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.
As I wrote in the comments, this can be proved with the Marcinkiewicz interpolation theorem. Here's the main steps:
It's well-known that $\mathscr{F}\colon L^1(\mathbb R^2)\to L^\infty(\mathbb R^2)$ and $\mathscr{F}\colon L^2(\mathbb R^2)\to L^2(\mathbb R^2)$ is bounded. Moreover, the function $x\mapsto|x|^{-1}$ lies in the weak $L^2$ space $L^{2,\rm w}(\mathbb R^2)$. For the operator $T$ defined by $$ Tf(x) := |x|^{-1}\mathscr{F}(f)(x) $$ we thus obtain that $T: L^1(\mathbb R^2)\to L^{2,\rm w}(\mathbb R^2)$ and $T: L^2(\mathbb R^2)\to L^{1,\rm w}(\mathbb R^2)$. In other words, $T\,$ is of weak type $(1,2)$ and of weak type $(2,1)$. Now the Marcinkiewicz interpolation theorem yields: $\,T\,$ is of strong type $(p,q)$ (i.e., $T: L^p(\mathbb R^2)\to L^q(\mathbb R^2)$ is bounded) if $p$ and $q$ are such that $p\le q$ and $$ \frac1p = \frac\theta1 + \frac{1-\theta}2 = \frac{1+\theta}2, \quad \frac1q = \frac\theta2 + \frac{1-\theta}1 = 1-\frac\theta2, $$ where $\theta\in(0,1)$. For $\theta=\frac12$ we obtain $p=q=\frac43$. Thus, in particular, $g=Tf \in L^{4/3}(\mathbb{R}^2)$ for all $f \in L^{4/3}(\mathbb{R}^2)$.