outside the $2/q$ power follows from this stronger estimate, but that weaker form alone cannot be iterated infinitely.

For completeness, the estimate also extends to nonnegative locally $H^1$ weak solutions without assuming they are already bounded. For $s\ge 0$ define $F_M(s)=s^{q/2}$ up to $M$ and continue it linearly with slope $(q/2)M^{q/2-1}$ above $M$; set ${\Psi }_M(s)={\int }_0^s{F}_M^{\prime }(r{)}^{2}dr$. Both have bounded derivatives for fixed $M$. Direct calculation gives ${\Psi }_M/F_M^{\prime }\le F_M$ and $F_M(s)\le (q/2)s^{q/2}$. Test the weak equation with ${\zeta }^2{\Psi }_M(u)$, using smooth approximations to these Sobolev chain rule functions if needed. The resulting weighted energy bound is $\int {\zeta }^2|\nabla F_M(u){|}^2\le 4R^2\int F_M(u{)}^2|\nabla \zeta {|}^2$. If $u\in L^q(B_{r_2})$, dominated convergence on the right and the Fatou lemma on the Sobolev left give the boxed gain as $M\to \infty$, with the same uniform constant. Starting with $q=2$, this justifies every subsequent iteration step.

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 nonnegative measurable functions $f_n$ on a measure space, the Fatou lemma statesIn particular $\mathbb{E}[\mathrm{liminf}{X}_n]\le \mathrm{liminf}\mathbb{E}{X}_n$ for nonnegative random variables; infinite values are allowed. Nonnegativity, or a common integrable lower bound after subtraction, is essential.

Since $\phi (0)=1$ and $\phi (h)+\phi (-h)=2\mathbb{E}\cos (hX)$, the expression in the hypothesis equals $\mathbb{E}{Z}_n$, with $Z_n=2[1-\cos (h_{n}X)]/h_n^2\ge 0$. For every finite real $X$, the Taylor expansion of cosine gives $Z_n\to X^2$. The Fatou lemma therefore givesNo differentiability of the characteristic function or prior integrability of $X^2$ was assumed; the nonnegative difference quotient establishes it.

Choose continuous representatives using Morrey's inequality. On the closed unit ball they are uniformly bounded by $CM$ and satisfyThe Arzela-Ascoli theorem therefore gives a uniformly convergent subsequence with a continuous limit. For $p=\infty$, use the uniform Lipschitz estimate instead; this part remains valid at that endpoint.

For weak lower semicontinuity of the Hilbert norm, weak convergence gives $⟨u_n-\varphi ,\varphi ⟩\to 0$, and henceTaking the lower limit proves $\boxed{\Vert \varphi \Vert \le \underset{n}{\mathrm{liminf}}\Vert u_n\Vert }$.

For the nonlinear integral, use local Sobolev compactness gives lower semicontinuity of a nonnegative integral. Select a subsequence attaining the lower limit of the integrals. The Rellich-Kondrachov compactness theorem on successively larger balls gives a diagonal subsequence converging strongly in local L2 space to $\varphi$. After a further subsequence it converges almost everywhere. Since $1-\cos s\ge 0$, the Fatou lemma givesEach integral is finite because $0\le 1-\cos s\le s^2/2$. The argument uses local compactness rather than convexity: $1-\cos s$ is not a convex function on the whole real line.

Write $B=-A=-{\partial }_x^2+1+x^2$. Smooth compactly supported functions lie in $D(A)$ and are dense in L2 space, so $A$ is densely defined.

For $\lambda >0$, use the formon the one-dimensional harmonic oscillator form domain $X$. It is bounded and coercive in the $X$ norm. The Lax-Milgram theorem gives a unique $u\in X$ satisfying $a_{\lambda }(u,v)=⟨g,v⟩$ for every $v\in X$. Thus $-u^{\prime \prime }+(1+\lambda +x^2)u=g$ in distributions.

We must verify the full specified operator domain, not just the form domain. The equation first gives $u\in H_{\mathrm{loc}}^2$. For compactly supported $H^2$ functions, integration by parts gives the graph estimate for the harmonic oscillator, obtained fromApply this to $v={\chi }_{R}u$, with ${\chi }_R=1$ on $[-R,R]$, supported in $[-2R,2R]$, and derivatives bounded by $C/R$ and $C/R^2$. Its right-hand source is ${\chi }_{R}g-2{\chi }_R^{\prime }{u}^{\prime }-{\chi }_R^{\prime \prime }u$, uniformly bounded in L2 space. The estimate and the Fatou lemma give $x^{2}u\in L^2$; the original equation then gives $u^{\prime \prime }\in L^2$, and hence $u\in H^2$. Thus $u\in D(A)$, proving that $\lambda -A$ is onto. The energy identity proves injectivity and the resolvent of the shifted harmonic oscillator estimateTo prove closedness, suppose $u_n\to u$ and $A{u}_n\to v$ in L2 space. For one fixed $\lambda >0$,Therefore $u$ belongs to the range of the resolvent, namely $D(A)$, and $Au=v$.

All hypotheses of the Hille-Yosida theorem now hold; the resolvent bound is stronger than $1/\lambda$. Let $S(t)$ be its contraction semigroup. The exponentially shifted semigrouphas generator $A+I={\partial }_x^2-x^2$. It is continuous in L2 space, takes the initial value ${\psi }_0$, and satisfies $\boxed{\Vert \psi (t){\Vert }_2\le e^t\Vert {\psi }_0{\Vert }_2}$. For ${\psi }_0\in D(A)$, its time derivative exists in L2 space and equals $(A+I)\psi (t)$, so the equation holds as a strong L2 space solution. For general initial data it is the semigroup solution, without an unproved pointwise differentiability claim.