Is it true that this function does not belong to $S(\mathbb{R}^d)$?

Question

Define $f: \mathbb{R}^d \to \mathbb{R}$ by:

$$\displaystyle f(x) = \frac{J_{d/2}(|x|)}{|x|^{d/2}},$$

where $|x|$ is the Euclidean norm of $x \in \mathbb{R}^d$, and $J_{\nu}$ denotes the Bessel function of the first kind. Is it correct to say that, for all $d$, we have $f \not \in S(\mathbb{R}^d)$?

My justification would be that it is immediate: from the definition of Schwartz space, all derivatives of this function would blow up to infinity whenever we take $x = 0$. That is, for some $a, b \in \mathbb{Z}_{+}$, we have:

$$\displaystyle \sup_{x \in \mathbb{R}^d} |x^aD^bf(x)| = \infty.$$

Is this correct?

Answer 1

You are correct, $J_{d/2}$ is a regular function, hence $f$ has a singularity at the origin, preventing $f\in\mathcal{S}(\mathbb{R}^d)$. As an alternative...

Assuming $f\in\mathcal{S}(\mathbb{R}^d)$ it follows that $J_{d/2}(\left|x\right|)$ belongs to $\mathcal{S}(\mathbb{R}^d)$, too. However, the Fourier transform of $J_{d/2}$ has a polynomial decay, hence $J_{d/2}$ and $f$ do not belong to such Schwartz space.

Answer 2

No, because $f(x)$ may be written as a power series in $\lvert x \rvert^2$ (recall that $J_{\nu}(r) = r^{\nu} \sum_{k=0}^{\infty} a_k (-r^{2})^{k}$ for coefficients that I can't be bothered to remember at the moment).

The problem actually lies in the large-$\lvert x \rvert$ region, because $J_{\nu}(r) \sim \sqrt{\frac{2}{\pi r}} \cos{(r-(2\nu+1)\pi/4)} $, so the decay is only that of $r^{-1/2}$.