Theorem: First-correction layer expansion

Let $S_{\delta }=(q+\delta z,q\mathbf{n})$. Uniformly for bounded $z$ and compact non-tip $(q,\mathbf{n})$,

For up to four $z$ derivatives, one tangential derivative in $(q,\mathbf{n})$, and $l$ additional smooth spectator derivatives,

The same estimate holds for

This operator is the logarithmic-scale derivative at fixed unscaled $S$.

Proof. We give an exact fixed-domain reduction before Taylor expansion.

Put $H=P+Q$, ${\mathbf{H}}_{\perp }=\mathbf{H}-(\mathbf{n}\cdot \mathbf{H})\mathbf{n}$, and $h=H^0-\mathbf{n}\cdot \mathbf{H}$. Define

This extends smoothly to $\delta =0$ on the fixed compact profile support, with

Exactly,

Make the change $u=h+\delta c_{\delta }$. Its temporal Jacobian is one. The retarded indicator becomes $\Theta (z-u)$, independent of $\delta$.

Thus (eq:translation) becomes an integral over $u\le z$ of the smooth integrand

where

A fixed compact interval in $u$ contains every nonzero integrand for the parameter range under consideration. The coefficient may be extended across $d=0$ using Lemma lem:regularity; on the actual integration domain for $\delta >0$, its argument is causal.

Taylor expansion in $\delta$, with this fixed domain, is therefore legitimate in the stated mixed norms. Two $\delta$ derivatives, four $z$ derivatives, and one tangential derivative use at most seven coefficient derivatives. Lemma lem:regularity uses at most fourteen weight derivatives. Spectator derivatives add $l$. Endpoint terms from differentiating the upper limit in $u$ have no larger derivative count. The integral Taylor remainder gives (eq:first-error).

It remains to calculate the first derivative. At $\delta =0$, the zeroth term is $\mathcal{G}[w]\mathcal{F}(z)$. The coefficient/weight derivative contributes

The density derivative from $h=u-\delta c_{\delta }$ contributes

The sum density of $H$ is radial in four Euclidean dimensions. Rotation symmetry and interchange of $P,Q$ give the density identities

No division by a marginal density is needed.

For

radial integration by parts gives

For completeness, choose the coordinate $t=h/\sqrt{2}$ and three perpendicular Euclidean coordinates $Y$. If the sum density is $a(t^2+|Y{|}^2)$, then

Compact support fixes the integration constant.

Along the null boundary,

Substitute these identities and (eq:first-moments) into (eq:first-integrand). Then add (eq:curvature-term) using (eq:transverse-moment). The result is precisely the first-order term in (eq:first-correction).

For direct scale control, repeat the same fixed-domain Taylor argument with the signed profile ${\rho }^{[1]}$. It retains rotation and interchange symmetries, so the same calculation applies. Its cumulative moments are

These are exactly the $\mathcal{S}$-derivatives of the zeroth and first-order terms in (eq:first-correction). The remainder for this signed profile has the same bound, proving the scale estimate without differentiating (eq:first-error).