Disposition: PASS on the assigned algebra; OPEN on the all-height sign.
Reviewed baseline: d93aaf6b8dbe1fcbebf6fff4f6d1f71d993f4208 (PR75).
Read the present review brief, derivation.md, applicable AGENTS.md
instructions, and PR75’s audits/theta-addition-admission-2026-10-05-v1/derivation.md.
This is an independent paper check of equations (7)-(10) and their boundary
premises, not a replay of every registered conclusion or certificate. The
parent supplied the register/dependency review; this reviewer did not
repeat it. No scan, symbolic computation, new sign candidate or certificate.
Wall check: Same open gap. Closest: NS100/101 and PR74/75. The subtraction repairs the ordinary-transform domain, while the complete first-Laguerre signed obligation is unchanged. The following identities do not provide an estimating method for that obligation.
Let u=s+t, v=s-t, and F(s,t)=f(u)f(v). Holding the other original
coordinate fixed gives
partial_u=(partial_s+partial_t)/2,
partial_v=(partial_s-partial_t)/2.
Since phi=(1/4-partial_u^2)f/4, the two negative signs in the product
cancel. Thus
phi(u)phi(v)
= [((partial_s+partial_t)^2-1)
*((partial_s-partial_t)^2-1)] F/256
= [partial_s^4-2*partial_s^2*partial_t^2+partial_t^4
-2*partial_s^2-2*partial_t^2+1] F/256.
In the polynomial multiplication, write A=partial_s^2+partial_t^2-1
and B=2*partial_s*partial_t; the commuting product is A^2-B^2.
This explicitly retains the mixed derivative. Integrability and vanishing
boundaries follow from the stated exponential bounds for all derivatives
of f. Two integrations by parts give integral s^2 F_ss=2*A_0; four
give integral s^2 F_ssss=0. Consequently equation (7) is exactly
256*C=(partial_t^2-1)^2*A_2-4*(partial_t^2+1)*A_0.
PASS. Neither the mixed term nor the old PR75 boundary issue is hidden in the factor 256.
The absolute determinant of (s,t)->(u,v) is 2, so
ds dt=du dv/2, s=(u+v)/2, and 2rt=r(u-v). Evenness and reality of f
make Y real and even on the real axis. Fubini is justified, also with
second moments. Therefore
integral A_0(t)*exp(2irt)dt=Y(r)*Y(-r)/2=Y(r)^2/2.
For A_2, the prefactor is 1/8, from the Jacobian and s^2.
The square moments are -Y''(r) and -Y''(-r), while the first moments
are -iY'(r) and -iY'(-r)=iY'(r). Hence
integral A_2(t)*exp(2irt)dt
=[-2*Y(r)*Y''(r)+2*Y'(r)^2]/8
=L1[Y](r)/4.
The imaginary parts vanish. In particular the cross moments have a plus
sign, not a minus sign. Applying the same derivation directly to phi gives
integral C(t)*cos(2rt)dt=L1[X](r)/4, independently checking the final
normalization. PASS on equation (8).
Set p=r^2+1/4. Transforming equation (7), using the legitimate decaying
A_j and their derivatives, gives
256*integral C(t)*cos(2rt)dt
=(4*r^2+1)^2*L1[Y]/4-2*(1-4*r^2)*Y^2
=4*p^2*L1[Y]+8*(r^2-1/4)*Y^2.
Multiplication by 4/256 therefore gives
16*L1[X]=p^2*L1[Y]+2*(r^2-1/4)*Y^2.
An independent product expansion verifies it: the two p*p'*Y*Y'
cross terms cancel, giving L1[pY]=p^2*L1[Y]+(p'^2-p*p'')*Y^2.
Here p'^2-p*p''=2*(r^2-1/4) and X=pY/4.
Since p is strictly positive for real r, rearranging yields exactly
equation (10). PASS. The multiplier correction is essential; dropping
it changes the target, especially for abs(r)<1/2.
r=0, equation (10) is L1[Y](0)>=8*Y(0)^2. This point is
consistent with strict positivity of the original moments. Let
F_j=integral u^j*f(u)du, m_j=integral u^j*phi(u)du, for j=0,2.
Integrating (1/4-partial_u^2)f=4*phi gives
F_0=16*m_0 and F_2=8*F_0+16*m_2. Since Y is even,
L1[Y](0)=F_0*F_2, so
L1[Y](0)-8*Y(0)^2=256*m_0*m_2>0.
This is a point check from the positive-kernel premises, not an
all-height estimate or a new research target.abs(r)=1/2, p=1/2 and the correction vanishes:
L1[X]=L1[Y]/64. Thus first-Laguerre positivity of the two transforms
is equivalent at exactly those two heights.X'=p*Y'/4 and
16*L1[X]=p^2*Y'^2>=0; multiple zeros may give equality. Equation
(10) remains regular without dividing by either transform.abs(r)>1/2, L1[Y]>=0 would suffice, but need not be necessary,
because the correction in (9) is nonnegative. For abs(r)<1/2, the
required positive margin cannot be dropped. Pointwise positivity of
A_2 does not supply its Fourier-transform sign in either region.PASS on all requested exceptional-point checks. The draft’s claims are correctly scoped; the r=0 moment calculation could be added as an optional normalization check, but no correction is required.
Put g(u)=2*cosh(u/2), so f=g-h. Then
F=f(u)f(v)=h(u)h(v)-g(u)h(v)-h(u)g(v)+g(u)g(v).
The full pair operator annihilates each displayed term containing a g
factor because (partial_u^2-1/4)g=0 (or the v analogue). Thus
P F=P H exactly. PR75 instead used R=H-2*cosh(s), with
P[2*cosh(s)]=0; hence P F=P R=P H.
These equalities concern the complete two-variable differential operator. They do not permit separately integrating or transforming growing cross terms. The advantage of the present subtraction is that the complete F and its required weighted derivatives are integrable in both variables. PR75’s M_j remain growing; nothing here retroactively justifies their separate ordinary transforms. Equation (9)’s correction is the exact remaining effect of recovering X from its decaying primitive’s transform.
No algebraic defect was found in the assigned formulas. This review supports publication of the boundary repair and exact corrected target as an audit. It supplies no independent all-height lower bound, no RH implication from first-Laguerre positivity, and no escape from the matched NS101 structural obstruction. The original nonlinear arithmetic estimate remains open.