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Coming from a physics background, I was taught the formula for the composition of a Dirac delta and a function. Indeed, if we consider a nice function $ f : \mathbb{R} \to \mathbb{R} $, one can write $$ \delta_{\rm D} (f(x)) = \sum_{x_{0} \in \mathcal{Z}_{f}} \frac{\delta_{\rm D} (x \!-\! x_{0})}{|f'(x_{0})|} \, , $$ where ${ \mathcal{Z}_{f} = \{ x_{0} \,|\, f(x_{0}) = 0 \} }$ is the set of all the zeroes of $f$, and where we assumed that all the zeroes of $f$ are simple, so that ${f'(x_{0}) \neq 0}$.

My question is now related to what would happen in higher dimensions. (Such formulas are often encountered in kinetic theory). Let's assume I want to compute the double integral $$ I = \iint dx \, dy \, \delta_{\rm D} (f (x,y)) \, g(x,y) \, , $$

My questions are then the following ones :

  • Let's assume that $f(x,y) = f(x)$, so that $y$ is absent from the Dirac delta. Provided that the zeroes of $f$ are simple, do we have ? $$ I = \sum_{x_{0} \in \mathcal{Z}{f}} \frac{1}{|f'(x_{0})|} \int dy \, g(x_{0} , y) $$

  • Let's assume that the set ${ \mathcal{Z}_{f} = \{ (x_{0},y_{0}) \,|\, f(x_{0},y_{0}) \} }$ is made of isolated points. Can we compute $I$ ? If yes, how can it be done ?

  • Let's assume that the set ${ \mathcal{Z}_{f} = \{ (x_{0},y_{0}) \,|\, f(x_{0},y_{0}) \} }$ can be parametrized under the form ${ \mathcal{Z}_{f} = \{ (x(\lambda),y(\lambda)) \,|\, \lambda \in \mathbb{R} \} }$, i.e. the zeroes of ${ f(x,y) }$ are along a given nice curve in $\mathbb{R}^{2}$. Is it possible to compute $I$ ?

  • Let's assume that the set ${ \mathcal{Z}_{f} = \{ (x_{0},y_{0}) \,|\, f(x_{0},y_{0}) \} }$ is even more filled than a line, so that the zeroes of $f$ constitute a closed region of $\mathbb{R}^{2}$ of non-zero volume. (I am not sure of the best way to phrase it...) For example, ${ \mathcal{Z}_{f} = \{ (x_{0},y_{0}) \,|\, x_{0}^{2} \!+\! y_{0}^{2} \leq 1 \} }$. Does the double integral $I$ have a meaning ?

  • Finally, are there any other sorts of others $\mathcal{Z}_{f}$ for which the integral $I$ could have a meaning and be computed ?

jibe
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  • Do you mean to write "zeroes" instead of "poles"? (Also, this is a great question!) – Micah Dec 18 '14 at 17:25
  • @Micah Yes, in my mind "zeroes" and "poles" were here identical concepts, i.e. $f(x_{0}) = 0$, but I guess I might not be using the best words. If "zeroes" is the correct term, I will of course edit my question accordingly. – jibe Dec 18 '14 at 17:33
  • At least in English, "pole" is generally synonymous with "singularity" not "zero". – Micah Dec 18 '14 at 17:44

1 Answers1

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Provided $\nabla f$ is non-vanishing, the formula is

$$ \int_{\Bbb R^n}\delta(f(x))g(x) \,dx = \int_{f^{-1}(0)} \dfrac{g(x)}{|\nabla f (x) |}\,d\sigma $$

where $d\sigma$ is the surface measure$^*$ on $f^{-1}(0)$. This is basically just the change of variables ($u$-substitution) for integration.

On a volume, you're going to have trouble. The gradient will vanish, and it's sort of intuitively clear that you shouldn't be able to do this anyway: for instance, if $g \equiv 1$

$$ \int\int \delta\left(\chi_{\{x^2 + y^2 \leq 1\}}\right) g(x,y) \,dx\,dy $$

ought to be infinite

*: or counting measure, if/where $f^{-1}(0)$ consists of isolated points.

BaronVT
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  • Thank you, I was missing this generalization. The notation $|\nabla f(x)|$ may be misleading. Do we agree that it is the euclidian norm of $\nabla f(x)$, so that $|\nabla f(x)| = \sqrt{(\partial f/\partial x_{1})^{2} + ... + (\partial f/\partial x_{2})^{2} }$ ?

    It also appeared to me that for nice function $f$, of non-vanishing gradient, the case where $\mathcal{Z}_{f}$ is made of isolated points can not occur, so that the surface element $d \sigma$ is always of dimension $(n-1)$, and $f^{-1} (0)$ is always a set of dimension $(n-1)$. Am I correct ?

     – jibe Dec 19 '14 at 09:14
  • Yes, we agree on the notation for gradient and norm (sorry for any confusion). A smooth $f$ can certainly have isolated zeros; for instance $f = x^2 + y^2$ is only zero at the origin. You could multiply shifted paraboloids to get zeros wherever you want: $f = ((x - x_1)^2 + (y - y_1)^2)((x - x_2)^2 + (y - y_2)^2)$, etc. – BaronVT Dec 19 '14 at 15:59
  • ... though of course those gradients vanish. Sorry for the hasty response, I'll think more about this, you may indeed be correct. – BaronVT Dec 19 '14 at 16:58
  • Ok, I think the Whitney extension theorem will produce examples ( http://en.wikipedia.org/wiki/Whitney_extension_theorem ). Basically, for any closed set $A$ of $\Bbb R^n$, you can cook up a $C^m$ function $f$ that has your choice of values and derivatives (up to order $m$) on $A$. – BaronVT Dec 19 '14 at 17:24
  • See also this: http://math.stackexchange.com/questions/48746/existence-of-non-constant-continuous-functions-with-infinitely-many-zeros – BaronVT Dec 19 '14 at 17:25
  • I do agree that Whitney extension theorem allows you to build up a nice function which cancels out in a point $x_0$ with non-vanishing gradient. However, it doesn't ensure that in the neighborhood of $x_0$, $f$ doesn't cancel out more. Indeed, if $\nabla f (x_0) \neq 0$, then thanks to the theorem of implicit functions, $f$ can locally be inverted, and it seems to imply that in the neighborhood of $x_0$, $f$ cancels out on a curve of the form ${(x_0 (\lambda), y_0 (\lambda))}$, so that having $\mathcal{Z}_f$ made of isolated points with non-vanishing gradients is impossible. Do you agree ? – jibe Dec 21 '14 at 12:48
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    I think you may be right. This may be suitable for a separate question if you have any doubts about your argument (something like "Can a $C^1 , \Bbb R^n \to \Bbb R, n \geq 2$ function have isolated zeros?") – BaronVT Dec 21 '14 at 17:34
  • I realize this question is old, but I was wondering if the formula still works when the zero set of the gradient has measure zero in $f^{-1}(0)$. For example $f(x,y)=x^2-y^2$ – TheEmptyFunction Apr 26 '23 at 19:55