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The prototypical example of a distribution is the Dirac delta function, defined as a linear functional taking a well behaved test function $\phi:\mathbb{R} \to \mathbb{R}$ and returning its value at the origin, $$(\delta,\phi) = \phi(0).$$ Intuitively, one can imagine $\delta$ as a sharply peaked function around the origin so that $$(\delta, \phi) = \int_{\mathbb{R}} \delta(x) \phi(x) dx.$$ A lot of properties of the delta functions are motivated by considering the above integral and performing various partial integrations etc. Then one uses the property that the test functions, together with all of their derivatives, vanish at the boundary of integration, which is infinity.

However, in physical applications, one often has to consider distributions for which the test functions are defined on a different domain $D$. For example:

  1. $D = [0, \infty)$: for spherical symmetric problems, physical quantities depend only on the radial variable $r$ which is inherently non-negative. Therefore, the test functions of the radial variable have positive reals as their domain, and the radial delta function $\delta_r$ should(?) exist with the fundamental property $$(\delta_r, \phi) = \int_0^\infty \delta_r (r) \phi(r) dr = \phi(0).$$ However, due to the non-trivial boundaries, the properties of the radial delta function should be different from the usual delta function. E.g. we cannot perform the partial integrations so easily. The similar problem persists in other domains, such as
  2. $D = [0,1]$: as in the case of zenith angle variable in spherical coordinates;
  3. $D = [0,2\pi)$: with $\lim_{x \to 2\pi} \phi(x) = \phi(0)$; the periodic boundary conditions, as in the case of azimuth angle variable in spherical coordinates.

How does one approach distributions on these domains? Can they be simply related to the usual distributions on unbounded domains?

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    $\begingroup$ Looks like you could just work with distributions (on $\mathbb R$) supported by $D$ in cases (1), (2), and I think in case (3) you're really dealing with $S^1$, so there are no problems here. (See here for the notion of support of a distribution: en.wikipedia.org/wiki/… ) $\endgroup$ Jun 10, 2018 at 1:19
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    $\begingroup$ Can the support of the distribution be a (semi) closed set? $\endgroup$
    – Fizikus
    Jun 10, 2018 at 9:50
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    $\begingroup$ It's a closed set by its definition. $\endgroup$ Jun 10, 2018 at 15:32
  • $\begingroup$ A simple way to define distributions on a closed or half closed interval is simply distributions on the whole line whose support lies in the interval. This works, because any compactly smooth function on the interval can be extended to one on the entire line. $\endgroup$
    – Deane Yang
    Jun 11, 2018 at 3:39
  • $\begingroup$ @DeaneYang: In the case of radial delta function, the support of the distribution is only one point - $r=0$ - the left boundary. This means that the usual formula for the derivatives does not work: $(\delta_r', \phi) \neq - (\delta_r, \phi')$. So I think that extending the domain, as you suggest, has some issues. $\endgroup$
    – Fizikus
    Jun 11, 2018 at 17:49

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