Asymptotic decay of two-dimensional Fourier transform

Question

As is well known, the smoothness of a function of one variable is related to the behaviour of its Fourier transform for arbitrarily large frequency. Specifically (see here), if a function $f(t)$ is in $C^m$ its Fourier transform $\hat f(\omega)$ is $O(\omega^{-(m+2)})$ for $\omega \rightarrow \infty$.

Is there a similar result for the two-dimensional Fourier transform?

I am only interested in radial functions, for which the two-dimensional Fourier transform can be expressed as a Hankel transform. So a result for the latter would be useful too; in fact more so. Also, my initial function has bounded support.

For example:

• The "unit cylinder" function (which equals $1$ in the unit circle and is $0$ outside) is discontinuous. Its Fourier transform is $2\pi J_1(\omega_\mathrm{r}) / \omega_\mathrm{r}$, where $\omega_\mathrm{r}$ is the "radial frequency", and so it decays like $O(\omega_\mathrm{r}^{-3/2})$.
• Replacing the cylinder by a cone, which makes the function continuous, gives a (more complicated) transform that seems to decay like $O(\omega_\mathrm{r}^{-5/2})$.

So, also in two dimensions there seems to be a faster asymptotic decrease rate of the transform as the original function is smoother.

Answer 1

There are some elementary estimates that give some insight. I use the convention $$f(t) = \int_{\mathbb{R}^d}\hat{f}(\omega)e^{i\omega t}\,d\omega,$$ $$\hat{f}(\omega) = \frac{1}{(2\pi)^d}\int_\mathbb{R^d}f(t)e^{-i\omega t}\,dt.$$ Let $\alpha = (\alpha_1, \dots, \alpha_d) \in \mathbb{N}_0^d$ be a multi-index. Let $\partial^{\alpha} = \partial_{x_1}^{\alpha_1}\dots\partial_{x_d}^{\alpha_d}$. For vector $x \in \mathbb{R}^d$, let $x^{\alpha} = x_1^{\alpha_1}\dots x_d^{\alpha_d}$. Since $\widehat{\partial^{\alpha}f} = (i\omega)^{\alpha}\hat{f}$, we get $$\|\omega^{\alpha}\hat{f}\|_{L^2} = (2\pi)^{-d/2}\|\partial^{\alpha}f\|_{L^2},$$ $$\|\omega^{\alpha}\hat{f}\|_{L^{\infty}} \leq (2\pi)^{-d}\|\partial^{\alpha}f\|_{L^1}.$$ Also if $\partial^{\alpha}f \in L^1$, then by the Riemann-Lebesgue lemma, $\omega^{\alpha}\hat{f}(\omega) \to 0$ as $|\omega| \to \infty$.