Hamiltonians for Polaron Models with Subcritical Ultraviolet Singularities

Abstract

We treat the ultraviolet problem for polaron-type models in nonrelativistic quantum field theory. Assuming that the dispersion relations of particles and the field have the same growth at infinity, we cover all subcritical (superrenormalisable) interactions. The Hamiltonian without cutoff is exhibited as an explicit self-adjoint operator obtained by a finite iteration procedure. The cutoff Hamiltonians converge to this operator in the strong resolvent sense after subtraction of a perturbative approximation for the ground-state energy.

Appendices

A Products in $\mathcal {K}_n$

In this appendix we derive in detail the quantitative bounds on $\kappa \star _\ell H_0^{-1} \kappa '$ that imply Theorem 3.2 and are used in the proof of Theorem 3.5.

Recall the Definition (62) of $\rho _{n,\lambda }$, ${\tilde{\rho }}_{n, \lambda }$ and note that

$$\begin{aligned} {\tilde{\rho }}_{n, \lambda }(R,E)=\rho _{n, \lambda }((r_n, \dots , r_1), E). \end{aligned}$$

(139)

We also have the property

$$\begin{aligned} \begin{aligned} \rho _{m, 1}(Q, E+\Omega (Q')) \rho _{m', \lambda }(Q', E)&= \rho _{m+m', \lambda }((Q, Q'), E) \\ {\tilde{\rho }}_{m, \lambda }(R,E) {\tilde{\rho }}_{m', 1}(R', E+\Omega (R))&= {\tilde{\rho }}_{m+m', \lambda }((R, R'), E). \end{aligned} \end{aligned}$$

(140)

We will frequently use an elementary bound on a class of integrals. The proof makes explicit the role of the parameters $\alpha , \gamma$.

Lemma A.1

Assume the Hypothesis 1.1. Let $s, t\ge 0$ with $s\ne 1+\delta /\gamma$ and $s+t>1+\delta /\gamma$. Then for $b>0$, $a\ge 0$

$$\begin{aligned} \int \frac{|v(\xi )|^2 \textrm{d}\xi }{(a+\omega (\xi ))^s(b+\omega (\xi ))^{t}} \lesssim (a+1)^{-(s-1-\delta /\gamma )_+} b^{-\min \{s+t -1 -\delta /\gamma , t\}}, \end{aligned}$$

(141)

Proof

By the hypothesis 1.1 the integral is bounded by

$$\begin{aligned} (141) \lesssim \int \frac{\textrm{d}\xi }{|\xi |^{2\alpha } (a+(c_1+\xi ^2)^{\gamma /2})^s(b+|\xi |^{\gamma })^{t}}. \end{aligned}$$

(142)

For $s>1+\delta /\gamma$, i.e. $2\alpha + s \gamma >d$, we drop $|\xi |^\gamma$ from the second factor in the denominator and obtain

$$\begin{aligned} (141) \lesssim b^{-t} \int \frac{\textrm{d}\xi }{|\xi |^{2\alpha } (a+(c_1+\xi ^2)^{\gamma /2})^s} \lesssim (a+1)^{-s -2\alpha /\gamma +d/\gamma }b^{-t}, \end{aligned}$$

(143)

which yields the claim as $-2\alpha /\gamma +d/\gamma =1+\delta /\gamma$. For $s<1+\delta /\gamma$, we have $2\alpha + s \gamma <d$ and

$$\begin{aligned} (141) \lesssim \int \frac{\textrm{d}\xi }{|\xi |^{2\alpha +s\gamma }(b+|\xi |^{\gamma })^{t} } \lesssim b^{-s-t-2\alpha /\gamma +d/\gamma }, \end{aligned}$$

(144)

which proves the claim. $\square$

The following Lemmas propagate bounds on $\kappa \in \mathcal {K}_n, \kappa '\in \mathcal {K}_{n'}$ to $\kappa H_0^{-1} \star _\ell \kappa '$. We treat the cases $\ell =\min \{n, n'\}=0$, $0<\ell <\max \{n, n'\}$, and $\ell =n=n'$ separately, starting with the case $\ell =\min \{n, n'\}=0$.

Lemma A.2

Let $n\in \mathbb {N}$ and $\kappa \in \mathcal {K}_n$, $\kappa '\in \mathcal {K}_{0}$. Suppose that for some $\mu \ge 0$, $0\le \lambda \le 1$, and $\lambda '\ge 0$ we have the bounds

$$\begin{aligned} |\kappa (Q,R,p,E)|&\lesssim E^{-\mu } \min _{s\in [\lambda -1, 1-\lambda ]} \rho _{n,\lambda +s}(Q,E){\tilde{\rho }}_{n,\lambda -s}(R,E) \\ |\kappa '(p,E)|&\lesssim H_0(p,E)E^{-\lambda '}. \end{aligned}$$

Then with

$$\begin{aligned} \sigma&=\min \{\lambda +\lambda ', 1\}, \\ \tau&=\mu +(\lambda +\lambda '-1)_+, \end{aligned}$$

we have

$$\begin{aligned} |\kappa H_0^{-1} \star _0 \kappa '|(Q,R,p,E)&\lesssim E^{-\tau } \min _{s\in [\sigma -1, 1-\sigma ]}\rho _{n,\sigma +s}(Q,E){\tilde{\rho }}_{n,\sigma -s}(R,E), \\ |\kappa ' H_0^{-1} \star _0 \kappa |(Q,R,p,E)&\lesssim E^{-\tau } \min _{s\in [\sigma -1, 1-\sigma ]}\rho _{n,\sigma +s}(Q,E){\tilde{\rho }}_{n,\sigma -s}(R,E). \end{aligned}$$

Proof

The kernel of $\kappa ' H_0^{-1} \star _0 \kappa$ is

$$\begin{aligned} \kappa ' H_0^{-1} \star _0 \kappa (Q,R,p,E) = \frac{\kappa '\Big (p+\sum _{\mu =1}^n q_\mu , E + \omega (Q)\Big )\kappa (Q,R,p,E)}{\Omega (p+\sum _{\mu =1}^n q_\mu ) + E+\omega (Q)}.\nonumber \\ \end{aligned}$$

(145)

We thus have for any $t\in [\lambda -1,1-\lambda ]$

$$\begin{aligned} |\kappa ' H_0^{-1} \star _0 \kappa (Q,R,p,E)|&\lesssim E^{-\mu }(E+\omega (Q))^{-\lambda '} \rho _{n,\lambda +t}(Q,E) {\tilde{\rho }}_{n,\lambda -t}(R,E) \nonumber \\&\lesssim E^{-\mu }\rho _{n,\lambda +2\lambda '+t}(Q,E) {\tilde{\rho }}_{n,\lambda -t}(R,E). \end{aligned}$$

(146)

Choosing $t=s-\sigma +\lambda$ for $s\in [\sigma -1,1-2\lambda +\sigma ]\supset [\sigma -1, 1-\sigma ]$, this becomes

$$\begin{aligned} |\kappa ' H_0^{-1} \star _0 \kappa (Q,R,p,E)|&\lesssim E^{-\mu }\rho _{n,2\lambda +2\lambda ' - \sigma +t}(Q,E) {\tilde{\rho }}_{n,\sigma -s}(R,E) \nonumber \\&\lesssim E^{-\tau } \rho _{n,\sigma +t}(Q,E) {\tilde{\rho }}_{n,\sigma -s}(R,E). \end{aligned}$$

(147)

The proof for the $\kappa H_0^{-1} \star _0 \kappa '$ is essentially the same, with the roles of Q, R reversed. $\square$

The next Lemma treats the general case of $\kappa \star H_0^{-1} \kappa '$, except for the special case $n=n'=\ell$.

