Proposition: Evaluated spectral remainders

The pair kernel $P$ is Hilbert–Schmidt, and $L$ defines a bounded compact operator ${\mathfrak{h}}_{\varphi }\to {\mathfrak{h}}_g$. There are finite constants $C_P,C_L$, independent of the choices of masks satisfying (eq:masks), such that

For $f\ne 0$, $L$ is not Hilbert–Schmidt; more precisely,

for all sufficiently large $R$, with $c,C>0$.

Proof. For pair signs the denominator in (eq:master2) contains $1+|p|+E_q$. Put $k=p+q$, and use ${E_q}^{-1}\le C_m⟨k⟩⟨p{⟩}^{-1}$. With arbitrarily large $n$,

The radial integral of the last factor is finite:

Its integral for $|p|<R^{-1}$ is $O(R^{-2})$, and for $|p|>R$ is $O(R^{-1})$.

For the scalar UV tail split $|p|<1$ and $|p|\ge 1$. On the latter, (eq:Pmajor) and $⟨p{⟩}^{-4}\le C⟨p+q{⟩}^4⟨q{⟩}^{-4}$ give a bound whose integral over $|q|>R$ is $O(R^{-1})$. On the former, spatial transfer decay gives $C_n|p{|}^{-1}⟨q{⟩}^{-n}$, with arbitrarily rapid integrated decay. Since

the pair estimate follows.

For the mixed signs, discard the bounded inverse frequency factor, but retain transfer decay. For $|p|\ge 1$, Peetre's inequality gives both bounds

For $|p|<1$,

The low-momentum part is Hilbert–Schmidt and the high part is bounded by the Schur test. The low gravitational tail has squared integral $O(R^{-2})$. For $|p|>R$ or $|q|>R$, the appropriate bound in (eq:Schur2) has both Schur integrals $O(R^{-1})$; the remaining low-$p$, high-$q$ part decays faster. This proves the operator-norm rate. Each $L_R$ is Hilbert–Schmidt, so its norm limit $L$ is compact.

For the last assertion, the squared mixed estimate gives

Integration over $|p|<2R$ proves the upper bound $CR$.

For a lower bound write $p=rn,q=rn+k$, with $n\in S^2$ and $k$ in a fixed compact set. Uniformly there,

The function $J$ is the Fourier transform of the compactly supported integrable function $I_{[0,\ell ]}(s)v_f(t_0+s,x)$; hence it is real analytic. It is nonzero if $f\ne 0$, because $({{\partial }_s}^2+{{\Omega }_m}^2)v_f=f$. It therefore cannot vanish on the open set $|\nu |<|k|$. Choose a nonzero value there and an $n$ with $n\cdot k=\nu$. The TT norm is strictly positive. Continuity gives open sets of positive measure in $n,k$ on which the limit in (eq:mixedasymptotic) is bounded below. For $r_0\le r\le R/2$, both masks equal one when $R$ is large. The Jacobian is $r^2\mathrm{d}r\mathrm{d}n\mathrm{d}k$, so integrating the lower bound ${c_0}^2{r}^{-2}$ gives a quantity bounded below by $cR$. The same region without radial truncation proves that $L$ is not Hilbert–Schmidt.