Boundedness of Fourier transform on weak $L^2$ spaces.

Question

It is well-known that Fourier transform $\mathcal{F}$ is isometry on $L^2(\mathbb{R}^d)$.

I would like to know whether $\mathcal{F}$ is bounded or not on weak space $L^{2,\infty}(\mathbb{R}^d)$.

Answer 1

In the following paper(http://matwbn.icm.edu.pl/ksiazki/cm/cm71/cm7129.pdf), authors proved the following results:

Theorem 1.

1. The largest rearrangement invariant Banach function space which is mapped by the Fourier transform into a space of locally integrable functions is the space $L^1+L^2$.
2. The only rearrangement invariant Banach function space $X$ on which the Fourier transform is bounded, $\Vert \mathcal{F} \phi \Vert_X \leq c \Vert \phi \Vert_X$ is the Hilbert space $L^2$.

It is well-known that $L^{2,\infty}$ is rearrangement-invariant Banach function space. Hence the answer to your question is negative.