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What is $\int_{\mathbb{R}^r}\delta\left(f(z)\right) g(z) dz$?

Here $\delta(\cdot)$ is a $d$-dimensional Dirac-delta function, $z$ is a $r$-dimensional variable ($r<d$). $f:\mathbb{R}^r\to\mathbb{R}^d$ is a differentiable function. $g(z)$ is a bounded. To make things simple, we assume that there exists only one $z$ such that $f(z)=0$.

I know this is going to be infinite but I want to know its divergence rate.

We can write $\delta\left(f(z)\right)$ as the limit of a Gaussian distribution: $$ \lim_{\gamma\to0}\mathcal{N}(x|f(z),\gamma I) $$ I guess $\int_{\mathbb{R}^r} \mathcal{N}(x|f(z),\gamma I) g(z)dz = \Theta(\gamma^{(d-r)/2})$, meaning they are the same order infinity. Am I correct? If so, how can I prove it?

user500250
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  • Your question doesn't make sense as it is. Did you mean $\int_{\mathbb{R}^d} $ ? Is $f$ a function $\mathbb{R}^d \to \mathbb{R}^d$ ? Is it differentiable, and is $\nabla f(z_0)$ inversible at the $z_0$ such that $f(z_0) = 0$ ? – reuns Dec 05 '17 at 10:54
  • @reuns Yes I mean $\int_{\mathbb{R}^d}$. $f$ is a function $\mathbb{R}^r\to\mathbb{R}^d$ and it is differentiable. $\bigtriangledown f(z_0)$ is not inversible since it is not a square matrix. Sorry for the confusion. I will change the question soon. Thanks. – user500250 Dec 05 '17 at 11:08
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    Related – Giuseppe Negro Dec 05 '17 at 11:31
  • If you don't give any clues I can't help more.. – reuns Dec 05 '17 at 12:39
  • Comment to the post (v2): The condition $r<d$ makes little sense. It should presumably be $r\geq d$. – Qmechanic Dec 12 '17 at 18:55

1 Answers1

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Your question doesn't make sense.

Do you see that it is natural to define the Dirac delta by $$\int_{-\infty}^\infty \delta(x) \varphi(x)dx = \lim_{a \to 0^+} \int_{-\infty}^\infty \frac{1_{x \in (-a,a)}}{2a} \varphi(x)dx = \lim_{a \to 0^+}\frac{1}{2a} \int_{-a}^a \varphi(x)dx = \varphi(0)$$ where the last equality assumes $\varphi$ is continuous at $x=0$.

Thus for $f: \mathbb{R} \to \mathbb{R}$ it is natural to define $\delta(f(x))$ by $$\int_{-\infty}^\infty \delta(f(x)) \varphi(x)dx = \lim_{a \to 0^+} \int_{-\infty}^\infty \frac{1_{f(x) \in (-a,a)}}{2a} \varphi(x)dx$$ If $f$ is differentiable, has only one zero at $x=x_0$ and $f'(x_0) \ne 0$ and $\varphi$ is continuous at $x_0$ then $$ \lim_{a \to 0^+} \int_{-\infty}^\infty \frac{1_{f(x) \in (-a,a)}}{2a} \varphi(x)dx =\lim_{a \to 0^+} \frac{1}{2a}\int_{x_0-\frac{a}{|f'(x_0)|}}^{x_0+\frac{a}{|f'(x_0)|}} \varphi(x)dx= \frac{\varphi(x_0)}{|f'(x_0)|}$$ This generalizes in dimension $d$ with $$\int_{\mathbb{R}^d} \delta(F(x)) \Phi(x)dx = \lim_{a \to 0^+} \int_{\mathbb{R}^d} \frac{1_{\|F(x)\| < a}}{a^d S_d} \varphi(x)dx=\frac{\Phi(x_0)}{|\det(\nabla F(x_0))|}$$ whenever $\Phi : \mathbb{R}^d \to \mathbb{R}^d $ is continuous at $x_0$ and $F : \mathbb{R}^d \to \mathbb{R}^d $ has only one zero at $x_0$, is differentiable and the matrix $\nabla F(x_0))$ is inversible and $S_d = \int_{\mathbb{R}^d} 1_{\|x\| < 1}dx$

reuns
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  • Thanks for your answer. Can I define the Dirac delta function as $\delta(x)=\lim_{\gamma\to0}\mathcal{N}(x|0,\gamma I)$ where $I$ is a $d$-dimensional identity matrix? Then the integration becomes $\lim_{\gamma\to0}\int_{\mathbb{R}^r}\mathcal{N}(f(z)|0,\gamma I)dz$. I know this is going to be infinite. But I want to know the divergence rate comparing to $\gamma^{(d-r)/2}$. Is it possible? If $d=r$, then it converges to the result you given. – user500250 Dec 05 '17 at 14:49
  • @daib09 No, you can define the Dirac delta differently only once the one I mentioned is clear to you. Thus, what do you get from the $\frac{1_{|x| < a}}{a^d}$ definition, can you make your question (and your assumptions about $f,g$) explicit from it ? – reuns Dec 05 '17 at 14:51
  • I though we can define a delta function as the limit of Gaussian distribution (http://functions.wolfram.com/GeneralizedFunctions/DiracDelta/09/). Actually I was deriving this: $\lim_{\gamma\to0}\log\int_{\mathbb{R}^r}\mathcal{N}(x|f(z),\gamma I_d)\mathcal{N}(z|0,I_r)dz+\frac{d-r}{2}\log\gamma-\frac{1}{2}\log|f^\prime(z_0)^\top f^\prime(z_0)|$. Here $I_d$ is a $d$-dimensional matrix and $f$ is a differentiable function. So I replaced the limit of Gaussian distribution with a delta function. If that was wrong, do you have any idea how to solve this one? Thanks a lot. – user500250 Dec 05 '17 at 15:07
  • and $z_0$ is the only solution for $f(z)=x$. – user500250 Dec 05 '17 at 15:08
  • @daib09 ..Can you prove that $\lim_{n \to \infty} \int_{-\infty}^\infty ne^{-\pi n^2 x^2} \varphi(x)dx = \varphi(0)$ whenever $\varphi$ is bounded and continuous at $x=0$ ? If $f : \mathbb{R}^r \to \mathbb{R}^r$ is differentiable and has only one zero where $\nabla f(z_0)$ is inversible then the result is in my answer. Otherwise the limit diverges and so it doesn't make sense. – reuns Dec 05 '17 at 15:25
  • OK. I will think more about it. Thanks a lot:) – user500250 Dec 05 '17 at 15:29