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In the Wikipedia page for the Dirac Delta function this formula appears under "Properties in $n$ dimension". $$ \int f(x) \delta(g(x)) dx = \int_{g^{-1}(0)} \frac{f(x)}{|\nabla g(x)|} d\sigma(x) $$ It is said that this is a consequence of the Co-Area formula but no proof is given and the only reference ("Hörmander (1983), The analysis of linear partial differential operators I") doesn't seem to have this formula in it.

I have a few questions, in order of importance.

  1. What is a proof of this statement?
  2. What other references are there about this statement and its generalizations to a function $g:\mathbb{R}^n\to\mathbb{R}^m$ with $n > m > 1$?
  3. In the above the author uses $\delta (g(x)) dx$ as if $\delta$ was a function, where in fact it is a Schwartz distribution or a measure. What did they mean? Especially because now it is concatenated with another function.
Definition of Dirac Distribution

It's a linear functional that maps test functions $\varphi$ to $$ \delta_x[\varphi] = \int \varphi(y) \delta_x^{\text{measure}}(dy) = \varphi(y) $$ where $\delta_x^{\text{measure}}$ is the Dirac Measure which for any measurable set $A$ is defined as $$ \delta_x^{\text{measure}}(A) = \begin{cases} 1 & x\in A \\ 0 & x\notin A \end{cases} $$

Co-Area Formula for Lipschitz Functions

If $g:\mathbb{R}^n\to\mathbb{R}^m$ with $n > m$ then $$ \int_{\mathbb{R}^n} f(x) dx = \int_{\mathbb{R}^m} \left[\int_{g^{-1}(y)} f(x) |J_g(x) J_g(x)^\top|^{-1/2} \mathcal{H}^{n-m}(dx) \right]dy $$ where $J_g(x)$ is the Jacobian matrix of $g$.

Euler_Salter
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  • Perhaps they are using an abuse of notation and what they mean is $$ \delta_x(g(A)) = \begin{cases} 1 & x\in g(A) \ 0 & x\notin g(A) \end{cases} $$ In which case $\delta(g(\cdot))$ is actually the pushforward measure of $\delta$ by $g^{-1}$ (i.e. pullback measure by $g$). If that's the case, then the LHS is actually $$ \int_{\mathbb{R}^n} f(x) \delta_0(g(dx)) = \int_{\mathbb{R}} f(g^{-1}(y)) \delta_0(dy) $$ using the change of variables formula – Euler_Salter Dec 28 '22 at 17:19
  • Although the last equation looks dodgy because technically $g^{-1}(y)$ should read $g^{-1}({y})$... – Euler_Salter Dec 28 '22 at 17:20
  • To be clear nobody understands my answer? – reuns Dec 28 '22 at 20:02
  • Related https://math.stackexchange.com/questions/2590419/integral-over-dirac-delta-of-multivariate-funciton – Euler_Salter Dec 29 '22 at 22:44
  • Related https://math.stackexchange.com/questions/56939/property-of-dirac-delta-function-in-mathbbrn?noredirect=1&lq=1 – Euler_Salter Jan 04 '23 at 10:11

1 Answers1

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They mean $$\int_{\Bbb{R}^N} f(x)\delta(g(x))dx=\lim_{n\to \infty} \int_{\Bbb{R}^N} f(x)\frac{n}2 1_{|g(x)|< 1/n}dx$$ where $f\in C^\infty_c(\Bbb{R}^N)$ and $g\in C^\infty(\Bbb{R}^N)$ with $\|\nabla g\| \ne 0$. What is $d\sigma(x)$? It is just the measure/distribution defined by $$\int_{Z(g)} f(x)d\sigma(x) = \lim_{n\to \infty} \frac{n}2\int_{dist(Z(g),x)<1/n} f(x)dx$$ where $Z(g)$ is the vanishing set of $g$ and $$dist(Z(g),x)=\inf_{a\in Z(g)} \|x-a\|$$

For $x$ close to $a\in Z(g)$ we have $g(x)\approx \nabla g(a) \cdot (x-a)$ from which $$dist(Z(g),x) \approx \frac{|g(x)|}{\|\nabla g(a)\|}$$

