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Multiply the representation in part (c) by ${\int }_0^{\infty }{\alpha }_{u}d{W}_u$ and take expectations. This Itô integral has mean zero, and the bilinear form of the Itô isometry givesEquating this with part (b), and writing both ordinary integrals with the same time variable, yieldsAll terms are integrable by the Cauchy-Schwarz inequality, the assumed square integrability of $\alpha$, and boundedness of ${\varphi }^{\prime }$. This is an orthogonality statement against predictable processes; its second term need not itself be predictable.

Let $V_t=⟨M{⟩}_t$, and first suppose $V_{\infty }=\infty$ almost surely in addition to strict increase. The Dambis-Dubins-Schwarz theorem says thatdefine a Brownian motion $B$ in the time-changed filtration and giveEach $T_u$ is a stopping time and is finite. Continuity and strict increase of $V$ make $u\mapsto T_u$ continuous, $V_{T_u}=u$ and $T_{V_t}=t$.

Here are the martingale details behind this inverse-clock proof of the Dambis-Dubins-Schwarz theorem. Stopping a continuous local martingale when its bracket reaches $n$ makes it an L2-bounded continuous martingale. This follows from the stopped Itô isometry or from the $p=2$ estimate proved in Question 1(a), applied after localization. In particular it is uniformly integrable, and optional sampling is valid even at an unbounded stopping time by taking limits. Applying this to $M$ stopped at $T_n$ shows that $(M_{T_{u\wedge n}}{)}_{u\ge 0}$ is a martingale. Thus $B$ is a continuous local martingale. Time-changing $M^2-V$ in the same way shows $B_u^2-u$ is a local martingale, so $⟨B{⟩}_u=u$.

For completeness, the Lévy characterization of Brownian motion follows directly from the Itô formula. If a continuous local martingale $N$, starting at zero, has bracket $u$, thenis a complex local martingale. Its modulus is bounded on each deterministic finite horizon, so it is a true martingale there. ConsequentlyConditional characteristic functions give Gaussian increments independent of the past. Iterating this identity gives independent increments, and continuity completes the Brownian characterization. The identical vector argument proves the Lévy characterization of multidimensional Brownian motion when the bracket matrix is $uI$.

The printed strict-increase hypothesis does not imply $V_{\infty }=\infty$. For example, $M_t={\int }_0^t{e}^{-s}d{W}_s$ has strictly increasing bracket $(1-e^{-2t})/2$. To state the theorem under exactly the printed hypothesis, allow an independent enlargement of the probability space if the terminal clock $L=V_{\infty }$ can be finite.

On $\{L<\infty \}$ the martingale $M$ has a finite terminal limit. Indeed, stopping at each bracket level $n$ gives an L2-bounded continuous martingale which converges; on $\{V_{\infty }<n\}$ the stopped process is the original one. This proves the finite-bracket convergence lemma. Set $T_u=\infty$ when $u\ge L$ and continue $N_u=M_{T_u}$ by that terminal limit. The optional-sampling argument just given makes $N$ a continuous local martingale with bracket $u\wedge L$. Moreover $L$ is a stopping time in ${\mathcal{G}}_u={\mathcal{F}}_{T_u}$.

On a product extension add an independent Brownian motion $\beta$ in clock time, and putThe two summands have zero quadratic covariation, and their brackets are $u\wedge L$ and $(u-L{)}^{+}$. Thus $⟨\tilde{B}{⟩}_u=u$; the proved characterization makes $\tilde{B}$ Brownian. Since $V_t<L$ at every finite $t$ when $L$ is finite, $M_t={\tilde{B}}_{V_t}$ still holds. This is the finite-lifetime extension of the Dambis-Dubins-Schwarz theorem. An infinite clock gives Brownian motion on the original space; a finite clock may require the independent extension.

