$d(\lambda)=d_{1}(\lambda)|{\varphi}_{1}\rangle\langle{\varphi}_{1}|,\quad
d_{1}(\lambda)=\frac{1|}{g_{1}(\lambda)(|\langle{\varphi}_{1},v_{1}\rangle|^{2}+cg_{1}(\lambda)^{-1})}\,.$
(9.5)

$d_{1}(\lambda)$ is a Mikhlin multiplier. Then, $B_{1}(\lambda)$ is invertible
by Lemma 6.1 and

$\displaystyle
B_{1}(\lambda)^{-1}=\lambda^{-2}({S_{2}(S_{2}{\mathcal{M}}_{1}S_{2})^{-1}S_{2}}+{d_{1}(\lambda)}Q)\,,$
(9.6) $\displaystyle
Q=\begin{pmatrix}|{\varphi}_{1}\rangle\langle{\varphi}_{1}|&-|{\varphi}_{1}\rangle\langle{\varphi}_{1}|{\tilde{a}}_{12}{\tilde{a}}_{22}^{-1}\\\
-|{\tilde{a}}_{22}^{-1}{\tilde{a}}_{21}{\varphi}_{1}\rangle\langle{\varphi}_{1}&|{\tilde{a}}_{22}^{-1}{\tilde{a}}_{21}{\varphi}_{1}\rangle\langle{\varphi}_{1}|{\tilde{a}}_{12}{\tilde{a}}_{22}^{-1}\end{pmatrix}.$
(9.7)

Note that $Q$ is $\lambda$-independent, ${\rm rank}\,Q=2$ and
$Q=({\varphi}_{1}\oplus{\tilde{\varphi}})\otimes({\varphi}_{1}\oplus{\tilde{\varphi}})$
with ${\tilde{\varphi}}=-{\tilde{a}}_{22}^{-1}{\tilde{a}}_{21}{\varphi}_{1}$.
Then, (9.4) and (9.6) implies
$B(\lambda)=(1+L_{2}(\lambda)B_{1}(\lambda)^{-1})B_{1}(\lambda)$ is invertible
and $B(\lambda)^{-1}$ is given by

$B_{1}(\lambda)^{-1}+L_{3}(\lambda),\
L_{3}(\lambda)=-B_{1}(\lambda)^{-1}L_{2}(\lambda)B_{1}(\lambda)^{-1}+{\mathcal{O}}^{(3)}(\lambda^{2}\langle\log\lambda\rangle^{4}).$
(9.8)

Hece ${\mathcal{M}}(\lambda)^{-1}$ exists by Lemma 6.2 and
${\mathcal{M}}(\lambda)^{-1}={\mathcal{L}}(\lambda)+{\mathcal{L}}(\lambda)S_{1}B(\lambda)^{-1}S_{1}{\mathcal{L}}(\lambda)$.

######  Lemma 9.2.

Modulo a good producer $M_{v}{\mathcal{M}}(\lambda)^{-1}M_{v}\equiv
M_{v}S_{1}B_{1}(\lambda)^{-1}S_{1}M_{v}$ .

###### Proof.

We have
$M_{v}{\mathcal{M}}(\lambda)^{-1}M_{v}=M_{v}{\mathcal{L}}(\lambda)M_{v}+M_{v}{\mathcal{L}}(\lambda)S_{1}B(\lambda)^{-1}S_{1}{\mathcal{L}}(\lambda)M_{v}$.
By Lemma 6.6 again $M_{v}{\mathcal{L}}(\lambda)M_{v}$ is a good producer.
Substituting $B(\lambda)^{-1}$ by (9.8), we see that the second term on the
right is equal to $E_{1}(\lambda)+E_{2}(\lambda)$ where

$E_{1}(\lambda)=M_{v}{\mathcal{L}}(\lambda)S_{1}B_{1}(\lambda)^{-1}S_{1}{\mathcal{L}}(\lambda)M_{v},\quad
E_{2}(\lambda)=M_{v}{\mathcal{L}}(\lambda)S_{1}L_{3}(\lambda)S_{1}{\mathcal{L}}(\lambda)M_{v}\,.$

