First of all, it is easy to check that for any cube $I$, $$\begin{aligned}
\langle |b_1|\rangle_{\Phi_0, I} \langle |b_2|\rangle_{\Phi_0,I}\le \Big\langle \Phi_0^{-1}\big((M1_{I_0}(x))^{-1}\big)^{-1} |b_2|\Big\rangle_{\Phi_0,I}\le 1.\end{aligned}$$ Then by the triangle inequality and general Hölder’s inequality we have $$\langle |b_1-\langle b_1\rangle_I|\rangle_{\Phi_0, I} \langle |b_2-\langle b_2\rangle_I|\rangle_{\Phi_0,I}\lesssim 1.$$ On the other hand, since $|b_1b_2|\le  1$, using triangle inequality and general Hölder’s inequality again we have $$\langle |b_1-\langle b_1\rangle_I|  |b_2-\langle b_2\rangle_I|\rangle_{\Phi_0,I}\lesssim 1.$$ Using Theorem \[upperbound\] we know that $C_bH$ is bounded on $L^2$, thanks to $\bar\Phi_0\in B_2$.

It remains to show that $S_{A,B}(b_1,b_2)=\infty$ and $T_C(b_1,b_2)=\infty,$ for all log -bumps $A,B,C$ with $\bar A,\bar B,\bar C\in B_2.$ Without loss of generality we can assume that $A(t)=t^2 \log(e+t)^{1+\alpha}$, $B(t)=t^2 \log(e+t)^{1+\beta}$ and $C(t)=t^2 \log(e+t)^{1+\gamma}$, where $\alpha, \beta, \gamma>0$. For $S_{A,B}(b_1, b_2)$ again we test with the interval $J_k=(-k,k)$ with $k\ge 2$. Since $b_1$ and $b_2$ are odd functions, we have $$\begin{aligned}
\langle |b_1-\langle b_1\rangle_{J_k}|\rangle_{A, J_k} \langle |b_2-\langle b_2\rangle_{J_k}|\rangle_{B,J_k}&=\langle |b_1|\rangle_{A, J_k} \langle |b_2|\rangle_{B,J_k}\\
&\simeq k^{-\frac 12}\log(e+k)^{\frac{1+\alpha}2} k^{\frac 12}\log(e+k)^{-\frac 12}\log\log(e^e+k)^{-\frac 34}\\
&\overset{k\rightarrow\infty}{\rightarrow}\infty\end{aligned}$$ For $T_C(b_1, b_2)$, we test with the cube $I_k=(0,k)$, $k\ge 100$, we have $$\begin{aligned}
    T_{C}(b_1, b_2)&\ge \lim_{k\rightarrow \infty}  {\big \langle}|b_1- \langle b_1\rangle_{I_k}| |b_2- \langle b_2\rangle_{I_k}| {\big \rangle}_{ C, I_k} \ge \lim_{k\rightarrow \infty}  {\big \langle}|b_1- \langle b_1\rangle_{I_k}| |b_2- \langle b_2\rangle_{I_k}| 1_{(0,1)}{\big 