For a finite list of real coefficients $(a_j{)}_{j\le m}$,A deterministic Itô integral is centered Gaussian: first verify it for step functions as a linear combination of independent Gaussian Brownian increments, then pass to an L2 approximation using the Itô isometry and characteristic functions. Its variance is the squared L2 norm of the integrand, which is ${\sum }_j{a}_j^2$ by the supplied orthonormality. ThereforeThis factors as the joint characteristic function of independent standard normals. Since every finite subfamily has this law, the entire sequence is independent and each ${\xi }_n$ has law N(0,1). This is the Gaussian coordinates of deterministic orthonormal Wiener integrands principle.

Multiply the linear stochastic differential equation by $e^{-at}$. The Itô product rule givesThe integrand is deterministic, so the Itô integral has a centered normal distribution. Its variance, by the Itô isometry, isHenceAt $a=0$ the correct continuous-limit formula is $N(z,b^{2}t)$. A zero variance, for example when $b=0$, denotes the deterministic distribution. For $a<0$ the variance is still positive because both numerator and denominator in its quotient are negative. This is the explicit Ornstein-Uhlenbeck solution, allowing either sign of the linear drift coefficient.

The angle is a bounded local martingale, hence a uniformly integrable martingale. The martingale convergence theorem gives an almost sure and $L^1$ limit ${\Theta }_{\infty }\in [-\pi /2,\pi /2]$ withIt must lie at an endpoint. Indeed, the Itô isometry and boundedness giveThus the bracket integral is finite almost surely. If the angle converged to an interior point, its squared cosine would eventually be bounded below by a positive constant, forcing this integral to be infinite. ConsequentlyThis proves endpoint convergence of a bounded angle diffusion.

Let $p_x=\mathbb{P}({\Theta }_{\infty }=\pi /2)$. Since $X_t=\tan {\Theta }_t$, this is exactly $\mathbb{P}(X_t\to +\infty )$, and the other endpoint means $X_t\to -\infty$. Taking expectations gives

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.

The Novikov condition holds on every horizon, sinceTherefore $Z=\mathcal{E}(M)$ is a true martingale and the Girsanov theorem applies through each finite $T$.

The deterministic bound also gives a genuine infinite-horizon density. For any $p>1$, the stochastic exponential identity isA nonnegative local martingale is a supermartingale, soThis $L^p$ bound gives uniform integrability. Also $M$ is a L2-bounded continuous martingale, since localization and the Itô isometry give $\mathbb{E}{M}_t^2=\mathbb{E}[M{]}_t\le C$. The L2 martingale convergence theorem gives a finite limit $M_{\infty }$; hence $Z_{\infty }=\exp (M_{\infty }-[M{]}_{\infty }/2)>0$ and $\mathbb{E}{Z}_{\infty }=1$. Weighting by $Z_{\infty }$ defines an equivalent probability measure on ${\mathcal{F}}_{\infty }$. This is the bounded-bracket criterion for a stochastic exponential.

The printed assertion is false as a supremum bound uniform over shrinking bandwidths. The supplied hint proves a pointwise stochastic order with constants uniform in $x$. Taking the supremum inside the probability changes the problem: a growing number of spatial windows produces a logarithmic cost. We first establish the valid consequence of the hint, then give a counterexample satisfying the printed assumptions.

For $I=I_{x,h}=[x-h,x+h]$, let $p=\mu (I)$, $A_T={\int }_0^T{1}_I(X_t)dt$ and $N_T={\int }_0^T{1}_I(X_t)\sigma (X_t)d{W}_t$. The Invariant distribution of an Itô diffusion has a probability density function bounded above and bounded away from zero on the required compact interval. Choosing $h_0<M/4$ therefore gives constants $c_0,C_0>0$ such that $c_{0}h\le p\le C_{0}h$, uniformly for $|x|\le M/2$ and $0<h<h_0$.

For the centered function $f=1_I-p$, $\int |f|d\mu =2p(1-p)\le 2p$. Outside $[-M,M]$, $f=-p$. The given second-moment estimate and the Chebyshev inequality yieldOn the event $A_T\ge Tp/2$, the drift coefficient contributes an average whose error is at most $R{h}^{\alpha }$, by Hölder continuity. The Itô isometry and stationarity give $\mathbb{E}{N}_T^2\le \Vert \sigma {\Vert }_{\infty }^{2}Tp$. Another use of the Chebyshev inequality provesThis also accounts for the zero-denominator convention through the first exceptional event. The bound proves pointwise $R{h}^{\alpha }+O_{\mathbb{P}}((Th{)}^{-1/2})$, uniformly in the location's tail probabilities. It supplies no bound for the probability of a supremum over all locations.

