2D Dirac delta function: how to prove $\delta(ax, by) = \frac{1}{|ab|} \delta(x,y)$

dirac deltadistribution-theory

Let $\delta(x)$ be a Dirac-delta function. I am wondering how to prove the following:
\begin{align}
\delta(ax, by) = \frac{1}{|ab|} \delta(x,y)
\end{align}
Because of the definition (if I understand it correctly),
\begin{align}
\delta(ax, by) = \begin{cases}
\infty & \text{if $ax = 0,\ by=0,$}\\
0 & \text{otherwise.}
\end{cases}
\end{align}
If this is true, the scale of the argument should not influence the value of the delta function itself? Correct me if I'm wrong.

Best Answer

Note that the dirac delta satisfies:

  1. case $ℝ→ℝ:\quad$ ${\displaystyle \delta (g(x))=\sum _{i}{\frac {\delta (x-x_{i})}{|g'(x_{i})|}}}$
  2. case $ℝ^n→ℝ:\quad$ ${\displaystyle \int _{\mathbf {R} ^{n}}f(\mathbf {x} )\,\delta (g(\mathbf {x} ))\,d\mathbf {x} =\int _{g^{-1}(0)}{\frac {f(\mathbf {x} )}{|\mathbf {\nabla } g|}}\,d\sigma (\mathbf {x} )}$
  3. case $ℝ^n→ℝ^n:\quad$ ${\displaystyle \int _{\mathbf {R} ^{n}}\delta (g(\mathbf {x} ))\,f(g(\mathbf {x} ))\left|\det g'(\mathbf {x} )\right|\,d\mathbf {x} =\int _{g(\mathbf {R} ^{n})}\delta (\mathbf {u} )f(\mathbf {u} )\,d\mathbf {u} }$

In Question Expression for Dirac delta $\delta(xy)$, we had an instance of case (2). Your question is an instance of case (3), with $g(\begin{smallmatrix}x\\ y\end{smallmatrix}) = (\begin{smallmatrix}ax\\ by\end{smallmatrix})$:

$$\begin{aligned} && δ(g(x, y))|\det g'(x)| &= \delta(x, y) \\&⟹& δ(ax, by) \Big|\det \pmatrix{a & 0 \\0 & b}\Big| &= \delta(x, y) \\&⟹&δ(ax, by)|ab| &= \delta(x, y) \\&⟹&δ(ax, by) &= \frac{1}{|ab|}\delta(x, y) \end{aligned}$$

For a proof of the theorem, see for example Rigorous proof of the change of coordinates formula for Dirac's delta..

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