# Decay of the Fourier transform of a function with a discontinuty at zero

Let $f \colon \mathbb R^2 \to \mathbb R^2$ be a function from the Schwartz class and $f(0) \neq 0$. Define it's projection $g(x) = \langle f(x), \frac{x}{|x|} \rangle \frac{x}{|x|}$, where $\langle a, b\rangle = a_1 b_1 + a_2 b_2$, $a = (a_1,a_2)$, $b=(b_1,b_2)$. Then $g$ has the same rate of decay at infinity as $f$ but it has discontinuity at zero of special kind. By the Riemann-Lebesgue lemma we conclude that $\hat g(\xi) \to 0$ as $\xi \to \infty$, where $$\hat g(\xi) = \int\limits_{\mathbb R^2} e^{- i \langle \xi, x \rangle} g(x) \, dx$$ is the Fourier transform of $g$. Is it possible to find an estimation of the decay rate of $\hat g$ in this particular case? Is it possible, for example, to say that for some small $\varepsilon > 0$ we have $(1+|\xi|^2)^{\frac \varepsilon 2} \hat g(\xi) \in L_1(\mathbb R^2)$? I've tried to find the answer in literature on the Fourier transform and to ask on MSE but without success.

If $D^2h\in L^1$, then we have $|\hat h(y)|\le C\frac{\|D^2h\|_{L^1}}{|y|^2}$. Now take $r>0$ and take a smooth $\psi$ supported on $B(0,2r)$ such that $\psi=1$ on $B(0,r)$ and $|D^2\psi|\le Cr^{-2}$. Split $g=g\psi+g(1-\psi)$. The first part is small in $L^1$ (about $r^2$) and the second part is $C^2$-smooth with the $L^1$-norm of the second differential (which grows as $|x|^{-2}$ near the origin until the cutoff kills it) bounded by $\log(1/r)$. Hence, we have $$|\hat g(y)|\le C[r^2+|y|^{-2}\log(1/r)]$$ Now just plug in $r=|y|^{-1}$ to get the decay $|y|^{-2}\log|y|$ at infinity. It is somewhat short of the correct order of magnitude ($|y|^{-2}$) but I think you got the idea.