(i) We first show that $E_{2}(\lambda)$ is a good producer. We obtain by
combining (9.4), (9.6) and (9.8) that
$S_{1}L_{3}(\lambda)S_{1}=\sum_{j,k=1}^{n}(\log\lambda)^{2}g_{jk}(\lambda)|{\varphi}_{j}\rangle\langle{\varphi}_{k}|$
with Mikhlin multipliers
$g_{jk}(\lambda)\in{\mathcal{O}}^{(3)}_{{\mathbb{C}}}(1)$, $j,k=1,\dots,n$. We
then recall (7.4) which implies that for $j=1,\dots,n$

$\displaystyle
M_{v}{\mathcal{L}}(\lambda){\varphi}_{j}=\psi_{j0}+g_{1}(\lambda)\lambda^{2}\psi_{j1}(x)+\lambda^{2}\psi_{j2}(x)+\psi_{j3}(\lambda,x),$
(9.9) $\displaystyle\psi_{j0},\,\psi_{j1},\,\psi_{j2}\in(L^{1}\cap
L^{2}),\quad\psi_{j3}\in{\mathcal{O}}^{(3)}_{L^{1}\cap
L^{2}}(\lambda^{4}(\log\lambda)^{2})\,.$ (9.10)

Since the integral kernel of ${\mathcal{L}}(\lambda)^{\ast}$ is the complex
conjugate of ${\mathcal{L}}(\lambda)$,
$M_{v}{\mathcal{L}}(\lambda)^{\ast}{\varphi}_{k}$, $k=1,\dots,n$ is expressed
similarly. Hence
$E_{2}(\lambda)=\sum(\log{\lambda})^{2}g_{jk}(\lambda)({\mathcal{L}}+{\mathcal{O}}^{(3)}_{{\mathcal{L}}}(h_{2}(\lambda)))$
and Lemma 3.7 for $(j,\ell)=(2,2)$ and Proposition 3.9 imply that
$E_{2}(\lambda)$ is a good producer.

ii) Define $B_{2}(\lambda)=S_{1}B_{1}(\lambda)^{-1}S_{1}$. By virtue of (9.6)
and (9.7) we have

$B_{2}(\lambda)=\sum_{j,k=1}^{n}\lambda^{-2}f_{jk}(\lambda)|{\varphi}_{j}\rangle\langle{\varphi}_{k}|,\quad
f_{jk}(\lambda)\in{\mathcal{O}}^{(3)}_{{\mathbb{C}}}(1).$ (9.11)

Substituting ${\mathcal{L}}(\lambda)$ by (6.6) and using
$D_{0}S_{1}=S_{1}D_{0}=S_{1}$, we express
$E_{1}(\lambda)=E_{11}(\lambda)+E_{12}(\lambda)+E_{13}(\lambda)+E_{14}(\lambda)$
where

$\displaystyle
E_{11}=M_{v}(D_{0}-D_{0}L_{1}(\lambda))B_{2}(\lambda)(D_{0}-D_{0}L_{1}(\lambda))M_{v},\
E_{12}=M_{v}{\mathcal{L}}(\lambda)B_{2}(\lambda)\tilde{L}_{1}(\lambda)M_{v},$
$\displaystyle
E_{13}=M_{v}D_{0}\tilde{L}_{1}(\lambda)B_{2}(\lambda){\mathcal{L}}(\lambda)M_{v},\quad
E_{14}=M_{v}D_{0}\tilde{L}_{1}(\lambda)B_{2}(\lambda)\tilde{L}_{1}(\lambda)M_{v}.$

(a) We first show that $E_{12}(\lambda)$ is a good producer.
$E_{12}=\sum\lambda^{-2}f_{jk}(\lambda)(M_{v}{\mathcal{L}}(\lambda){\varphi}_{j})\otimes(M_{v}\tilde{L}_{1}(\lambda)^{\ast}{\varphi}_{k})$
by (9.11). We can apply (9.9) and (9.10) to
$M_{v}{\mathcal{L}}(\lambda){\varphi}_{j}$ and we have
$M_{v}\tilde{L}_{1}(\lambda)^{\ast}{\varphi}_{k}\in{\mathcal{O}}^{(3)}_{{\mathcal{H}}\cap{\mathcal{L}}}(\lambda^{4}(\log\lambda)^{2})$
by virtue of (6.5) and (6.6). It follows
$E_{12}(\lambda)\in{\mathcal{O}}^{(3)}_{{\mathcal{L}}}(\lambda^{2}(\log\lambda)^{2})$
and $E_{12}(\lambda)$ is a good producer by virtue of Proposition 3.9. Similar
argument implies $E_{13}(\lambda)$ and $E_{14}(\lambda)$ are both good
producers.

