Finite part of the regulated Cauchy-reflection triangle A strong numerical distributional limit for the VII-C family

For the insertion-kernel regularization fixed in VII-C, we prove existence of the strong distributional finite part of its numerical reflection-triangle coefficient. The proof applies to arbitrary compact joint tests. An explicit normalization ledger identifies the leading reflected contribution with the prescribed light-ray residue, including its density and every Fourier factor. A joint spatial Fourier argument gives a regulator-independent ultraviolet decomposition: the radial coefficient differs from that residue by $O(r^{-1})$ in a fixed finite-order distribution seminorm, and the remaining regulator-placement error is $O(\epsilon (1+|\log \epsilon |))$. The estimates retain the full phase and are uniform through the intersection of reflected collinearity with the first–third observation diagonal. For two admissible normalized radial mollifiers, the complete conversion is $c_{ab}r$, with an explicit real coefficient, as a corollary of the ultraviolet decomposition. The result is a numerical subtraction prescription, not an algebra construction.

Figure fig:split-VII-D records one view of the same separation. Figure fig:chain-VII-D records one view of the same separation. Figure fig:pair-VII-D records one view of the same separation. Figure fig:gate-VII-D records one view of the same separation. The drawings add no hypothesis.

Figure fig:scope-VII-D records that separation. The drawing adds no hypothesis.

Let $M={\mathbb{R}}^{1+3}$, $M_{+}=\{t>0\}$, and $X=M_{+}^3$. We ask whether

exists strongly in ${\mathcal{D}}^{\prime }(X)$, with the counterterm class fixed to

The family ${T_{\epsilon }}^{\rho }$ is precisely the numerical VII-C family: both slots of each insertion are regularized with the same fixed mollifier $\rho$, while the single middle observable is unsmeared.

The formulas on the following convention sheet are from 085T_Landmark_VIIC_v1.tex VIIC. These displayed formulas define the family studied in this paper. The proof requires no companion methods manuscript and no unspecified normalization.

We reserve $\mu$ for the subtraction scale and write $\vartheta$ for the mollifier profile denoted by $\mu$ in VII-C. The symbol $g$ denotes the middle test; $s=y^0$ is a time coordinate.

{Insertion and placement (eq:B, eq:C, eq:Beps in VII-C).}

{Even covariance (eq:Hhh, eq:Hscalar, eq:positive).}

{Zero-Cauchy mode and mixed covariance (eq:K).}

The factors $P^{-1}$ and $P^{-2}$ are locally integrable in three momentum dimensions. No graviton positivity is assumed.

{Profiles and spectral identity (eq:F, eq:spectral, lem:spectral).}

There is one graph type. Its factor $2$ counts the assignments of the two middle legs. The middle observable is not regularized.

{Locked residue (eq:ray, eq:residue, thm:log).}

Here $\mathrm{d}\Omega$ is ordinary surface area measure: ${\int }_{S^2}\mathrm{d}\Omega =4\pi$.

{Off-shell identity (eq:reconstruction).}

The source is retained. Equation (sc:covariance) does not set a classical field to zero.

Write

The scalar mass $m\ge 0$ and each admissible mollifier are fixed independently of $\epsilon$.

Our Fourier convention is

The dot product in this convention is the coordinate pairing, not the Lorentzian metric pairing. Radiality of $\rho$ is Euclidean.

For a compact $K\subset X$, let

Tests are extended by zero to the ambient Euclidean space when taking Fourier transforms. Their supports are separated from the boundary of $X$.

All estimates below can be taken with the single test order $64$. Its size is inessential; fixing it makes independence from the cutoff and the radial parameter explicit.

Strong convergence in ${\mathcal{D}}^{\prime }(X)$ means uniform convergence on bounded subsets of $\mathcal{D}(X)$. A bounded subset of $\mathcal{D}(X)$ is contained and bounded in some fixed-support test space. Thus

on every compact implies strong convergence.

Define

The field regularization and the mixed-kernel convolution in (sc:placement) define the same insertion. Their numerical coefficient is

For every $\epsilon >0$, this is a smooth function on $X$, and

Its pairing on real product tests is exactly (sc:spectral).

