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.

  • 1
    $\begingroup$ The best way to think about the Dirac delta is not to think of it as a function on numbers, but to think of it as a functional taking in functions in a function space with an integral norm and outputting the value of that function at 0 $\endgroup$
    – Alan
    Oct 5, 2021 at 16:17
  • 2
    $\begingroup$ That is not a good definition of $\delta.$ A close to correct definition is that $$\iint_{\mathbb{R}^2}\delta(x,y)\,\phi(x,y)\,dx\,dy=\phi(0,0)$$ for all smooth functions $\phi: \mathbb{R}^2\to\mathbb{R}.$ $\endgroup$
    – md2perpe
    Oct 5, 2021 at 16:18
  • 2
    $\begingroup$ And to follow that up, consider the change of variables $t = ax$ and $s=by$, which gives $dx = dt/a$ and $dy=ds/b$. $\endgroup$
    – bonsoon
    Oct 5, 2021 at 16:19

2 Answers 2


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..


Simple answer: this is a simple change of variable. Heuristically, since the Dirac delta $\delta$ is defined by its action on smooth functions $\varphi$ by $$ \langle\delta,\varphi\rangle = \iint_{\Bbb R^2} \varphi(x,y)\,\delta(x,y)\,\mathrm d x\,\mathrm d y = \varphi(0,0), $$ it follows by the change of variable $x'=a\,x$ and $y'=b\,y$ that $$ \iint_{\Bbb R^2} \varphi(x,y)\,\delta(ax,by)\,\mathrm d x\,\mathrm d y = \frac{1}{|ab|}\iint_{\Bbb R^2} \varphi(\tfrac{x'}{a},\tfrac{y'}{a})\,\delta(x',y')\,\mathrm d x'\,\mathrm d y' = \frac{1}{|ab|} \,\varphi(0,0), $$ or in other words $\langle\delta(ax,ay),\varphi\rangle = \frac{1}{|ab|}\langle\delta(x,y),\varphi\rangle$ for any $\varphi$, which implies $\delta(ax,ay) = \frac{1}{|ab|}\,\delta(x,y)$

Digging a bit more, now why can we do this change of variable in the integral? Actually this is because the composition of distributions is defined so that the change of variable formula works as if everything was done with classical integrals.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .