Then $$\label{ml1}
                    {\Big|{\dashint_{Q} {{\dashint_{Q} {(b_1(x)-b_1(y))(b_2(x)-b_2(y))\overline{b_1(x)}\overline{b_2(y)} {\,\mathrm{d}}y {\,\mathrm{d}}x}}}}\Big|_{}} \geq {\dashint_{Q} {{|b_1|}^2}}{\dashint_{Q} {{|b_2|}^2}}$$ and $$\label{ml2}
                    {\dashint_{Q} {{\dashint_{Q} {(b_1(x)-b_1(y))(b_2(x)-b_2(y)\overline{b_1(x)}\overline{b_2(x)} {\,\mathrm{d}}y {\,\mathrm{d}}x}}}} = \dashint_Q{|b_1b_2|}^2 + {\Big|\dashint_Qb_1b_2\Big|_{}}^2,$$ where we have replaced the latter occurrance of $b_2(y)$ with $b_2(x).$

Multiplying out shows that $$\begin{split}
                    &\dashint_Q\dashint_Q(b_1(x)-b_1(y))(b_2(x)-b_2(y))\overline{b_1(x)}\overline{b_2(y)}{\,\mathrm{d}}y {\,\mathrm{d}}x \\
                    &= \dashint_Qb_1(x)b_2(x)\overline{b_1(x)}{\,\mathrm{d}}x \dashint_Q\overline{b_2(y)}{\,\mathrm{d}}y - \dashint_Q b_1(x)\overline{b_1(x)}{\,\mathrm{d}}x\dashint_Q b_2(y)\overline{b_2(y)}{\,\mathrm{d}}y\\
                    &- \dashint_Q b_2(x)\overline{b_1(x)}{\,\mathrm{d}}x\dashint_Qb_1(y)\overline{b_2(y)}{\,\mathrm{d}}y + \dashint_Q\overline{b_1(x)}{\,\mathrm{d}}x\dashint_Qb_1(y)b_2(y)\overline{b_2(y)}{\,\mathrm{d}}y \\
                    &= -\dashint_Q{|b_1(x)|}^2{\,\mathrm{d}}x \dashint_Q{|b_2(y)|}^2{\,\mathrm{d}}y - {\Big|\dashint_Q b_1(x)\overline{b_2(x)}{\,\mathrm{d}}x\Big|_{}}^2,
                \end{split}$$ whence $$\begin{split}
                        & {\Big|\dashint_Q\dashint_Q(b_1(x)-b_1(y))(b_2(x)-b_2(y)\overline{b_1(x)}\overline{b_2(y)}{\,\mathrm{d}}y {\,\mathrm{d}}x \Big|_{}}  \\
                        &=  \dashint_Q{|b_1(x)|}^2{\,\mathrm{d}}x \dashint_Q{|b_2(y)|}^2{\,\mathrm{d}}y + {\Big|\dashint_Q b_1(x)\overline{b_2(x)}{\,\mathrm{d}}x\Big|_{}}^2 \ge  \dashint_Q{|b_1(x)|}^2{\,\mathrm{d}}x \dashint_Q{|b_2(y)|}^2{\,\mathrm{d}}y.
                    \end{split}$$ As for we compute: $$\begin{split}
        &   \dashint_Q\dashint_Q\left(b_1(x)-b_1(y)\right)\left(b_2(x)-b_2(y)\right)\overline{b_1(x)b_2(x)}{\,\mathrm{d}}y{\,\mathrm{d}}x \\
        &= \dashint_Q{|b_1b_2|}^2-\dashint_Qb_2\dashint_Q{|b_1|}^2\overline{b_2}-\dashint_Qb_1\dashint_Q\overline{b_1}{|b_2|}^2+\dashint_Qb_1b_2\dashint_Q\overline{b_1b_2} \\
        &= \dashint_Q{|b_1b_2|}^2 + {\Big|\dashint_Qb_1b_2\Big|_{}}^2.
\end{split}$$

The lower bounds now follow by combining lemmas \[rieszlemma\] and \[magiclemma\].

\[lowerbound\] Let $R_j$, $j= 1,\dots,d,$ be the Riesz transforms, $b_1b_2\in L^2_{loc},$ $b_1,b_2\in L^3_{loc}.$ Then $$S_2(b_1,b_2)+T_2(b_1,b_2) \leq C_d\sum_{j=1}^{d}{\big\|C_bR_j\big\|_{L^2\to L^2}}.$$

Denote $\psi_i = b_i-\langle b_i\rangle_Q,$ $i=1,2.$ Then $\int_Q\psi_i$ = 0 and the assumptions of Lemma \[magiclemma\] are satisfied by which by and lemma \[rieszlemma\] we get the necessary condition $S_2$ $$\begin{split}
        {\dashint_{Q} {{|\psi_1(x)|}^2{\,\mathrm{d}}x}}{\dashint_{Q} {{|\psi_2(y)|}^2{\,\mathrm{d}}y}} &\leq {\Big|{\dashint_{Q} {{\dashint_{Q} {(\psi_1(x)-\psi_1(y))(\psi_2(x)-\psi_2(y)\overline{\psi_2(y)}\overline{\psi_1(x)} {\,\mathrm{d}}y {\,\mathrm{d}}x}} {}}}\Big|_{}}  \\
        &\leq C_d\sum_{i=1}^{k}{\big\|C_bR_i\big\|_{L^2\to L^2}} \left( {\dashint_{Q} {{|\psi_1|}^2}} \right) ^{1/2}\left( {\dashint_{Q} {{|\psi_2|}^{2}}} \right) ^{1/2}.
        \end{split}$$ For the condition $T_2$, we apply lemma \[rieszlemma\] with $f_2 = 1_Q,$ $f_1 = \overline{\psi_1\psi_2}$ and lemma \[magiclemma\] by with $b_i = 