Appendix B 4 October 2026

Exact moment bound for the nonlinear witness

Jeremy Rodgers · Independent Researcher
doi:10.5281/zenodo.23131064

Reading position 12 of 14

B Exact moment bound for the nonlinear witness

Set $\tau=R^{-2}$ , $S=2x^2+4y^2$ in (6.1). Since $g_R=yf_R$ , the product rule gives

b_R=\bigl(0,e^{-\tau S}P_\tau\bigr),\qquad P_\tau=x^2-\tau x^4+\tau^2x^6+4\tau^2x^4y^2 .

All moments needed for $\|b_R-b_\infty\|_{L^2(r)}^2$ are finite Gaussian integrals:

\mathbb E_r[x^{2p}y^{2q}e^{-j\tau S}] =\frac{(2p-1)!!(2q-1)!!} {2^p4^q(1+2j\tau)^{p+q+1}},\qquad (-1)!!=1.

Expanding the two finite polynomials yields

\begin{aligned}\|b_R-b_\infty\|_{L^2(r)}^2 &=A(\tau)-2B(\tau)+\frac34,\\ A(\tau)&= \frac{48+528\tau+3228\tau^2+8148\tau^3+17883\tau^4} {64(1+4\tau)^7},\\ B(\tau)&=\frac{12+18\tau+123\tau^2} {16(1+2\tau)^5}. \end{aligned}

At $\tau=1/256$ this is the rational value $E$ displayed in section 6. The strict estimate used there is certified by

\frac1{576}-\frac{4E}{3} =\frac{261360546794830111543} {747189504563051511000000}>0 .

This calculation is an exact continuum integral, not a spectral cutoff or numerical trajectory approximation.