$\limsup_{h\searrow 0} \left(\frac{1}{\psi(\mathrm{e}^{\frac{1}{h}})} \int_{\textbf{T}\times \textbf{N}} |f *W_n)(e^{i\theta})|^{p+h} d\nu(\theta, n)\right)^{\frac{p}{p+h}} <\infty;$

4.  $\limsup_{t\to \infty} \frac1{\psi(t)} \int_0^t \Phi_f(s)^p ds < \infty;$

5.  $H_{\overline{f}} \in \mathcal{M}_\psi^{(p)}.$

The quantities in (1)-(4) are all equivalent to $\mathrm{d}_{\mathcal{M}_\psi}(|H_{\overline{f}}|^p,\mathcal{M}_{\psi,0})$.

The proof of this result is a straightforward repetition of that of [@EZ Theorem 1] with the replacement of $\log t$, $p-1$ and the use of [@EZ Proposition 7] by $\psi(t)$, $\psi(e^\frac1{h})$ and the use of Proposition \[prop7\], respectively. Again, as in Proposition \[prop7\] we can reduce the proof to $p=1$.

\[prop8\] Let $\psi$ satisfy . Let $H=H^*\in L^q(0,\infty)$ for all $1<q<1+\delta$ for some $\delta>0$. Let $\omega$ be an exponentiation invariant extended limit on $L^\infty(0,\infty)$ and $\hat{\omega}:=\omega \circ {\rm exp}$.

\(a) For every $\alpha>1$ and sufficiently large $t>0$ one has $\mu_H(1/t) \le t^\alpha;$

\(b) One has $$\lim_{t\to \omega} \frac1{\psi(t)} \int_0^t H(s) ds = \lim_{t\to \omega} \frac1{\psi(t)} \int_0^{\mu_H(1/t)} H(s) ds;$$

\(c) One has $$\lim_{r\to \hat{\omega}} \frac{\|H\|_{1+1/r}}{\psi(\mathrm{e}^{r})}= \lim_{t\to \omega} \frac1{\psi(t)} \int_0^t H(s) ds.$$

\(a) Denote for brevity $a:=\mu_H(1/t).$ For sufficiently large $t>0$ we have $$c_H := \sup_{t>2} \frac1{\psi(t)} \int_0^t H(s) ds \ge \frac1{\psi(a)} \int_0^a H(s) ds \ge \frac{a H(a)}{\psi(a)}= \frac{a}{t \psi(a)},$$ since $H$ is nonincreasing and $H(\mu_H(1/t))=1/t$. Since the function $\psi$ is slowly varying, it follows that for every $0<\varepsilon 