Proof. and kernel convolution. For $\check{\rho }(z)=\rho (-z)$, transposition of convolution gives

Since $\rho$ is even, this is the kernel convolution displayed in (sc:placement). The identity holds on arbitrary smooth configurations: the original mixed kernel has compact support, and the convolution kernels are compactly supported.

-wave multiplier identities. The coefficients of $C$ are constant, so

for every smooth $h$. For any constant tensor $v$,

This applies in particular to the de Donder polarizations in (sc:covariance). Applying the fixed operator ${\mathsf{E}}_{0}C$ to this equality retains the scalar multiplier $\vartheta (\epsilon P)$. This proves the displayed formula for $K_{\epsilon }$, without commuting $S_{\epsilon }$ through ${\mathsf{E}}_0$.

For the scalar leg, Euclidean radiality gives

The two scalar insertion slots therefore supply ${M_{\epsilon }}^2$. The derivative factors are

multiplying the scalar spectral density $1/(2{\omega }_m(q))$ gives (eq:Deps). The two insertion signs also multiply to $+1$. The middle observable is untouched.

and the product-test identity. For fixed positive $\epsilon$, every coordinate derivative of the momentum integrands in $K_{\epsilon }$ and $D_{\epsilon }$ is integrable, locally uniformly in the observations. At zero momentum the only graviton denominators are $P^{-1}$ and $P^{-2}$, locally integrable in three dimensions. The scalar density is locally integrable also for $m=0$. At infinity the profiles provide arbitrary polynomial decay.

Thus (eq:Texact) is a product of smooth functions. Its compact-test bound follows by a supremum on a compact neighborhood of $K$. Integrating against $f\otimes g\otimes f^{\prime }$ and using (eq:Deps) yields (sc:spectral).

tests. The distribution (eq:Texact) is complex-linear in its joint test. On complex product tests, the spectral factors are

The formula with $\overline{F_{f^{\prime },\epsilon }}$ alone applies to real $f^{\prime }$. Kernel conjugation in (eq:Texact) does not conjugate the test.

On $\mathbf{x}\ne \mathbf{y}$, define

The chart domain has $a>0$ and $a+\sigma >0$. The inverse spatial formulas are

Consequently,

For fixed times, the linear transformation to $(\mathbf{y},\mathbf{v},\mathbf{w})$ has determinant of absolute value one. Replacing $\mathbf{y}$ by $\mathbf{z}=\mathbf{y}+s\mathbf{n}$ is triangular with unit determinant. Polar coordinates in $\mathbf{v}$ therefore give

The full transformation, including the times, has the same Jacobian because its time block is the identity. Replacing $t^{\prime }$ by $d=t-t^{\prime }$ has absolute Jacobian one. There is no geometric $1/d$.

Let

with coordinates $(\mathbf{z},\mathbf{n},t,s,t^{\prime })$, and set

The subtraction distribution is

Equivalently, for an arbitrary joint test,

The map $\Psi$ is a proper embedding over compact subsets of $X$. Its inverse on its image is given by the observation times and

On the preimage of a compact in $X$, the time coordinates remain bounded and positive, the spatial coordinate $\mathbf{z}$ is bounded, and $\mathbf{n}\in S^2$. The preimage is closed in this bounded parameter region and hence compact.

It follows directly that $r$ is a positive order-zero distribution, with support exactly $\mathrm{Rel}=\Psi (Z)$. Its product-test pairing is (sc:residue). The separated-support nonzero witness is already established in VII-C, lem:separated.

Relative to the ambient density (eq:density),

Indeed, multiplying this expression by $(a+\sigma {)}^2$ and integrating in $\sigma ,\mathbf{w}$ gives exactly (eq:rjoint). There is no $\delta (d)$ in this measure.

If the same embedding and parameter density are written as $r_{\Sigma }=c_{\Sigma }{\Psi }_{\ast }\nu$, their normalization comparison is

This paper always subtracts (eq:rdef). The fact that the actual leading graph coefficient equals this prescribed distribution is proved in Proposition prop:ledger; it is not built into the definition of the radial family.

For fixed $(t,s,t^{\prime },\mathbf{y})$, define

This preserves all correlations of the joint test.

