fa.functional analysis - Is the Fourier-Transform a bounded operator on Lorentz spaces L(2,q)?

It is well known that the Fourier transform $mathcal{F}$ maps $L^1(mathbb{R}^n)$ continuously into $L^infty(mathbb{R}^n)$ and $L^2(mathbb{R}^n)$ continuously into $L^2(mathbb{R}^n)$.

Then, by interpolation theorems such as Marcinkiewicz' Theorem one can deduce that $mathcal{F}$ maps as well the Lorentz spaces $L^{p,q}(mathbb{R}^n)$ into $L^{p',q}(mathbb{R}^n)$ if $1 < p < 2$, $1 leq q leq infty$. (Here, $p' = frac{p}{p-1}$ is the Hölder-conjugate of $p$).

On the other hand, for $p > 2$ it can be shown that the Fourier Transform $mathcal{F}$ is defined on $L^p(mathbb{R}^n)$ only in the sense of distributions, and does not map $L^p$ into $L^{p'}$, and in particular it does not map Lorentz spaces $L^{p,q}$ continuously into $L^{p',q}$ for $p > 2$, $1 leq q leq infty$.

So my question is, what happens in the in-between spaces $L^{2,q}$? Is still an isomorphism as in the case $L^{2,2} = L^2$?