Let $S(\mathbb{R}^n)$ be the Schwartz space of $\mathbb{R}^n$. Assume that $f \in S(\mathbb{R}^n)$ and $u \in S'(\mathbb{R}^n)$ is a tempered distribution. Here, I define for all $\phi \in S(\mathbb{R}^n)$, $\partial_i u [\phi] = -u [\partial_i \phi]$ and $(f \cdot \partial_i u) [\phi] = \partial_i u[f \cdot \phi]$. Also, Fourier transform is defined as $\widehat{u}[\phi] = u[\widehat{\phi}]$.
My (not so well defined question) is that I would like to see how can one simplify calculation of the Fourier transform of the distribution $f \cdot \partial_i u$. In particular, in the physics context, one would want to argue that "$\widehat{f \cdot \partial_i u}(k) = k_i g(|k|)$" for some distribution $g$. I believe if we have $f, u \in S(\mathbb{R}^n)$ and both radial then this result can indeed be shown. I am now interested in understanding this in terms of distributions.
If $f, u \in S(\mathbb{R}^n)$ and both are radial, then I believe the following holds. First, use my convention for Fourier transform, where integral exists as $f \cdot \partial_i u \in S(\mathbb{R}^n)$. Then, use chain rule as $u$ is radial and $S(\mathbb{R}^n) \subset C^{\infty}(\mathbb{R}^n)$. If somehow I could argue that $u'(|x|)/|x|$ is again in $S(\mathbb{R}^n)$ then I could proceed with the following equalities.
$$ \widehat{f \cdot \partial_i u}(k) = \int \limits_{\mathbb{R}^n} (f \cdot \partial_i u)(x)e^{i\langle k,x \rangle}\mathrm{d}^nx = \int \limits_{\mathbb{R}^n} f(|x|)x_i \frac{u'(|x|)}{|x|}e^{i\langle k,x \rangle}\mathrm{d}^nx = \int \limits_{\mathbb{R}^n} -i f(|x|) \frac{u'(|x|)}{|x|} \frac{\partial}{\partial k_i}\left( e^{i\langle k,x \rangle} \right) \mathrm{d}^nx $$ $$ = -i \frac{\partial}{\partial k_i} \int \limits_{\mathbb{R}^n} f(|x|) \frac{u'(|x|)}{|x|} e^{i\langle k,x \rangle} \mathrm{d}^nx$$
Question 1: It is possible to justify that $u'(|x|)/|x|$ is again in $S(\mathbb{R}^n)$?
To continue with the computation, use that in the last step we obtain Fourier transform of a radial function which makes Fourier transform radial as well. Then, one can use chain rule to compute $\partial/\partial k_i$. This then shows the required result.
$$ \widehat{f \cdot \partial_i u}(k) = -i \frac{k_i}{|k|} \frac{\partial}{\partial |k|} \int \limits_{\mathbb{R}^n} f(|x|) \frac{u'(|x|)}{|x|} e^{i\langle k,x \rangle} \mathrm{d}^nx $$
I wanted to mimic the calculation for the case when $u$ is a tempered distribution. Let me think of distribution $u$ being approximated by $u_{\varepsilon} \in S(\mathbb{R}^n)$ such that for all $\phi \in S(\mathbb{R}^n)$ we have $\langle u_{\varepsilon}, \phi \rangle \to u[\phi]$ as $\varepsilon \to 0^+$.
In that case, I would have the following for all $\varepsilon > 0$.
$$ \widehat{f \cdot \partial_i u_{\varepsilon}}(k) = -i \frac{k_i}{|k|} \frac{\partial}{\partial |k|} \int \limits_{\mathbb{R}^n} f(|x|) \frac{u_{\varepsilon}'(|x|)}{|x|} e^{i\langle k,x \rangle} \mathrm{d}^nx $$
However, I don't see why the limit $\varepsilon \to 0^+$ should exist and whether it should be equal to the $(-i k_i/|k|) \partial_{|k|} \widehat{(f \cdot u'/|\cdot|)}[\phi]$ for some distribution $f \cdot u'/|\cdot|$. Moreover, why does $\partial_{|k|}$ make sense, and so on.
Question 2: Is it possible to make sense of $\varepsilon \to 0^+$ in a sense of tempered distributions?
As an alternative calculation, I wanted to go back to the definition of tempered distributions, where one "passes" all of the manipulations to the Schwartz functions instead. I considered the following in a sense of tempered distributions. First, use product rule (which I am comfortable with) and then use properties of Fourier transform that it is linear and that $\widehat{\partial_i u} = -ik_i \widehat{u}$.
$$ \widehat{f \cdot \partial_i u} = FT[\partial_i (f \cdot u) - (\partial_i f) \cdot u] = -ik_i \widehat{f \cdot u} - \widehat{\partial_i f \cdot u}$$
Here, the first term is well defined, and the problem is only with the second term. However, repeating previous calculations and assuming that if $f \in S(\mathbb{R}^n)$ is radial then $f'(|x|)/|x| \in S(\mathbb{R}^n)$, I would have the following for all $\varepsilon > 0$.
$$ \widehat{u_{\varepsilon} \cdot \partial_i f}(k) = -i \frac{k_i}{|k|} \frac{\partial}{\partial |k|} \int \limits_{\mathbb{R}^n} u_{\varepsilon}(|x|) \frac{f'(|x|)}{|x|} e^{i\langle k,x \rangle} \mathrm{d}^nx $$
Question 3: Can I show that this expression is well defined as $\varepsilon \to 0^+$? I understand why the integral expression would be defined in this limit as it would just be Fourier transform of a tempered distribution $u \cdot f'(|\cdot|)/|\cdot|$. However, it is not clear to me how the derivative $\partial_{|k|}$ is defined in such limit and whether one can exchange differentiation and $\varepsilon \to 0^+$ limits. If this could be done, I would be done with the problem as I could use that Fourier transform of a radial distribution is radial as well.
I appreciate any other suggestions and ideas to solve this problem.