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Econ-34-A3 · revision r1 · on Econ-34

An L^1 counterexample at critical logarithmic BSDE growth

Full disproof claimed · LLM-generated · Submitted by Ji Ho Bae · Ji Ho Bae · gpt-6 astra, claude opus 5.5, gpt-6 astra pro

Unverified claim of a full disproof. Automated screening checks form, honesty and scope, not correctness.

Claim: full disproof claimed

For the exact critical logarithmic generator, a nonnegative terminal variable of expectation one has no class-(D) solution on every finite positive horizon and in every Brownian dimension.

Limitations: This is a constructed integrable-data nonexistence result, not a claim that all terminal data fail.; No independent community or formal verification is recorded here.; Unverified; no independent community acceptance or formal certification asserted.

Contributor’s completion estimate: 100%. A negative answer to the recorded critical-growth L1 existence question (Fan-Hu-Tang Problem 5.1). Not a probability of correctness.

Manuscript

Contents
  1. An $L^1$ counterexample at critical logarithmic BSDE growth
  2. Bounded Gaussian lower solutions
  3. Divergence of the initial values
  4. One integrable terminal value
  5. Scope.
  6. Use of AI tools
  7. References

An $L^1$ counterexample at critical logarithmic BSDE growth

Ji Ho Bae
LawAll AI

10 October 2026

\begin{abstract} For the generator $g(z)=|z|/\sqrt{\log(e+|z|)}$, we construct a nonnegative terminal value of expectation one for which the scalar BSDE has no class-$(D)$ solution. The construction works on every finite positive horizon and in every Brownian dimension. It gives a negative answer to Fan, Hu and Tang's Problem 5.1. \end{abstract}

Let $B$ be a $d$-dimensional Brownian motion, $d\geq1$, on the Brownian basis of [FHT], and fix $T>0$. Class $(D)$ means that $\{Y_\tau:\tau\leq T\text{ a stopping time}\}$ is uniformly integrable. A solution has continuous adapted $Y$, predictable $Z$, and $\int_0^T|Z_t|^2dt<\infty$ almost surely, with the BSDE identity below. All logarithms are natural; vector norms are Euclidean.

Theorem: There is an explicitly specified $\mathcal F_T$-measurable $\xi\geq0$ with $\mathbb E\xi=1$ for which no class-$(D)$ solution satisfies \begin{equation}\label{wsref-0-62-73-64-65} Y_t=\xi+\int_t^T g(Z_s)\,ds-\int_t^T Z_s\cdot dB_s, \qquad g(z)=\frac{|z|}{\sqrt{\log(e+|z|)}}. \end{equation}

We first record why bounded lower solutions can be compared with a putative class-$(D)$ solution despite a change of measure.

Lemma: If $(Y,Z)$ solves (bsde) with $\xi\geq0$ and $Y$ of class $(D)$, then $Y\geq0$. Suppose a continuous adapted process $u$ satisfies $0\leq u\leq C$ for a deterministic finite constant $C$, and \[ du_t=-f_t\,dt+z_t\cdot dB_t, \quad \int_0^T|z_t|^2dt<\infty, \quad f_t\leq g(z_t),\quad u_T\leq\xi. \] Then $u\leq Y$.
Proof: Since $g\geq0$, localization of (bsde) gives $Y_t\geq\mathbb E[Y_{\tau_n}\mid\mathcal F_t]$ for stopping times $\tau_n\uparrow T$. The stopped stochastic integrals may be chosen square-integrable. Class $(D)$ and continuity give $Y_{\tau_n}\to\xi$ in $L^1$, whence $Y_t\geq\mathbb E[\xi\mid\mathcal F_t]\geq0$. Taking rational $t$ and using continuity makes this assertion simultaneous in time.

For $h(r)=r/\sqrt{\log(e+r)}$, $0\leq h'(r)\leq1$ on $r>0$. Thus $g$ is globally $1$-Lipschitz, also in $\mathbb R^d$. Put $\Delta=u-Y$ and choose the predictable vector \[ \theta_t=\begin{cases} \displaystyle\frac{g(z_t)-g(Z_t)}{|z_t-Z_t|^2}(z_t-Z_t),&z_t\ne Z_t,\\ 0,&z_t=Z_t. \end{cases} \] Then $|\theta_t|\leq1$. Novikov's condition holds, so under the equivalent measure with density $\exp(\int_0^T\theta_t\cdot dB_t-\frac12\int_0^T|\theta_t|^2dt)$, the process $B^{\theta}=B-\int_0^{\cdot}\theta_tdt$ is Brownian. Writing $r_t=g(z_t)-f_t\geq0$, we obtain \[ d\Delta_t=r_tdt+(z_t-Z_t)\cdot dB_t^{\theta}. \] Tanaka's formula makes $\Delta^+$ a local submartingale. It is bounded by $C$, because $Y\geq0$, hence is a true submartingale. Its terminal value is zero, so it vanishes identically. This argument needs no integrability of $Y$ or $\xi$ under the changed measure. ∎

Bounded Gaussian lower solutions

Write $\Phi$ and $\phi$ for the standard normal distribution function and density. Set \[ S=\min\{T,(2\pi)^{-1}\},\qquad H_N=\sum_{k=1}^N\frac1k,\qquad \delta_k=\frac{S}{kH_N},\qquad t_k=\sum_{j=1}^k\delta_j. \] For each integer $N\geq1$, define \begin{equation}\label{wsref-0-74-65-72-6d-69-6e-61-6c-4e} X_N=2^N\prod_{k=1}^N \Phi\left(\frac{B^1_{t_k}-B^1_{t_{k-1}}}{\sqrt{\delta_k}}\right). \end{equation} The standardized increments are independent standard normals, and $\mathbb E\Phi(G)=1/2$ by symmetry. Consequently $\mathbb E X_N=1$.

