Let the function $\Phi : [0,\infty) \rightarrow [0,\infty)$ be a convex bijection, $$ \|f\|_{L^{\Phi}}:=\inf\left\{\lambda>0:\int_{\mathbb{R}^n}\Phi\Big(\frac{|f(x)|}{\lambda}\Big)dx\leq 1\right\}, $$ and $$ \Vert f\Vert_{WL^{\Phi}}:=\inf\left\{\lambda>0\ :\ \sup_{t>0}\Phi(t)|\{x\in\mathbb{R}^n: \frac{|f(x)|}{\lambda}>t\}|\ \leq 1\right\}. $$ I want to show that $$ \Vert f\Vert_{WL^{\Phi}}=\sup_{t>0}\|t\chi_{\{|f|>t\}}\|_{L^{\Phi}}, $$ where $\chi$ denotes the characteristic function.
My attempt: First note that $$ \Vert f\Vert_{WL^{\Phi}}=\inf\left\{\lambda>0\ :\ \sup_{t>0}\Phi(\frac{t}{\lambda})|\{x\in\mathbb{R}^n: |f(x)|>t\}|\ \leq 1\right\}=:\inf B. $$ If we write $\sup_{t>0}\|t\chi_{\{|f|>t\}}\|_{L^{\Phi}}$ explicitly, we get $$ \sup_{t>0}\|t\chi_{\{|f|>t\}}\|_{L^{\Phi}}=\sup_{t>0}\inf\left\{\lambda>0:\Phi\Big(\frac{t}{\lambda}\Big)\int_{\{x\in\mathbb{R}^n: |f(x)|>t\}}dx\leq 1\right\}=:\sup_{t>0}\inf A_t. $$
If $\lambda\in B$ then $\sup_{t>0}\Phi(\frac{t}{\lambda})|\{x\in\mathbb{R}^n: |f(x)|>t\}|\ \leq 1$. Therefore $\Phi(\frac{t}{\lambda})|\{x\in\mathbb{R}^n: |f(x)|>t\}|\ \leq 1,\, \forall t>0$ which means $\lambda \in A_t$ for all $t>0$. Hence $B\subset A_t,\, \forall t>0$ and this imply $\inf B\geq \inf A_t,\, \forall t>0$ and consequently $\inf B\geq \sup_{t>0}\inf A_t$. Now let $t$ be fixed and $\lambda \in A_t$. Then $\Phi(\frac{t}{\lambda})|\{x\in\mathbb{R}^n: |f(x)|>t\}|\ \leq 1$. I think that i can not write $\sup_{t>0}\Phi(\frac{t}{\lambda})|\{x\in\mathbb{R}^n: |f(x)|>t\}|\ \leq 1$ since $t$ is not arbitrary? Am i right? How can i continue? I am very confused.