For an explicit counterexample, take $\sigma \equiv 1$ and the bounded globally Lipschitz functionIt has Lipschitz constant $R=1$, exponent $\alpha =1$, and satisfies the inward-drift conditions with $M=2$, $\gamma =2$. The stochastic differential equation has a unique strong solution of a stochastic differential equation. Its Invariant distribution of an Itô diffusion has densityThe exponent is zero on $[-1,1]$, equals $-(|x|-1{)}^2$ when $1<|x|<2$, and equals $3-2|x|$ outside $[-2,2]$. This verifies integrability, positivity and a constant density ${\rho }_0=Z^{-1}$ on the central interval. Start the Itô diffusion in this Invariant distribution of an Itô diffusion to obtain the required stationary process.

Set $h=T^{-1/3}$ and choose $N=\lfloor (4h{)}^{-1}\rfloor$ windows with centers $x_j=-1/2+(4j-2)h$, $1\le j\le N$. Their intervals $I_j=[x_j-h,x_j+h]$ are disjoint and lie in $[-1,1]$, where the drift is zero. WriteThen ${\hat{b}}_T(x_j,h)=M_j(T)/A_j(T)$ whenever $A_j(T)>0$. The same occupation bound as above gives $\mathbb{E}(A_j(T)-q{)}^2\le C_{1}T{h}^2$ for $T\ge 1$. With $\delta =T^{-1/6}$, a union bound and the Chebyshev inequality implyThe quadratic variations are $⟨M_j⟩=A_j$, and the cross quadratic covariations vanish since the windows are disjoint. Each clock tends to infinity almost surely: for any fixed window, the occupation estimate at times $2^k$, followed by the Borel-Cantelli lemmas, gives $A_j(2^k)/(2^k\mu (I_j))\to 1$. The Knight theorem for orthogonal martingales therefore represents $M_j(t)=B_j(A_j(t))$ using independent standard Brownian motions $B_1,\dots ,B_N$. This theorem is applied to the finite collection for each $T$; no growing-dimensional central limit theorem is being assumed.

On the preceding clock event, compare each time-changed Brownian motion with its value at deterministic time $q$. The Brownian reflection principle and a union bound show that, for every fixed $\epsilon >0$,This bound does not require the Brownian motions to be independent of their clocks. In particular, ${\max }_j|M_j(T)-B_j(q)|/\sqrt{q}\to 0$ in probability. The variables $B_j(q)/\sqrt{q}$ are independent $N(0,1)$ random variables. Their Gaussian maximum exceeds $\sqrt{\log N}$ with probability tending to one: for $r\ge 1$, integration of the normal density over $[r,r+1/r]$ gives $\mathbb{P}(|Z|>r)\ge c{r}^{-1}{e}^{-r^2/2}$, and henceSince $A_j(T)\le (1+\delta )q$ on the clock event, it follows that, for some $c_3>0$,But $\sqrt{Th}Rh=1$ for this bandwidth. Thus the error after subtraction of $Rh$ and multiplication by $\sqrt{Th}$ is not bounded in probability, contradicting the printed stochastic order. The valid pointwise bound above is provable; the uniform shrinking-bandwidth assertion is not. This counterexample shows why a uniform rate needs at least a $\sqrt{\log (1/h)}$ enlargement of the noise scale. If $h$ is instead held fixed and constants may depend on $h$, this counterexample does not contradict that different interpretation.

