Theorem 3.8].

We shall now state the first principal result of this section, in which we characterize the optimal range partner to a given space with respect to the operator $M$.

\[T:maximal-operator\] Let $X$ be an r.i. space over $\mathbb{R}^{n}$ such that $$\begin{aligned}
\label{E:psi-condition}
\psi \in X'(0,\infty ),\end{aligned}$$ where $\psi (t)=\chi _{(0,1)}(t)\log \tfrac{1}{t}$, $t\in (0,\infty )$. Define the functional $\sigma $ by $$\begin{aligned}
\sigma (f)=\left \|  \int _{t}^{\infty }f^{*}(s)\frac{{\mathrm{d}}s}{s}\right \|  _{X'(0,\infty )}, \quad f\in \mathcal{M}_{+}(\mathbb{R}
^{n}).\end{aligned}$$ Then $\sigma $ is an r.i. norm and $$\begin{aligned}
\label{E:M-bounded}
M\colon X\to Y,\end{aligned}$$ where $Y=Y(\sigma ')$. Moreover, $Y$ is the optimal smallest r.i. space for which  holds.

Conversely, if  is not true, then there does not exist an r.i. space $Y$ for which  holds.

We now turn our attention to the question of the optimal domain space when the target space is prescribed. This situation is considerably simpler than the reverse one as no associate norms need to be involved.

\[T:maximal-operator-domain\] Let $Y$ be an r.i. space over $\mathbb{R}^{n}$ such that $$\begin{aligned}
\label{E:psi-condition-domain}
\psi \in Y(0,\infty ),\end{aligned}$$ where $\psi (t)=\min \{1,\frac{1}{t}\}$ for $t\in (0,\infty )$. Define the functional $\varrho $ by $$\begin{aligned}
\varrho (f)=\|f^{**}\|_{Y(0,\infty )}, \quad f\in \mathcal{M}_{+}(
\mathbb{R}^{n}).\end{aligned}$$ Then $\varrho $ is an 