$p\geq 1$. Indeed condition is readily verified for $\psi^{1/p}$ and condition for $\psi^{1/p}$ follows from that $\psi$ has regular variation of index $0$ and [@GS Proposition 2.17 and 2.23]. We can therefore replace $f$ by $H:=|f|^p\geq 0$ and $\psi$ by $\psi^p$, and thusly assume that $p=1$. For any $C>\|H\|_{1,\limsup}$ there is $q_0>0$ such that $$\frac{\|H\|_{1+h}}{\psi(\mathrm{e}^\frac{1}{h})} <C, \quad\mbox{for}\; 0<h<q_0.$$

Using the Hölder inequality, for any $0<q<q_0$ we obtain $$\begin{aligned}
\int_0^t H(s) \mathrm{d} s &\le \left( \int_0^t H(s)^{1+q} \mathrm{d}s \right)^\frac1{1+q} \left( \int_0^t  \mathrm{d}s \right)^\frac{q}{1+q}\\
&\le C \cdot \psi(\mathrm{e}^\frac1{q})\cdot t^\frac{q}{1+q} \le C \cdot\psi(\mathrm{e}^\frac1{q})\cdot t^q.\end{aligned}$$

If $t>\mathrm{e}^{1/q_0}$ (that is, $q_0>1/\log t$), one can take $q=1/\log t$. Thus, $$\int_0^t H(s) \mathrm{d}s \le C \mathrm{e} \cdot\psi(t), \quad\mbox{for}\;  t>\mathrm{e}^{1/q_0}.$$ Therefore, $$\|H\|_{1,\lim \psi} \le \mathrm{e} \|H\|_{1,\limsup}.$$

Conversely for $C> \|H\|_{1,\lim \psi}$ there exists $t_0>0$ such that $$\label{eq1}
\frac1{\psi(t)} \int_0^t H(s) \mathrm{d}s \le C, \quad\forall t\ge t_0.$$

Equivalently, $$\int_0^t H(s) \mathrm{d}s \le \int_0^t C \psi'(s) \mathrm{d}s, \quad \forall t\ge t_0.$$

For the function $$G(t):=\begin{cases}
H(t), &t\ge t_0\\
\min\{H(t), C \psi'(t)\}, &t<t_0.
\end{cases}$$ we clearly have $$\int_0^t G(s) \mathrm{d}s \le \int_0^t C \psi'(s) \mathrm{d}s,\quad \forall t> 0.$$

This means that the function $G$ is submajorised by the function $C \psi'$ (in the sense of Hardy-Littlewood). Thus, for every $h>0$ one 