Let $\varphi:\mathbb{R}^n \to [0,\infty)$ be compactly supported and $C^{\infty}$. Define $k_s(x) = |x|^{-\alpha}$ for $x \in \mathbb{R}^n$, where $0 < \alpha < n$. I know that, as tempered distributions,
$$
\mathcal{F}(k_\alpha) = k_{n-\alpha}
$$
and
$$
\mathcal{F}(\varphi \ast k_\alpha) = \mathcal{F}(\varphi)\mathcal{F}(k_\alpha).
$$
Here $\mathcal{F}$ denotes the Fourier transform.
Question Is the following true? If yes, how can it be proved? \begin{align*} \int_{\mathbb{R}^n} | \mathcal{F}(\varphi \ast k_\alpha) |^2 dx = \int_{\mathbb{R}^n} | \mathcal{F}(\varphi)\mathcal{F}(k_\alpha) |^2 dx \end{align*}
In my application, we can assume $2(\alpha-n) > -n$ so that the integral on the right is is finite, in case that helps.