GPT 6 Astra Ultra attempt on Econ-38
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Oscillatory BSDE uniqueness: one controlled integrand is sufficient
Problem ID:C65-37. Research attempt; independent review pending.Target and scope
On the usual $d$-dimensional Brownian basis, fix $p>1$ and $\xi\in L^p$. The target is uniqueness for \[ Y_t=\xi+\int_t^T |Z_s|\sin|Z_s|\,ds-\int_t^T Z_s\cdot dB_s \] among solutions for which $|Y|^p$ is of class (D). Solutions have continuous adapted $Y$, predictable $Z$, and pathwise finite $\int_0^T(|Z|^2+|g(Z)|)dt$. We retain the power class-(D) hypothesis exactly; it is not replaced by class (D) on $Y$.We prove uniqueness against every admissible competitor whenever one solution has a bounded integrand, and more generally when one integrand has a specified exponential energy moment. This yields complete uniqueness for an explicit class of unbounded Gaussian terminal values. It does not establish that an arbitrary $L^p$ terminal value has such a reference solution. The source question and its missing uniform continuity assumption were checked in [FHT, Problem 5.3 and Theorem 3.14(iv)] on 9 October 2026. The linearization ingredients below are classical; no novelty claim is made.
The stochastic-calculus results used below (Itô integration, Brownian martingale representation, and Girsanov change of measure) use their standard forms in [RY]; their additional integrability requirements are checked explicitly in the proofs.
A secant estimate that uses only one endpoint
Lemma: For $g(z)=|z|\sin|z|$ and all $z,z'\in{\mathbb{R}}^d$, \begin{equation}\label{wsref-0-73-65-63-61-6e-74-62-6f-75-6e-64}
|g(z)-g(z')|\leq
2(1+\min\{|z|,|z'|\})|z-z'|.
\end{equation} The function $g$ is nevertheless not uniformly continuous on ${\mathbb{R}}^d$.
Proof: Put $m=\min(|z|,|z'|)$ and $d_0=|z-z'|$. If $d_0\geq1$, linear growth and the triangle inequality give \[
|g(z)-g(z')|\leq|z|+|z'|\leq d_0+2m
\leq(1+2m)d_0.
\] If $d_0<1$, the two radii lie in $[m,m+1]$. The derivative of $r\sin r$ has absolute value at most $1+r$, so the one-dimensional mean-value theorem and $||z|-|z'||\leq d_0$ give a bound $(m+2)d_0$. Both bounds are at most $2(1+m)d_0$. For the final assertion take $z_n=2\pi n e_1$ and $z'_n=(2\pi n+(2\pi n)^{-1})e_1$. Their distance tends to zero, while $g(z_n)=0$ and $g(z'_n)=(2\pi n+(2\pi n)^{-1})\sin((2\pi n)^{-1})\to1$. ∎
Lemma (A density moment and class-(D) transfer): Let $r>1$ and let $\beta$ be predictable with $A=\int_0^T|\beta_s|^2ds<\infty$ almost surely. Put $\kappa_r=2r^2-r$. If ${\mathbb{E}} e^{\kappa_r A}<\infty$, then $D=\mathcal E(\int\beta\,dB)$ is a uniformly integrable martingale, $D_T>0$, and \[
\sup_{\tau\leq T}{\mathbb{E}} D_\tau^r\leq({\mathbb{E}} e^{\kappa_r A})^{1/2}.
\] If $p=r/(r-1)$ and $|Y|^p$ is class (D) under ${\mathbb{P}}$, then $Y$ is class (D) under $d{\mathbb{Q}}=D_Td{\mathbb{P}}$.
Proof: For $M=\int\beta\,dB$, the identity \[
D_\tau^r=\mathcal E(2rM)_\tau^{1/2}
\exp\left((r^2-r/2)\langle M\rangle_\tau\right)
\] and Cauchy--Schwarz imply the moment bound: the expectation of the nonnegative local-martingale exponential on the right is at most one. The bound applies to all localizing times for $D$ and implies uniform integrability of its stopped values. Thus $D$ is a true uniformly integrable martingale, with terminal expectation one. Finite quadratic variation makes its terminal value strictly positive. For the last claim, Hölder's inequality gives \[
{\mathbb{E}}_{\mathbb{Q}}[|Y_\tau|{\mathbf 1}_{\{|Y_\tau|>R\}}]
\leq ({\mathbb{E}} D_\tau^r)^{1/r}
({\mathbb{E}}[|Y_\tau|^p{\mathbf 1}_{\{|Y_\tau|>R\}}])^{1/p}.
