and is inspired by work of Engliš-Zhang [@EZ].

\[allthethingsGSsaid\] Let $\psi: [0,\infty)\to [0,\infty)$ be an increasing concave function satisfying and , $\psi(0)=0$, $\lim_{t\to\infty} \psi(t)=\infty$. We set $k_\psi := \log(A_{\psi}(e))$.

1.  For any exponentiation invariant extended limit $\omega\in (L^\infty)^*$ and $T\in \mathcal{M}_\psi^{(p)}$, the formula $$\mathrm{Tr}_{\omega,\psi}(|T|^p)=\frac1{\Gamma(1+k_\psi)}\cdot\lim_{h\to  \tilde{\omega}} \frac{1}{\psi(\mathrm{e}^{1/h})}\|T\|_{p+h}^{p+h}$$ holds where $\tilde{\omega}\in (L_\infty)^*$ is the extended limit at $0$ given by $$\label{omegatildedef}
    \lim_{h\to  \tilde{\omega}} x(t) := \lim_{t\to \omega} x(\frac1{\log(t)}).$$

2.  For any $T\in \mathcal{N}$, $$\|T\|_{ \mathcal{M}_\psi^{(p)}}\sim  \sup_{h>0} \frac{1}{\psi(\mathrm{e}^{1/h})}\|T\|_{p+h}^{p+h}.$$ In particular, $T\in  \mathcal{M}_\psi^{(p)}$ if and only if $\|T\|_{p+h}^{p+h}=O(\psi(\mathrm{e}^{1/h}))$.

3.  Assume that $\mathcal{N}$ is atomic. For any $T\in \mathcal{M}_\psi^{(p)}$, we have that $$\begin{aligned}
    \limsup_{h\searrow 0} \frac{1}{\psi(\mathrm{e}^{1/h})}\|T\|_{p+h}^{p+h}&\leq \mathrm{d}_{\mathcal{M}_\psi}(|T|^p,\mathcal{M}_{\psi,0})\leq \mathrm{e} \limsup_{h\searrow 0} \frac{1}{\psi(\mathrm{e}^{1/h})}\|T\|_{p+h}^{p+h}.\end{aligned}$$

Before proving the third statement of this theorem, we need two lemmas. The following result is an extension of [@EZ Proposition 7].

\[prop7\] Let $\psi: [0,\infty)\to [0,\infty)$ be an increasing concave function satisfying the conditions and and moreover that $\psi(0)=0$, $\lim_{t\to\infty} \psi(t)=\infty$. For a function $f\in \cap_{0<h<\delta}L^{p+h}(0,\infty)$ for some $\delta>0$ we define the quantities $$\|f\|_{p,\limsup} := \limsup_{h\searrow0} \frac{\|f\|_{p+h}^{1+h}}{\sqrt[p]{\psi(\mathrm{e}^{\frac{1}{h}})}}\quad
\mbox{and} \quad 
\|f\|_{p, \lim \psi} := \limsup_{t\to \infty} \sqrt[p]{\frac{1}{\psi(t)}\int_0^t |f(s)|^p \mathrm{d}s}$$ It then holds that $$\|f\|_{p, \lim \psi}\leq \|f\|_{p,\limsup}\leq \mathrm{e} \|f\|_{p, \lim \psi}.$$

If $\psi$ satisfies the conditions and , then so does $\psi^{1/p}$ for any 