Lemma A.3

Let $n, n'\in \mathbb {N}$ and $\kappa \in \mathcal {K}_n$, $\kappa '\in \mathcal {K}_{n'}$. Suppose that for some $\mu , \mu '\ge 0$ and $0\le \lambda , \lambda '\le 1$ with $\lambda +\lambda '\ne 1+\delta /\gamma$ we have the bounds

$$\begin{aligned} |\kappa (Q,R,p,E)|&\lesssim E^{-\mu } \min _{s\in [\lambda -1, 1-\lambda ]} \rho _{n,\lambda +s}(Q,E){\tilde{\rho }}_{n,\lambda -s}(R,E) \\ |\kappa '(Q',R',p,E)|&\lesssim E^{-\mu '} \min _{s\in [\lambda '-1, 1-\lambda ']} \rho _{n',\lambda '+s}(Q',E){\tilde{\rho }}_{n',\lambda '-s}(R',E). \end{aligned}$$

Then for all $0\le \ell \le \min \{n, n'\}$ with $\ell <\max \{n, n'\}$ and

$$\begin{aligned} \sigma&=\min \{\lambda +\lambda '+\ell (1-\delta /\gamma ), 1\}, \\ \tau&=\mu +\mu '+(\lambda +\lambda '+\ell (1-\delta /\gamma )-1)_+, \end{aligned}$$

we have

$$\begin{aligned} |\kappa H_0^{-1} \star _\ell \kappa '|(Q,R,p,E) \lesssim E^{-\tau } \min _{s\in [\sigma -1, 1-\sigma ]}\rho _{n+n'-\ell ,\sigma +s}(Q,E){\tilde{\rho }}_{n+n'-\ell ,\sigma -s}(R,E). \end{aligned}$$

Proof

For $\ell =0$, we then have from (140) (taking $s=1-\lambda$ and $s'=\lambda '-1$)

$$\begin{aligned}&|\kappa \star _0 H_0^{-1}\kappa '|((Q,Q'), (R,R'),p,E) \nonumber \\&\quad \lesssim E^{-\mu -\mu '}\frac{{\tilde{\rho }}_{n, 2\lambda -1}(R, E+\omega (Q')) {\tilde{\rho }}_{n', 1}(R',E+\omega (R))}{E+\omega (Q')+\omega (R)} \nonumber \\&\qquad \times \rho _{n, 1}(Q, E+\omega (Q'))\rho _{n', 2\lambda '-1}(Q',E+\omega (R)) \nonumber \\&\quad \le E^{-\mu -\mu '}\frac{\rho _{n+n',2\lambda '-1}((Q,Q'),E) {\tilde{\rho }}_{n+n', 2\lambda -1}((R,R'),E)}{E+\omega (q'_{n'})+\omega (r_1)} \nonumber \\&\quad \le E^{-\mu -\mu '} \rho _{n+n',2\lambda '+t}((Q,Q'),E) {\tilde{\rho }}_{n+n', 2\lambda -t}((R,R'),E), \end{aligned}$$

(148)

for any $-1\le t\le 1$. Setting $s=\lambda '-\lambda +t$ and bounding any inverse powers of $E+\omega (r_1)$, $E+\omega (q_{n+n'})$ in excess of one by inverse powers of E gives the desired inequality.

Now let $\ell >0$. In view of (66), we need to integrate in $\Xi =(\xi _1, \dots , \xi _\ell )$ the quantity

$$\begin{aligned}&\frac{ \Big |\kappa \Big (Q_1^n, S,p+\sum \nolimits _{\mu =1}^{n'-\ell } q_{n+\mu }, E+\omega (Q_{n+1}^{n+n'-\ell })\Big ) \Big |}{E+\omega (\Xi )+ \omega (Q_{n+1}^{n+n'-\ell })+\omega (R_1^{n-\ell })}\nonumber \\&\quad \times \Bigg |\kappa '\Bigg (U, R_{n-\ell +1}^{n+n'-\ell },p+\sum \limits _{\mu =1}^{n-\ell } r_\mu , E+\omega (R_1^{n-\ell })\Bigg )\Bigg | \end{aligned}$$

(149)

evaluated at $S_I=\Xi =U_J$, $S_{I^c}=R_1^{n-\ell }$, $U_{J^c}=Q_{n+1}^{n+n'-\ell }$, where $I=(i_1, \dots , i_\ell )$ with $1\le i_1< \dots < i_\ell \le n$ and $J=(j_1, \dots j_\ell )$ with pairwise different $j_1, \dots , j_\ell \in \{1, \dots , n'\}$.

We first restrict to $\ell <\min \{n,n'\}$. As for $\ell =0$, we then use hypothesis with $s=1-\lambda$, $s'=\lambda '-1$ to obtain

$$\begin{aligned} (149)&\le \frac{{\tilde{\rho }}_{n, 2\lambda -1}(S,E+\omega (Q_{n+1}^{n+n'-\ell }))\rho _{n',2\lambda '-1}(U, E+\omega (R_1^{n-\ell }))}{E+\omega (\Xi )+ \omega (Q_{n+1}^{n+n'-\ell })+\omega (R_1^{n-\ell })} \end{aligned}$$

(150)

$$\begin{aligned}&\quad \times E^{-\mu -\mu '}\rho _{n, 1}(Q_1^n, E+\omega (Q_{n+1}^{n+n'-\ell }) \tilde{\rho }_{n',1}(R_{n-\ell +1}^{n+n'-\ell }, E+\omega (R_1^{n-\ell })). \end{aligned}$$

(151)