Define deterministic constants backwards, starting with $k=N$: \begin{equation}\label{wsref-0-62-61-63-6b-77-61-72-64-73} D_k=(2\pi\delta_k)^{-1/2},\quad L_k=2^ND_k\prod_{j>k}p_j,\quad q_k=[\log(e+L_k)]^{-1/2},\quad p_k=\Phi(q_k\sqrt{\delta_k}/\sqrt2). \end{equation} This recursion uses only constants already defined. For $t_{k-1}\leq t\leq t_k$, put \[ A_k=2^N\prod_{j<k}\Phi(\Delta B^1_j/\sqrt{\delta_j}) \prod_{j>k}p_j, \qquad w_t=\frac{B^1_t-B^1_{t_{k-1}}+q_k(t_k-t)}{\sqrt{\delta_k+t_k-t}}, \] where $\Delta B^1_j=B^1_{t_j}-B^1_{t_{j-1}}$, and define \[ u^N_t=A_k\Phi(w_t),\qquad z^N_t=\frac{A_k\phi(w_t)}{\sqrt{\delta_k+t_k-t}}\,e_1. \] At $t_k$, the next stage starts with its Gaussian factor equal to $p_{k+1}$, so the values of $u^N$ agree. After $S$, set $u^N=X_N$ and $z^N=0$. Direct Itô differentiation on each interval gives \begin{equation}\label{wsref-0-69-74-6f} du^N_t=-q_k|z^N_t|dt+z^N_t\cdot dB_t,\qquad 0\leq |z^N_t|\leq L_k. \end{equation} The finite collection of deterministic bounds gives square-integrability of $z^N$. Moreover $0\leq u^N\leq2^N$, $u^N_T=X_N$, and \[ q_k|z^N_t|\leq g(z^N_t),\qquad u^N_0=\prod_{k=1}^N2p_k. \] Thus $u^N$ satisfies the bounded lower-solution conditions of Lemma Lemma.

Divergence of the initial values

Put $x_k=q_k\sqrt{\delta_k}/\sqrt2$ and $\ell_k=\log(2p_k)$. Since $D_k\geq1$ and $p_j\geq1/2$, we have $L_k\geq2^k$ and $q_k\leq(k\log2)^{-1/2}$. Also $q_k\leq1$, so $0\leq x_k\leq1$. On this interval, \begin{equation}\label{wsref-0-6e-6f-72-6d-61-6c-62-6f-75-6e-64} \phi(1)x\leq\log(2\Phi(x))\leq\sqrt{2/\pi}\,x. \end{equation} Indeed the derivative is $\phi(x)/\Phi(x)$, between $\phi(1)$ and $2\phi(0)$. It follows that \begin{equation}\label{wsref-0-74-61-69-6c} \sum_{j>k}\ell_j \leq\sqrt{\frac{S}{\pi\log2}}\,\sqrt{H_N}. \end{equation} Set $a=\log(1/(2\pi S))\geq0$. For $k\geq\lceil H_N\rceil$ and $H_N\geq\max\{8,a\}$, use \[ \log L_k=k\log2+\tfrac12\log(kH_N)+\tfrac12a+\sum_{j>k}\ell_j. \] The coefficient in (tail) is at most one. Since $L_k\geq2^k$, $\log(e+L_k)\leq1+\log L_k$, and hence \[ \log(e+L_k)\leq1+k\log2+\log k+k/2+\sqrt{k}\leq5k. \] The lower bound in (normalbound) now gives \begin{align} \log u^N_0 &\geq\phi(1)\sqrt{\frac{S}{10H_N}} \sum_{k=\lceil H_N\rceil}^{N}\frac1k\notag\\ &\geq\frac{\phi(1)}{2\sqrt{10}}\sqrt{SH_N}.\label{wsref-0-64-69-76-65-72-67-65-6e-63-65} \end{align} For the last inequality, the omitted harmonic sum is at most $1+\log H_N\leq H_N/2$ when $H_N\geq8$.

One integrable terminal value

For $m\geq1$, choose the explicit integers \[ N_m=\left\lceil\exp\left(8+a+ \frac{160m^2(\log2)^2}{S\phi(1)^2}\right)\right\rceil, \qquad \xi=\sum_{m=1}^{\infty}2^{-m}X_{N_m}. \] Tonelli's theorem and (terminalN) give $\mathbb E\xi=1$, so this sum is finite almost surely. The summands need not be independent. Since $H_N\geq\log(N+1)$, (divergence) gives $u^{N_m}_0\geq2^{2m}$.

Suppose a class-$(D)$ solution with terminal value $\xi$ existed. For $0<c\leq1$, $g(cz)\geq c g(z)$, because the logarithmic denominator decreases under scaling. Hence $2^{-m}u^{N_m}$ is still a bounded nonnegative lower solution, with terminal value at most $\xi$. Lemma Lemma yields \[ Y_0\geq2^{-m}u^{N_m}_0\geq2^m\qquad(m\geq1). \] These countably many inequalities contradict the finiteness of $Y_0$. This proves Theorem Theorem and the negative answer to [FHT, Problem 5.1, equation $5.1$].

Scope.
The generator, exponent, Brownian basis and class-$(D)$ condition are unchanged. Only an admissible terminal value is chosen to refute the universal existence claim.

Use of AI tools

The author used research AI agents that the author built and configured (GPT-6 Astra max, Claude Opus 5.5 max and GPT-6 Astra Pro) during the proof work, for proof search and construction, proof checking, translation, Python computation and similar tasks. The final manuscript was written, and the proofs were checked, by the human author. All responsibility for this work rests with the author personally.

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, §1 and §5, Problem 5.1. Primary text.