Use the Radon-Nikodym derivative at $T_1$ and the discounted bond price martingale from part (b):For the variance, put $b_t^i={\int }_t^{T_i}\sigma (t,u)du$ and $d_t=b_t^2-b_t^1={\int }_{T_1}^{T_2}\sigma (t,u)du$, for $t\le T_1$. The ratio $R_t=P(t,T_2)/P(t,T_1)$ has, by the Itô formula,Under the T-forward measure ${\mathbb{Q}}_{T_1}$, $d{W}_t=d{W}_t^{T_1}-b_t^{1}dt$, so its drift becomes $-d_t^2/2$. Since $R_{T_1}=P(T_1,T_2)$,The first two terms are deterministic; the Itô isometry supplies the variance of the last. The two requested answers areIn fact the terminal zero-coupon bond price has a log-normal distribution under this forward measure, with the deterministic negative half-variance correction ensuring the displayed mean.

A simple predictable process has the form $H_s={\sum }_{j=0}^{m-1}{h}_j{1}_{(t_j,t_{j+1}]}(s)$, where $0=t_0<\cdots <t_m<\infty$ are deterministic and $h_j$ is bounded and ${\mathcal{F}}_{t_j}$-measurable. A possible ${\mathcal{F}}_0$-measurable value at zero is irrelevant to this Itô integral. DefineEach coefficient is known before the corresponding increment. The conditional expectation of each subsequent increment is zero, so the result is a continuous martingale starting at zero. It is an L2-bounded continuous martingale because it is a finite sum of bounded coefficients times stopped increments of an L2-bounded continuous martingale. This definition is independent of the chosen subdivision: splitting an interval simply splits its increment into a telescoping sum. General admissible predictable processes are then integrated by completion using the Itô isometry; unbounded step coefficients are allowed when the weighted $L^2$ condition in the quadratic-variation measure in the next part holds.

Fix a L2-bounded continuous martingale $M$. Its quadratic-variation measure on the predictable sigma-algebra isIt is finite, since ${\nu }_M(\Omega \times [0,\infty ))=\mathbb{E}[M{]}_{\infty }=\mathbb{E}{M}_{\infty }^2-\mathbb{E}{M}_0^2$. The source Hilbert space is $L^2(M)=L^2(\Omega \times [0,\infty ),\mathcal{P},{\nu }_M)$: predictable processes with finite $\mathbb{E}\int H^{2}d[M]$, identified when equal ${\nu }_M$-almost everywhere. Its norm is $\Vert H{\Vert }_{L^2(M)}^2=\mathbb{E}\int H^{2}d[M]$.

The target ${\mathcal{M}}_0^2$ consists of L2-bounded continuous martingales starting at zero, identified up to indistinguishability of stochastic processes, with norm $\Vert K{\Vert }_{{\mathcal{M}}_0^2}=\Vert K_{\infty }{\Vert }_2$. It is a Hilbert space: terminal values belong to the closed linear subspace of $L^2({\mathcal{F}}_{\infty })$ whose conditional expectation given ${\mathcal{F}}_0$ is zero and whose associated martingales have continuous versions. Closure of that continuous-version subspace follows from the Doob L2 maximal inequality and an almost surely uniformly convergent subsequence.

The Itô isometry says that integration extends uniquely from simple predictable processes to a linear isometry $L^2(M)⟶{\mathcal{M}}_0^2$, withFor every finite $t$, the same identity holds with the integral restricted to $[0,t]$ and the left side $\mathbb{E}|(H\cdot M{)}_t{|}^2$. No claim of surjectivity onto all of ${\mathcal{M}}_0^2$ is needed: that would require an additional Martingale representation theorem.

Put $S=H\cdot M$ and $A_t={\int }_0^t{H}_s^{2}d[M{]}_s$. The candidate $A$ is continuous, adapted, nondecreasing, starts at zero and has integrable terminal value. We prove $S^2-A$ is a martingale, rather than assuming the desired quadratic variation of a stochastic integral.

The assumed boundedness of $M$ also makes $M^2-[M]$ a true martingale on every finite horizon. Indeed, apply its local martingale identity at a localizing sequence; bounded stopped squares and the Fatou lemma show that $[M{]}_t$ is integrable. The stopped expressions are then dominated by ${\sup }_s|M_s{|}^2+[M{]}_t$, so the dominated convergence theorem gives the unrestricted conditional identity on that horizon.