We expand $\rho$, ${\tilde{\rho }}$ in (150) using the Definition (62) and evaluate the variables S, U according to the prescription above. For any pair with $i_\nu \ne 1$, $j_\nu \ne n'$, $\nu =1,\dots , \ell$, we group the two factors containing $v(s_{i_\nu })=v(\xi _\nu )=v(u_{j_\nu })$ together and drop $\omega (\xi _\nu )$ from all other factors, which gives an upper bound. The integral over $\xi _\nu$ is then given by

$$\begin{aligned}&\int \frac{|v(\xi _\nu )|^2 \textrm{d}\xi _\nu }{(E+\omega (\xi _\nu ) + \omega (R_1^{i_\nu -1})+ \omega (Q_{n+1}^{n+n'-\ell }))(E+\omega (\xi _\nu )+\omega (R_1^{n-\ell })+\omega (Q_{j_\nu +1}^{n+n'-\ell }))} \nonumber \\&\quad {\mathop {\lesssim }\limits ^{(141)}} (E+\omega (r_1)+\omega (q_{n+n'-\ell }))^{\delta /\gamma -1}. \end{aligned}$$

(152)

If $\nu =1$ and $i_1=1$, $j_1\ne n'$, we include the factor with $v(r_1)$ (here we use that $\ell <n$ and thus $r_1=s_a$ for some $a>1$) before dropping $\omega (\xi _1)$, which gives an upper bound on the $\xi _1$-integral by

$$\begin{aligned}&| v(r_1)| \int \frac{|v(\xi _1)|^2 \textrm{d}\xi _1}{(E+\omega (\xi _1))^\lambda (E +\omega (\xi _1) + \omega (r_1)+\omega (q_{n+n'-\ell }))^2} \nonumber \\&\quad {\mathop {\lesssim }\limits ^{(141)}} \frac{| v(r_1)|}{(E+\omega (r_1)+\omega (q_{n+n'-\ell }))^{\lambda +1-\delta /\gamma }}, \end{aligned}$$

(153)

since $\lambda \le 1<1+\delta /\gamma$. If for some $\nu \in \{1, \dots , \ell \}$, $j_\nu =n'$ and $i_\nu \ne 1$ the argument is the same with $v(r_1)$ replaced by $v(q_{n+n'-\ell })$ and $\xi _1$ by $\xi _\nu$.

This gives us the inequality

$$\begin{aligned}&\int (150)\Bigg \vert {\begin{array}{c} S_I=\Xi \\ S_{I^c}=R_1^{n-\ell } \\ U_J=\Xi \\ U_{J^c}=Q_{n+1}^{n+n'-\ell } \end{array}} \textrm{d}\Xi \nonumber \\&\quad \lesssim \frac{\rho _{n'-\ell , 2\lambda '-1}(Q_{n+1}^{n+n'-\ell },E){\tilde{\rho }}_{n-\ell ,2\lambda -1 }(R_1^{n-\ell }, E)}{(E+\omega (r_1)+\omega (q_{n+n'-\ell }))^{1+\ell (1-\delta /\gamma )}} \end{aligned}$$

(154)

Combining this with (151) and splitting the denominator as in the case $\ell =0$ gives a bound by

$$\begin{aligned} E^{-\tau } \rho _{n+n'-\ell ,\sigma + s}(Q,E){\tilde{\rho }}_{n+n'-\ell , \sigma -s}(R,E), \end{aligned}$$

(155)

with $2\lambda '-1-\sigma \le s\le \sigma +1-2\lambda$, which is a weaker condition than $\sigma -1\le s\le 1-\sigma$. This proves required bound for the terms with $j_\nu \ne n'$.

If I, J are such that $i_1=1$ and $j_1=n'$ we include both the factor with $v(r_1)$ and $v(q_{n+n'-\ell })$ in the $\xi _1$-integral, leading to

$$\begin{aligned}&|v(r_1)||v(q_{n+n'-\ell })| \int \frac{|v(\xi _1)|^2 \textrm{d}\xi _1 }{(E+\omega (\xi _1))^{\lambda +\lambda '}(E +\omega (\xi _1) + \omega (r_1)+\omega (q_{n+n'-\ell }))^2} \nonumber \\&\quad {\mathop {\lesssim }\limits ^{(141)}} \frac{|v(r_1)||v(q_{n+n'-\ell })|}{(E+\omega (r_1)+\omega (q_{n+n'-\ell }))^{\min \{\lambda +\lambda '+1-\delta /\gamma ,2\}}} E^{-(\lambda +\lambda '-1-\delta /\gamma )_+} \nonumber \\&\quad \lesssim \frac{|v(r_1)||v(q_{n+n'-\ell })|}{(E+\omega (r_1)+\omega (q_{n+n'-\ell }))^{\sigma - (\ell -1)(1-\delta /\gamma )}} E^{-(\lambda + \lambda '+\ell (1-\delta /\gamma )-1)_+} \end{aligned}$$

(156)

where we used that $1+\delta /\gamma \ne \lambda + \lambda '$. From here we conclude as before, and this proves the claim for $\ell <\min \{n, n'\}$.

The remaining case is $\ell =\min \{n, n'\}$. Let $\ell =n$, $\ell <n'$. We then use the hypothesis differently, keeping the freedom of choosing the value of $s'\in [\lambda '-1, 1-\lambda ']$. Instead of (150), (151), this gives for the case at hand

$$\begin{aligned} (149)&=\frac{\Big |\kappa \big (Q_1^n, \Xi ,p+\sum _{\mu =1}^{n'-\ell } q_{n+\mu }, E+\omega (Q_{n+1}^{n+n'-\ell })\big ) \kappa '\big (U, R,p, E)\Big |}{E+\omega (\Xi )+ \omega (Q_{n+1}^{n+n'-\ell })} \nonumber \\&\lesssim \frac{{\tilde{\rho }}_{n, 2\lambda -1}(\Xi ,E+\omega (Q_{n+1}^{n+n'-\ell }))\rho _{n',\lambda '+s'}(U, E)}{E+\omega (\Xi )+ \omega (Q_{n+1}^{n+n'-\ell })} \end{aligned}$$

(157)

$$\begin{aligned}&\qquad \times E^{-\mu -\mu '} \rho _{n, 1}(Q_1^n, E+\omega (Q_{n+1}^{n+n'-\ell })) {\tilde{\rho }}_{n',\lambda '-s'}(R, E)). \end{aligned}$$

(158)

The integral over $\xi _1=\xi _{i_1}$ is then bounded using the denominator in (157), which gives for $j_1\ne n'$

$$\begin{aligned}&\int \frac{|v(\xi _1)|^2 \textrm{d}\xi _1}{(E +\omega (\xi _1) +\Omega (Q_{n+1}^{n+n'-\ell }))^{1+\lambda }(E +\omega (\xi _1) +\Omega (Q_{n+j_1+1}^{n+n'-\ell }))} \nonumber \\&\quad {\mathop {\lesssim }\limits ^{(141)}} (E+\omega (q_{n+n'-\ell }))^{-\lambda -1+\delta /\gamma }. \end{aligned}$$

