have used the fact that $M(1_{I_0})(x)\sim (1+|x|)^{-1}$. To conclude notice that immediately by definition $\lim_{t\rightarrow \infty}\bar A^{-1}(t)=\infty$ and $$\lim_{t\rightarrow \infty} \frac{\bar A(t)}{t^2}\le \frac 4{\log 2}\lim_{t\rightarrow \infty} \int_{t}^{2t} \frac{\bar A(s)}{s^2}\frac{ds}{s}=0.$$ On the other hand with $I_k=(0, k)$, $k\ge 100,$ we have $$\begin{aligned}
    T_{C}(b_1, b_2)&\ge \lim_{k\rightarrow \infty}  {\big \langle}|b_1- \langle b_1\rangle_{I_k}| |b_2- \langle b_2\rangle_{I_k}| {\big \rangle}_{ C, I_k} \ge \lim_{k\rightarrow \infty}  {\big \langle}|b_1- \langle b_1\rangle_{I_k}| |b_2- \langle b_2\rangle_{I_k}| 1_{(0,1)}{\big \rangle}_{ C, I_k}\\
    &=\lim_{k\rightarrow \infty} (1-k^{-1})  {\big \langle}|b_2- \langle b_2\rangle_{I_k}| 1_{(0,1)}{\big \rangle}_{ C, I_k}.
    \end{aligned}$$ Since for $x>1$, $b_2(x)=(\frac{x+1}{2})^{\frac 12}$ and for $0<x\le 1$, $b_2(x)=1,$ another direct calculation shows that $$\langle b_2\rangle_{I_k}= \frac{\sqrt 2}{3k}\big[ (k+1)^{\frac 32}-2\sqrt 2 \big]$$ by which and the assumption $k\ge 100$ we see that $$|b_2- \langle b_2\rangle_{I_k}| 1_{(0,1)}\ge c k^{\frac 12}.$$ Hence $$\begin{aligned}
    T_{C, 2}(b_1, b_2)\ge\lim_{k\rightarrow \infty}  c(1-k^{-1})k^{\frac 12}\langle 1_{(0,1)}\rangle_{C, I_k}=\lim_{k\rightarrow \infty}  c(1-k^{-1})\frac{k^{\frac 12}}{C^{-1}(k)}=\infty.
    \end{aligned}$$ Next, we show that $C_bH: L^2\not\to L^2.$ To see this, let $$f(x)=x^{-\frac 12}(\log x)^{-1}1_{[100,\infty)}(x) \in L^2({\mathbb{R}}).$$ We claim that $|C_bHf(x)|=\infty$ for all $x\in I_0$, and in showing this, hence conclude the unboundedness of $C_bH$ on $L^2$. Indeed, since for $y\in [100,\infty)$ $$b_2(y)=(M(1_{I_0}))^{-\frac 12}=\sqrt{ \frac {y+1}2},$$ and $b_1(x)=b_2(x)=\operatorname{sgn}(x)$, for any $x\in I_0,$ we have $$\begin{aligned}
    |C_bHf(x)|&=\Big| \int_{100}^\infty (b_1(x)-b_1(y)) (b_2(x)-b_2(y))\frac{f(y)}{x-y}dy\Big| =\Big| \int_{100}^\infty   (\operatorname{sgn}(x)-\sqrt{\frac{y+1}2})\frac{f(y)}{x-y}dy\Big|\\
    &\sim\int_{100}^\infty   \sqrt{\frac{y+1}2}\frac{f(y)}{y-x}dy\sim \int_{100}^\infty 