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 RiemannLebesgue 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.
1 Answer
You can always do something cheap like isolating singularity by a cutoff with small support, or, if you want to get true asymptotics, subtract something with known Fourier transform and the same singularity so that the difference behaves better. Let me show a (crude) way to apply the first method.
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 $y^{2}\logy$ at infinity. It is somewhat short of the correct order of magnitude ($y^{2}$) but I think you got the idea.
Any decent textbook in beginning Fourier analysis uses this trick several times so I'm a bit perplexed about your claim concerning unsuccessful literature search.

$\begingroup$ Thank you, and in which book have you found this trick? $\endgroup$– AppliquéOct 20, 2013 at 21:37