(159)

If $j_1=n'$, then we additionally include the factor with $v(q_{n+n'-\ell })$ as we did for $\ell <\min \{n, n'\}$ and obtain (with $(1+\lambda '-s')/2\le 1$)

$$\begin{aligned}&|v(q_{n+n'-\ell })| \int \frac{|v(\xi _1)|^2 \textrm{d}\xi _1}{(E +\omega (\xi _1) +\omega (q_{n+n'-\ell }))^{2+\lambda }(E +\omega (\xi _1))^{(1+\lambda '-s')/2}} \nonumber \\&\quad \lesssim \frac{|v(q_{n+n'-\ell })|}{(E+\omega (q_{n+n'-\ell }))^{1-\delta /\gamma + (1+2\lambda +\lambda '-s')/2}}. \end{aligned}$$

(160)

This gives us a bound on the integral by

$$\begin{aligned} \int (157) \bigg |{\begin{array}{c} \scriptscriptstyle U_J=\Xi \\ U_{J^c}=Q_{n+1}^{n+n'-\ell } \end{array}} \; \textrm{d}\Xi \lesssim \rho _{n'-\ell ,2\lambda +\lambda '+2\ell (1-\delta /\gamma )+ t}(Q_{n+1}^{n+n'-\ell }, E). \end{aligned}$$

(161)

Combining with (158) and setting $s=s'+\sigma -\lambda '$ (with the resulting restriction $s\in [\sigma -1, 1+\sigma -2\lambda ']\supset [\sigma -1,1-\sigma ]$) gives the claim. $\square$

We now turn to the remaining case $\ell =n=m$.

Lemma A.4

Let $n\in \mathbb {N}$ and $\kappa , \kappa ' \in \mathcal {K}_n$. Suppose that for some $\mu , \mu '\ge 0$ and $0\le \lambda , \lambda '\le 1$ we have the bounds

$$\begin{aligned} |\kappa (Q,R,p,E)|&\lesssim E^{-\mu } \min _{s\in [\lambda -1, 1-\lambda ]} \rho _{n,\lambda + s}(Q,E){\tilde{\rho }}_{n,\lambda -s}(R,E) \\ |\kappa '(Q,R,p,E)|&\lesssim E^{-\mu '} \min _{s\in [\lambda '-1, 1-\lambda ']} \rho _{n,\lambda ' + s}(Q,E){\tilde{\rho }}_{n,\lambda '-s}(R,E) \end{aligned}$$

for $E\ge 1$. Then for all $0\le \sigma \le 1$ satisfying

$$\begin{aligned} \max \{\lambda , \lambda '\}\le \sigma&<\lambda +\lambda '+n(1-\delta /\gamma ) \end{aligned}$$

and

$$\begin{aligned} \tau&=\mu +\mu '+(\lambda +\lambda '+n(1-\delta /\gamma )-1)_+ \end{aligned}$$

we have for $E\ge 1$

$$\begin{aligned} |\kappa \star _n H_0^{-1}\kappa '|(Q,R,p,E) \lesssim E^{-\tau }\min _{s\in [\sigma -1, 1-\sigma ]} \rho _{n,\sigma _\varepsilon +s}(Q,E){\tilde{\rho }}_{n,\sigma -s}(R,E). \end{aligned}$$

Proof

This is the case $n=\ell =n'$ of the previous lemma and we adopt the notation from there. In the present case we have $i_\nu =\nu$, and $j_1, \dots , j_n$ is just a permutation of $1, \dots , n$. The integral then simplifies to

$$\begin{aligned} \int \frac{\Big |\kappa \big (Q, \Xi , E)\big ) \kappa '\big (\Xi _J, R, E)\Big |}{E+\Omega (\Xi )} \textrm{d}\Xi&\lesssim E^{-\mu -\mu '} \rho _{n, \lambda + t}(Q, E) {\tilde{\rho }}_{n,\lambda '-t'}(R,E)\nonumber \\&\quad \times \int \frac{{\tilde{\rho }}_{n, \lambda -t}(\Xi , E)\rho _{n, \lambda '+t'}(\Xi _J, E)}{E+\omega (\Xi )} \textrm{d}\Xi , \end{aligned}$$

(162)

where $\Xi _J=(\xi _{j_1}, \dots , \xi _{j_n})$ are the permuted variables. Hence we only need to prove that the integral (162) is bounded by $E^{-(\lambda +\lambda '+n(1-\delta /\gamma )-1)_+}$ for appropriate choices of $t, t'$.

Let us first consider the case $j_1=n$. Then the integral is bounded by

$$\begin{aligned} (162)&\le \int \frac{|v(\xi _1)|^2}{(E+\omega (\xi _1))^{2+(\lambda - t+\lambda '+t')/2}} \prod _{j=2}^n \frac{|v(\xi _j)|^2}{(E+\omega (\xi _j))(E+\omega (\Xi _1^j))} \textrm{d}\Xi \nonumber \\&{\mathop {\lesssim }\limits ^{(141)}} \int \frac{|v(\xi _1)|^2\textrm{d}\xi _1}{(E+\omega (\xi _1))^{2+(\lambda - t+\lambda '+t')/2+(n-1)(1-\delta /\gamma )}}. \end{aligned}$$

(163)

For the final integral to be finite, we need

$$\begin{aligned} 2+(\lambda + \lambda '- t+t')/2+(n-1)(1-\delta /\gamma )>1+\delta /\gamma . \end{aligned}$$

(164)

Now let $\sigma \le 1$ as in the statement, and set for $\sigma -1\le s\le 1-\sigma$

$$\begin{aligned} t=s+\sigma -\lambda , \quad t'=s - \sigma + \lambda '. \end{aligned}$$

(165)

These choices are admissible since $\sigma \ge \lambda , \lambda '$. Because $\sigma < \lambda + \lambda ' + n(1-\delta /\gamma )$, we have

$$\begin{aligned} (162)&\lesssim \int \frac{|v(\xi )|^2\textrm{d}\xi }{(E+\omega (\xi ))^{2+(\lambda +\lambda '+(n-1)(1-\delta /\gamma )-\sigma )}} {\mathop {\lesssim }\limits ^{(141)}} E^{-(\lambda +\lambda '+n(1-\delta /\gamma )-\sigma )}. \end{aligned}$$

(166)

As $\tau -\mu -\mu '\le (\lambda +\lambda '+n(1-\delta /\gamma )-\sigma )$, this yields

$$\begin{aligned} |\kappa \star _n H_0^{-1}\kappa '|(Q,R,p,E) \lesssim E^{-\tau } \min _{s\in [\sigma -1, 1-\sigma ]} \rho _{n, \sigma +s}(Q,E){\tilde{\rho }}_{n, \sigma -s}(R,E). \end{aligned}$$

(167)

This proves the claim for the case $j_1=n$.