First let $H$ be a bounded simple predictable process. On each interval with coefficient $h$, an increment of $S$ is $h(M_t-M_a)$. The defining quadratic variation identity for $M$, together with the martingale property of $M$, makesa martingale for $t\ge a$. Expanding $S_t^2-S_a^2$ gives a term $2S_{a}h(M_t-M_a)$ and the square term above; their conditional expectations yield the desired identity on that interval. Integrability follows because the variables are square-integrable; concatenating the finitely many intervals gives $S^2-A$ a martingale.

For general bounded predictable $H$, choose bounded simple predictable processes $H^n\to H$ in $L^2(M)$, which is possible since they generate the predictable sigma-algebra. Set $S^n=H^n\cdot M$ and $A_t^n={\int }_0^t(H_s^n{)}^{2}d[M{]}_s$. The Itô isometry and Doob L2 maximal inequality imply $S_t^n\to S_t$ in $L^2$; the Cauchy-Schwarz inequality gives $(S_t^n{)}^2\to S_t^2$ in $L^1$ andPass the conditional expectation identity for $(S^n{)}^2-A^n$ to the $L^1$ limit. Thus $S^2-A$ is a martingale. Uniqueness in the defining quadratic variation property now gives

Set $V(x)=(1+x^2{)}^{-1}$ and $A_t={\int }_0^{t}V(B_s)ds$. On a fixed horizon $T$, the Itô formula and Itô product rule giveThe drift vanishes by the partial differential equation. This makes $M$ a continuous local martingale. Since $0\le A_t\le T$,The right side is a finite deterministic bound under the stated boundedness assumption, so the dominated convergence theorem removes a localizing sequence and $M$ is a true martingale. If the derivatives are also bounded, the Itô isometry alternatively proves the stochastic integral is square-integrable because $\mathbb{E}{\int }_0^T{e}^{2A_s}|u_x(T-s,B_s){|}^{2}ds\le T{e}^{2T}\Vert u_x{\Vert }_{\infty }^2$.

For the bounded parabolic differentiability class, distinguish global bounds from bounds on each finite time slab. Only the latter are needed here; a single global bound on the classical solution $u$ would conflict with its requested long-time growth in part (c).

The post-quench Ornstein-Uhlenbeck process has $r_1(q)=\Gamma (a_1+\kappa q^2)$. Its future noise is independent of the initial field. Consequently the cross term between the initial Fourier mode and its future noise integral has zero expectation. The Itô isometry, or the given white-noise covariance, then yieldsThe initial Gaussian Boltzmann distribution, with $a=a_0$, gives $S_q(0)=k_{B}T/(a_0+\kappa q^2)$ in the same unitary normalization. Thus the Gaussian Model A quench hasThe static structure factor interpolates monotonically mode by mode between its two positive equilibrium values. The covariance relaxes at twice the amplitude decay rate. These statements assume that $T$, $\Gamma$ and $\kappa$ are unchanged by the quench; both $a_0,a_1>0$ exclude a spinodal instability.

For deterministic step functions, define directly from Brownian motion incrementswith $s_0$ the left endpoint of the first interval and $f=0$ outside the listed intervals. The printed endpoint list omits $s_0$; the intended convention is $0\le s_0<s_1<\cdots <s_n$, usually $s_0=0$.

Given finitely many observation times, refine the deterministic partition of an interval to include all observation times and all $s_i$. Every $Z_{t_j}$ is a linear combination of the same finite family of independent centered normal Brownian motion increments. Thus every linear combination of $(Z_{t_1},\dots ,Z_{t_m})$ is normal, including possibly a degenerate constant: $Z$ is a Gaussian process. By independence, only increments in the common interval $[0,s\wedge t]$ contribute to the covariance, givingThe same calculation for two step functions $f,g$ proves, rather than assumes, the Itô isometryFor $f\in L^2({\mathbb{R}}_{+})$, choose deterministic step functions $f_n\to f$ in the L2 space. This identity makes $Z_t(f_n)$ a Cauchy sequence in the L2 space; the Riesz-Fischer theorem defines $Z_t(f)$ by mean-square convergence, independently of the approximation. The covariance formula passes to the limit by the Cauchy-Schwarz inequality. For finitely many times and coefficients $a_j$, the characteristic functions converge toso the extension remains a Gaussian process.

