What gets broken if we assign a value to the square of Delta function?

Question

Particularly, what gets broken if $$\pi^2\delta^2(x)=2i\pi\delta'(x)-\frac1{12}$$?

Answer 1

There is the famous impossibility theorem of Schwartz. You can have a look at the original paper (in French)

Benci and Baglini provide a more modern writeup here with an exact list of conditions. So no matter how you want to define $\delta^2$, of of these will break. Many different suggestions have been made on what the "proper" generalization should be. (for example Colombeau Algebra) However from my experience with the topic one doesn't come across any "satisfying" answer.

Answer 2

"Things" get broken when trying to define an "intrinsic" multiplication of distributions, i.e. when trying to define the multiplication of two distributions $\alpha, \gamma \in \mathcal{D}^{'}(\mathbb{R})$ as an element of $\mathcal{D}^{'}(\mathbb{R})$. The following example is from "Multiplication of Distributions and Applications to Partial Differential Equations" by M. Oberguggenberger.

Let's try to define the multiplication of two Dirac deltas as an element of $\mathcal{D}^{'}(\mathbb{R})$. We do this by regularising the dirac delta, squaring it, and then trying to take its limit. Let $\phi_\varepsilon$ satisfy:
$$\text{ supp } (\phi_\varepsilon)={0} \text{ as } \varepsilon \to 0$$ $$ \int \phi_\varepsilon (x) dx =1 \; \forall \varepsilon >0$$ $$ \int |\phi_\varepsilon (x)| dx =1 \; \text{ is bounded independently of } \varepsilon $$ $$\phi_\varepsilon \text{ is real valued}$$ Let $\psi \in \mathcal{D}(\mathbb{R})$ be such that $\psi \equiv 1$ in a neighbourhood of the support of $\phi_\varepsilon$. Thus, we can write: $$ \langle \phi_\varepsilon ^2, \psi \rangle = \langle \phi_\varepsilon, \phi_\varepsilon \psi \rangle = \int \phi_\varepsilon ^2 (x) dx$$ If this sequence converges weakly in $\mathcal{D}^{'}(\mathbb{R})$, then it's bounded in $L^2(\mathbb{R})$ which is a Hilbert space, thus by this theorem it has a weakly convergent subsequence in $L^2(\mathbb{R})$. This implies that $\delta \in L^2(\mathbb{R})$ wich is not correct.

However, according to the same book, complex-valued regularisations of $\delta$ can be constructed, such that their squares converge to various distributions.