The case $j_1\ne n$ arises only for $n\ge 2$. We choose $t,t'$ as before, and we then group the denominator in (162) with the $\xi _{j_1}$-integral. This is then bounded by

$$\begin{aligned}&\int \frac{|v(\xi _{j_1})|^2\textrm{d}\xi _{j_1}}{(E+\omega (\xi _{j_1}))^{(1+2\lambda '+s-\sigma )/2}(E+\omega (\xi _1) + \omega (\xi _{j_1}))^2} \nonumber \\&\quad {\mathop {\lesssim }\limits ^{(141)}} (E+\omega (\xi _1))^{-(1-\delta /\gamma )-(1+2\lambda '+s-\sigma )/2}, \end{aligned}$$

(168)

since $2\lambda '+s-\sigma \le \lambda '+s \le 1$. Treating the integrals over $\xi _\nu$, $\nu =2, \dots , n$, $\nu \ne j_1$, in the same way as before, we are left with the $\xi _1$-integral

$$\begin{aligned} (162)&\lesssim \int \frac{|v(\xi )|^2\textrm{d}\xi }{(E+\omega (\xi ))^{2+(\lambda +\lambda '+(n-1)(1-\delta /\gamma )-\sigma )}}. \end{aligned}$$

(169)

This proves the claim by the same argument as for $j_1=1$. $\square$

In the proof of Theorem 3.5 we additionally need bounds on $a(v)H_0^{-1}\star _{\ell } \kappa H_0^{-1}\star _{\ell '} a^*(v)$.

Lemma A.5

Let $n\in \mathbb {N}$, $\kappa \in \mathcal {K}_n$ and $\ell , \ell '\in \{0,1\}$ with $\ell +\ell '\le n$. Suppose that for some $\mu \ge 0$ and $0\le \lambda \le 1$ we have

$$\begin{aligned} |\kappa (Q,R,p,E)| \lesssim E^{-\mu } \min _{s\in [\lambda -1, 1-\lambda ]} \rho _{n,\lambda +s}(Q,E){\tilde{\rho }}_{n,\lambda -s}(R,E). \end{aligned}$$

Then

$$\begin{aligned} \kappa _{\ell , \ell '}:= \Big (a(v)H_0^{-1}\star _{\ell } \kappa \Big ) H_0^{-1}\star _{\ell '} a^*(v) \in \mathcal {K}_{n+1-\ell -\ell '}, \end{aligned}$$

and for

$$\begin{aligned} \sigma&= \min \{ \lambda + (\ell +\ell ')(1-\delta /\gamma ),1\} \\ \tau&= \mu + (\lambda + (\ell +\ell ')(1-\delta /\gamma )-1)_+ \end{aligned}$$

we have

$$\begin{aligned} |\kappa _{\ell , \ell '}(Q,R,p,E)|\lesssim E^{-\tau } \min _{s\in [\sigma -1, 1-\sigma ]} \rho _{n,\sigma +s}(Q,E){\tilde{\rho }}_{n,\sigma -s}(R,E). \end{aligned}$$

Proof

The kernel of $\kappa _{0,0}$ is

$$\begin{aligned}&\kappa _{0,0}(Q,R,p,E) \nonumber \\&\quad = \frac{v(r_1) v(q_{n+1}) \kappa \Big (Q_1^n,R_1^n, p+r_1+q_{n+1}, E+ \omega (r_1)+\omega (q_{n+1})\Big )}{(E+ \omega (Q)+\omega (r_1))(E+\omega (R)+\omega (q_{n+1}))}. \end{aligned}$$

(170)

Using the hypothesis, it thus satisfies

$$\begin{aligned}&|\kappa _{0,0}(Q,R,p,E)| \nonumber \\&\quad \lesssim \frac{|v(r_1)| |v(q_{n+1})| \rho _{n, 1}(Q_1^n, E+\omega (q_{n+1})){\tilde{\rho }}_{n, 2\lambda -1}(R_2^{n+1}, E+ \omega (r_1)+\omega (q_{n+1})) }{(E+ \omega (q_{n+1})+\omega (r_1))^{1+\mu +\lambda }(E+\omega (R))^{1-\lambda }} \nonumber \\&\quad = \frac{ \rho _{n+1, -1}(Q_1, E)) {\tilde{\rho }}_{n+1, -1}(R, E) }{(E+ \omega (q_{n+1})+\omega (r_1))^{1+\lambda +\mu }} \nonumber \\&\quad \le E^{-\mu } \rho _{n+1, \lambda + s}(Q, E)) {\tilde{\rho }}_{n+1, \lambda -s}(R, E) \end{aligned}$$

(171)

for any $-1-\lambda \le s \le 1+\lambda$.

For the case $\ell +\ell '=1$, we give the details only for $\kappa _{0,1}$ (the proof for $\kappa _{1,0}$ is similar, but the term corresponding to $i=1$ below does not occur). The kernel of the first parenthesis, where there is no contraction, satisfies

$$\begin{aligned}&\left| a(v)H_0^{-1} \star _0 \kappa (Q,R,p,E) \right| \nonumber \\&\quad \lesssim E^{-\mu } \frac{|v(r_1)| \rho _{n, \lambda + s}(Q, E) {\tilde{\rho }}_{n, \lambda - s}(R_2^{n+1}, E+\omega (r_1))}{E+\omega (Q)+\omega (r_1)}. \end{aligned}$$

(172)

Distinguishing the contraction with the first variable from the remaining ones, where we take $s=\lambda -1$, we obtain

$$\begin{aligned}&|\kappa _{0,1} (Q,R,p,E)| \nonumber \\&\quad =\left| \sum _{i=1}^{n+1} \int \frac{\big (a(v)H_0^{-1} \star _0 \kappa \big )(Q,S,p,E) v(\xi )}{E+\omega (\xi ) + \omega (R)}\bigg \vert \begin{array}{c} s_i=\xi \\ S_{\{i\}^c}=R \end{array}\, \textrm{d}\xi \right| \nonumber \\&\quad \lesssim E^{-\mu } \rho _{n, \lambda +s}(Q, E){\tilde{\rho }}_{n, \lambda -s}(R,E) \int \frac{|v(\xi )|^2 \textrm{d}\xi }{(E+\omega (\xi ) + \omega (Q))(E+\omega (\xi ) + \omega (R))} \nonumber \\&\qquad + \frac{| v(r_1)| {\tilde{\rho }}_{n-1, 1}(R_2^n, E+\omega (r_1)) \rho _{n, 2\lambda -1}(Q, E)}{E+\omega (Q)+\omega (r_1)} \nonumber \\&\qquad \times \sum _{i=2}^{n+1} \int \frac{|v(\xi )|^2 \textrm{d}\xi }{(E+\omega (\xi ) + \omega (R_1^{i-1}))(E+\omega (\xi ) + \omega (R))} \nonumber \\&\quad {\mathop {\lesssim }\limits ^{(141)}} E^{-\mu } \frac{\rho _{n, \lambda +s}(Q, E){\tilde{\rho }}_{n,\lambda -s}(R,E)}{ (E+\omega (r_1)+\omega (q_{n}))^{1-\delta /\gamma }} \nonumber \\&\qquad + nE^{-\mu } \frac{{\tilde{\rho }}_{n, -1 +2 (1-\delta /\gamma )}(R, E) \rho _{n, 2\lambda - 1}(Q, E)}{E+\omega (r_1)+\omega (q_n)} \nonumber \\&\quad \lesssim E^{-\mu - (\lambda -\delta /\gamma )_+}\rho _{n, \sigma +s}(Q, E){\tilde{\rho }}_{n, \sigma -s}(R,E) \end{aligned}$$

