Theorem: Newton–wave form and cutoff removal

The fully Wick-ordered expression defines a continuous sesquilinear form on ${\mathcal{D}}_{\varphi }$, with

Its regulated forms satisfy

The limit is joint and independent of the bounded profiles with the stated support properties. For real $f$ the form is symmetric. No distribution supported at transfer zero is added.

Proof. For a kernel $A$, write

The majorant in (eq:cubicmajorant) is integrable in this norm. In transverse and six tangential variables $y$, the singular integral is bounded by

Integration in $y$ gives a multiple of $(1+|\mathbf{q}{|}^{-2})⟨\mathbf{q}{⟩}^{-9}$, which is integrable.

The ultraviolet error is supported where $R\ge \Lambda$; take out ${\Lambda }^{-1}$ and use the same integrability with exponent $-14$ instead of $-15$. For transfer removal integrate over $|\mathbf{q}|\le 2\delta$; the resulting bound is $C({\delta }^3+\delta )$. The edge strip has length at most $2ϵ$, and the pre-time-integration majorant gives $Cϵ$. Hence

In (eq:cubicsector), the two pointwise Schwartz bounds give a factor $⟨z{⟩}^{-60}$ for every spectator and ${\prod }_{\mathrm{external}}⟨p{⟩}^{-30}\le R^{-30}$. The spectator integrals give $I_{\ast }^{n-s}$. Also

The weights in $p$ dominate these factors. Applying the weighted-kernel bound and then summing occupied sectors proves both estimates.

Complex conjugation reverses all mode signs, and the classical maps are real. Thus $\overline{{K_f}^{\sigma }}={K_f}^{-\sigma }$ for real $f$. Together with full signed permutation symmetry this interchanges the $(r,s)$ and $(s,r)$ blocks and proves symmetry. Ordinary weighted integral convergence gives the locally integrable distribution itself, with no extra zero-supported term.