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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}

where

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

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.

Make42
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1 Answers1

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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 $0$.

reuns
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  • 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? – Make42 Nov 29 '20 at 20:56
  • 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 $0$. – reuns Nov 29 '20 at 21:25
  • 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.) – Make42 Nov 29 '20 at 21:30
  • 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 $1$. – reuns Nov 29 '20 at 21:36
  • 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? – Make42 Nov 29 '20 at 21:43
  • What I need in the end is the "distribution-or-something-similar-1" that, when convoluted with a Gaussian, results in a function which marginal in some of the coordinates is a Gaussian and the rest of the coordinates is a "distribution-or-something-similar-2" which is a probability density that is uniform but with infinite support. (I would also be happy with finite support, but I don't think I would find a "distribution-or-something-similar-1" that could provide this.) – Make42 Nov 29 '20 at 21:46
  • The smooth functions that vanish for $|x|$ large enough. But we often enlarge it to add the Schwartz functions. The convolution of a distribution and a test function is well defined, it sends it to a smooth function. No you don't need a pdf uniform with support $\Bbb{R}$ because you don't have an underlying $\sigma$-algebra. – reuns Nov 29 '20 at 21:46
  • Let us continue this discussion in chat. – Make42 Nov 29 '20 at 21:48