# 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 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 $L^{2,2} = L^2$?

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I know very little about Lorentz spaces, but I think isomorphism is the wrong word to use; $\mathcal{F}$ maps some $L^{p}$ spaces into others, but very rarely onto ($L^2$ 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 would guess that $\mathcal{F}$ 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
Also, try considering functions just on $[0,1]$, this is technically much easier (we have genuine Lebesgue integrals rather than densely defined operators, etc.) – Zen Harper Jun 22 '10 at 10:38
Note that for Lorentz spaces we have the following inclusion relation $L^{p,q_1}(\mathbb{R}^n) \subset L^{p,q_2}(\mathbb{R}^n)$ if $1 < p < \infty$ and $1 \leq q_1 \leq q_2 \leq \infty$. This implies in particular that the Fourier transform $\mathcal{F}$ maps $L^{2,q}(\mathbb{R}^n)$ into a proper subspace of $L^2(\mathbb{R}^n)$ if $1 \leq q < 2$. – Armin Jun 23 '10 at 6:42
As a (typical?) example which belongs to $L^{2,\infty}(\mathbb{R}^n)$ but not to $L^{2}(\mathbb{R}^n)$ one can take $\vert x \vert^{-\frac{n}{2}}$. Its Fourier transform (in the distributional sense) is up to a constant again $\vert x \vert^{-\frac{n}{2}}$, so at least in this very special case the Fourier transform maps $L^{2,\infty}(\mathbb{R}^n)$ into $L^{2,\infty}(\mathbb{R}^n)$. – Armin Jun 23 '10 at 6:50
But this might be a bad example to get intuition from, because in general $\mathcal{F}$ maps $\vert x \vert^{-\frac{n}{p}} \in L^{p,\infty}$ into (up to a constant) $\vert x\vert ^{-\frac{n}{p'}} \in L^{p',\infty}$ where $p'$ is the Hölder conjugate, but if $p > 2$ the Fourier transform does not map $L^{p,\infty}$ into $L^{p',\infty}$. – Armin Jun 23 '10 at 6:51