\bar C\in B_2$. Moreover, $C_bH: L^2\not\to L^2.$

We prove the result via the following example. Let $I_0=[-1,1]$ and $$\sigma= 1_{I_0},\qquad w=M(\sigma)^{-1},$$ notice that both $\sigma$ and $w$ are even functions. It is immediate to see that $$\label{eee}
    \sup_I \langle \sigma\rangle_I \langle w \rangle_I \le \sup_I \inf_{x\in I} M(\sigma)(x) \langle w \rangle_I\le \sup_I\langle  M(\sigma)w \rangle_I =1.$$ Now define $$\begin{aligned}
    b_1(x):= \operatorname{sgn}(x)\sigma(x),\qquad b_2(x):= \operatorname{sgn}(x) w^{\frac 12}(x),
    \end{aligned}$$ and notice that immediately $b_1, b_2\in L_{\rm{loc}}^{\infty}.$ By we see that $$\label{end1}
    S_2(b_1, b_2)^2=\sup_I {\big \langle}|b_1-\langle b_1\rangle_I|^2{\big \rangle}_I {\big \langle}|b_2-\langle b_2\rangle_I|^2{\big \rangle}_I\le 16 \sup_I {\big \langle}|b_1|^2{\big \rangle}_I {\big \langle}|b_2|^2{\big \rangle}_I \le 16< \infty.$$ We also have $$\begin{aligned}
    |b_1-\langle b_1\rangle_I|^2|b_2-\langle b_2\rangle_I|^2 & \le 4(|b_1|^2+ |\langle b_1\rangle_I|^2)(|b_2|^2+ |\langle b_2\rangle_I|^2),
    \end{aligned}$$ and by $|b_1b_2|\le 1$, direct calculations give us $$\label{end2}
    T_2(b_1, b_2)^2=\sup_I {\big \langle}|b_1-\langle b_1\rangle_I|^2|b_2-\langle b_2\rangle_I|^2{\big \rangle}_I \le 4+ 12{\big \langle}|b_1|^2{\big \rangle}_I {\big \langle}|b_2|^2{\big \rangle}_I\le 16<\infty.$$ However, for $J_k=(-k,k)$, $k\ge 2$, since $b_1$ and $b_2$ are odd functions, $$\begin{aligned}
    S_{A, B}(b_1, b_2) &\ge \lim_{k\rightarrow \infty}{\big \langle}|b_1 |{\big \rangle}_{ A, J_k}{\big \langle}|b_2 |  {\big \rangle}_{B, J_k} \gtrsim \lim_{k\rightarrow \infty}{\big \langle}|b_1 |{\big \rangle}_{ A, J_k}{\big \langle}|b_2 |^2  {\big \rangle}_{ J_k}^{\frac 12} \\
    &\sim \lim_{k\rightarrow \infty}\frac{k^{\frac 12}}{ A^{-1}(k)}\sim \lim_{k\rightarrow \infty}\frac{\bar A^{-1}(k)}{k^{\frac 12}}= \lim_{k\rightarrow \infty}\Big(\frac{(\bar A^{-1}(k))^2}{\bar A(\bar A^{-1}k) }\Big)^{\frac12},
    \end{aligned}$$ where we 