we consider the setup in the present corollary. (Since $\frac{2}{p}=\theta\in(0,1)$ if and only if $p>2$). With this adjustment in mind, the conclusion to theorem \[eqn.interpolate-indices\] becomes $\|{\hat{f}}\|^*_{p',p}\leq B\|f\|^*_{p,p}=B\|f\|_p$.

As in the proof of corollary \[cor.HY-Lorentz\], we obtain the following extension immediately from the interpolation theorem \[thm.lorentz-inter\].

The Heckman–Opdam transform is a continuous mapping from $L^p(A,d\mu)^W$ to $L^{p',p}(i{\mathfrak{a}}^*,d\nu)^W$ whenever $1<p<2$.

The preceding two corollaries are stronger than their respective standard forms since $L^{p^\prime,p}$ is continuously and properly embedded in $L^{p^\prime}$.

The last result on Lorentz spaces that we will need is due to R. O’Neil, [@Oneil-convolution], and concerns the pointwise product of two functions.

\[thm.Oneil\] Let $q\in(2,\infty)$ and set $r=\frac{q}{q-2}$. For $g\in L^q(X)$ and $h\in L^{r,\infty}(X)$ it holds that $gh$ belongs to $L^{q',q}(X)$ with $\|gh\|^*_{q',q}\leq\|g\|_q\|h\|_{r,\infty}^*$.

The Hardy–Littlewood inequalities {#sec.HY-ineqs}
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The first part of the present section generalizes the inequality . We have decided to treat the rank one case separately as an illustrative example of the interpolation arguments that will be used throughout the section.

Assume $\mathrm{dim}\,{\mathfrak{a}}=1$, $m_\alpha+m_{2\alpha}\geq 1$, and define $Tf(\lambda)=|\lambda|^2\mathcal{F}f(\lambda)$. Since $-m_\alpha-m_{2\alpha}+1\leq 0$, it follows that $(1+|\lambda|)^{-(m_\alpha+m_{2\alpha})+1}\leq 1$ for all $\lambda\in i{\mathfrak{a}}^*$. Moreover, $$\begin{split}
\|Tf\|_2^2 &=\int_{i{\mathfrak{a}}^*}|Tf(\lambda)|^2|W|^{-1}(1+|\lambda|)^{-(m_\alpha+m_{2\alpha})+1}|\lambda|^{-4}|{\mathbf{c}}(\lambda)|^{-2}\,d\lambda\\
&= |W|^{-1}\int_{i{\mathfrak{a}}^*}|\mathcal{F}f(\lambda)|^2(1+|\lambda|)^{-(m_\alpha+m_{2\alpha})+1}|{\mathbf{c}}(\lambda)|^{-2}\,d\lambda\\
&\leq |W|^{-1}\int_{i{\mathfrak{a}}^*}|\mathcal{F}f(\lambda)|^2|{\mathbf{c}}(\lambda)|^{-2}\,d\lambda = \|f\|_2^2,
\end{split}$$ so $T$ is of strong 