Let $\Psi(t)=\int_0^t\psi(s)$ be a Young's function. Then, $\Psi$ satisfies the Lorentz-Shimogaki condition if $$ \int_0^{\infty}\frac{\Psi(st)}{v(t)^2}\psi(t)dt< \infty. $$ Denote $\rho_{\Psi}=\frac{1}{\Psi^{-1}(t^{-1})}$ by $Q(t)=\int_0^tg$. One has $$ L_{\Psi}(\mathbf R_+)=M(g)=\left\{f \in M_{+}(\mathbf R_+): \int_0^tf^*\leq C\int_0^tg=CQ(t)\right\}. $$ Hence, if $\Phi(t)=\int_0^t\phi$ is a Young's function and $T_k: L_{\Psi}(\mathbf R^n)\to L_{\Phi}(\mathbf R^n)$ if $$ \int_{R_+}\Phi\big(s(Qk^{**}+S_{k^*}g)\big)<\infty. \label{1}\tag{1} $$
Given $\psi(s)=e^{-1/s}$, when $s<<1$ and $\psi(s)=e^s$, when $s>>1$, I want to investigate when \eqref{1} holds for $\Phi(t)=\int_0^t\log(1+s)ds$
