Is the Function g(x)=|x|^{-1}\mathscr{F}(f)(x) in L^{4/3}(\mathbb{R}^2)?

[feynman456]

Suppose $f \in L^{4/3}(\mathbb{R}^2)$ and denote its Fourier transform by $\mathscr{F}(f).$ Is it true that the function $g:\mathbb{R}^2 \rightarrow \mathbb{C}$ defined by
$$g(x)=|x|^{-1}\mathscr{F}(f)(x)$$ is in $L^{4/3}(\mathbb{R}^2)$ also?

Simply appealing to Hausdorff-Young and Hölder's inequality doesn't suffice.

Edit: It turns out that this can be proved using the Marcinkiewicz interpolation theorem[\url], as described [url=http://math.stackexchange.com/questions/47951/fourier-transform-of-function-in-l4-3]here[\url].

 

[jvicens]

Thank you for bringing up this interesting question. It is true that the function g defined in the post is also in L^{4/3}(\mathbb{R}^2). However, as you mentioned, simply using Hausdorff-Young and Hölder's inequality does not suffice to prove this. Instead, we can use the Marcinkiewicz interpolation theorem to prove this statement.

The Marcinkiewicz interpolation theorem states that if a function f is in L^{p_0}(\mathbb{R}^n) and in L^{p_1}(\mathbb{R}^n), where p_0 < p < p_1, then f is also in L^p(\mathbb{R}^n). In our case, we have f \in L^{4/3}(\mathbb{R}^2) and g(x)=|x|^{-1}\mathscr{F}(f)(x). Using the fact that the Fourier transform is an isometry on L^{p}(\mathbb{R}^n), we can rewrite g as g(x)=|x|^{-1/2} \mathscr{F}(f)(x) and thus, g \in L^{4/3}(\mathbb{R}^2) by the Marcinkiewicz interpolation theorem.

I hope this helps to answer your question. If you have any further questions or concerns, please do not hesitate to ask.