\frac {f(y)}{\sqrt y}dy=\infty.
        \end{aligned}$$

If we take $A(t)=B(t)=C(t)=t^{2+\varepsilon}$, where $\varepsilon>0$, we immediately have the following

\[JNCEE\] The conditions $S_2,T_2$ holding simultaneously does not improve to $S_{2+\varepsilon}$ or $T_{2+\varepsilon}$ for any $\varepsilon > 0$.

For our next example, we note that functions $\Phi:[0,\infty)\to[0,\infty)$ of the form $$\Phi(t) = t^p\log(e+t)^{p-1+\delta},\qquad p\in (1,\infty), \ \delta\in (0,\infty)$$ are called log -bumps. These are Young functions, and we recall some facts from [@CMP10 Chapter 5]:

i)  If $\Phi(t) = t^p\log(e+t)^{p-1+\delta},$ then $$\Phi^{-1}(t) \sim t^{1/p}\log(e+t)^{-\frac{1}{p'}-\frac \delta p} \mbox{ and } \bar \Phi(t) \sim t^{p'}[\log(e+t)]^{-1-(p'-1)\delta}\in B_{p'}.$$

ii) If $\Phi(t) = t^p\log(e+t)^{p-1}\log\log(e^e+t)^{p-1+\delta}$(which is referred to as a loglog-bump), then $$\Phi^{-1}(t) \sim t^{\frac 1 p}\log(e+t)^{-\frac1{p'}}\log\log(e^e+t)^{-\frac{1}{p'}-\frac \delta p}$$ and $$\bar \Phi(t) \sim t^{p'}\log(e+t)^{-1}[\log\log(e^e+t)]^{-1-(p'-1)\delta}\in B_{p'}.$$

\[lastexample\] There exist functions $b_1,b_2 \in L^{\infty}_{loc}$ such that $C_bH:L^2\to L^2$ boundedly, but $S_{A,B}(b_1,b_2)=\infty$ and $T_C(b_1,b_2)=\infty,$ for all log -bumps $A,B,C$ with $\bar A,\bar B,\bar C\in B_2.$

The idea is to construct a pair of functions $(b_1, b_2)$ such that it satisfies the assumption in Theorem \[upperbound\] so that we can conclude the boundedness of $C_bH$ directly, meanwhile, the related bump function increases slower than log-bumps. Let $\Phi_0=t^2\log(e+t)\log\log(e^e+t)^{3/2}$, and define $$b_1(x)=\operatorname{sgn}(x)1_{I_0}(x),\qquad b_2(x) = \operatorname{sgn}(x) \Phi_0^{-1}\big((M1_{I_0}(x))^{-1}\big),\ \ I_0=[-1,1].$$We will show that $(b_1, b_2)$ is what we need. 