Lemma: Integrable transfer singularities

Let

There is $C_f$ such that

The same bound holds before the outer response-time integration. For a chosen stress pair, equal signs give locally bounded transfer behavior. Opposite signs can retain an inverse-square transfer term.

Proof. Use coordinates $p_2={\sigma }_2(q-{\sigma }_1{p}_1)$, of unit absolute Jacobian. For equal signs, Taylor expansion of ${\omega }_{-p+\sigma q}$ in (eq:numerators) gives

Both initial amplitudes are locally bounded, and the source amplitude is smooth.

For opposite signs, at $q=0$,

Thus ${Z_{00}}^{\mathrm{in}}$ has a generally nonzero $|q{|}^{-2}$ numerator. The spatial initial term is at worst $|q{|}^{-1}$, while the source term is smooth. Away from the other transfers the cyclic sum cannot universally cancel that pole.

For the global bound, finite time gives

Stress numerators cost at most $C{R}_p^2$, spatial initial numerators at most $C{R}_p^3$, and the scalar Hessian contraction at most $C{R}_p^2$. Equation (eq:dcancel) bounds the separate divergence terms. The massive mode normalizations are bounded, and $|F_f(t,w)|\le {\Vert f(t,\cdot )\Vert }_{L^1}$. This proves (eq:kernelbound) term by term.

All time integrals remain finite integrals. No artificial resonance denominator is introduced.