Product of Distinct $\delta$ functions

Question

Let $\delta_a$ and $\delta_b$ be delta functions where $a,b \in \mathbb{R}$ and $a < b$. Is it the case that $\delta_a \delta_b = 0$? My idea is that if we take $\varepsilon = |b-a| > 0$, then for every $\phi \in C_0^\infty(\mathbb{R})$ we have the following: \begin{align*} \langle \delta_a \delta_b , \phi \rangle &= \int_{-\infty}^\infty \delta_a \delta_b \phi(x) \ dx \\ &= \int_{-\infty}^{a + \frac{\varepsilon}{2}} \delta_a \cdot\left( 0 \cdot \phi(x) \right) dx \ + \int_{a + \frac{\varepsilon}{2}}^{\infty} \delta_b \cdot \left( 0 \cdot \phi(x) \right) \\ &= 0 + 0 \\ &= 0, \end{align*} and hence, $\delta_a \delta_b \equiv 0$. Is this correct reasoning or is multiplying distinct $\delta$ functions just as nonsensical as squaring a single $\delta$ function?

Answer 1

Given a sufficiently nice approximate identity $g_n$ e.g. $g_n(x)=\frac{n}{\sqrt{2 \pi}} e^{-n^2 x^2/2}$, $g_n(x-a) g_n(x-b)$ will converge in the sense of distributions to zero if $a \neq b$. In this sense "$\delta_a \delta_b=0$", since we formally identify the limit of one factor with $\delta_a$ and the limit of the other factor with $\delta_b$.

But in the strict sense of distribution theory $\delta_a \delta_b$ is not defined.