(173)

for $\sigma -1\le s \le 1-\sigma$ (in fact, the range can be chosen larger here). This shows the bound as claimed.

For $\ell =\ell '=1$ recall that we suppose that $n\ge 2$. The kernel we need to bound is

$$\begin{aligned}&|\kappa _{1,1}(Q,R,p,E)| \nonumber \\&\quad = \left| \sum _{i, j=1}^n \int \frac{v(\xi _1) }{E+\omega (\xi _1)+ \omega (Q)} \frac{v(\xi _2)}{E+\omega (\xi _2) + \omega (R)} \kappa (U,S,p,E)\bigg \vert \begin{array}{c} u_j=\xi _1 \\ U_{\{j\}^c}=Q \\ s_i=\xi _2 \\ S_{\{i\}^c}=R \end{array} \, \textrm{d}\xi _1 \textrm{d}\xi _2\right| \nonumber \\&\quad \lesssim E^{-\mu }\bigg (\sum _{j=1}^n \int \frac{|v(\xi _1)| \rho _{n, \lambda +s}(U, E) }{E+\omega (\xi _1) + \omega (q_{n-1})} \bigg \vert \begin{array}{c} u_j=\xi _1 \\ U_{\{j\}^c}=Q \end{array}\,\textrm{d}\xi _1\bigg ) \end{aligned}$$

(174)

$$\begin{aligned}&\qquad \times \bigg (\sum _{i=1}^n \int \frac{|v(\xi _2)| {\tilde{\rho }}_{n, \lambda -s}(S, E) }{E+\omega (\xi _2) + \omega (r_1)} \bigg \vert \begin{array}{c} s_i=\xi _2 \\ S_{\{i\}^c}=R \end{array} \, \textrm{d}\xi _2\bigg ) \end{aligned}$$

(175)

To bound the integral (175), we expand $\tilde{\rho }$ as a product of fractions with numerator $|v(s_\nu )|$ using its definition. For $i\ge 2$, we drop $\omega (\xi _2)$ from the denominators of all factors except the one of $v(\xi _2)=v(s_i)$, which gives the bound

$$\begin{aligned} \int \frac{|v(\xi )| {\tilde{\rho }}_{n, \lambda -s}(S, E) }{E+\omega (\xi ) + \omega (r_1)} \bigg \vert \begin{array}{c} s_i=\xi \\ S_{\{i\}^c}=R \end{array} \, \textrm{d}\xi&\le {\tilde{\rho }}_{n-1, \lambda -s}(R, E) \int \frac{|v(\xi )|^2}{(E+\omega (\xi ) + \omega (r_1))^2} \nonumber \\&{\mathop {\lesssim }\limits ^{(141)}} {\tilde{\rho }}_{n-1, \lambda -s+2(1-\delta /\gamma )}(R, E). \end{aligned}$$

(176)

For $i=1$ we do not drop $\omega (\xi )$ in the factor of $v(r_1)=v(s_2)$. This leads to (keeping in mind that $1+\lambda -s \le 2$)

$$\begin{aligned}&\int \frac{|v(\xi )| {\tilde{\rho }}_{n, \lambda -s}((\xi , R), E) }{E+\omega (\xi ) + \omega (r_1)} \, \textrm{d}\xi \nonumber \\&\quad \le {\tilde{\rho }}_{n-2,1}(R_2^{n-1}, E+\omega (r_1)) \int \frac{|v(r_1)| |v(\xi )|^2 \textrm{d}\xi }{(E+\omega (\xi ))^{(1+\lambda -s)/2}(E+\omega (\xi ) + \omega (r_1))^2} \nonumber \\&\quad {\mathop {\lesssim }\limits ^{(141)}} {\tilde{\rho }}_{n-1, \lambda -s+2(1-\delta /\gamma )}(R,E)\le E^{-(\lambda -\delta /\gamma )_+} {\tilde{\rho }}_{n-1, \sigma -s}(R,E). \end{aligned}$$

(177)

Arguing in the same way for the other integral (174) proves the claim. $\square$

B Operator Bounds

We first give a well known Lemma on the boundedness of $a(v)\textrm{d}\Gamma (\omega )^{-s}$.

Lemma B.1

For $s>\tfrac{1}{2}(1+\delta /\gamma )$

$$\begin{aligned} \Vert a(v) d\Gamma (\omega )^{-s} \Vert \le \Vert v\omega ^{-s}\Vert _{L^2}. \end{aligned}$$

Proof

Let $n\in \mathbb {N}_0$ and $\Psi \in {\mathcal {F}}^{(n+1)}$. Then, using Cauchy–Schwarz inequality, the symmetry of $\Psi$, and the fact that $2s\ge 1$, we have

$$\begin{aligned}&\Vert a(v) d\Gamma (\omega )^{-s} \Psi \Vert ^2_{{\mathcal {F}}^{(n)}} \nonumber \\&\quad =(n+1) \int _{\mathbb {R}^{dn}} \bigg | \int _{\mathbb {R}^d} \frac{v(\xi )}{\omega ^s(\xi )} \frac{\omega ^s(\xi )\Psi (K, \xi )}{(\omega (\xi )+\sum _{j=1}^{n} \omega (k_j))^{s}} d \xi \bigg |^2 d K\nonumber \\&\quad \le \Vert v\omega ^{-s}\Vert ^2 (n+1) \int _{\mathbb {R}^{(n+1)d}} \omega ^{2s}(k_{n+1}) \frac{|\Psi (K)|^2}{\big (\sum _{j=1}^{n+1} \omega (k_j)\big )^{2s}} dK \nonumber \\&\quad = \Vert v\omega ^{-s}\Vert ^2 \int _{\mathbb {R}^{(n+1)d}} \frac{\sum _{\ell =1}^{n+1} \omega ^{2s}(k_{\ell })}{(\sum _{j=1}^{n+1} \omega (k_j))^{2s}} |\Psi (K)|^2 dK \nonumber \\&\quad \le \Vert v\omega ^{-s} \Vert ^2_{L^2}\Vert \Psi \Vert ^2_{{\mathcal {F}}^{(n+1)}}. \end{aligned}$$

(178)

This proves the claim. $\square$

The following Lemmas provide bounds on elements of $\mathcal {K}_{n, \lambda }$, as operators respectively quadratic forms. Similar bounds (with $n=1$) appear in the literature on contact interactions [5, 9, 23].