\] The second factor tends to zero uniformly by the assumed class (D) of $|Y|^p$, proving the claim. ∎
Uniqueness against an unrestricted admissible competitor
Theorem: Fix $p>1$, set $r=p/(p-1)$ and $\kappa_r=2r^2-r$. Let $(Y,Z)$ and $(\widetilde Y,\widetilde Z)$ solve the target BSDE with the same terminal value and both power transforms of class (D). If the first solution satisfies \begin{equation}\label{wsref-0-6f-6e-65-65-6e-65-72-67-79}
{\mathbb{E}}\exp\left(8\kappa_r\int_0^T|Z_s|^2ds\right)<\infty,
\end{equation} then the two solutions are equal. No exponential energy hypothesis on $\widetilde Z$ is required. In particular the conclusion holds if $|Z_t|\leq R$ for a deterministic $R<\infty$ almost everywhere.
Proof: Put $U=Y-\widetilde Y$, $V=Z-\widetilde Z$, and set \[
\beta_t={\mathbf 1}_{\{V_t\ne0\}}
\frac{g(Z_t)-g(\widetilde Z_t)}{|V_t|^2}V_t.
\] This is predictable and satisfies $g(Z)-g(\widetilde Z)=\beta\cdot V$. By Lemma Lemma, $|\beta|\leq2(1+|Z|)$, hence $\int|\beta|^2\leq8T+8\int|Z|^2$. Hypothesis (oneenergy) verifies Lemma Lemma (A density moment and class-(D) transfer) at the exact conjugate exponent $r$. Thus $D=\mathcal E(\int\beta\,dB)$ defines an equivalent measure ${\mathbb{Q}}$, both $Y$ processes are class (D) under ${\mathbb{Q}}$, and $B^{\mathbb{Q}}=B-\int\beta ds$ is Brownian by Girsanov's theorem. The difference equation reduces to $dU=V\cdot dB^{\mathbb{Q}}$. Its stochastic integral is locally defined because both candidate energies are finite pathwise. A local martingale of class (D) is a true martingale: stop along a localizing sequence and pass to conditional expectations using uniform integrability. Since $U_T=0$, $U$ vanishes. The quadratic variation $\int|V|^2dt$ vanishes as well. Equality at each time extends to indistinguishability by continuity. If $|Z|\leq R$, the expectation in (oneenergy) is bounded by $\exp(8\kappa_rR^2T)$. ∎
Corollary (A complete unbounded-terminal subcase): Let $a:[0,T]\to{\mathbb{R}}^d$ be deterministic, measurable, and essentially bounded, and let $c\in{\mathbb{R}}$. For $\xi=c+\int_0^T a_s\cdot dB_s$, the target BSDE has exactly one solution with $|Y|^p$ of class (D), for every fixed $p>1$. It is \[
Z_t=a_t,\qquad
Y_t=c+\int_0^t a_s\cdot dB_s+\int_t^T|a_s|\sin|a_s|\,ds.
\]
Proof: The displayed pair has the prescribed terminal value and differentiating its finite-variation term verifies the BSDE. Its pathwise integrability is immediate from bounded $a$. The Gaussian martingale $M_t=\int_0^t a_s\cdot dB_s$ has terminal moments of every order. For any $q>p$, Doob's $L^q$ inequality gives ${\mathbb{E}}\sup_{t\leq T}|M_t|^q<\infty$; the deterministic drift term is bounded by $T\|a\|_\infty$. Therefore $\sup_t|Y_t|^p$ is integrable, and it dominates all stopped powers, proving class (D). The reference integrand is bounded, so Theorem Theorem proves uniqueness against every competitor in the full prescribed solution class. ∎
What remains unresolved
The estimate uses linear growth in a stronger way than the local derivative bound $1+|z|$: a secant is controlled by its smaller endpoint. Nevertheless, for arbitrary $\xi\in L^p$ we have not established a solution whose integrand satisfies (oneenergy). The target's power class-(D) hypothesis does not itself assert such a moment. The counterexample to uniform continuity in Lemma Lemma shows why a bounded generator-to-control ratio cannot simply replace a Lipschitz secant estimate. Thus the unrestricted uniqueness question remains unresolved, while Corollary Corollary (A complete unbounded-terminal subcase) is a fully verified class of unbounded data within it.References
- [FHT] S. Fan, Y. Hu and S. Tang, 1D nonlinear backward stochastic differential equations: a unified theory and applications, arXiv:2309.06233v2, 11 February 2026. https://arxiv.org/html/2309.06233v2.
- [RY] D. Revuz and M. Yor, Continuous Martingales and Brownian Motion, 3rd edition, Springer, 1999. Chapters IV, V and VIII cover stochastic integration, Brownian martingale representation and Girsanov's theorem. https://doi.org/10.1007/978-3-662-06400-9.