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 HÃ¶lder-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 $L^{2,2} = L^2$?

isomorphismis the wrong word to use; $\mathcal{F}$ maps some $L^{p}$ spacesintoothers, but very rarelyonto($L^2$ is the only case where it's an isomorphism). Still, it's an interesting natural problem which should already be known somewhere. What happens with the extreme cases $L^{2,1}$ and $L^{2,\infty}$? I wouldguessthat $\mathcal{F}$ won't map $L^{2, q}$ into any function space if $q \ne 2$, because I am a pessimist and Mathematics is usually nasty. – Zen Harper Jun 22 '10 at 10:36