from which (ii) follows.

In order to generalize theorem \[thm.Ray-Sarkar\], we must therefore choose suitable measure spaces, define convenient sublinear operators $T$ and finally choose a good Young function.

\[lemma-J\] Let $({\mathfrak{a}},\Sigma,m)$ be a fixed root datum. The function $$\psi(x)=J(x)=\prod_{\alpha\in\Sigma^+}|e^{\alpha(x)}-e^{-\alpha(x)}|^{m_\alpha}$$ is a Young function.

Since $|e^{\alpha(x)}-e^{-\alpha(x)}|\leq 2e^{\|\alpha\|\|x\|}$ for every $\alpha\in\Sigma^+$, $x\in{\mathfrak{a}}$, it follows that $|J(x)|\leq C e^{2\|\rho\|\|x\|}$ for all $x\in{\mathfrak{a}}$, where $\rho=\frac{1}{2}\sum_{\alpha\in\Sigma^+}m_\alpha\alpha$. Since $\int_0^\infty s^{n-1}e^{-2\|\rho\|s}ds<\infty$ for all $n\in{\mathbb{N}}$, it follows as in the discussion succeeding example \[example.young\] that $\psi$ is a Young function.

The following result is a direct analogue of [@Ray-Sarkar-trans Theorem 4.11], and our proof follows theirs closely. The main addition is that we once again have to choose the underlying measure spaces and the operator $T$ in such a way that the weak type $(1,1)$ estimate – now with respect to a weighted measure in the $n$-dimensional vector space $i{\mathfrak{a}}^*$. As in the proof of theorem \[thm.HY-qlarge\] this is essentially achieved by working our way backwards from the desired weak type $(1,1)$ estimate, modifying $T$ accordingly. We shall not work with an arbitrary Young function but rather the choice in lemma \[lemma-J\]. Accordingly $L^{(q)}({\mathfrak{a}})$, $q>2$ denotes the space of measurable $W$-invariant functions $f:{\mathfrak{a}}\to{\mathbb{C}}$ for which $$\|f\|_{(q)}:=\Bigl(\int_{\mathfrak{a}}|f(x)|^qJ(x)^{q-2}d\mu(x)\Bigr)^{1/q}<\infty.$$

\[thm.HL-ver3\]

