$$\begin{split}
        {\Big|\langle S_1(f),\psi\rangle\Big|_{}} &\leq \sum_{Q\in S} \int_Q{|b_1-\langle b_1\rangle_Q|}{|\psi|}\langle{|b_2-\langle b_2\rangle_Q|}{|f|}\rangle_Q \\
        &\lesssim \sum_{Q\in S} {|Q|}\langle {|b_1-\langle b_1\rangle|}\rangle_{A,Q}\langle{|\psi|} \rangle_{\bar A, Q}\langle{|b_2-\langle b_2\rangle_Q|}\rangle_{B,Q}\langle{|f|} \rangle_{\bar B, Q} \\
        &\leq S_{A,B}(b_1,b_2) \sum_{Q\in S} {|Q|}\langle{|\psi|} \rangle_{\bar A,Q}\langle{|f|} \rangle_{\bar B, Q} \leq \gamma^{-1}S_{A,B}(b_1,b_2) \sum_{Q\in S} \int_{E_Q}\langle{|\psi|} \rangle_{\bar A,Q}\langle{|f|} \rangle_{\bar B, Q} \\
        &\leq \gamma^{-1}S_{A,B}(b_1,b_2) \sum_{Q\in S} \int_{E_Q}  {M}_{\bar A}\psi {M}_{\bar B}f\leq  \gamma^{-1}S_{A,B}(b_1,b_2) {\big\| {M}_{\bar A}\psi\big\|_{L^2}}{\big\| {M}_{\bar B}f\big\|_{L^2}} \\
        &\lesssim S_{A,B}(b_1,b_2)  {\big\|\psi\big\|_{L^2}}{\big\|f\big\|_{L^2}},
        \end{split}$$ where we have used Proposition \[PE95\] in the last step.

Next we use the condition $T_C$ to control the term $S_3$: $$\begin{aligned}
{\Big|\langle S_3(f),\psi\rangle\Big|_{}} &\leq \sum_{Q\in S} {|Q|}\langle {|\psi|} \rangle_Q  \langle{|b_1-\langle b_1\rangle_Q|}{|b_2-\langle b_2\rangle_Q|}{|f|}\rangle_Q \\
&\leq  \sum_{Q\in S} {|Q|}\langle {|\psi|} \rangle_Q \langle {|f|} \rangle_{\bar C, Q} \langle{|b_1-\langle b_1\rangle_Q|} {|b_2-\langle b_2\rangle_Q|} \rangle_{C,Q} \\
&\leq T_C(b_1,b_2)\sum_{Q\in S} {|Q|}\langle {|\psi|} \rangle_Q \langle {|f|} \rangle_{\bar C, Q}  \leq \gamma^{-1} T_C(b_1,b_2){\big\|M\psi\big\|_{L^{2}}}{\big\|M_{\bar C}f\big\|_{L^2}}\\
&\lesssim  T_C(b_1,b_2){\big\|\psi\big\|_{L^{2}}}{\big\|f\big\|_{L^2}}.\end{aligned}$$

Since with $A(t)=t^p,$ $\bar A \in B_{2},$ for $p>2,$ we immediately get:

\[rmk:p\] Let $T$ be as before and assume that a pair of functions $b_1,b_2\in L^{2p}_{loc}({\mathbb{R}}^d)$ satisfy the conditions $T_{p}$ and $S_{p}$ for some $p>2.$ Then we have $$\begin{split}
    C_bT\colon L^2({\mathbb{R}}^d)\longrightarrow L^2({\mathbb{R}}^d)
    \end{split}$$ boundedly.

We close this section with some remarks.

For Theorem \[upperbound\] the difference in the case $p\not=2$ is that we need to introduce 