I decided to ask this question here, since I did not get any answer from MSE and perhaps this topic is somewhat far from MSE's topics. I am following the paper here. I am trying to understand how to make sense of products of distributions, using Heaviside function as an example (it is my first time as a physics PhD student trying to fight functional-analytic subtleties). This is particularly relevant for me when trying to work with stuff related to algebraic framework in quantum field theory (AQFT).
The crucial thing about Heaviside function $\Theta(x)$ that I am interested in is the fact that naively, we have in "calculus sense" $\Theta(x)^2 = \Theta(x)$. After all, it is discontinuous only at zero. So it looks like since this function is discontinuous at the origin, I have no real expectation of it obeying Leibniz rule, something like $(\Theta^2)'=2\Theta(x)\Theta(x)'\,\, ``=" 2\Theta(x)\delta(x)$, where the last "equality" is obtained by cheating (using weak derivative). But in general one cannot really make sense of $\Theta(x)\delta(x)$ as products of distributions, so I don't expect Leibniz rule to hold.
However, one (clever) way I learnt from the short review paper above is that we define products in slightly roundabout way: we consider product of distribution in terms of the Fourier transform: that is, if I have two distributions $u,v\in \mathcal{D}'(\mathbb{R}^n)$, then one way to do it is to use the fact that the Fourier transform of products is convolution of the Fourier transforms: \begin{align} \widetilde{uv} = \widetilde{u}\star\widetilde{v}\,. \end{align} Hence the product, whenever it makes sense, is defined to be the inverse Fourier transform of the $\widetilde{u}\star \widetilde{v}$.
A simple try (using Mathematica) shows that \begin{align} \tilde\Theta(k) = \pi \delta (k)+\frac{i}{k} \end{align} where I used the same convention for Fourier transform \begin{align} \tilde{u}(k) = \int dx\,e^{ikx}u(x)\,. \end{align} However, using this precription via inverse Fourier transform of the convolution in this case give (see Eq.~(1) in the paper) \begin{align} \Theta^{2}(x) = \frac{1}{2}\Theta\,. \end{align} So this disagrees with the intuitive picture I got earlier.
I will not go into the wavefront set yet (I am still learning it), but it seems that Hörmander's result is supposed to give a prescription that respects Leibniz rule, while both naive and Fourier transform methods do not obey Leibniz rule. From this alone, it seems that essentially Heaviside function falls outside Hörmander's requirement. However, I cannot see why the Fourier transform method gives different prescription from naive one.
Question: what is the underlying reason behind the difference in Fourier method vs the naive one, and any other approaches (e.g., wavefront sets)? I expect that this shouldn't be "too ambiguous" because I am not multiplying ill-defined objects like $\delta(x)\Theta(x)$ or $\delta(x)\delta(x)$, but something that could at least looks sensible in calculus. Perhaps one way to ask is this: in what sense is $\theta^2=\Theta$ rigorous as a product of distributions?
One possibility I can think of is that this is the analog of ambiguity in prescribing the value of an asymptotic (divergent) series: sums like $\sum_{n=1}^\infty (-1)^n$ can be assigned different finite values depending on which method you use to sum them (e.g., Cesaro gives $1/2$, while naive rearrangement can give $0,\pm 1$, or more generally any rational number.)
Any clarification (even if it is only sensible in terms of more advanced tools like microlocal analysis) would be appreciated, since I can then try to look in the right direction.
UPDATE: As mentioned earlier, I was experimenting this with Mathematica, and as Willie Wong pointed out it cannot be right since $i/k$ is not locally integrable so the convolution does not work. From my perspective it means something worse: the naive calculus version does not fit Fourier transform method either (which cannot be done because of $i/k$). How should I understand these different approaches? I have updated the question to account for this stupid miss on my part.
