Let $\mathcal{D}$ denote the space of $C^{\infty}$, compactly supported functions on $\mathbb{R}^{d}$, and let $\mathcal{D}'$ denote its dual (i.e. the space of distributions). In volume II of Reed and Simon's Methods of Mathematical Physics, the authors introduce the following definition of the product of distributions (pg. 91), which I have reproduced below modulo some minor notational changes.
Definition. Let $T,S\in\mathcal{D}'$. We say that $W\in\mathcal{D}'$ is the product of $T$ and $S$ if and only if for each $x\in\mathbb{R}^{d}$, there exists some $f\in \mathcal{D}$, with $f\equiv 1$ near $x$ so that for each $\xi\in\mathbb{R}^{d}$, $$\mathcal{F}(f^{2}W)(\xi) = (2\pi)^{-d/2}\int_{\mathbb{R}^{d}}\mathcal{F}(fT)(\eta) \mathcal{F}(f S)(\xi-\eta)d\eta, \tag{*}$$ where the integral in the RHS is absolutely convergent. If such a $W$ exists, we say the product of $T$ and $S$ exists.
Above, $\mathcal{F}$ denotes the Fourier transform on the space of tempered distributions.
The authors go on to claim in Theorem IX.43 that the product is well-defined and that there is at most one $W$ satisfying the definition. I am having trouble seeing that the uniqueness of the product. Looking at Reed and Simon's sketch of the proof, they claim without proof that if $W$ is a product for $T$ and $S$, then for any $g\in\mathcal{D}$,
$$\mathcal{F}(gf^{2}W) = (2\pi)^{-d/2}\mathcal{F}(gf T) \ast \mathcal{F}(f S) = (2\pi)^{-n/2} \mathcal{F}(f T)\ast \mathcal{F}(fg S), \tag{**}$$
citing associativity, which is valid because of the absolute convergence of the integral in the RHS of (*).
I do not see why absolute convergence of the integral alone is sufficient to warrant the change of variables needed to prove (**). It seems like one also needs some condition of the form $$\xi \mapsto \int_{\mathbb{R}^{d}}|\mathcal{F}(fT)(\eta)| |\mathcal{F}(fS)(\xi-\eta)|d\eta$$ is polynomially bounded, but this condition isn't assumed, nor do I see why it follows from (*). Am I missing something obvious?