Lemma B.2

Let $\kappa \in \mathcal {K}_n$ with kernel satisfying

$$\begin{aligned}&|\kappa (Q,R,p,E)| \lesssim \rho _{n,1}(Q,E){\tilde{\rho }}_{n,2\lambda -1}(R,E), \end{aligned}$$

for some $0\le \lambda \le 1$. Then for any non-negative $s>1-\lambda -\tfrac{1}{2} n(1-\delta /\gamma )$, the Formula (61) defines a bounded operator

$$\begin{aligned} \kappa :D\Big (\textrm{d}\Gamma (\omega )^{s}\Big )\rightarrow \mathcal {F}. \end{aligned}$$

Proof

Let $\Phi , \Psi \in \mathcal {F}$ be finite linear combinations of compactly supported functions in $L^2(\mathbb {R}^{dn})$ and note that the set of such elements is dense in $\mathcal {F}$. We have for any function h on $\mathbb {R}^d$

$$\begin{aligned}&|\langle \Phi , \kappa \Psi \rangle | \nonumber \\&\quad \le \int \limits _{\mathbb {R}^{dn}\times \mathbb {R}^{dn}} \left| \left\langle \left( \prod _{i=1}^n a_{q_i}\right) \Phi , \kappa (Q,R,\textrm{d}\Gamma (k)-P, \textrm{d}\Gamma (\omega )+E_0) \left( \prod _{i=1}^n a_{r_i}\right) \Psi \right\rangle \right| \textrm{d}Q \textrm{d}R \nonumber \\&\quad \lesssim \int \limits _{\mathbb {R}^{dn}\times \mathbb {R}^{dn}} \left\| \rho _{n,1}(Q, \textrm{d}\Gamma (\omega ))\left( \prod _{i=1}^n a_{q_i}\right) \Phi \right\| \left\| {\tilde{\rho }}_{n,2\lambda -1}(R, \textrm{d}\Gamma (\omega ))\left( \prod _{i=1}^n a_{r_i}\right) \Psi \right\| \textrm{d}Q \textrm{d}R \nonumber \\&\quad \le \left( \int \limits \left( \prod _{i=1}^n \frac{h(q_i)}{h(r_i)}\right) \left\langle \left( \prod _{i=1}^n a_{q_i}\right) \Phi , \rho _{n,1}(Q, \textrm{d}\Gamma (\omega ))^2 \left( \prod _{i=1}^n a_{q_i}\right) \Phi \right\rangle \textrm{d}R \textrm{d}Q \right) ^{1/2} \end{aligned}$$

(179)

$$\begin{aligned}&\quad \quad \times \left( \int \left( \prod _{i=1}^n \frac{h(r_i)}{h(q_i)}\right) \left\langle \left( \prod _{i=1}^n a_{r_i}\right) \Psi , {\tilde{\rho }}_{n,2\lambda -1}(R, \textrm{d}\Gamma (\omega ))^2 \left( \prod _{i=1}^n a_{r_i}\right) \Psi \right\rangle \textrm{d}Q \textrm{d}R \right) ^{1/2}. \end{aligned}$$

(180)

We choose $h(q)=\omega (q)^t/|v(q)|^2$ with $t=1+\delta /\gamma + \varepsilon /n$, where $\varepsilon >0$ is such that $t\le 2$. Then $\int \frac{\textrm{d}q}{h(q)}<\infty$ and, using the pull-through Formula (64),

$$\begin{aligned} (180)&\lesssim \left\langle \Psi ,\int \limits _{\mathbb {R}^{dn}}\left( \prod _{i=1}^n h(r_i) a_{r_i}^*\right) {\tilde{\rho }}_{n,2\lambda -1}(R, \textrm{d}\Gamma (\omega ))^2 \left( \prod _{i=1}^n a_{r_i}\right) \Psi \textrm{d}R \right\rangle ^{1/2} \nonumber \\&= \Bigg \langle \Psi ,\int \limits _{\mathbb {R}^{dn}}\left( \prod _{i=1}^n a_{r_i}^*\right) \frac{\omega (r_1)^t}{(\textrm{d}\Gamma (\omega ) +\omega (r_1))^{2\lambda }}\nonumber \\&\quad \times \bigg (\prod _{j=2}^{n}\frac{\omega (r_j)^t}{(\textrm{d}\Gamma (\omega ) + \omega (R_1^j))^2} \bigg ) \left( \prod _{i=1}^n a_{r_i}\right) \Psi \textrm{d}Q \Bigg \rangle ^{1/2} \nonumber \\&= \Bigg \langle \Psi ,\int \limits _{\mathbb {R}^{dn}}\left( \prod _{i=2}^{n} a_{r_i}^*\right) \omega (r_1)^t a^*_{r_1} a_{r_1}\textrm{d}\Gamma (\omega )^{-2\lambda } \nonumber \\&\qquad \times \bigg (\prod _{j=2}^{n}\frac{\omega (r_j)^t}{(\textrm{d}\Gamma (\omega ) + \omega (R_2^j))^2} \bigg ) \left( \prod _{i=2}^{n} a_{r_i}\right) \Psi \textrm{d}Q \Bigg \rangle ^{1/2} \nonumber . \end{aligned}$$

Now since $t\ge 1$,

$$\begin{aligned} \int _{\mathbb {R}^d}\omega (r_i)^t a^*_{r_1} a_{r_1}\textrm{d}\Gamma (\omega )^{-2\lambda } \textrm{d}q_n = \textrm{d}\Gamma (\omega ^{t})\textrm{d}\Gamma (\omega )^{-2\lambda } \le \textrm{d}\Gamma (\omega )^{t-2\lambda }. \end{aligned}$$

(181)