For completeness, the construction has a continuous version. Each approximating $Z(f_n)$ is a martingale, directly by the centered independent increments of Brownian motion; the Doob L2 maximal inequality givesTo prove the inequality used here without assuming Itô integral properties, a continuous martingale $N$ satisfies $\lambda \mathbb{P}(N_T^{\ast }\ge \lambda )\le \mathbb{E}(|N_T|1_{\{N_T^{\ast }\ge \lambda \}})$, by conditioning at the first crossing on finite deterministic grids and then using continuity; integrating in $\lambda$ and using the Cauchy-Schwarz inequality yields $\Vert N_T^{\ast }{\Vert }_2\le 2\Vert N_T{\Vert }_2$. A sufficiently fast subsequence therefore converges uniformly almost surely on every compact time interval, giving the continuous version of the stochastic integral.

Multiplication by the integrating factor $e^{t/2}$ and the Itô product rule give the Ornstein-Uhlenbeck processThe two summands are independent centered normal random variables, because $X_0$ is independent of the Brownian motion and the integrand is deterministic. By the Itô isometry, their variances sum to $e^{-t}+{\int }_0^t{e}^{-(t-u)}du=1$. ThusFor $t\ge s$, split the Itô integral at $s$ to obtain $X_t=e^{-(t-s)/2}{X}_s+{\int }_s^t{e}^{-(t-u)/2}d{B}_u$, where the latter term is centered and independent of $X_s$. ConsequentlyIn particular this initialization makes the Gaussian process a stationary process.

Put $\Delta B_h=B_{t+h}-B_t$ and $R_h={\int }_t^{t+h}(H_s-H_t)d{B}_s$. The constant-in-time coefficient $H_t$ on $(t,t+h]$ is previsible, so the ratio minus $H_t$ is $R_h/\Delta B_h$. Since $|H|\le C$, continuity and the dominated convergence theorem implyThe Itô isometry gives $\mathbb{E}{R}_h^2\le h{\eta }_h$. For $\epsilon ,\delta >0$, use the Chebyshev inequality and the normal distribution of $\Delta B_h/\sqrt{h}$ to obtainFirst let $h\downarrow 0$, then $\delta \downarrow 0$; the normal distribution has no atom at zero. This provesNo independence between $R_h$ and the denominator is needed.

The first estimate suggested by the hint is valid: Brownian scaling and the Gamma function give $\mathbb{E}|\Delta B_h{|}^{-1/2}=h^{-1/4}{2}^{-1/4}\Gamma (1/4)/\sqrt{\pi }$. The second hint requires care about the placement of its expectation: with $V_h={\int }_t^{t+h}(H_s-H_t{)}^{2}ds$, the interpretation $\mathbb{E}|R_h{|}^{1/2}\le (\mathbb{E}{V}_h{)}^{1/4}$ is correct by the Jensen inequality and the Itô isometry, whereas the interpretation $\mathbb{E}|R_h{|}^{1/2}\le \mathbb{E}(V_h^{1/4})$ is false with constant $1$. For the latter interpretation, the Burkholder-Davis-Gundy inequality supplies a constant $C_{1/2}$ instead. One can see the obstruction with a Brownian motion stopped on exiting $(-1,1)$: if $\tau$ is that Brownian exit time, then $|B_{\tau }|=1$, $\mathbb{E}\tau =1$, and strict Jensen inequality gives $\mathbb{E}{\tau }^{1/4}<1=\mathbb{E}|B_{\tau }{|}^{1/2}$.

This obstruction also survives the continuity condition on $H$. On a long finite interval approximate $1_{\{s<\tau \}}$ by a continuous adapted process equal to $1$ until $\tau$ and decreasing linearly to $0$ during the next $ϵ$ units of time, and take its initial value to be $0$ by inserting a short continuous ramp before starting the exit experiment. The Itô isometry controls these approximations, and letting the interval length tend to infinity recovers the strict inequality above; hence some bounded continuous adapted example also violates the latter interpretation with constant $1$. The direct proof avoids relying on this ambiguous hint.