The multivariate Dirac delta distribution can be - more or less intuitively - be expressed as

\begin{align} \delta(\mathbf x) = \begin{cases} \lim\limits_{a\rightarrow0} \quad \dfrac{1}{a^n} & \forall x_i \in [-\frac a2,\frac a2], 1\le i\le n \\[6pt] \quad 0 & \text{otherwise} \end{cases} \end{align}


$$ \int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty} \delta(\mathbf x) \text{ d}\mathbf x = 1 $$

Is there an "opposite" of that, which can be expressed as

\begin{align} \epsilon(\mathbf x) = \begin{cases} \lim\limits_{a\rightarrow\infty} \quad \dfrac{1}{a^n} & \forall x_i \in [-\frac a2,\frac a2], 1\le i\le n \\[6pt] \quad 0 & \text{otherwise} \end{cases} \end{align}

where also

$$ \int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty} \epsilon(\mathbf x) \text{ d}\mathbf x = 1 $$


Is there a name for this distribution and/or a symbol?

For context: I am planning to use them in convolutions and I am treating them as probability densities.


Both limits $$\lim_{a\to 0} a^{-n} 1_{x\in [-a/2,a/2]^n}, \qquad \lim_{a\to \infty} a^{-n} 1_{x\in [-a/2,a/2]^n}$$ are perfectly rigorous definitions of distributions, the first one converges in the sense of distributions to $\delta$ and the second one to $0$.

  • $\begingroup$ Maybe I was not clear in my question: Is there a name for this (second) distribution and/or a symbol? Or maybe I was clear and I do not understand your answer: Is there a distribution called "0" = "zero" which integral over the domain is 1? $\endgroup$
    – Make42
    Nov 29 '20 at 20:56
  • $\begingroup$ We don't integrate a distribution $T$ on $\Bbb{R}^n$, we only look at $\int_{\Bbb{R}^n} T(x)\phi(x)dx$ for each $\phi\in C^\infty_c(\Bbb{R}^n)$ (alternatively for $\phi$ in the Schwartz space). The $0$ distribution gives $0$ for all $\phi$. If you want to add the constants to your test function space you can but it won't be called distributions, it will be the dual of $V=C^\infty_c(\Bbb{R}^n)+\Bbb{C}$ and $\lim_{a\to \infty} a^{-n} 1_{x\in [-a/2,a/2]^n}$ will still converge in $V'$ but not to $0$. $\endgroup$
    – reuns
    Nov 29 '20 at 21:25
  • $\begingroup$ Well, I want to combine the 0 distribution with the Dirac delta distribution, e.g. like $\delta(x_1) \cdot \epsilon(x_2)$. As far as I understand the Dirac delta distribution does integrate to 1 - I need this, because I use it as a probability density and I want to use the $\epsilon$ / 0 distribution as well as a probability density. I have not found any article by web search that mentions a "zero distribution" or "0 distribution". Can you link to something? (Long term, my test function is going to be a Gaussian - just for sake of giving more context.) $\endgroup$
    – Make42
    Nov 29 '20 at 21:30
  • $\begingroup$ A distribution is just a linear map $C^\infty_c(\Bbb{R}^n)\to \Bbb{C}$ and $\delta$ is the one sending $\phi$ to $\phi(0)$. You can add the constant functions to $C^\infty_c(\Bbb{R}^n)$ but it won't be called distributions. And I doubt that you need something which sends $1+\phi$ to $1$. $\endgroup$
    – reuns
    Nov 29 '20 at 21:36
  • $\begingroup$ I don't know what any of that means. Can you tell me what $C_c^{\infty}(\mathbb R^n)$ is? What do you mean by "sending" to? If I convolute a Gaussian with a Dirac, I get the Gaussian - where is the "sending to" in this context? $\endgroup$
    – Make42
    Nov 29 '20 at 21:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.