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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.

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    I'm pretty sure one can do it with the Marcinkiewicz interpolation theorem. (Or is this overkill?)2011-06-27
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    @Hendrik: I am not quite seeing it. What are you proposing to use as the end points?2011-06-27
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    @Willie: $4/3-\varepsilon$ and $4/3+\varepsilon$, for a suitable $\varepsilon>0$. The point is that $x\mapsto|x|^{-1}$ is in the _weak_ $L_2$ space.2011-06-27
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    (Egads, my interpolation theory is rusty.) Of course: $f\mapsto g$ is $L^1\to L^{2,w}$ and $L^2 \to L^{1,w}$. I was looking at the scale the wrong way and forgot to dualize.2011-06-27
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    @Willie: I took the freedom to use your end points in my answer `:-)`2011-06-27

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