Assume first that $t\ge 2\lambda$. Then $\textrm{d}\Gamma (\omega )^{t-2\lambda } \le (\textrm{d}\Gamma (\omega )+ \omega (r_2))^{t-2\lambda }$, and we can iterate this argument to obtain

$$\begin{aligned}&\int \limits _{\mathbb {R}^{d\nu }}\Big (\prod _{i=1}^{\nu } a_{r_i}^*\Big )\frac{\omega (r_1)^t}{(\textrm{d}\Gamma (\omega ) +\omega (r_1))^{2\lambda }} \left( \prod _{j=2}^{n}\frac{\omega (r_j)^t}{(\textrm{d}\Gamma (\omega ) + \omega (R_1^j))^2} \right) \left( \prod _{i=1}^\nu a_{r_i}\right) \nonumber \\&\quad \le (\textrm{d}\Gamma (\omega )+\omega (r_{\nu +1}))^{t\nu -2(\nu -1)-2\lambda }\left( \prod _{j=\nu +1}^{n}\frac{\omega (r_j)^t}{(\textrm{d}\Gamma (\omega ) + \omega (R_{\nu +1}^j))^2} \right) ,\qquad \end{aligned}$$

(182)

as long as

$$\begin{aligned} t\nu -2(\nu -1)-2\lambda = \nu (\delta /\gamma -1)+2-2\lambda + \frac{\varepsilon \nu }{n}\ge 0. \end{aligned}$$

(183)

Now if $\lambda +\tfrac{1}{2} n(1-\delta /\gamma )\le 1$ this holds true up to $\nu =n$ and we obtain

$$\begin{aligned} (180) \lesssim \left\langle \Psi , \textrm{d}\Gamma (\omega )^{2-2\lambda +n(\delta /\gamma -1)+\varepsilon )} \Psi \right\rangle ^{1/2} = \Vert \textrm{d}\Gamma (\omega )^{1-\lambda -n(1-\delta /\gamma )/2 +\varepsilon /2} \Psi \Vert .\nonumber \\ \end{aligned}$$

(184)

If $\lambda +\tfrac{1}{2}n(1-\delta /\gamma )> 1$, let $\nu _0$ be the smallest $\nu \ge 1$ (which exists for small enough $\varepsilon$) such that $t\nu -2\nu +2-\lambda +\varepsilon \nu /n\le 0$. Then we proceed as before, but bound the negative power of $\textrm{d}\Gamma (\omega )$ by a constant, which gives

$$\begin{aligned}&\int \limits _{\mathbb {R}^{d\nu _0}}\Big (\prod _{i=1}^{\nu _0} a_{r_i}^*\Big )\frac{\omega (r_1)^t}{(\textrm{d}\Gamma (\omega ) +\omega (r_1))^{2\lambda }} \left( \prod _{j=2}^{n}\frac{\omega (r_j)^t}{(\textrm{d}\Gamma (\omega ) + \omega (R_1^j))^2} \right) \Big (\prod _{i=1}^{\nu _0} a_{r_i}\Big ) \nonumber \\&\quad \lesssim \prod _{j=\nu _0+1}^{n}\frac{\omega (r_j)^t}{(\textrm{d}\Gamma (\omega ) + \omega (R_{\nu _0+1}^j))^2} , \end{aligned}$$

(185)

These remaining factors lead to a bounded operator by the same reasoning, because $t\le 2$, so we obtain in this case

$$\begin{aligned} (180) \lesssim \Vert \Psi \Vert . \end{aligned}$$

(186)

By the same argument, up to renaming of R, Q, we also have

$$\begin{aligned} (179) \lesssim \Vert \Phi \Vert , \end{aligned}$$

(187)

and thus

$$\begin{aligned}&|\langle \Phi , \kappa \Psi \rangle | \lesssim \Vert \Phi \Vert \Vert \textrm{d}\Gamma (\omega )^{(1-\lambda -n(1-\delta /\gamma )/2 +\varepsilon /2)_+} \Psi \Vert , \end{aligned}$$

(188)

which proves the claim. $\square$

Lemma B.3

Let $\kappa \in \mathcal {K}_n$ with kernel satisfying

$$\begin{aligned}&|\kappa (Q,R,p,E)| \lesssim \rho _{n,\lambda }(Q,E){\tilde{\rho }}_{n,\lambda }(R,E), \end{aligned}$$

for some $0\le \lambda \le 1$. Then for any non-negative $s>1-\lambda -n(1-\delta /\gamma )$, $\kappa$ defines a quadratic form on $D\Big (\textrm{d}\Gamma (\omega )^{s/2}\Big )$ satisfying

$$\begin{aligned} |\langle \Phi , \kappa \Psi \rangle | \lesssim \Vert \textrm{d}\Gamma (\omega )^{s}\Phi \Vert \Vert \textrm{d}\Gamma (\omega )^{s}\Psi \Vert . \end{aligned}$$

In particular, if $\lambda +n(1-\delta /\gamma )>1$ then $\kappa$ defines a bounded operator on $\mathcal {F}$.

Proof

As in the proof of Lemma B.2, we take $\Phi , \Psi$ as finite combinations of compactly supported functions and obtain

$$\begin{aligned}&|\langle \Phi , \kappa \Psi \rangle |^2 \nonumber \\&\quad \le \int \limits _{\mathbb {R}^{dn}\times \mathbb {R}^{dn}} \Bigg (\prod _{i=1}^n \frac{h(q_i)}{h(r_i)}\Bigg ) \left\langle \Bigg (\prod _{i=1}^n a_{q_i}\Bigg ) \Phi , \rho _{n,\lambda }(Q, \textrm{d}\Gamma (\omega ))^2 \Bigg (\prod _{i=1}^n a_{q_i}\Bigg ) \Phi \right\rangle \textrm{d}R \textrm{d}Q \nonumber \\&\qquad \times \int \limits _{\mathbb {R}^{dn}\times \mathbb {R}^{dn}} \Bigg (\prod _{i=1}^n \frac{h(r_i)}{h(q_i)}\Bigg ) \left\langle \Bigg (\prod _{i=1}^n a_{r_i}\Bigg ) \Psi , {\tilde{\rho }}_{n,\lambda }(R, \textrm{d}\Gamma (\omega ))^2 \Bigg (\prod _{i=1}^n a_{r_i}\Bigg ) \Psi \right\rangle \textrm{d}Q \textrm{d}R .\nonumber \\ \end{aligned}$$

(189)

With the identical choice for h, we conclude using the arguments of Lemma B.3 that

$$\begin{aligned}&|\langle \Phi , \kappa \Psi \rangle | \lesssim \Vert \textrm{d}\Gamma (\omega )^{(1-\lambda -n(1-\delta /\gamma ) +\varepsilon )/2} \Phi \Vert \Vert \textrm{d}\Gamma (\omega )^{(1-\lambda -n(1-\delta /\gamma ) +\varepsilon )/2} \Psi \Vert , \end{aligned}$$

(190)

if $\lambda + n(1-\delta /\gamma ) \le 1$, and

$$\begin{aligned} |\langle \Phi , \kappa \Psi \rangle | \lesssim \Vert \Phi \Vert \Vert \Psi \Vert \end{aligned}$$

(191)

if $\lambda + n(1-\delta /\gamma )>1$. This proves the claim. $\square$