(x(n))^p}{\psi(t_n)}\mathop{\longrightarrow}\limits_{n\to\infty} 0,$$ since $x=\mu(x)$ and $x(n)\to 0$ as $n\to\infty$. Hence, $\mathrm{d}_{\mathcal{M}_\psi(0,\infty)}(x^p,\mathcal{M}'_{\psi,0}(0,\infty))=0\le a.$

3\. If $0 = \liminf_{n\to\infty} t_n \le \limsup_{n\to\infty} t_n<\infty$, then $$\frac{1}{\psi(t_n)} \int_n^{t_n+n} (x(s))^p \ \mathrm{d}s \le \frac{t_n (x(n))^p}{\psi(t_n)}\mathop{\longrightarrow}\limits_{n\to\infty} 0,$$ since $x$ is bounded and $\frac{t}{\psi(t)}\to 0$ as $t\to0$. Hence, $\mathrm{d}_{\mathcal{M}_\psi(0,\infty)}(x^p,\mathcal{M}'_{\psi,0}(0,\infty))=0\le a.$

On the other hand, for every $x=\mu(x) \in \mathcal{M}_\psi^{(p)}(0,\infty)$ and $y\in \mathcal{M}'_{\psi,0}(0,\infty)$ by [@KPS Theorem II.3.1] we have $$\begin{aligned}
\mathrm{d}_{\mathcal{M}_\psi(0,\infty)}(x^p,\mathcal{M}'_{\psi,0}(0,\infty)) &=\|x^p-y\|_{\mathcal{M}_\psi(0,\infty)}\ge \|\mu(x^p)-\mu(y)\|_{\mathcal{M}_\psi(0,\infty)}\\
&\ge \limsup_{t\to \infty}  \frac1{\psi(t)}\int_0^t (\mu(x^p)-\mu(y))(s) \mathrm{d}s\\
&= \limsup_{t\to \infty}  \frac1{\psi(t)}\int_0^t \mu(x^p)(s) \mathrm{d}s,\end{aligned}$$ since $y\in \mathcal{M}'_{\psi,0}(0,\infty)$. This proves the assertion.

Set $f=\mu_T$. Assuming that $\mathcal{N}$ is atomic, Lemma \[prop7.5\] ensures that $\mathrm{d}_{\mathcal{M}_\psi}(|T|^p,\mathcal{M}_{\psi,0})= \|f\|_{p,\lim \psi}^p$. By definition, $$\limsup_{h\searrow 0} \frac{1}{\psi(\mathrm{e}^{1/h})}\|T\|_{p+h}^{p+h}=\|f\|_{p,\limsup}^p.$$ We conclude the inequality stated in the third statement of Theorem \[allthethingsGSsaid\] from Lemma \[prop7\].

The aspect of Theorem \[allthethingsGSsaid\] relevant to this paper lies in its implications on Hankel operators. To formalize this, we state an immediate corollary of Theorem \[allthethingsGSsaid\]. If $(X_h)_{h\in [0,1]}$ is a family of Banach spaces with $X_h\subseteq X_{h'}$ continuously for $h<h'$, we define the extrapolation space $X_\psi\subseteq \cap_{h\in (0,1]}X_h$ to be the set of all elements $x\in\cap_{h\in (0,1]}X_h$ for which $$\|x\|_{X_\psi}:=\sup_{h>0} \frac{1}{\psi(\mathrm{e}^{1/h})}\|x\|^{1+h}_{X_h}<\infty.$$

\[equivalencescor\] 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} 