Why the radial delta function is not equal to the product of two 1D delta functions

Question

The first proof that comes to my mind is that the units of left side and right side does not match. Delta functions take an input and spit out an output of units = 1/[units of input].

What would be the rigorous proof of this?

Statement to be proven: (also given below):

$$\delta \left ( r \right )\space \space \neq \space \space \delta \left ( x \right )\delta \left ( y \right )$$

Answer 1

Since$$\iint f(r,\,\theta)\delta(r)\delta(\theta)drd\theta=f(O)=\iint f(x,\,y)\delta(x)\delta(y)\underbrace{dxdy}_{rdrd\theta},$$we have the dimensionally homogeneous result $\delta(r)\delta(\theta)=r\delta(x)\delta(y)$. @MarkViola's comment has noted a more general finding we ca get by writing $f(p)$ for some point $p$ as two different double integrals.

Answer 2

Delta Function in Two Dimensions

The problem with this question is that the radial delta function is not $\delta(r)$, where $\delta$ is the usual $1$-dimensional delta function.

An approximation of $\delta(x)$ is $ne^{-\pi n^2x^2}$, so $$ \begin{align} \delta(x)\,\delta(y) &=ne^{-\pi n^2x^2}ne^{-\pi n^2y^2}\tag{1a}\\ &=n^2e^{-\pi n^2r^2}\tag{1b} \end{align} $$ Now, let's see what the integral of $n^2e^{-\pi n^2r^2}$ is (in polar coodinates) $$ \begin{align} \int_0^{2\pi}\int_0^\infty n^2e^{-\pi n^2r^2}\,r\,\mathrm{d}r\,\mathrm{d}\theta &=\int_0^{2\pi}\int_0^\infty\frac1{2\pi}\,e^{-u}\,\mathrm{d}u\,\mathrm{d}\theta\tag{2a}\\ &=1\tag{2b} \end{align} $$ Explanation:
$\text{(2a)}$: $u=\pi n^2r^2$
$\text{(2b)}$: evaluate the integrals

So, it seems that $n^2e^{-\pi n^2r^2}$ is an approximation of the radial delta function. Taking the limit says that $\delta(x)\,\delta(y)$ is the radial delta function.

---

The Radial Delta Function is not $\boldsymbol{\delta(r)}$

However, if we use $\delta(r)$ in place of $\delta(x)\,\delta(y)$, we get $$ \begin{align} \int_0^{2\pi}\int_0^\infty ne^{-\pi n^2r^2}\,r\,\mathrm{d}r\,\mathrm{d}\theta &=\int_0^{2\pi}\int_0^\infty\frac1{2\pi n}\,e^{-u}\,\mathrm{d}u\,\mathrm{d}\theta\tag{2a}\\ &=\frac1n\tag{2b} \end{align} $$ which tends to $0$.

---

The Radial Dirac Delta Function $$ \begin{align} \int_0^{2\pi}\int_0^\infty\overbrace{\frac{n}{\pi r}e^{-\pi n^2r^2}}^{\large\frac1{\pi r}\delta(r)}\,r\,\mathrm{d}r\,\mathrm{d}\theta &=\int_0^{2\pi}\int_0^\infty \frac1{\pi}e^{-\pi r^2}\,\mathrm{d}r\,\mathrm{d}\theta\tag{3a}\\ &=1\tag{3b} \end{align} $$ Thus, the radial delta function is $\frac1{\pi r}\delta(r)$.