(b) We have $E_{11}(\lambda)=M_{v}B_{2}(\lambda)M_{v}+E_{3}(\lambda)$ where
$E_{3}(\lambda)$ is defined by

$E_{3}(\lambda)=-M_{v}B_{2}(\lambda)L_{1}(\lambda)M_{v}-M_{v}D_{0}L_{1}(\lambda)B_{2}(\lambda)M_{v}+M_{v}D_{0}L_{1}(\lambda)B_{2}(\lambda)L_{1}(\lambda)M_{v}.$

We prove $E_{3}(\lambda)$ is a good producer to finish the proof. Recalling
(6.5) and that $v{\varphi}\in{\langle x\rangle}^{-2-\delta}(L^{1}\cap L^{4})$
for ${\varphi}\in S_{1}{L^{2}}$, we obtain as previously that, for
$j=1,\dots,n$,

$\displaystyle\lambda^{-2}M_{v}D_{0}L_{1}(\lambda){\varphi}_{j}(x)=g_{1}(\lambda)\tilde{\psi}_{j1}(x)+\tilde{\psi}_{j2}(x)+\tilde{\psi}_{j3}(\lambda,x),$
$\displaystyle\tilde{\psi}_{j1},\ \tilde{\psi}_{j2}\in(L^{1}\cap L^{2}),\
\tilde{\psi}_{3}(\lambda)\in{\mathcal{O}}^{(3)}_{L^{1}\cap
L^{2}}(h_{2}(\lambda)).$

An obvious modification of the argument shows that similar expressions are
satisfied by $\lambda^{-2}M_{v}L_{1}(\lambda)^{\ast}{\varphi}_{k}$ for
$k=1,\dots,n$. Combining these with (9.11) produces the expression for
$E_{3}(\lambda)$:

$E_{3}(\lambda)=\sum_{j,k=1}^{n}(\log\lambda)^{2}h_{jk}(\lambda){\mathcal{L}}_{jk}+{\mathcal{O}}^{(3)}_{{\mathcal{L}}}(\lambda^{2}\langle\log\lambda\rangle^{2})$

with ${\mathcal{L}}_{jk}\in{\mathcal{L}}$ and
$h_{jk}(\lambda)\in{\mathcal{O}}_{{\mathbb{C}}}^{(3)}(1)$ for $1\leq j,k\leq
n$. Thus, $E_{3}(\lambda)$ is a good producer by Lemma 3.7 and Proposition
3.9. ∎

#### Proof of Theorem 1.4 (2c)

In view of Lemma 9.2, it suffices to prove that the operator $Z$ defined by

$Zu(x)=\int_{0}^{\infty}G_{0}(-\lambda)M_{v}S_{1}B_{1}(\lambda)^{-1}S_{1}M_{v}\Pi(\lambda)u(x)\chi_{\leq
a}(\lambda)\lambda d\lambda$ (9.12)

is bounded in $L^{p}$ for $1<p<2$ and unbounded for $2<p<\infty$. We
substitute (9.6) for $B_{1}(\lambda)^{-1}$, which makes $Z=Z_{1}+Z_{2}$ where
$Z_{1}$ and $Z_{2}$ are produced by
$\lambda^{-2}S_{2}(S_{2}{\mathcal{M}}_{1}S_{2})^{-1}S_{2}$ and
$\lambda^{-2}d_{1}(\lambda)(v({\varphi}_{1}+{\tilde{\varphi}}))\otimes(v({\varphi}_{1}+{\tilde{\varphi}}))$.

We may repeat the argument of the previous section §8 to $Z_{1}$ with obvious
modifications, which proves that $Z_{1}$ is bounded in $L^{p}$ for $1<p<4$ and
unbounded for $4<p<\infty$ in general and, it becomes a good operator if all
${\varphi}\in S_{2}{L^{2}}$ satisfy the extra cancellation property $\langle
v,y_{j}{\varphi}\rangle=0$, $j=1,\dots,4$.

The operator $Z_{2}$ is the same as the one defined by (7.5) if ${\varphi}$
and $\mu(\lambda)$ are replaced by ${\varphi}_{1}+{\tilde{\varphi}}$ and
$d_{1}(\lambda)$ respectively. Then, the repetition of the argument below
(7.5) implies that $Z_{1}$ is bounded in $L^{p}$ for $1<p<2$ and is unbounded
for $2<p<\infty$. This completes the proof of Theorem 1.4. ∎

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