Whence for $f_a\in C^\infty_c(\Bbb{R}^N)$ supported on a small ball around $a$ we have $$\lim_{n\to \infty} \int_{\Bbb{R}^N} f_a(x)\frac{n}2 1_{|g(x)|< 1/n}dx\approx \int_{Z(g)} \frac{f_a(x)}{\|\nabla g(a)\|}d\sigma(x)$$ Writing $f$ as a sum of smooth functions supported on very small balls we get the formula $$\int_{\Bbb{R}^N} f(x)\delta (g(x))dx=\int_{Z(g)} \frac{f(x)}{\|\nabla g(x)\|}d\sigma(x)$$

reuns
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  • Thanks! How do you know the first identity? I am desperately looking for a proof of something similar, but no one seems to make it rigurously. I think that rescaled indicator function is what some people call "approximation to the identity" or "smoothing kernel" but no one shows that (in the limit) the Dirac Delta Distribution and the Mollifier Distribution act the same on test functions – Euler_Salter Dec 28 '22 at 17:52
  • This is the definition of $\delta$ – reuns Dec 28 '22 at 17:52
  • Do you have a reference for this definition? I can only find the definition I gave above – Euler_Salter Dec 28 '22 at 17:53
  • Forget about the definitions you saw, $\lim_{n\to \infty} \frac{n}2 1_{|t|<1/n}$ is the definition of $\delta(t)$ (convergence in the sense of distributions). – reuns Dec 28 '22 at 17:54
  • Do you have a reference though? – Euler_Salter Dec 28 '22 at 18:19
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    No why do you want some, ie. what is unclear in my answer? I realize that I wrote $\Delta$ instead of $\nabla$, I corrected. – reuns Dec 28 '22 at 18:56
  • I’m a bit confused. What definition of $d\sigma$ are you using? I thought it was the Hausdorff measure, but don’t recognise your definition of it – Euler_Salter Dec 28 '22 at 22:42
  • I gave the definition in my answer. See instead https://en.wikipedia.org/wiki/Minkowski_content – reuns Dec 28 '22 at 22:53
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    The definition of $\delta_0$ as a distribution is, of course, $\delta_0(\varphi)=\varphi(0)$ for $\varphi\in\mathscr D(\mathbb R^N)$. That it is equal to $\lim\limits_{n\to\infty} c_{n,N} 1_{{|t|<1/n}}$ where $c_{n,N}^{-1}$ is the volume of the $N$-dimensional ball with radius $1/n$ is a simple fact (whose proof just uses the continuity of test functions). The definition of reuns thus works only in dimension $N=1$. – Jochen Dec 29 '22 at 10:04
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    @Jochen Here it is the 1-dimensional Dirac delta. Please take a look at https://math.stackexchange.com/questions/4604827/how-can-i-interpret-expression-lx-t-where-l-is-a-distribution-in-a-single/4605036#4605036 to understand why defining $\delta$ as a limit of a sequence of functions is the only right definition. Obviously if I insist on it it is because we are dealing with $\delta(g(x)),g\in C^\infty(\Bbb{R}^N)$ which doens't make much sense otherwise. – reuns Dec 29 '22 at 10:41
  • I think I understand the proof intuitively, but it seems very hand-wavey – Euler_Salter Dec 30 '22 at 11:21
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    @Euler_Salter Which part? For now you only said that you were affraid of the definitions (of the distributions $\delta(g(x))$ and $d\sigma(x)$), for bad reasons, as the ones I gave are the only right ones in the context of your question. Then the formula doesn't have anything strange/surprising/difficult: for $x$ close to $Z(g)$, $|g(x)|$ is a good approximation to $|\nabla g(x)| dist(Z(g),x)$ – reuns Dec 30 '22 at 11:36
  • I understand it’s a good approximation, but sometimes the limit of a good approximation is still biased – Euler_Salter Dec 30 '22 at 12:12
  • @Euler_Salter It is a $O(dist(Z(g),x)^2)$ approximation (uniform on the support of $f$) and this term disappears as $n\to \infty$ – reuns Dec 30 '22 at 12:18
  • @reuns how do you know that $\text{dist}(Z(g), x) \approx \frac{|g(x)|}{|\nabla g(a)|}$ – Euler_Salter Jan 04 '23 at 10:10