We shall use the fixed momentum-decay exponent $N=12$. Applying $(1-{\Delta }_{\mathbf{x}}{)}^6(1-{\Delta }_{{\mathbf{x}}^{\prime }}{)}^6$ under the integral gives

Only $24+j+j^{\prime }\le 32$ derivatives are needed for this estimate; order $64$ also accommodates the fixed localizations used below. The variables $(t,s,t^{\prime },\mathbf{y})$ range over a fixed compact projection of $K$.

Let $r\ge 1$, $\mathbf{\nu }\in S^2$, $\ell \in \{1,2\}$, $k\in \{0,1,2\}$, and $N>k+3$. Then

The constants are uniform in $r,\mathbf{\nu }$. For $0\le r\le 1$, the first integral is uniformly bounded. In particular, all these assertions apply with $N=12$.

Proof. On $|\alpha |\le r/2$, the denominator is at least $(r/2{)}^{\ell }$, and the remaining weight is integrable.

On $|\alpha |>r/2$, first consider $|\alpha -r\mathbf{\nu }|\le r/2$. Here $|\alpha |$ is comparable to $r$, and

This part is bounded by $C{r}^{3-\ell +k-N}$. In the remaining part the denominator gives $C{r}^{-\ell }$, while

This proves the tail bound. Since $N>k+3$, the tail also satisfies the full-integral bound.

For $0\le r\le 1$, split into a fixed-size neighborhood of $\alpha =r\mathbf{\nu }$ and its complement. Local integrability of $|p{|}^{-\ell }$ treats the first region, and the fixed polynomial weight treats the second.

. These bounds include all momentum ratios and all directions. They use no time separation, regulator decay, or angular stationary-phase hypothesis.

Set

Throughout the momentum calculation we use the single substitution

For $A,B\in \{D,R\}$, define

For $\epsilon =0$, the two profile factors are replaced by one. For each fixed $q$, the integral is absolutely convergent after taking the joint spatial transform.

The exact positive-cutoff identity is

The spatial signs follow from

Define ${S_{\epsilon }}^{\rho }$ by taking all four branches in (eq:jointbranches) on $|q|\le 1$, and only $DD,DR,RD$ on $|q|>1$. Define $S$ by the same prescription with all multipliers replaced by one. The high-$|q|$ integrals defining $S$ are understood after performing the compact time integrations in (eq:Qbranches).

The distribution $S$ exists, is independent of $\rho$, and satisfies

Proof. The profiles are bounded and globally Lipschitz, with value one at zero. Telescoping the product of multipliers gives

For $|q|\le 1$, (eq:FTbound) and the bounded-$r$ part of Lemma lem:convolution give an integrable bound. The extra powers in (eq:multdifference) are harmless: $P\le |\alpha |+1$ and $P^{\prime }\le |\beta |+1$. Thus this sector has a cutoff-free limit with $O(\epsilon )$ error.

For $r=|q|\ge 1$, suppose the left branch is direct. Its $t$-frequency is $-(P+{\omega }_m(q))$. Integrating eight times by parts in $t$ gains

The amplitude is at most linear in $t$; its differentiated coefficients are still bounded by $C_K(P^{-1}+P^{-2})$. The other amplitude is bounded by $C_K(P^{\prime -1}+P^{\prime -2})$. If only the right branch is direct, use $t^{\prime }$ instead.

The shifted bounds show that the two momentum integrations contribute at most $C{r}^{-2}$. Since $r^2{\omega }_m(r)\le C_m{r}^3$, the unregulated radial majorant is $C{r}^{-7}$. For the regulated difference,

and (eq:multdifference) costs at most $C\epsilon r$ after the weighted integrations. The resulting radial majorant is $C\epsilon r^{-6}$. Both are integrable.

All integrations by parts are performed against the full compact joint test. No restriction to an observation diagonal is used.

. This lemma holds on every compact in $X$, including ordinary diagonals and the reflected intersection. It covers bounded scalar momentum and every sector with a direct branch. The test order is $64$, independent of $\epsilon$; the error is $O_{K,\rho ,m}(\epsilon )$, without a lower bound on $|t-t^{\prime }|$.

For $q=r\mathbf{\nu }$, $r\ge 1$, set

in accordance with (eq:substitution).

The radial families are defined intrinsically from the exact branch integrals:

No residue constant has been inserted in these definitions. They give the exact decomposition

The use of $r$ as the radial integration variable is distinct from its use for the fixed residue distribution. In expressions such as $A(r)-r$, the second term is the distribution (eq:rdef).

The leading reflected–reflected radial coefficient derived from (eq:Aintrinsic) equals the prescribed distribution (eq:rdef) on all joint tests. In particular, its constant is $c_0=1/(64(2\pi {)}^3)$, with the parameter density in (eq:nu).

Proof. The calculation is recorded in six identities.

1: exact pointwise kernel. Proposition prop:positive gives

2: reflected amplitudes and conjugation. The reflected terms of the two mixed covariances have amplitudes

The corresponding time phases are, separately, $e^{iP(t+s)}$ and $e^{-i{P}^{\prime }(t^{\prime }+s)}$. The right spatial phase is $e^{-i{p}^{\prime }\cdot ({\mathbf{x}}^{\prime }-\mathbf{y})}$. These are conjugations of the kernel, not of $\Phi$.

3: scalar density. The scalar factor is

4: combined numerical prefactor. Consequently the reflected–reflected integrand has prefactor

The assignment factor in L1 has now been exhausted. No further factor of two occurs in $A(r)$.

5: polar measure and radial normalization. With ordinary sphere area measure,

This explains the extra $r^3$ in (eq:Aintrinsic), rather than adding a normalization by hand.

6: translation and Fourier inversion. For fixed $q$, (eq:substitution) is a translation and has Jacobian one. The two mixed-covariance measures become exactly ${\mathrm{d}}^3\alpha {\mathrm{d}}^3\beta /(2\pi {)}^6$. The Fourier inversion identity is

In particular, the two factors $(2\pi {)}^{-3}$ supplied by the mixed covariances are the normalization of the two Fourier inversions. They cancel the factors $(2\pi {)}^3$ from the unnormalized Fourier delta identities; they are not canceled a second time against another Fourier measure.

For the leading coefficient, replace the amplitude factors $P^{-2},P^{\prime -2}$ by $r^{-2},r^{-2}$ and $\omega (r)$ by $r$. In the phase, however, retain the first-order expansions

Substituting $P=P^{\prime }=r$ into the phase itself would be incorrect. The retained time phase is exactly the inversion phase in L6.

The only remaining numerical factor is therefore

After L6, the remaining measure is $\mathrm{d}t\mathrm{d}s\mathrm{d}{t}^{\prime }{\mathrm{d}}^3\mathbf{y}\mathrm{d}\Omega (\mathbf{\nu })$. Changing $\mathbf{y}$ to $\mathbf{z}=\mathbf{y}+s\mathbf{\nu }$ has unit Jacobian, so this is precisely the pushforward density (eq:nu). We have proved

The cutoff factors $M_{\epsilon }(r{)}^2\vartheta (\epsilon P)\vartheta (\epsilon P^{\prime })$ have not been used in this leading identification. They are treated by the replacement estimate below. Direct-branch amplitudes do not enter this calculation.

As a consequence of the ledger, the intrinsic definition (eq:Aintrinsic) can now be written

The formula for ${A_{\epsilon }}^{\rho }(r)$ has the two additional factors $\vartheta (\epsilon P)\vartheta (\epsilon P^{\prime })$.

For every compact $K\subset X$,

In particular,

The estimate holds without a positive lower bound on $|t-t^{\prime }|$.

Proof. We retain the exact phase, form an exact difference, and only then estimate its terms.

1. Exact phase. Set

The temporal phase in (eq:Adef) is exactly

Before the spatial integration defining ${\mathcal{H}}_{\Phi }$, the exact full phase in the incidence chart is

Thus both angular variables and all of $(\sigma ,\mathbf{w},d)$ are retained. The phase ${\phi }_0$ is the leading temporal phase in Cartesian Fourier space; the spatial phase is already inside ${\mathcal{H}}_{\Phi }$. One does not delete the first line of (eq:intersectionphase) in an incidence-coordinate integral.

2. Exact central–tail decomposition. Let

and abbreviate, within this proof,

All integrals over ${\mathcal{C}}_r$ or its complement below include the remaining variables, with measure

Write ${\mathcal{H}}_{\Phi }$ at its full arguments from (eq:jointFT). Define

These definitions precede all absolute-value estimates. The algebraic identity behind them is